Category: Class 11 Mathematics

NCERT Exemplar Class 11 Maths Chapter 14 Mathematical Reasoning are part of NCERT Exemplar Class 11 Maths. Here we have given NCERT Exemplar Class 11 Maths Chapter 14 Mathematical Reasoning.

NCERT Exemplar Class 11 Maths Chapter 14 Mathematical Reasoning

Q1. Which of the following sentences are statements? Justify.
(i) A triangle has three sides.
(ii) O is a complex number.
(iii) Sky is red,
(iv) Every set is an infinite set.
(V) 15 +8>23
(vi) y+9=7
(vii) Where is your bag?
(viii) EVery square is a rectangle.
(ix) Sum of opposite angles of a cyclic quadrilateral is 1800.
(x) sin² x+cos² x=O
Sol: As we know, a statement is a sentence which is either true or false but not
both simultaneously.
(j) It is true statement.
(ii) It is true statement.
(iii) It is false statement.
(iv) It is false statement.
(y) It is false statement.
(vi) y +9 = 7
It is not considered as a statement, since the value of y is not given.
(vii) It is a question, so it is not a statement.
(viii) It is a true statement.
(ix) It is a true statement.
(x) It is a false statement.

Q2. Find the component statements of the following compound statements.
(i) Number 7 is prime and odd.
(ii) Chennai is in India and is the capital of Tamil Nadu.
(iii) The number loo is divisibLe by 3, 11 and 5.
(iv) Chandigarh is the capital of Irlaryana and U.P.
(y) √7 is a rational number or an irrational number.
(vi) O is less than every positive integer and every negative integer.
(vii) Plants use sunlight, water and carbon dioxide for photosynthesis.
(viii) Two lines in a plane either intersect at one point or they are parallel.
(ix) A rectangle is a quadrilateral or a 5-sided polygon.
Sol: (i) p: Number 7 is prime.
q: Number 7 is odd.
(ii) P: Chennai is in India.
q: Chennai is capital of Tamil Nadu.
(iii) p. 100 is divisible by 3.
q: 100 isdivisibleby 11.
r: 100 ¡s divisible by 5.
(iv) p. Chandigarh is capital of Haryana.
q: Chandigarb is capital of UP
(v) p: √7 is a rational number.
q: √7 is an irrational number.
(vi) p: 0 is less than every positive integer.
q: O is less than every negative integer.
(vii) p: Plants use sunlight for photosynthesis.
q: Plants use water for photosynthesis.
q:- Plants use carbon dioxide for photosynthesis.
(viii) p: Two lines in a plane intersect at one point.
q: Two lines ¡n a plane are parallel.
(ix) p: A rectangle is a quadrilateral.
q. A rectangle is a 5-sided polygon.

Q3. Write the component statements of the following compound statements and
check whether the compound statement is true or false.
(i) 57 is divisible by 2 or 3.
(ii) 24 is a multiple of 4 and 6.
(iii) All living things have two eyes and two legs.
(iv) 2 is an even number and a prime number.
Sol: (i) Here component statements are:
p: 57 is divisible by 2. [false]
q: 57 is divisible by 3. [true]
Given compound statement is of the form ‘pvq’.
Since, the statement ‘pvq’ has the truth value T whenever either p or q
or both have the truth value T.
So, it is true statement as 57 is divisible by 3.
(ii) Here component statements are: p: 24 is multiple of 4. q: 24 is multiple of 6.
Given compound statement is of the form ‘p ^ q’
Since, the statement ‘p A q’ has the truth value T whenever bothp and q have the truth value T.
So, it is a true statement as 24 is divisible by 4 and 6.
(iii) Here component statements are:
p: All living things have two eyes. [false]
q: All living things have two legs. [false]
Given compound statement is of the form ‘p ^q’
It is a false statement. Since ‘p ^ q’ has truth value F whenever either p or q or both have the truth value F
(iv) Here component statements are:
p: 2 is an even number. [true]
q: 2 is a prime number. [true]
Given compound statement is of the form ‘p ^ q’.
It is a true statement. Since ‘p ^ q’ has truth value T whenever both p and q or both have the truth value T.

Q4. Write the negative on the following simple statements.
(i) The number 17 is prime.
(ii) 2 + 7 = 6.
(iii) Violets are blue.
(iv) √5 is a rational number.
(v) 2 is not a prime number.
(vi) Every real number is an irrational number.
(vii) Cow has four legs.
(viii) A leap year has 366 days.
(ix) All similar triangles are congruent.
(x) Area of a circle is same as the perimeter of the circle.
Sol: (i) The number 17 is not prime.
(ii) 2 + 7≠6.
(iii) Violets are not blue.
(iv) √5 is not a rational number.
(v) 2 is a prime number.
(vi) Every real number is not an irrational number.
(vii) Cow does not have four legs.
(viii) A leap year does not have 366 days.
(ix) There exist similar triangles which are not congruent.
(x) Area of a circle is not same as the perimeter of the circle

Q5. Translate the following statements into symbolic form.
(i) Rahul passed in Hindi and English.
(ii) x and y are even integers.
(iii) 2, 3 and 6 are factors of 12.
(iv) Either x or x + 1 is an odd integer.
(v) A number is either divisible by 2 or 3.
(vi) Either x = 2 or x = 3 is a root of 3x² – x – 10 = 0
(vii) Students can take Hindi or English as an optional paper.
Sol: (i) p: Rahul passed in Hindi.
q: Rahul passed in English. p ∧ q: Rahul passed in Hindi and English.
(ii) p: x is even integers. . q: y is even integers.
p∧q: x andy are even integers.
(iii) p: 2 is factor of 12. q: 3 is factor of 12. r: 6 is factor of 12.
p ∧ q ∧ r: 2, 3 and 6 are factors of 12
(iv) p: x is an odd integer.
q: (x + 1) is an odd integer. p v q: Either x or (x + 1) is an odd integer.
(v) p: A number is divisible by 2. q: A number is divisible by 3.
pv q: A number is either divisible by 2 or 3.
(vi) p: x = 2 is a root of 3×2 – x – 10 = 0. q: x = 3 is a root of 3×2 – x – 10 = 0
p v q: Either x = 2orx = 3isa root of 3×2 – x – 10 = 0
(vii) p: Students can take Hindi as an optional paper. q: Students can take English as an optional paper.
p v q: Students can take Hindi or English as an optional paper.

Q6. Write down the negation of following compound statements.
(i) All rational numbers are real and complex.
(ii) All real numbers are rationals or irrationals.
(iii) x = 2 and x = 3 are roots of the quadratic equation x² -5x +6 = 0
(iv) A triangle has either 3-sides or 4-sides.
(v) 35 is a prime number or a composite number.
(vi) All prime integers are either even or odd.
(vii) |x| is equal to either x or -x.
(viii) 6 is divisible by 2 and 3.
Sol. (i) Let p: All rational numbers are real.
q: All rational numbers are complex.
~ p: All rational numbers are not real.
~ q ; All rational numbers are not complex.
Then, the negation of the given compound statement is:
~ (p ∧ q): All rational numbers are not real or not complex.
[~(p ∧ q) = ~p v ~q]
(ii) Let p: All real numbers are rationals. q: All real numbers are irrationals.
Then, the negation of the given compound statement is:
~ (p v q): All real numbers are not rational and all real numbers are not irrational. [~(p v q) = ~p ∧ ~ q]
(iii) Let p ; x = 2 is root of quadratic equation x² – 5x + 6 = 0. q: x = 3 is root of quadratic equation x² – 5x + 6 = 0.
Then, the negation of the given compound statement is:
~ (p ∧  q) : x = 2 is not a root of quadratic equation x²– 5x + 6 = 0 or x = 3 is not a root of the quadratic equationx² – 5x + 6 = 0.
(iv) Let p: A triangle has 3-sides. q: A triangle has 4-sides.
Then, the negation of the given compound statement is:
~ (p v q): A triangle has neither 3-sides nor 4-sides.
(v) Let p: 35 is a prime number. q: 35 is a composite number.
Then, the negation of the given compound statement is:
~ (p v q): 35 is not a prime number and it is not a composite number.
(vi) Let p: All prime integers are even. q: All prime integers are odd.
Then, the negation of the given compound statement is given by
~(p v q): All prime integers are not even and all prime integers are not odd.
(vii) Let p:|x| is equal to x. q: |x| is equal to —x.
Then, the negation of the given compound statement is:
~ (p v q): |x| is not equal to JC and it is not equal to —x.
(viii) Let p: 6 is divisible by 2. q: 6 is divisible by 3.
Then, the negation of the given compound statement is:
~ (p∧q): 6 is not divisible by 2 or it is not divisible by 3

Q7. Rewrite each of the following statements in the form of conditional statements.
(i) The square of an odd number is odd.
(ii) You will get a sweet dish after the dinner.
(iii) You will fail, if you will not study.
(iv) The unit digit of an integer is 0 or 5, if it is divisible by 5.
(v) The square of a prime number is not prime.
(vi) 2b = a + c, if a, b and c are in AP.
Sol: (i) If the number is odd number, then its square is odd number.
(ii) It you take the dinner, then you will get sweet dish.
(iii If you will not study, then you will fail.
(iv) If an integer is divisible by 5, then its unit digits are 0 or 5.
(v) If the number is prime, then its square is not prime.
(vi) If a, b and c are in AP, then 2b = a + c.

Q8. Form the biconditional statement p⟷q, where
(i) p: The unit digits of an integer is zero.
q: It is divisible by 5.
(ii) p: A natural number is odd.
q: Natural number is not divisible by 2.
(iii) p: A triangle is an equilateral triangle.
q: All three sides of a triangle are equal.
Sol:(i) p ⟷ q: The unit digit of on integer is zero, if and only if it is divisible by 5.
(ii) p ⟷ q: A natural number is odd if and only if it is not divisible by 2.
(iii) p ⟷q: A triangle is an equilateral triangle if and only if all three sides of triangle are equal.

Q9. Write down the contra positive of the following statements.
(î) lf x =y and y=3,then x = 3.
(ii) If n is a natural number, then n is an integer.
(iii) If all three sides ola triangle are equal, then the triangle is equilateral.
(iv) If x andy are negative integers, then .ty is positive.
(v) If natural number n is divisible by 6, then n is divisible by 2 and 3.
(vi) Jf it snows, then the weather will be cold.
(vii) lix is a real number such that O <x < 1, then x2 < 1
Sol: (i) If x ≠ 3, then x ≠y or y ≠ 3.
(ii) 1f n is not an integer, then it is not a natural number.
(iii) If the triangle is not equilateral, then all three sides of the triangle are not equal.
(iv) If xv is not positive integer, then either x or y is not negative integer.
(v) If natural number n is not divisible by 2 or 3, then n is not divisible by 6.
(vi) The weather will not be cold, if it does not snow.
(vii) lf x² is not less than I, thenx is not a real number such that O <x <1.

Q10. Write down the converse of following statements.
(i) If a rectangle lR’ is a square, then R is a rhombus.
(ii) If today is Monday, then tomorrow is Tuesday.
(iii) If you go to Agra, then you must visit Taj Mahal.
(iv) If sum of squares of two sides of a triangle is equal to the square of third side of a triangle, then the triangle is right angled.
(v) If all three angles of a triangle are equal, then the triangle is equilateral.
(vi) If x : y = 3 : 2, then 2x = 3y.
(vii) If S’ is a cyclic quadrilateral, then the opposite angles of S are supplementary.
(viii) If x is zero, then x is neither positive nor negative.
(ix) If two triangles are similar, then the ratio of their corresponding sides are equal.
Sol: (i) If the rectangle ‘S’ is rhombus, then it is square.
(ii) It tomorrow is Tuesday, then today is Monday.
(iii) If you must visit Taj Mahal, you go to Agra.
(iv) If the triangle is right angle, then the sum of squares of two sides of a triangle is equal to the square of third side.
(v) If the triangle is equilateral, then all three angles of triangle are equal.
(vi) If 2x = 3y, thenx :y = 3:2
(vii) If the opposite angles of a quadrilateral are supplementary, then S is cyclic.
(viii) If x is neither positive nor negative, then x is 0.
(ix) If the ratio of corresponding sides of two triangles are equal, then triangles are similar

Q11. Identify the quantifiers in the following statements.
(i) There exists a triangle which is not equilateral.
(ii) For all real numbers x and y, xy= yx.
(iii) There exists a real number which is not a rational number.
(iv) For every natural number x, x + 1 is also a natural number.
(v) For all real numbers x with x > 3, x² is greater than 9.
(vi) There exists a triangle which is not an isosceles triangle.
(vii) For all negative integers x, x³ is also a negative integers.
(viii) There exists a statement in above statements which is not true.
(ix) There exists an even prime number other than 2.
(x) There exists a real number x such that x² + 1 = 0.
Sol: Quantifier are the phrases like ‘There exists’ and ‘For every1, ‘For all’ etc.
(i) There exists
(ii) For all
(iii) There exists
(iv) For every
(v) For all
(vi) There exists
(vii) For all (viii) There exists
(ix) There exists
(x) There exists

Q12. Prove by direct method that for any integer ‘n’ ,n³ – n is always even.

Q13. Check validity of the following statements.
(i) p: 125 is divisible by 5 and 7.
(ii) q: 131 is a multiple of 3 or 11.
Sol: (i) We have,P : 125 is divisible by 5 and 7.
Let q: 125 is divisible by 5.
r: 125 is divisible by 7. q is true, r is false.
=> q ⋀ r is false.
[since, p ⋀ q has the truth value F (false) whenever eitherp or q or both have the truth value F]
Hence, p is not valid.
(ii) We have,  p: 131 is a multiple of 3 or 11.
Let q: 131 is multiple of 3.
r: 131 is a multiple of 11.
p is true, r is false.
=> p v r is true.
[since, p v q has the truth value T (true) whenever either p or q or both have the truth value T]
Hence, q is valid.

Q14. Prove the following statement by contradiction method.
p: The sum of an irrational number and a rational number is irrational.

Q15. Prove by direct method that for any real number x, y if x = y, then x ²=y²

Sol: Let p: x = y; x, y∈ R
On squaring both sides, we get
x² =y² :q
p ⟹q
Hence, proved.

Q16. Using contra positive method prove that, if n² is an even integer, then n is also an even integer.
Sol: Let p: n² is an even integer. q: n is also an even integer.
Let ~p is true i.e., n is not an even integer.
=> n² is not an even integer. [Since square of an odd integer is odd]
=> ~ p is true.
Therefore, ~q is true which provides that ~p is true.
Hence proved.

Objective Type Questions
Q17. Which of the following is a statement?
(a) x is a real number (b) Switch off the fan
(c) 6 is a natural number (d) Let me go
Sol: (c) As we know that a statement is a sentence which is either true or false.
6 is a natural number; this is true.
Hence, it is a statement.

Q18. Which of the following is not a statement.
(a) Smoking is injurious to health
(b) 2 + 2 = 4
(c) 2 is the only even prime number
(d) Come here
Sol: (d) No sentence can be called a statement, if it is an order. So, ‘Come here’ is not a statement.

Q19. The connective in the statement ‘2 + 7>9or2 + 7<9’is
(a) and
(b) or
(c) >
(d) <
Sol: (b) In ‘2 + 7 > 9 or 2 + 7 < 9’, or is the connective.

Q20. The connective in the statement “Earth revolves round the Sun and Moon is a satellite of earth” is
(a) or
(b) Earth
(c) Sun
(d) and
Sol: (d) Connective word is ‘and’.

Q21. The negation of the statement “A circle is an ellipse” is
(a) An ellipse is a circle
(b) An ellipse is not a circle
(c) A circle is not an ellipse
(d) A circle is an ellipse
Sol: (c) Let p: A circle is an ellipse.
~p: A circle is not an ellipse.

Q22. The negation of the statement “7 is greater than 8” is
(a) 7 is equal to 8
(b) 7 is not greater than 8
(c) 8 is less than 7
(d) None of these
Sol: (b) Letp: 7 is greater than 8.
~p: 7 is not greater than 8

Q23. The negation of the statement “72 is divisible by 2 and 3” is
(a) 72 is not divisible by 2 or 72 is not divisible by 3
(b) 72 is not divisible by 2 and 72 is not divisible by 3
(c) 72 is divisible by 2 and 72 is not divisible by 3
(d) 72 is not divisible by 2 and 72 is divisible by 3
Sol: (a) We have, p: 72 is divisible by 2 and 3.
Let q: 72 is divisible by 2.
r: 72 is divisible by 3.
~q: 72 is not divisible by 2.
~r. 72 is not divisible by 3.
~{q ⋀f ) -~q v ~r
⟹ 72 is not divisible by 2 or 72 is not divisible by 3.

Q24. The negation of the statement “Plants take in C0₂ and give out 0₂” is
(a) Plants do not take in C0₂ and do not given out 0₂
(b) Plants do not take in C0₂ or do not give out 0₂
(c) Plants take is C0₂ and do not give out 0₂
(d) Plants take in C0₂ or do not give out 0₂
Sol: (b) Now, p: Plants take in C0₂ and give out 0₂.
Let q: Plants take in C0₂.
r: Plants give out 0₂.
~q: Plants do not take in C0₂.
~r: Plants do not give out 0₂.
~(q ∧ r): Plants do not take in C0₂ or do not give out 0₂.

Q25. The negative of the statement “Rajesh or Rajni lived in Bangalore” is
(a) Rajesh lives in Bangalore and Rajni did not live in Bangalore
(b) Rajesh did not live in Bangalore and Rajni did not live in Bangalore
(c) Rajesh did not live in Bangalore or Rajni did not live in Bangalore
Sol: (c) We have, p: Rajesh or Rajni lived in Bangalore.
and                q: Rajesh lived in Bangalore.
r: Rajni lived in Bangalore.
~q: Rajesh did not live in Bangalore.
~r. Rajni did not live in Bangalore.
~ (q v r): Rajesh did not live in Bangalore and Rajni did not live in Bangalore.

Q26. The negation of the statement “101 is not a multiple of 3” is
(a) 101 is a multiple of 3
(b) 101 is a multiple of 2
(c) 101 is an odd number                     
(d) 101 is an even number
Sol: (a) Let p: 101 is not a multiple of 3.
~p: 101 is a multiple of 3,

Q27. The contra positive of the statement
“If 7 is greater than 5, then 8 is greater than 6” is
(a) If 8 is greater than 6, then 7 is greater than 5
(b) If 8 is not greater than 6, then 7 is greater than 5
(c) If 8 is not greater than 6, then 7 is not greater than 5
(d) If 8 is greater than 6, then 7 is not greater than 5
Sol: (c) Letp: 7 is greater than 5.
and q: 8 is greater than 6.
∴P→ q
~p: 7 is not greater than 5.
~q: 8 is not greater than 6.
(~q) → (~p) i.e., if 8 is not greater than 6, then 7 is not greater than 5.

Q28. The converse of the statement “If x > y, then x + a > y + a” is
(a)    If x <y, then x + a <y + a      
(b) If x + a >y + a, then x>y
(c)    If x <y, then x+ a <y + a        
(d) If x >y, then x + a<y+ a
Sol: (b) Let p: x >y
q:x + a>y + a
P ⟶ q
Converse of the above statement is:
q⟶P
i.e., If x + a > y + a, then x>y

Q29. The converse of the statement “If sun is not shining, then sky is filled with clouds” is
(a) If sky is filled with clouds, then the Sun is not shining
(b) If Sun is shining, then sky is filled with clouds
(c) If sky is clear, then Sun is shining
(d) If Sun is not shining, then sky is not filled with clouds
Sol: (a) Let p: Sun is not shining.
and q:Sky is filled with clouds.
Converse of the above statement p → q is q → p.
If sky is filled with clouds, then the Sun is not shining.

Q30. The contra positive of the statement “If p, then q” is
(a) if q, then p                                        
(b) if p, then ~q
(c) if ~q, then ~p                                  
(d) if ~p, then ~q
Sol:(c) p → q                                           ‘
If p, then q
Contra positive of the statement p → q is (~q) →(~ p).
If ~q, then ~p.

Q31. The statement “If x² is not even, then x is not even” is converse of the statement
(a) If x² is odd, then x is even
(b) If x is not even, then x² is not even
(c) If x is even, then x² is even
(d) If x is odd, then x² is even
Sol: (b) Let p: x² is not even.
and q: x is not even.
Converse of the statement p →q is q → p. i.e.,
If x is not even, then x² is not even.

Q32. The contra positive of statement ‘If Chandigarh is capital of Punjab, then Chandigarh is in India’ is
(a) If Chandigarh is not in India, then Chandigarh is not the capital of Punjab
(b) If Chandigarh is in India, then Chandigarh is Capital of Punjab
(c) If Chandigarh is not capital of Punjab, then Chandigarh is not capital of India
(d) If Chandigarh is capital of Punjab, then Chandigarh is not is India
Sol: (a) Let p: Chandigarh is capital of Punjab.
and q: Chandigarh is in India.
~ p: Chandigarh is not capital of Punjab.
~q: Chandigarh is not in India.
Contra positive of the statement p → q
if (~q), then (~p).
It Chandigarh is not in India, then Chandigarh is not the capital of Punjab.

Q33. Which of the following is the conditional p → q?
(a) q is sufficient for p                         
(b) p is necessary for q
(c) p only if q                                         
(d) if q then p
Sol: (c) ‘p → q is same as ‘p only if q’.

Q34. The negation of the statement “The product of 3 and 4 is 9” is
(a) it is false that the product of 3 and 4 is 9
(b) the product of 3 and 4 is 12
(c) the product of 3 and 4 is not 12
(d) it is false that the product of 3 and 4 is not 9
Sol: (a) The negation of the above statement is ‘It is false that the product of 3 and 4 is 9’.

Q35. Which of the following is not a negation of “A nature number is greater than zero”
(a) A natural number is not greater than zero
(b) It is false that a natural number is greater than zero
(c) It is false that a natural number is not greater than zero
(d) None of the above
Sol: (c) The false negation of the given statement is “It is false that a natural number is not greater than zero”.

Q36. Which of the following statement is a conjunction?
(a) Ram and Shyam are friends
(b) Both Ram and Shyam are tall
(c) Both Ram and Shyam are enemies
(d) None of the above
Sol: (d) If two simple statements p and q are connected by the word ‘and’, then the resulting compound statement p and q is called a conjunction ofp and q. Here, none of the given statement is conjunction.

Q37. State whether the following sentences are statements or not.
(i) The angles opposite to equal sides of a triangle are equal.
(ii) The moon is a satellites of Earth.
(iii) May God bless you.
(iv) Asia is a continent.
(v) How are you? ,
Sol: (i) It is a statement.
(ii) It is a statement,
(iii) It is not a statement, since it is an exclamations.
(iv) It is a statement.
(v) It is not a statement, since it is a question.

NCERT Exemplar ProblemsMathsPhysicsChemistryBiology

We hope the NCERT Exemplar Class 11 Maths Chapter 14 Mathematical Reasoning help you. If you have any query regarding NCERT Exemplar Class 11 Maths Chapter 14 Mathematical Reasoning, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Class 11 Maths Chapter 1 Sets are part of NCERT Exemplar Class 11 Maths. Here we have given NCERT Exemplar Class 11 Maths Chapter 1 Sets.

NCERT Exemplar Class 11 Maths Chapter 1 Sets

Short Answer Type Questions
Q1. Write the following sets in the roaster from

Q2. Write the following sets in the roaster form:

Q3. If Y = {x\x is a positive factor of the number 2^P(2^P – 1), where 2^P – 1 is a prime number}. Write Y in the roaster form.

Sol: Y- {x | x is a positive factor of the number 2^P-1 (2^P – 1), where 2^P – 1 is a prime number}.
So, the factors of 2^P-1 are 1,2,2²,2³,…, 2^{P- 1}.
Y= {1,2,2²,2³, …,2^p-1,2^p-1}

Q4. State which of the following statements are true and which are false. Justify your answer.
(i) 35 ∈ {x | x has exactly four positive factors}.
(ii) 128 e {y | the sum of all the positive factorsofy is 2y}
(iii) 3∉{x|x⁴-5x³ + 2jc²-112x + 6 = 0}
(iv) 496 ∉{y | the sum of all the positive factors of y is 2y}.
Sol: (i) The factors of 35 are 1, 5, 7 and 35. So, 35 is an element of the set. Hence, statement is true.

(ii) The factors of 128 hre 1,2,4, 8, 16, 32, 64 and 128.
Sum of factors = 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 = 255 * 2 x 128 Hence, statement is false.

(iii) We have, x⁴ – 5x³ + 2x² – 1 12jc + 6 = 0 Forx = 3, we have
(3)⁴ – 5(3)³ + 2(3)² – 112(3) + 6 = 0
=> 81 – 135 + 18-336 + 6 = 0
=>    -346 = 0, which is not true.
So 3 is not an element of the set
Hence, statement is true.

(iv) 496 = 2⁴ x 31
So, the factors of 496 are 1,2,4, 8, 16,31,62, 124,248 and 496.
Sum of factors = 1 +2 + 4 + 8+ 16 + 31+62+124 + 248 + 496 = 992 = 2(496)
So, 496 is the element of the set Hence, statement is false

Q5. Given L, = {1,2, 3,4},M= {3,4, 5, 6} and N= {1,3,5}
Verify that L-(M⋃N) = (L-M)⋂(L-N)
Sol: Given L,= {1,2, 3,4}, M= {3,4,5,6} and N= {1,3,5}
M⋃N= {1,3,4, 5,6}
L – (M⋃N) = {2}
Now, L-M= {1, 2} and L-N= {2,4}
{L-M) ⋂{L-N)= {2}
Hence, L-{M⋃N) = {L-M) ⋂ (L-N).

Q6. If A and B are subsets of the universal set U, then show that
(i) A⊂ A∪ B
(ii) A⊂B⟺A∪B = B
(iii) (A∩B)⊂ A



Q7. Given that N= {1,2,3, …, 100}. Then write
(i) the subset of N whose elements are even numbers.
(ii) the subset of N whose element are perfect square numbers.
Sol: We have, N= {1,2, 3,4,…, 100}
(i) subset of N whose elements are even numbers = {2,4, 6, 8,…, 100}
(ii) subset of N whose elements are perfect square = {1,4, 9, 16, 25, 36,49, 64, 81, 100}

Q8. If X= {1, 2, 3}, if n represents any member of X, write the following sets containing all numbers represented by
(i) 4n                     
(ii) n + 6                 
(iii) n/2
(iv) n-1 

Q9.If Y= {1,2,3,…, 10}, and a represents any element of Y, write the following sets, containing all the elements satisfying the given conditions.

Q10. A, B and C are subsets of Universal Set If A = {2, 4, 6, 8, 12, 20}, B= {3,6,9,12,15}, C= {5,10,15,20} and U is the set of all whole numbers, draw a Venn diagram showing the relation of U, A, B and C.

Q11. Let U be the set of all boys and girls in a school, G be the set of all girls in the school, B be the set of all boys in the school, and S be the set of all students in the school who take swimming. Some, but not all, students in the school take swimming. Draw a Venn diagram showing one of the possible interrelationship among sets U, G, B and S.

Q12. For all sets A, B and C, show that (A – B) ∩ (C – B) = A – (B ∪ C)

Instruction for Exercises 13-17: Determine whether each of the statements in these exercises is true or false. Justify your answer.

 

Q13. For all sets A and B, (A – B)∪ (A∩ B) = A
Sol: True
L.H.S. = (A-B) ∪ (A∩B) = [(A-B) ∪A] ∩ [(A – B) ∪B]
= A∩ (A-B) = A= R.H.S.
Hence, given statement is true.

Q14. For all sets A, B and C, A – (B-C) = (A- B)-C
Sol: False 

Q15. For all sets A, B and C, if A ⊂ B, then  A ∩C<⊂B ∩C

Sol: True
Let          x ∈A∩C
=> x ∈ A and x∈ C

Q16. For all sets A, B and C, if A⊂ B, then A ∪ C⊂ B ∪ C
Sol: True

Q17. For all sets A, B and C, if A⊂ C and B ⊂ C,then A∪ B ⊂ C

Instruction for Exercises 18-22: Using properties of sets prove the statements given in these exercises.

Q18. For all sets A and B, A ∪ (B -A) = A ∪ B

Q19. For all sets A and B, A – (A – B) = A ∩ B

Q20. For all sets A and B, A – (A ∩ B) = A – B

Q21. For all sets A and B,(A ∪ B)- B = A-B

Long Answer Type Questions

Q23. Let A, B and C be sets. Then show that A ∩ (B ∪ C) = (A ∩ B)∪ (A ∩ C).

From (i) and (ii), we get .                              .
A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

Q24. Out of 100 students; 15 passed in English, 12 passed in Mathematics, 8 in Science,6 in English and Mathematics, 7 in Mathematics and Science; 4 in English and Science; 4 in all the three. Find how many passed
(i) in English and Mathematics but not in Science
(ii) in Mathematics and Science but not in English
(iii) in Mathematics only
(iv) in more than one subject only
Sol. Let M be the set of students who passed in Mathematics, E be the set of students who passed in English and S be the set of students who passed in Science.

Given n (U) = 100,
n(E) = 15, n(M) = 12, n(S) = 8,
n(E ∩ M) = 6, n(M ∩S) = 7, n(E ∩ S) — 4, and n(E ∩M ∩ S) = 4,

Number of students passed in English and Mathematics but not in Science = b = 2
(ii) Number of students passed in Mathematics and Science but not in English = d = 3
(iii) Number of students passed in Mathematics only = e = 3
(iv) Number of students passed in more than one subject = a + b + c + d =4+2+0+3=9

Q25. In a class of 60 students, 25 students play cricket and 20 students play tennis, and 10 students play both the games. Find the number of students who play neither.
Sol: Let C be the set of students who play cricket and T be the set of students who play tennis.
n(U) = 60, n(C) = 25, n(T) = 20, and n(C ∩ T) = 10
n(C ∪ T) = n(C) + n(T) – n(C n T) = 25 + 20 – 10 = 35

Q26. In a survey of 200 students of a school, it was found that 120 study Mathematics, 90 study Physics and 70 study Chemistry, 40 study Mathematics and Physics, 30 study Physics and Chemistry, 50 study Chemistry and Mathematics and 20 none of these subjects. Find the number of students who study all the three subjects.
Sol: Let M be the set of students who study Mathematics, P be the set of students who study E Physics and C be the set of students who study Chemistry

Q27. In a town of 10,000 families it was found that 40% families buy newspaper A, 20% families buy newspaper B, 10% families buy newspaper C, 5% families buy A and B, 3% buy B and C and 4% buy A and C. If 2% families buy all the three newspapers. Find
(a) The number of families which buy newspaper A only.
(b) The number of families which buy none of A, B and C.
Sol: Let A be the set of families which buy newspaper A, B be the set of families which buy newspaper B and C be the set of families which buy newspaper C. The

Number of families which buy none of A, B and C = 10000 x (40/100)

Q28. In a group of 50 students, the number of students studying French, English, Sanskrit were found to be as follows: French = 17, English = 13, Sanskrit = 15, French and English = 09, English and Sanskrit = 4,French and Sanskrit = 5, English, French and Sanskrit = 3. Find the number of students who study
(i) French only 
(ii) English only 
(iii) Sanskrit only 
(iv) English and Sanskrit 
(v) French and  Sanskrit but not English
(vi) French and  English but not Sanskrit
(vii)     at least one  of the three languages
(viii) none of the  three languages but not French

Sol: Let F be the set of students who study French, E be the set of students who study English and S be the set of students who study Sanskrit.
Then, n{U) = 50, n(F) =17, n{E) = 13, and n{S) = 15,
n(F ∩ E) = 9, n(E ∩ S) = 4, n(F ∩ S) = 5, n(F ∩ E ∩ S) = 3

(i) Number of students studying French only = e = 6
(ii) Number of students studying English only = g = 3
(iii) Number of students studying Sanskrit only =f= 9
(iv) Number of students studying English and Sanskrit but not French = c = 1
(v) Number of students studying French and Sanskrit but not English = d = 2
(vi) Number of students studying French and English but not Sanskrit = b = 6
(vii) Number of students studying at least one of the three languages = a + b + c + d + e+f+g = 30
(viii) Number of students studying none of the three languages but not French = 50-30 = 20

Q30. Two finite sets have m and n elements. The number of subsets of the first set is 112 more than that of the second set. The values of m and n are, respectively, (a) 4,7 (b) 7,4 (c) 4,4 (d) 7, 7

Q32. Let F₁ be the set of parallelograms, F₂ the set of rectangles, F₃ the set of rhombuses, F₄ the set of squares and F₅ the set of trapeziums in a plane. Then F₁may be equal to
(a) F₂ ∩F₃                                              
(b) F₃ ∩F₄
(c) F₂ u F_s
(d) F₂ ∪ F₃ ∪ F₄ ∪ F₁
Sol: (d) Every rectangle, rhombus, square in a plane is a parallelogram but every trapezium is not a parallelogram.
F₁ = F₂ ∪ F₃ ∪ F₄ ∪ F₁

Q33. Let S = set of points inside the square, T = the set of points inside the triangle and C = the set of points inside the circle. If the triangle and circle intersect each other and are contained in a square, then

Q34. Let R be set of points inside a rectangle of sides a and b (a, b> 1) with two sides along the positive direction of x-axis andy-axis. Then

Q35. In a class of 60 students, 25 students play cricket and 20 students play tennis, and 10 students play both the games. Then, the number of students who play neither is
(a) 0 (b) 25 (c) 35 (d) 45
Sol: Let C be the set of students who play cricket and T be the set of students who play tennis.
n(U) = 60, n(C) = 25, n(T) = 20, and n(C ∩ T) = 10
n(C ∪ T) = n(C) + n(T) – n(C n T) = 25 + 20 – 10 = 35

Q36. In a town of 840 persons, 450 persons read Hindi, 300 read English and 200 read both. Then the number of persons who read neither is
(a) 210 (b) 290 (c) 180 (d) 260
Sol: (b) Let H be the set of persons who read Hindi and E be the set of persons who read English.

NCERT Exemplar ProblemsMathsPhysicsChemistryBiology

We hope the NCERT Exemplar Class 11 Maths Chapter 1 Sets help you. If you have any query regarding NCERT Exemplar Class 11 Maths Chapter 1 Sets, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Class 11 Maths Chapter 2 Relations and Functions are part of NCERT Exemplar Class 11 Maths. Here we have given NCERT Exemplar Class 11 Maths Chapter 2 Relations and Functions.

NCERT Exemplar Class 11 Maths Chapter 2 Relations and Functions

Relations and Functions

Short Answer Type Questions
Q1. If A = {-1, 2, 3 } and B = {1, 3}, then determine
(i) AxB (ii) BxC (c) BxB (iv) AxA
Sol: We have A = {-1,2,3} and B = {1,3}
(i) A x B = {(-1, 1), (-1, 3), (2, 1), (2, 3), (3,1), (3, 3)}
(ii) BxA = {( 1, -1), (1, 2), (1,3), (3,-1), (3,2), (3, 3)}
(iii) BxB= {(1,1), (1,3), (3,1), (3, 3)}
(iv) A xA = {(-1, -1), (-1, 2), (-1, 3), (2, -1), (2, 2), (2, 3), (3, -1), (3, 2), (3,3)}

Q2. If P = {x : x < 3, x e N}, Q= {x : x≤2,x ∈ W}. Find (P∪ Q) x (P∩ Q), where W is the set of whole numbers.
Sol: We have, P={x: x<3,x ∈ N} = {1,2}
And Q = {x :x≤ 2,x∈ W] = {0,1,2}
P∪Q= {0, 1,2} and P ∩ Q= {1,2}
(P ∪ Q) x (P ∩ Q) = {0,1, 2} x {1,2}
= {(0,1), (0, 2), (1,1), (1,2), (2,1), (2, 2)}

Q3. lfA={x:x∈ W,x < 2}, 5 = {x : x∈N, 1 <.x < 5}, C= {3, 5}. Find
(i) Ax(B∩Q) (ii) Ax(B∪C)
Sol: We have, A = {x :x∈ W,x< 2} = {0, 1};
B = {x : x ∈ N, 1 <x< 5} = {2, 3,4}; and C= {3, 5}

(i) B∩ C = {3}
A x (B ∩ C) = {0, 1} x {3} = {(0, 3), (1, 3)}

(ii) (B ∪ C) ={2,3,4, 5}
A x (B ∪ C) = {0, 1} x {2, 3,4, 5}
= {(0,2), (0,3), (0,4), (0,5), (1,2), (1,3), (1,4), (1, 5)}

Q4. In each of the following cases, find a and b. (2a + b, a – b) = (8, 3) (ii) {a/4, a – 2b) = (0, 6 + b)
Sol: (i) We have, (2a + b,a-b) = (8,3)
=> 2a + b = 8 and a – b = 3
On solving, we get a = 11/3 and b = 2/3

 

Q5. Given A = {1,2,3,4, 5}, S= {(x,y) :x∈ A,y∈ A}.Find the ordered pairs which satisfy the conditions given below
x+y = 5 (ii) x+y<5 (iii) x+y>8
Sol: We have, A = {1,2, 3,4, 5}, S= {(x,y) : x ∈ A,y∈ A}
(i) The set of ordered pairs satisfying x + y= 5 is {(1,4), (2,3), (3,2), (4,1)}
(ii) The set of ordered pairs satisfying x+y < 5 is {(1,1), (1,2), (1,3), (2, 1), (2,2), (3,1)}
(iii) The set of ordered pairs satisfying x +y > 8 is {(4, 5), (5,4), (5, 5)}.

Q6. Given R = {(x,y) : x,y ∈ W, x² + y² = 25}. Find the domain and range of R
Sol: We have, R = {(x,y):x,y∈ W, x² + y² = 25}
= {(0,5), (3,4), (4, 3), (5,0)}
Domain of R = Set of first element of ordered pairs in R = {0,3,4, 5}
Range of R = Set of second element of ordered pairs in R = {5,4, 3, 0}

Q7. If R₁ = {(x, y)| y = 2x + 7, where x∈ R and -5 ≤ x ≤ 5} is a relation. Then find the domain and range of R₁.
Sol: We have, R₁ = {(x, y)|y = 2x + 7, where x∈ R and -5 ≤x ≤ 5}
Domain of R₁ = {-5 ≤ x ≤ 5, x ∈ R} = [-5, 5]
x ∈ [-5, 5]
=> 2x ∈ [-10,10]
=>2x + 7∈ [-3, 17]
Range is [-3, 17]

Q8. If R₂ = {(x, y) | x and y are integers and x² +y² = 64} is a relation. Then find R₂
Sol: We have, R₂ = {(x, y) | x and y are integers and x² + y² – 64}
Clearly, x² = 0 and y² = 64 or x² = 64 andy² = 0
x = 0 and y = ±8
or x = ±8 and y = 0
R₂ = {(0, 8), (0, -8), (8,0), (-8,0)}

Q9. If R₃ = {(x, |x|) | x is a real number} is a relation. Then find domain and range
Sol: We have, R₃ = {(x, |x)) | x is real number}
Clearly, domain of R₃ = R
Now, x ∈ R and |x| ≥ 0 .
Range of R₃ is [0,∞)

Q10. Is the given relation a function? Give reasons for your answer.
(i) h={(4,6), (3,9), (-11,6), (3,11)}
(ii) f = {(x, x) | x is a real number}
(iii) g = {(n, 1 In)| nis a positive integer}
(iv) s= {(n, n²) | n is a positive integer}
(v) t= {(x, 3) | x is a real number}
Sol: (i) We have, h = {(4,6),(3,9), (-11,6), (3,11)}.
Since pre-image 3 has two images 9 and 11, it is not a function.
(ii) We have, f = {(x, x) | x is a real number}
Since every element in the domain has unique image, it is a function.
(iii) We have, g= {(n, 1/n) | nis a positive integer}
For n, it is a positive integer and 1/n is unique and distinct. Therefore,every element in the domain has unique image. So, it is a function.
(iii) We have, s = {(n, n²) | n is a positive integer}
Since the square of any positive integer is unique, every element in the domain has unique image. Hence, ibis a function.
(iv) We have, t = {(x, 3)| x is a real number}.
Since every element in the domain has the image 3, it is a constant function.

Q11. If f and g are real functions defined byf( x) = x² + 7 and g(x) = 3x + 5, find each of the following

Q12. Let f and g be real functions defined by f(x) = 2x+ 1 and g(x) = 4x – 7.
(i) For what real numbers x,f(x)= g(x)?
(ii) For what real numbers x,f (x) < g(x)?
Sol: We have,f(x) = 2x + 1 and g(x) = 4x-7
(i) Now f (x) = g(x)
=> 2x+l=4x-7
=> 2x = 8 =>x = 4
(ii) f (x) < g(x)
=> 2x + 1 < 4x – 7
=> 8 < 2x
=> x > 4

Q13. If f and g are two real valued ftmctions defined as f(x) = 2x + 1, g(x) = x² + 1, then find.

Q14. Express the following functions as set of ordered pairs and determine their range.
f:X->R,f{x) =  x³ + 1, where X= {-1,0, 3, 9, 7}
Sol: We have, f:X→ R,flx) = x³ + 1.
Where X = {-1, 0, 3, 9, 7}
Now f (-l) = (-l)³+1 =-l + 1 =0
f(0) = (0)³+l=0+l = l
f(3) = (3)³ + 1 = 27 + 1 = 28
f(9) = (9)³ + 1 = 729 + 1 = 730
f(7) = (7)³ + 1 = 343 + 1 = 344
f= {(-1, 0), (0, 1), (3, 28), (9, 730), (7, 344)}
Range of f= {0, 1, 28, 730, 344}

Q15. Find the values of x for which the functions f(x) = 3x² -1 and g(x) = 3+ x are equal.
Sol: f(x) = g(x)
=> 3x²-l=3+x => 3x²-x-4 = 0 => (3x – 4)(x+ 1) – 0
x= -1,4/3

Q16. Is g = {(1, 1), (2, 3), (3, 5), (4, 7)} a function? Justify. If this is described by the relation, g(x) = x +, then what values should be assigned to and ?
Sol:We have, g = {(1, 1), (2, 3), (3, 5), (4,7)}
Since, every element has unique image under g. So, g is a function.
Now, g(x) = x + For (1,1), g(l) = a(l) + P
=>             l = +            (i)
For (2, 3), g(2) = (2) +
=>             3 = 2 +         (ii)
On solving Eqs. (i) and (ii), we get = 2, = -l
f(x) = 2x-1
Also, (3, 5) and (4, 7) satisfy the above function.

Q17. Find the domain of each of the following functions given by

Q18. Find the range of the following functions given by

 

Q19. Redefine the function f(x) = |x-2| + |2+x| , -3 ≤x ≤3

Q21. Let f (x) = √x and g(x) = xbe two functions defined in the domain R⁺ ∪ {0}. Find
(i) (f+g)(x)        
(ii) (f-g)(x)
(iii) (fg)(x)
(iv) f/g(x)

 

Q23. If f( x)= y = ax-b/ cx-a then prove that f (y) = x

Objective Type Questions
Q24. Let n(A) = m, and n(B) = n. Then the total number of non-empty relations that can be defined from A to B is
(a) mⁿ                     
(b) n^m– 1                  
(c) mn – 1               
(d) 2^mn– 1

Sol: (d) We have, n(A) = m and n(B) = n
n(A xB) = n(A). n(B) = mn
Total number of relation from A to B = Number of subsets of AxB = 2^mn
So, total number of non-empty relations = 2^mn – 1

Q25. If [x]² – 5[x] + 6 = 0, where [. ] denote the greatest integer function, then
(a) x ∈ [3,4]
(b) x∈ (2, 3]            
(c) x∈ [2, 3]           
(d) x ∈ [2, 4)
Sol: (d) We have [x]² – 5[x] + 6 = 0 => [(x – 3)([x] – 2) = 0
=> [x] = 2,3 .
For [x] = 2, x ∈ [2, 3)
For [x] = 3, x ∈ [3,4)
x ∈ [2, 3) u [3,4)
Or x ∈ [2,4)

Q29. If fx) ax+ b, where a and b are integers,f(-1) = -5 and f(3) – 3, then a and b are equal to
(a) a = -3, b =-1
(b) a = 2, b =-3
(c) a = 0, b = 2
(d) a = 2, b = 3

Fill in the Blanks Type Questions

Q36. Let f and g be two real functions given by  f= {(0, 1), (2,0), (3,.-4), (4,2), (5, 1)}
g= {(1,0), (2,2), (3,-1), (4,4), (5, 3)}  then the domain of f  x g is given by________ .
Sol: We have, f = {(0, 1), (2, 0), (3, -4), (4, 2), (5,1)} and g= {(1, 0), (2, 2), (3, 1), (4,4), (5, 3)}
Domain of  f = {0,2, 3, 4, 5}
And Domain of g= {1, 2, 3,4, 5}
Domain of (f x g) = (Domain of f) ∩ (Domain of g) = {2, 3,4, 5}

Matching Column Type Questions

Q37. Let f= {(2,4), (5,6), (8, -1), (10, -3)} andg = {(2, 5), (7,1), (8,4), (10,13), (11, 5)} be two real functions. Then match the following:

True/False Type Questions

Q38. The ordered pair (5,2) belongs to the relation R ={(x,y): y = x – 5, x,y∈Z}
Sol: False
We have, R = {(x, y): y = x – 5, x, y ∈ Z}
When x = 5, then y  = 5-5=0 Hence, (5, 2) does not belong to R.

Q39. If P = {1, 2}, then P x P x P = {(1, 1,1), (2,2, 2), (1, 2,2), (2,1, 1)}
Sol:False
We have, P = {1, 2} and n(P) = 2
n(P xPxP) = n(P) x n(P) x n(P) = 2 x 2 x 2
= 8 But given P x P x P has 4 elements.

Q40. If A= {1,2, 3}, 5= {3,4} and C= {4, 5, 6}, then (A x B) ∪ (A x C) = {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3,4), (3, 5), (3,6)}.
Sol: True
We have.4 = {1,2, 3}, 5= {3,4} andC= {4,5,6}
AxB= {(1, 3), (1,4), (2, 3), (2,4), (3, 3), (3,4)}
And A x C =  {(1,4), (1, 5), (1, 6), (2,4), (2, 5), (2, 6), (3,4), (3, 5), (3, 6)}
(A x B)∪(A xC)= {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3,3), (3,4), (3, 5), (3,6)}

Q42. If Ax B= {(a, x), (a, y), (b, x), (b, y)}, thenM = {a, b},B= {x, y}.
Sol: True
We have, AxB= {{a, x), {a, y), (b, x), {b, y)}
A = Set of first element of ordered pairs in A x B = {a, b}
B = Set of second element of ordered pairs in A x B = {x, y}

NCERT Exemplar ProblemsMathsPhysicsChemistryBiology

We hope the NCERT Exemplar  Class 11 Maths Chapter 2 Relations and Functions help you. If you have any query regarding NCERT Exemplar Class 11 Maths Chapter 2 Relations and Functions, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Class 11 Maths Chapter 3 Trigonometric Functions are part of NCERT Exemplar Class 11 Maths. Here we have given NCERT Exemplar Class 11 Maths Chapter 3 Trigonometric Functions.

NCERT Exemplar Class 11 Maths Chapter 3  Trigonometric Functions

Class 11 Trigonometry Questions Chapter 3

Trigonometry Class 11 Questions Chapter 3

Trigonometry Class 11 Extra Questions Chapter 3

NCERT Exemplar Class 11 Trigonometry Chapter 3

Q4. If cos (α + ) =4/5 and sin (α- )=5/13 , where α lie between 0 and π/4, then find the value of tan 2α.

Solution:

Trigonometry Questions For Class 11 Chapter 3

Trigonometry Questions Class 11 Chapter 3

Q6. Prove that cos cos /2- cos 3 cos 9/2 = sin 7/2 sin 4 .

Trigonometry Class 11 Important Questions Chapter 3

Q7. If a cos θ + b sin θ =m and a sin θ -b cosθ = n, then show that a² + b²-m² + n²

Sol: We have, a cos θ + b sin θ = m (i)
and a sin θ -bcos θ = n (ii)

Q8. Find the value of tan 22°30′

NCERT Exemplar Class 11 Trigonometry Solutions Chapter 3

Q9. Prove that sin 4A = 4 sin A cos³A – 4 cos A sin³ A.

Sol: L.H.S. = sin 4A
= 2 sin 2A- cos 2A = 2(2 sin A cosA)(cos² A – sin² A)
= 4 sin A • cos³ A – 4 cos A sin³ A = R.H.S.

Q10. If tan + sin = m and tan – sin = n, then prove that m²-n² = 4 sin tan

Sol:We have, tan + sin = m   (i)
And tan -sin =n  (ii)
Now,         m + n = 2 tan
And          m – n = 2 sin.
(m + n)(m -n) = 4 sin 6
tan m² -n² = 4 sin -tan

Trigonometric Functions Class 11 Extra Questions Chapter 3

Q11. If tan (A + B) =p and tan (A – B) = q, then show that tan 2A = p+q / 1 – pq

Sol: We have tan (A + B) =p and tan (A – B) = q
tan2A = tan [(A + B) + (A-B)]

Q12. If cos + cos = 0 = sin + sin β, then prove that cos 2 + cos 2β = -2 cos (α + ).

NCERT Exemplar Class 11 Maths Trigonometry Solutions Chapter 3

Q15.  If sin θ+ cos θ =1, then find the general value of θ

Q16. Find the most general value of θ satisfying the equation tan θ = -1 and cos θ = 1/√2 .
Sol: We have tan θ = -1 and cos θ =1/√2 .
So, θ lies in IV quadrant.
θ = 7/4
So, general solution is θ = 7π/4 + 2 n π, n∈ Z

Q17. If cot θ + tan θ = 2 cosec θ, then find the general value of θ
Sol: Given that, cot θ + tan θ = 2 cosec θ

Q18. If 2 sin² θ =3 cos θ, where O≤θ≤2, then find the value of θ

Q19. If sec x cos 5x + 1 = 0, where 0 < x <π/2 , then find the value of x.

Long Answer Type Questions

Q20. If sin(θ + α) = a and sin(θ + β) = b , then prove that cos2(α – β) – 4abcos(α – β) = 1-2a² -2b²

Sol: We have sin(θ + α) = a —(i)
sin(θ + β) = b ——-(ii)

Q22. Find the value of the expression

Q23. If a cos 2+b sin 2 = c has α and β as its roots, then prove that tan α +tan β = 2b/a+c

Q24. If x = sec ϕ-tanϕandy = cosec ϕ + cot ϕ then show that xy + x -y +1=0.

Q25. If lies in the first quadrant and cos =8/17 , then find the value of cos (30° + ) + cos (45° – ) + cos (120° – ).

Q26. Find the value of the expression cos⁴ π/8 + cos⁴ 3π/8  + cos⁴ 5π/8  + cos⁴7π/8

Q27. Find the general solution of the equation 5 cos² +7 sin² -6 = 0.

Q28. Find the general solution of‘the equation sin x – 3 sin 2x + sin 3x = cos x – 3 cos 2x + cos 3x.
Sol: We have, (sin x + sin 3x) – 3 sin 2x = (cos x + cos 3x) – 3 cos 2x
=> 2 sin 2x cos x – 3 sin 2x = 2 cos 2x.cos x – 3 cos 2x
=> sin 2x(2 cos x – 3) = cos 2x(2 cos x – 3)
=> sin 2x = cos 2x (As cos x ≠ 3/2)
=>              tan 2x = 1    => tan 2x = tan π/4
=>              2x = nπ + π/4 , n∈Z
x = nπ/2 +π/8 , n∈Z

Q29. Find the general solution of the equation (√3- l)cos + (√3+ 1)sin = 2.

Objective Type Questions

Q30. If sin + cosec =2, then sin² + cosec² is equal to
(a) 1
(b) 4                          
(c) 2                         
(d) None of these

Q31. If f(x) = cos² x + sec² x, then ‘
(a) f(x) <1             
(b) f(x) = 1              
(c) 2 <f(x) < 1      
(d) fx) ≥ 2

Q32. If tan θ = 1/2 and tan ϕ = 1/3, then the value of θ + ϕ is

Q33. Which of the following is not correct?

(a) sin θ = – 1/5 (b) cos θ = 1                 (c) sec θ = -1/2         (d) tan θ = 20
Sol: (c)
We know that, the range of sec θ is R – (-1, 1).
Hence, sec θ cannot be equal to -1/2

Q34. The value of tan 1° tan 2° tan 3° … tan 89° is
(a) 0
(b) 1
(c) 1/2
(d) Not defined

Sol: (b)
tan 1° tan 2° tan 3° … tan 89°
= [tan 1° tan 2° … tan 44°] tan 45°[tan (90° – 44°) tan (90° – 43°)… tan (90° – 1°)]
= [tan 1° tan 2° … tan 44°] [cot 44° cot 43°……. cot 1°]
= 1-1… 1-1 = 1

Q36. The value of cos 1° cos 2° cos 3° … cos 179° is
(a) 1/√2
(b) 0
(c) 1
(d) -1

Sol: (b)
Since cos 90° = 0, we have
cos 1° cos 2° cos 3° …cos 90°… cos 179° = 0

Q37. If tan θ = 3 and θ lies in the third quadrant, then the value of sin θ is

Q38. The value of tan 75° – cot 75° is equal to

Q39. Which of the following is correct?
(a) sin 1° > sin 1                                     
(b) sin 1° < sin 1
(c) sin l° = sin l
(d) sin l° = π/18° sin 1

 

Sol: We know that, in first quadrant if θ is increasing, then sin θ is also increasing.
∴sin 1° < sin 1 [∵ 1 radian = 57◦30′]

Q41. The minimum value of 3 cos x + 4 sin x + 8 is
(a) 5
(b) 9
(c) 7
(d) 3
Sol: (d)
3 cos x + 4sin x + 8 = 5 (3/5 cos x + 4/5sin x) + 8
= 5(sin α cos x + cos α sin x) + 8
= 5 sin(α + x) + 8, where tan α = 3/4

Q42. The value of tan 3A – tan 2A – tan A is
(a) tan 3A . tan 2A . tan A
(b) -tan 3A .tan 2A . tan A
(c) tan A . tan 2A – tan 2A . tan 3A – tan 3A . tan A
(d) None of these
Sol: (a)
3A= A+ 2A
=> tan 3A = tan (A + 2A)
=> tan 3 A = tanA + tan2A/ 1 – tan A . tan 2A
=> tan A + tan 2A = tan 3A – tan 3A• tan 2A . tan A
=> tan 3 A – tan 2A – tan A = tan 3A . tan 2A . tan A

Q43. The value of sin (45° + )- cos (45° – ) is
(a) 2 cos              
(b) 2 sin              
(c) 1                         
(d) 0
Sol: (d)
sin (45° + ) – cos (45° – ) = sin (45° + ) – sin (90° – (45° – ))
= sin (45° + ) – sin (45°+ ) = 0

Q44. The value of (π/4+ ) cot (π/4- ) is
(a) -1                       
(b)  0  
(c)  1                     
(d)   Not defined

Q46. The value of cos 12° + cos 84° + cos 156° + cos 132° is
(a) 1/2            
(b) 1                       
(c) -1/2            
(d) 1/8

Q47. If tan A = 1/2 and tan B = 1/3 then tan (2A + B) is equal to
(a) 1
(b) 2
(c) 3
(d) 4

Q49. The value of sin 50° – sin 70° + sin 10° is equal to
(a) 1                       
(b) 0                       
(c) 1
(d) 2

Q50. If sin + cos =1, then the value of sin 2 is
(a) 1                      
(b) 1   
(c) 0                        
(d) -1

NCERT Exemplar ProblemsMathsPhysicsChemistryBiology

We hope the NCERT Exemplar Class 11 Maths Chapter 3 Trigonometric Functions help you. If you have any query regarding NCERT Exemplar Class 11 Maths Chapter 3 Trigonometric Functions, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Class 11 Maths Chapter 4 Principle of Mathematical Induction are part of NCERT Exemplar Class 11 Maths. Here we have given NCERT Exemplar Class 11 Maths Chapter 4 Principle of Mathematical Induction.

NCERT Exemplar Class 11 Maths Chapter 4 Principle of Mathematical Induction

Short Answer Type Questions
Q1. Give an example of a statement P(n) which is true for all n≥ 4 but P(l), P(2) and P(3) are not true. Justify your answer.

Sol. Consider the statement P(n): 3n < n!

For n = 1, 3 x 1 < 1!, which is not true
For n = 2, 3 x 2 < 2!, which is not true
For n = 3, 3 x 3 < 3!, which is not true
For n = 4, 3 x 4 < 4!, which is true
For n = 5, 3 x 5 < 5!, which is true

Q2. Give an example of a statement P(n) which is true for all Justify your answer.

Q3. 4ⁿ – 1 is divisible by 3, for each natural number
Sol: Let P(n): 4ⁿ – 1 is divisible by 3 for each natural number n.
Now, P(l): 4¹ – 1 = 3, which is divisible by 3 Hence, P(l) is true.
Let us assume that P(n) is true for some natural number n = k.
P(k): 4^k – 1 is divisible by 3
or               4^k – 1 = 3m, m∈ N  (i)
Now, we have to prove that P(k + 1) is true.
P(k+ 1): 4^k+1 – 1
= 4^k-4-l
= 4(3m + 1) – 1  [Using (i)]
= 12 m + 3
= 3(4m + 1), which is divisible by 3 Thus, P(k + 1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction P(n) is true for all natural numbers n.

Q4. 2³ⁿ – 1 is divisible by 7, for all natural numbers
Sol: Let P(n): 2³ⁿ – 1 is divisible by 7
Now, P( 1): 2³ — 1 = 7, which is divisible by 7.
Hence, P(l) is true.
Let us assume that P(n) is true for some natural number n = k.
P(k): 2^3k – 1 is divisible by 7.
or               2^3k -1 = 7m, m∈ N  (i)
Now, we have to prove that P(k + 1) is true.
P(k+ 1): 2^3(k+1)– 1
= 2^3k.2³– 1
= 8(7 m + 1) – 1
= 56 m + 7
= 7(8m + 1), which is divisible by 7.
Thus, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for all natural numbers n.

Q5. n³ – 7n + 3 is divisible by 3, for all natural numbers
Sol: Let P(n): n³ – 7n + 3 is divisible by 3, for all natural numbers n.
Now P(l): (l)³ – 7(1) + 3 = -3, which is divisible by 3.
Hence, P(l) is true.
Let us assume that P(n) is true for some natural number n = k.
P(k) = K³ – 7k + 3 is divisible by 3
or K³ – 7k + 3 = 3m, m∈ N         (i)
Now, we have to prove that P(k + 1) is true.
P(k+ 1 ):(k + l)³ – 7(k + 1) + 3
= k³ + 1 + 3k(k + 1) – 7k— 7 + 3 = k³ -7k + 3 + 3k(k + l)-6
= 3m + 3[k(k+l)-2]  [Using (i)]
= 3[m + (k(k + 1) – 2)], which is divisible by 3 Thus, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for all natural numbers n.

Q6. 3²ⁿ – 1 is divisible by 8, for all natural numbers
Sol: Let P(n): 3²ⁿ – 1 is divisible by 8, for all natural numbers n.
Now, P(l): 3² – 1 = 8, which is divisible by 8.
Hence, P(l) is true.
Let us assume that, P(n) is true for some natural number n = k.
P(k): 3^2k – 1 is divisible by 8
or               3^2k -1 = 8m, m ∈ N  (i)
Now, we have to prove that P(k + 1) is true.
P(k+ 1): 3^2(k+1)– l
= 3^2k • 3² — 1
= 9(8m + 1) – 1     (using (i))
= 72m + 9 – 1
= 72m + 8
= 8(9m +1), which is divisible by 8 Thus P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for all natural numbers n.

Q7. For any natural number n, 7ⁿ – 2ⁿ is divisible by 5.
Sol: Let P(n): 7ⁿ – 2ⁿ is divisible by 5, for any natural number n.
Now, P(l) = 7¹-2¹ = 5, which is divisible by 5.
Hence, P(l) is true.
Let us assume that, P(n) is true for some natural number n = k.

.’.  P(k) = 7^k -2^k is divisible by 5
or  7^k – 2^k = 5m, m∈ N                                                                           (i)
Now, we have to prove that P(k + 1) is true.
P(k+ 1): 7^k+1 -2^k+1
= 7^k-7-2^k-2
= (5 + 2)7^k -2^k-2
= 5.7^k + 2.7^k-2-2^k
= 5.7^k + 2(7^k – 2^k)
= 5 • 7^k + 2(5 m)     (using (i))
= 5(7^k + 2m), which divisible by 5.
Thus, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for all natural numbers n.

Q8. For any natural number n, xⁿ -yⁿ is divisible by x -y, where x and y are any integers with x ≠y
Sol: Let P(n) : xⁿ – yⁿ is divisible by x – y, where x and y are any integers with x≠y.
Now, P(l): x¹ -y¹ = x-y, which is divisible by (x-y)
Hence, P(l) is true.
Let us assume that, P(n) is true for some natural number n = k.
P(k): x^k -y^k is divisible by (x – y)
or   x^k-y^k = m(x-y),m ∈ N …(i)
Now, we have to prove that P(k + 1) is true.
P(k+l):x^k+l-y^k+l
= x^k-x-x^k-y + x^k-y-y^ky
= x^k(x-y) +y(x^k-y^k)
= x^k(x – y) + ym(x – y)  (using (i))
= (x -y) [x^k+ym], which is divisible by (x-y)
Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n.

Q9. n³ -n is divisible by 6, for each natural number n≥
Sol: Let P(n): n³ – n is divisible by 6, for each natural number n> 2.
Now, P(2): (2)³ -2 = 6, which is divisible by 6.
Hence, P(2) is true.
Let us assume that, P(n) is true for some natural number n = k.
P(k): k³ – k is divisible by 6
or    k³ -k= 6m, m∈ N       (i)
Now, we have to prove that P(k + 1) is true.
P(k+ 1): (k+ l)³-(k+ 1)
= k³+ 1 +3k(k+ l)-(k+ 1)
= k³+ 1 +3k² + 3k-k- 1 = (k³-k) + 3k(k+ 1)
= 6m + 3 k(k +1)  (using (i))
Above is divisible by 6.   (∴ k(k + 1) is even)
Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n,n≥ 2.

 

Q10. n(n² + 5) is divisible by 6, for each natural number
Sol: Let P(n): n(n² + 5) is divisible by 6, for each natural number.
Now P(l): 1 (l² + 5) = 6, which is divisible by 6.
Hence, P(l) is true.
Let us assume that P(n) is true for some natural number n = k.
P(k): k( k² + 5) is divisible by 6.
or K (k²+ 5) = 6m, m∈ N         (i)
Now, we have to prove that P(k + 1) is true.
P(K+l):(K+l)[(K+l)² + 5]
= (K + l)[K² + 2K+6]
= K³ + 3 K² + 8K + 6
= (K² + 5K) + 3 K² + 3K + 6 =K(K² + 5) + 3(K² + K + 2)
= (6m) + 3(K² + K + 2)        (using (i))
Now, K² + K + 2 is always even if A is odd or even.
So, 3(K² + K + 2) is divisible by 6 and hence, (6m) + 3(K² + K + 2) is divisible by 6.
Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n.

Q11. n² < 2ⁿ, for all natural numbers n ≥
Sol: Let P(n): n² < 2ⁿ for all natural numbers n≥ 5.
Now P(5): 5² < 2⁵ or 25 < 32, which is true.
Hence, P(5) is true.
Let us assume that P(n) is true for some natural number n = k.
∴ P(k): k² < 2^k  (i)
Now, to prove that P(k + 1) is true, we have to show that P(k+ 1): (k+ l)² <2^k+1
Using (i), we get
(k + l)² = k² + 2k + 1 < 2^k + 2k + 1         (ii)
Now let, 2^k + 2k + 1 < 2^k+1     (iii)
∴ 2^k + 2k + 1 < 2 • 2^k
2k + 1 < 2^k, which is true for all k > 5 Using (ii) and (iii), we get (k + l)² < 2^k+1 Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n,n≥ 5.

Q12. 2n<(n + 2)! for all natural numbers
Sol: Let P(n): 2n < (n + 2)! for all natural numbers n.
P( 1): 2 < (1 + 2)! or 2 < 3! or 2 < 6, which is true.
Hence,P(l) is true.
Let us assume that P(n) is true for some natural number n = k.
P(k) :2k<(k + 2)!  (i)
To prove that P(k + 1) is true, we have to show that
P(k + 1): 2(k+ 1) < (k + 1 + 2)!
or 2(k+ 1) < (k + 3)!
Using (i), we get
2(k + 1) = 2k + 2<(k+2) !  +2  (ii)
Now let, (k + 2)! + 2 < (k + 3)!  (iii)
=>  2 < (k+ 3)! – (k+2) !
=> 2 < (k + 2) ! [k+ 3-1]
=>2<(k+ 2) ! (k + 2), which is true for any natural number.
Using (ii) and (iii), we get 2(k + 1) < (k + 3)!
Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n.

Q14. 2 + 4 + 6+… + 2n = n² + n, for all natural numbers
Sol: Let P(n) :2 + 4 + 6+ …+2 n = n² + n
P(l): 2 = l² + 1 = 2, which is true
Hence, P(l) is true.
Let us assume that P(n) is true for some natural number n = k.
∴ P(k): 2 + 4 + 6 + .,.+2k = k² + k  (i)
Now, we have to prove that P(k + 1) is true.
P(k + l):2 + 4 + 6 + 8+ …+2k+ 2 (k +1)
= k² + k + 2(k+ 1)  [Using (i)]
= k² + k + 2k + 2
= k² + 2k+1+k+1
= (k + 1)² + k+ 1
Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n.

Q15. 1 + 2 + 2² + … + 2ⁿ = 2^{n +1} – 1 for all natural numbers
Sol: Let P(n): 1 + 2 + 2² + … + 2ⁿ = 2^{n +1} – 1, for all natural numbers n
P(1): 1 =2^{0 + 1} — 1 = 2 — 1 = 1, which is true.
Hence, ,P(1) is true.
Let us assume that P(n) is true for some natural number n = k.

P(k): l+2 + 2²+…+2^k = 2^k+1-l              (i)

Now, we have to prove that P(k + 1) is true.

P(k+1): 1+2 + 2²+ …+2^k + 2^k+1
= 2^{k +1} – 1 + 2^k+1  [Using (i)]
= 2.2^k+l– 1 = 1
₌ 2^(k+1)+1-1
Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n.

Q16. 1 + 5 + 9 + … + (4n – 3) = n(2n – 1), for all natural numbers
Sol: Let P(n): 1 + 5 + 9 + … + (4n – 3) = n(2n – 1), for all natural numbers n.
P(1): 1 = 1(2 x 1 – 1) = 1, which is true.
Hence, P(l) is true.
Let us assume that P(n) is true for some natural number n = k.
∴ P(k):l+5 + 9 +…+(4k-3) = k(2k-1)  (i)
Now, we have to prove that P(k + 1) is true.
P(k+ 1): 1 + 5 + 9 + … +  (4k- 3) + [4(k+ 1) – 3]
= 2k² -k+4k+ 4-3
= 2k² + 3k + 1
= (k+ 1)( 2k + 1)

= (k+l)[2(k+l)-l]

Hence, P(k + 1) is true whenever P(k) is true.

So, by the principle of mathematical induction P(n) is true for any natural number n.

Long Answer Type Questions
Q17. A sequence a_x, a₂, a₃, … is defined by letting a₁=3 and a_k = 7a_k–₁ for all natural numbers k≥ Show that a_n = 3 • 7 ^n-1 for all natural numbers.

Q18. A sequence b₀, b₁, b₂, … is defined by letting b₀ = 5 and b_k = 4 + b_k–₁, for all natural numbers Show that b_n = 5 + 4n, for all natural number n using mathematical induction.
Sol. We have a sequence b₀, b₁, b₂,… defined by letting b₀ = 5 and b_k = 4 + b_k–₁,, for all natural numbers k.

So, by the principle of mathematical induction P(n) is true for any natural number rt,n> 1.

Q25. Prove that number of subsets of a set containing n distinct elements is 2″, for all n ∈
Sol: Let P(n): Number of subset of a set containing n distinct elements is 2″, for all ne N.
For n = 1, consider set A = {1}. So, set of subsets is {{1}, ∅}, which contains 2¹ elements.
So, P(1) is true.
Let us assume that P(n) is true, for some natural number n = k.
P(k): Number of subsets of a set containing k distinct elements is 2^k To prove that P(k + 1) is true,
we have to show that P(k + 1): Number of subsets of a set containing (k + 1) distinct elements is 2^k+1
We know that, with the addition of one element in the set, the number of subsets become double.
Number of subsets of a set containing (k+ 1) distinct elements = 2×2^k = 2^k+1
So, P(k + 1) is true. Hence, P(n) is true.

NCERT Exemplar ProblemsMathsPhysicsChemistryBiology

We hope the NCERT Exemplar Class 11 Maths Chapter 4 Principle of Mathematical Induction help you. If you have any query regarding NCERT Exemplar Class 11 Maths Chapter 4 Principle of Mathematical Induction, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations are part of NCERT Exemplar Class 11 Maths. Here we have given NCERT Exemplar Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations.

NCERT Exemplar Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations

Complex Numbers Class 11 Questions And Answers NCERT

Short Answer Type Questions

NCERT Exemplar Class 11 Complex Numbers

Questions On Complex Numbers Class 11 NCERT

NCERT Exemplar Class 11 Maths Ch 5

Q6. If a = cos θ + i sin θ, then find the value of (1+a/1-a) 
Sol:  a = cos θ + i sin θ


Q10. Show that the complex number z, satisfying the condition arg lies on arg (z-1/z+1) = π/4 lies on a  circle.

Sol: Let z = x + iy

Q11. Solve the equation |z| = z + 1 + 2i.
Sol: We have |z| = z + 1 + 2i
Putting z = x + iy, we get
|x + iy| = x + iy + 1+2i

Long Answer Type Questions
Q12. If |z + 1| = z + 2( 1 + i), then find the value of z.
Sol: We have |z + 11 = z + 2(1+ i)
Putting z = x + iy, we get
Then, |x + iy + 11 = x + iy + 2(1 + i)
⟹|x + iy + l|=x + iy + 2(1 +i)

Q13. If arg (z – 1) = arg (z + 3i), then find (x – 1) : y, where z = x + iy.
Sol: We have arg (z – 1) = arg (z + 3i), where z = x + iy
=>  arg (x + iy – 1) = arg (x + iy + 3i)
=> arg (x – 1 + iy) = arg [x + i(y + 3)]

Q14. Show that | z-2/z-3| = 2 represents a circle . Find its center and radius .
Sol: We have | z-2/z-3| = 2
Puttingz=x + iy, we get

 

Q15. If z-1/z+1 is a purely imaginary number (z ≠1), then find the value of |z|.

Sol: Let   z = x + iy

Q17. If |z₁ | = 1 (z₁≠ -1) and z₂ = z₁ – 1/ z₁ + 1 , then show that real part of z₂ is zero .

Q18. If Z₁, Z₂ and Z₃, Z₄ are two pairs of conjugate complex numbers, then find arg (Z_1/ Z₄) + arg (Z_2/ Z₃)
Sol. It is given that z₁ and z₂ are conjugate complex numbers.

Q20. If for complex number z₁ and z₂, arg (z₁) – arg (z₂) = 0, then show that |z₁ – z₂| = | z₁|- |z₂ |

Q21. Solve the system of equations Re (z²) = 0, |z| = 2.

Sol: Given that, Re(z²) = 0, |z| = 2

Q22. Find the complex number satisfying the equation z + √2 |(z + 1)| + i = 0.

Fill in the blanks 

True/False Type Questions

Q26. State true or false for the following.
(i) The order relation is defined on the set of complex numbers.
(ii) Multiplication of a non-zero complex number by -i rotates the point about origin through a right angle in the anti-clockwise direction.
(iii) For any complex number z, the minimum value of |z| + |z – 11 is 1.
(iv) The locus represented by |z — 11= |z — i| is a line perpendicular to the join of the points (1,0) and (0, 1).
(v) If z is a complex number such that z ≠ 0 and Re(z) = 0, then Im (z²) = 0.
(vi) The inequality |z – 4| < |z – 2| represents the region given by x > 3.
(vii) Let Z₁ and Z₂ be two complex numbers such that |z, + z₂| = |z₁ j + |z₂|, then arg (z₁ – z₂) = 0.
(viii) 2 is not a complex number.

Sol:(i) False
We can compare two complex numbers when they are purely real. Otherwise comparison of complex numbers is not possible or has no meaning.

Matching Column Type Questions
Q24. Match the statements of Column A and Column B.

Column A			Column B
(a)	The polar form of i + √3 is	(i)	Perpendicular bisector of segment joining (-2, 0) and (2,0)
(b)	The amplitude of- 1 + √-3 is	(ii)	On or outside the circle having centre at (0, -4) and radius 3.
(c)	It |z + 2| = |z – 2|, then locus of z is	(iii)	2/3
(d)	It |z + 2i| = |z – 2i|, then locus of z is	(iv)	Perpendicular bisector of segment joining (0, -2) and (0,2)
(e)	Region represented by |z + 4i| ≥ 3 is	(v)	2(cos /6 +I sin /6)
(0	Region represented by |z + 4| ≤ 3 is	(Vi)	On or inside the circle having centre (-4,0) and radius 3 units.
(g)	Conjugate of 1+2i/1-I  lies in	(vii)	First quadrant
(h)	Reciprocal of 1 – i lies in	(viii)	Third quadrant

Q28. What is the conjugate of 2-i / (1 – 2i)²

Q29. If |Z₁| = |Z₂|, is it necessary that Z₁ = Z₂?
Sol: If |Z₁| = |Z₂| then z₁ and z₂ are at the same distance from origin.
But if arg(Z₁) ≠arg(z₂), then z₁ and z₂ are different.
So, if (z₁| = |z₂|, then it is not necessary that z₁ = z₂.
Consider Z₁ = 3 + 4i and Z₂ = 4 + 3i

Q30.If  (a²+1)² / 2a –i = x + iy, then what is the value of x² + y²?
Sol: (a²+1)² / 2a –i = x + iy

Q31. Find the value of z, if |z| = 4 and arg (z) = 5π/6

Q34. Where does z lies, if | z – 5i / z + 5i  |  = 1?
Sol: We have | z – 5i / z + 5i  |

Instruction for Exercises 35-40: Choose the correct answer from the given four options indicated against each of the Exercises.

Q35. sin x + i cos 2x and cos x – i sin 2x are conjugate to each other for

Question 40.

Q41. Which of the following is correct for any two complex numbers z₁ and z₂?

NCERT Exemplar ProblemsMathsPhysicsChemistryBiology

We hope the NCERT Exemplar Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations help you. If you have any query regarding NCERT Exemplar Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Class 11 Maths Chapter 6 Linear Inequalities are part of NCERT Exemplar Class 11 Maths. Here we have given NCERT Exemplar Class 11 Maths Chapter 6 Linear Inequalities.

NCERT Exemplar Class 11 Maths Chapter 6 Linear Inequalities

Linear Inequalities Class 11 Extra Questions NCERT

Q1. \(\frac { 4 }{ x+1 } \le 3\le \frac { 6 }{ x+1 } \)

NCERT Exemplar Class 11 Maths Chapter 6 Solutions

Q2. \(\frac { |x-2|-1 }{ |x-2|-2 } \le 0  \)

Sol:

Linear Inequalities Class 11 Questions And Answers NCERT

Q3. \(\frac { 1 }{ |x|-3 } \le \frac { 1 }{ 2 }   \)
Sol: 

NCERT Exemplar Linear Inequalities Class 11

Q4. |x-1|≤ 5, |x| ≥ 2
Sol:

Linear Inequalities Class 11 Exemplar Solutions NCERT

Q5. \(-5\le \frac { 2-3x }{ 4 } \le 9   \)
Sol:

Linear Inequalities Class 11 Important Questions NCERT

Q6. 4x + 3≥2x + 17, 3x – 5 < -2.
Sol: 

NCERT Exemplar Class 11 Maths Linear Inequalities

Q7. A company manufactures cassettes. Its cost and revenue functions are C(x) = 26000 + 30x and R(x) = 43x, respectively, where x is the number of cassettes produced and sold in a week How many cassettes must be sold by the company to realise some profit?

Sol. Cost function: C(x) = 26000 + 3Ox Revenue function: R(x) = 43x For profit, R(x) > C(x)
⟹    26000 + 30x < 43x
⟹43x – 30x > 26000
⟹ 13x> 26000
⟹     x > 2000
Hence, more than 2000 cassettes must be produced to get profit.

Linear Inequalities Class 11 Questions NCERT

Q8. The water acidity in a pool is considered normal when the average pH reading of three daily measurements is between 8.2 and 8.5. If the first two pH readings are 8.48 and 8.35, find the range of pH value for the third reading that will result in the acidity level being normal.

Sol: Given, first pH value = 8.48
And second pH value = 8.35
Let third pH value be x.
Since it is given that average pH value lies between 8.2 and 8.5, we get

Class 11 Maths Chapter 6 Extra Questions NCERT

Q9. A solution of 9% acid is to be diluted by adding 3% acid solution to it. The resulting mixture is to be more than 5% but less than 7% acid. If there is 460 litres of the 9% solution, how many litres of 3% solution will have to be added?

Sol:

Extra Questions Of Linear Inequalities Class 11 NCERT

Q10. A solution is to be kept between 40°C and 45°C. What is the range of temperature in degree Fahrenheit, if the conversion formula is F= 9/5 C + 32?
Sol.

Linear Inequalities Class 11 Exemplar NCERT

Q11. The longest side of a triangle is twice the shortest side and the third side is 2 cm longer than the shortest side. If the perimeter of the triangle is more than 166 cm then find the minimum length of the shortest side.
Sol: Let the length of shortest side be x cm.
According to the given information,
Longest side = 2 x Shortest side = 2x cm
And third side = 2 + Shortest side = (2 + x) cm
Perimeter of triangle = x + 2x + (x + 2) = 4x + 2
But it is given that,
Perimeter > 166 cm
=> 4x + 2 > 166 => 4x> 166-2 => 4x>164
x>164/4 =41 cm
Hence, the minimum length of shortest side is 41 cm.

Q12. In drilling world’s deepest hole it was found that the temperature T in degree Celsius, jc km below the earth’s surface was given by T = 30 + 25 (x – 3), 3 ≤ x≤ At what depth will the temperature be between 155°C and 205°C?

Q13. \(\frac { 2x+1 }{ 7x-1 } >5,\frac { x+7 }{ x-8 } >2 \)
Solution:

Q14. Find the linear inequalities for which the shaded region in the given figure is the solution set.

Sol: We observe that the shaded region and the origin are on the same side of the line 3x + 2y = 48.
For (0, 0), we have 3(0) + 2(0) – 48 < 0. So, the shaded region satisfies the inequality 3x + 2y≤ 48.
Also, the shaded region and the origin are on the same side of the line x+y = 20.
For (0,0), we have 0 + 0 – 20 < 0. So, the shaded region satisfies the inequality x +y ≤ 20.
Also, the shaded region lies in the first quadrant. So, x ≥0,y≥0.
Thus, the linear inequation corresponding to the given solution set are 3x + 2y ≤ 48, x + y ≤ 20 and x ≥ 0, y ≥0.

Q15Find the linear inequalities for which the shaded region in the given figure is the solution set.

Sol: We observe that the shaded region and the origin are on the same side of the linex+y = 8.
For (0,0), we have 0 + 0 – 8 < 0. So, the shaded region satisfies the inequality x + 2≤8.
The shaded region and the origin are on the opposite side of the line x+y = 4.
For (0,0), we have 0 + 0 – 4 < 0. So, the shaded region satisfies the inequality x + 2≥4.
Further, the shaded region and the origin are on the same side of the lines x = 5 andy = 5.
So, it satisfies the inequality x ≤ 5 andy < 5.
Also, the shaded region lies in the first quadrant. So, x > 0, y > 0.
Thus, the linear inequation comprising the given solution set are: x+y≥4;x+y≤ 8;x≤ 5;y < 5;x≥ 0 andy ≥ 0.

Q16. Show that the following system of linear inequalities has no solution: x + 2y≤3, 3x + 4y> 12,x≥0,y≥ 1
Sol: We have x + 2y ≤ 3, 3x + 4y > 12, x > 0, y ≥ 1
Now let’s plot lines x + 2y = 3, 3x + 4y = 12, x = 0 and y = 1 in coordinate plane.
Line x + 2y = 3 passes through the points (0, 3/2) and (3, 0).
Line 3jc + 4y = 12 passes through points (4, 0) and (0, 3).
For (0, 0), 0 + 2(0) – 3 < 0.
Therefore, the region satisfying the inequality x + 2y ≤ 3 and (0,0) lie on the same side of the line x + 2y = 3.
For (0, 0), 3(0) + 4(0)- 12 ≤0.
Therefore, the region satisfying the inequality 3x + 4y ≥ 12 and (0, 0) lie on the opposite side of the line 3x + 4y = 12.
The region satisfying x > 0 lies to the right hand side of the y-axis.
The region satisfying y > 1 lies above the line y = 1.
These regions are plotted as shown in the following figure

It is clear from the graph that the Shaded portions do not have common region. So, solution set is null set.

Q17. Solve the following system of linear inequalities:
3x+2y≥24,3x+y≤ 15,x≥4
Sol: We have, 3x + 2y ≥24,
3x +y ≤ 15, x ≥ 4
Now let’s plot lines 3x + 2y = 24, 3x + y = 15 and x = 4 on the coordinate plane.
Line 3x + 2y = 24 passes through the points (0, 12) and (8, 0).
Line 3x+y = 15 passes through points (5,0) and (0, 15).
Also line x = 4 is passing through the point (4, 0) and vertical.
For (0, 0), 3(0) + 2(0) – 24 < 0.
Therefore, the region satisfying the inequality 3x + 2y≥ 24 and (0, 0) lie on the opposite of the line 3x + 2y = 24.
For (0), 3(0) + (0) – 15 ≤ 0.
Therefore, the region satisfying the inequality 3x +y ≤ 15 and (0,0) lie on the same side of the line 3x +y = 15.
The region satisfying x ≥ 4 lies to the right hand side of the line x = 4.
These regions are plotted as shown in the following figure

It is clear from the graph that there is no common region corresponding to these inequalities.

Hence, the gi ven system of inequalities has no solution.

Q18. Show that the solution set of the following system of linear inequalities is an unbounded region:
2x +y ≥ 8, x + 2y > 10, x ≥ 0, y ≥ 0
Sol: We have 2x+y≥8,x + 2y> 10, x ≥ 0, y ≥ 0
Line 2x + y = 8 passes through the points (0, 8) and (4, 0).
Line x + 2y = 10 passes through points (10, 0) and (0, 5).\
For (0, 0), 2(0) + (0) – 8 < 0.
Therefore, the region satisfying the inequality 2x+y ≥ 8 and (0, 0) lie on the opposite side of the line 2x +y = 8.
For (0,0), (0) + 2(0)- 10 <0.
Therefore, the region satisfying the inequality x + 2y≥ 10 and (0,0) lie on the opposite side of the line x + 2y = 10.
Also, for x ≥ 0, y ≥0, region lies in the first quadrant.
The common region is plotted as shown in the following figure.

NCERT Exemplar ProblemsMathsPhysicsChemistryBiology</

We hope the NCERT Exemplar Class 11 Maths Chapter 6 Linear Inequalities help you. If you have any query regarding NCERT Exemplar Class 11 Maths Chapter 6 Linear Inequalities, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar  Class 11 Maths Chapter 7 Permutations and Combinations are part of NCERT Exemplar Class 11 Maths. Here we have given NCERT Exemplar Class 11 Maths Chapter 7 Permutations and Combinations.

NCERT Exemplar Class 11 Maths Chapter 7 Permutations and Combinations

NCERT Exemplar Class 11 Maths Permutations And Combinations

Short Answer Type Questions

Q1. Eight chairs are numbered 1 to 8. Two women and 3 men wish to occupy one chair each. First the women choose the chairs from amongst the chairs 1 to 4 and then men select from the remaining chairs. Find the total number of possible arrangements.
Sol. First the women choose the chairs from amongst the chairs numbered 1 to 4.

Permutations And Combinations Class 11 Extra Questions NCERT

Q2. If the letters of the word RACHIT are arranged in all possible ways as listed in dictionary, then what is the rank of the word RACHIT?

Sol: The alphabetical order of the letters of the word RACHIT is: A, C, H, I, R, T. Number of words beginning with A = 5!
Number of words beginning with C = 5!
Number of words beginning with H = 5!
Number of words beginning with 1 = 5!
Clearly, the first word beginning with R is RACHIT.
.•. Rank of the word RACHIT in dictionary = 4×5! + 1= 4 x120+1= 481

Permutation And Combination Class 11 Extra Questions With Answers

Q3. A candidate is required to answer 7 questions out of 12 questions, which are divided into two groups, each containing 6 questions. He is not permitted to attempt more than 5 questions from either group. Find the number of different ways of doing questions.

Sol: Since the candidate cannot attempt more than 5 questions from either group, he is able to attempt minimum two questions from either group.
The possible number of questions attempted from each group will be as given in the following table:

Group I	5	4	3	2
Group II	2	3	4	5

NCERT Exemplar Class 11 Maths Chapter 7 Solutions

Q4. Out of 18 points in a plane, no three are in the same line except five points which are collinear. Find the number of lines that can be formed joining the point.
Sol: There are 18 point in a plane, of which 5 points are collinear.

Q5. We wish to select 6 persons from 8 but, if the person A is chosen, then B must be chosen. In how many ways can selections be made?
Sol:

Permutation And Combination Class 11 Important Questions NCERT 

Q6. How many committee of five persons with a chairperson can be selected form 12 persons?
Sol: Total number of persons =12
Number of persons to be selected = 5

Q7. How many automobile license plates can be made, if each plate contains two different letters followed by three different digits?
Sol: There are 26 English alphabets and 10 digits (0 to 9).
It is given that each plate contains two different letters followed by three different digits.

Class 11 Maths Chapter 7 Extra Questions NCERT

Q8. A bag contains 5 black and 6 red balls. Determine the number of ways in • which 2 black and 3 red balls can be selected from the lot.

Q9. Find the number of permutations of n distinct things taken r together, in which 3 particular things must occur together.

Important Questions Of Permutation And Combination Class 11 NCERT

Q10. Find the number of different words that can be formed from the letters of the word TRIANGLE, so that no vowels are together.
Sol: Given word is: TRIANGLE Consonants are: T, R, N, G, L Vowels are: I, A, E
Since we have to form words in such a way that no two vowels are together, we first arrange consonants.
Five consonants can be arranged in 5! ways.

Q11. Find the number of positive integers greater than 6000 and less than 7000 which are divisible by 5, provided that no digit is to be repeated.
Sol: We have to form 4-digit numbers which are greater than 6000 and less than 7000.
We know that a number is divisible by 5, if at the unit place of the number there is 0 or 5.
So, unit digit can be filled in 2 ways.
The thousandth place can be filled by ‘6’ only.
The hundredth place and tenth place can be filled together in 8 x 7 = 56 ways. So, total number of ways = 56 x 2 = 112

Class 11 Permutations And Combinations Important Questions NCERT

Q12. Thereare 10persons named P₁,P₂,P₃,…,P_!0.Outof 10 persons, 5 persons are to be arranged in a line such that in each arrangement P, must occur whereas P₄ and P₅ do not occur. Find the number of such possible arrangements.
Sol. Given that, P₁, P₂, …, P₁₀, are 10 persons, out of which 5 persons are to be arranged but P, must occur whereas P₄ and P₅ never occurs.
As P, is already occurring w’e have to select now 4 out of 7 persons.
.•. Number of selections = ⁷C₄ = 35 Number of arrangements of 5 persons = 35 x 5! = 35 x 120 = 4200

Q13. There are 10 lamps in a hall each one of them can be switched on independently. Find the number of ways in which the hall can be illuminated.
Sol: There are 10 lamps in a hall.
The hall can be illuminated if at least one lamp is switched.
.•. Total number of ways =        ¹⁰C₁+ ¹⁰C₂ + ^l0C₃… + ¹⁰C_]0
= 2¹⁰– 1 = 1024- 1 = 1023

Permutation And Combination Class 11 Questions NCERT

Q14. A box contains two white, three black and four red balls. In how many ways can three balls be drawn from the box, if at least one black ball is to be included in the draw?
Sol: There are two white, three black and four red balls.
We have to draw 3 balls, out of these 9 balls in which at least one black ball is included.
So we have following possibilities:

Black balls	1	2	3
Other than black	2	1	0

.’. Number of selections = ³C₁ x ⁶C₂ + ³C₂ x ⁶C, + ³C₃ x ⁶C₀
= 3×15+ 3×6+1= 45+ 18 + 1= 64

Q15. If ⁿC_r-1=  36 ⁿC_r = 84 and ⁿC_r+1= 126, then find the value of ^rC₂.

Permutation And Combination Questions Class 11 NCERT

Q16. Find the number of integers greater than 7000 that can be formed with the digits 3, 5, 7, 8 and 9 where no digits are repeated.
Sol: We have to find the number of integers greater than 7000 with the digits 3,5, 7, 8 and 9.
So, with these digits, we can make maximum five-digit numbers because repetition is not allowed.
Since all the five-digit numbers are greater than 7000, we have Number of five-digit integers = 5x4x3x2x1 = 120 A four-digit integer is greater than 7000 if thousandth place has any one of 7, 8 and 9.
Thus, thousandth place can be filled in 3 ways. The remaining three places can be filled from remaining four digits in ⁴P₃ ways.
So, total number of four-digit integers = 3x ⁴P₃ = 3x4x3x2 = 72 Total number of integers = 120 + 72 = 192

 

Q17. If 20 lines are drawn in a plane such that no two of them are parallel and no three are concurrent, in how many points will they intersect each other?
Sol: It is given that no two lines are parallel which means that all the lines are intersecting and no three lines are concurrent.
One point of intersection is created by two straight lines.
Number of points of intersection = Number of combinations of 20 straight lines taken two at a time

Permutation And Combination Extra Questions NCERT

Q18. In a certain city, all telephone numbers have six digits, the first two digits always being 41 or 42 or 46 or 62 or 64. How many telephone numbers have all six digits distinct?
Sol: If first two digit is 41, the remaining 4 digits can be arranged in ⁸P₄ = 8 x 7 x 6×5 = 1680 ways.
Similarly, if first two digit is 42, 46, 62, or 64, the remaining 4 digits can be arranged in ⁸P₄ ways i.e., 1680 ways.
.’. Total number of telephone numbers having all six digits distinct = 5x 1680 = 8400

Q19. In an examination, a student has to answer 4 questions out of 5 questions, questions 1 and 2 are however compulsory. Determine the number of ways in which the student can make the choice.
Sol: It is given that 2 questions are compulsory out of 5 questions.
So, the other 2 questions can be selected from the remaining 3 questions in ³C₂ = 3 ways.

Q20. A convex polygon has 44 diagonals. Find the number of its sides.
[Hint: Polygon of n sides has (ⁿC₂ – n) number of diagonals.]
Sol: Let the convex polygon has n sides.
Number of diagonals=Number of ways of selecting two vertices – Number of sides = ⁿC₂ – n
It is given that polygon has 44 diagonals.

Long Answer Type Questions
Q21. 18 mice were placed in two experimental groups and one control group with all groups equally large. In how many ways can the mice be placed into three groups?
Sol: It is given that 18 mice were placed equally in two experimental groups and one control group i.e., three groups.
Each group is of 6 mice.

Q22. A bag contains six white marbles and five red marbles. Find the number of ways in which four marbles can be drawn from the bag, if (i) they can be of any colour (ii) two must be white and two red and (iii) they must all be of the same colour.
Sol:Total number of marbles = 6 white +- 5 red = 11 marbles

(a) If they can be of any colour means we have to select 4 marbles out of 11
∴ Required number of ways = ¹¹C₄
(b) Two white marbles can be selected in ⁶C₂
Two red marbles can be selected in ⁵C₂ ways.
∴ Total number of ways = ⁶C₂ x ⁵C₂ = 15 x 10 = 150
(c) If they all must be of same colour,
Four white marbles out of 6 can be selected in ⁶C₄ ways.
And 4 red marbles out of 5 can be selected in ⁵C₄ ways.
∴ Required number of ways = ⁶C₄ + ⁵C₄ = 15 + 5 = 20

Q23. In how many ways can a football team of 11 players be selected from 16 players? How many of them will

• include 2 particular players?
• exclude 2 particular players?

Sol: Total number of players = 16
We have to select a team of 11 players
So, number of ways = ¹⁶C₁₁
(i) If two particular players are included then more 9 players can be selected from remaining 14 players in ¹⁴C₉
(ii) If two particular players are excluded then all 11 players can be selected from remaining 14 players in ¹⁴C₁₁
Q24.  sports team of 11 students is to be constituted, choosing at least 5 from class XI and at least 5 from class XII. If there are 20 students in each of these classes, in how many ways can the team be constituted?
Sol: Total number of students in each class = 20
We have to select at least 5 students from each class.
So we can select either 5 students from class XI and 6 students from class XII or 6 students from class XI and 5 students from class XII.
∴ Total number of ways of selecting a team of 11 players = ²⁰C₅ x ²⁰C₆ + ²⁰C₆ x ²⁰C₅ = 2 x ²⁰C₅ x ²⁰C₆

Q25. A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected, if the team has
(i) no girls
(ii) at least one boy and one girl
(iii) at least three girls
Sol: Number of girls = 4;
Number of boys = 7
We have to select a team of 5 members provided that

Objective Type Questions
Q26. If ⁿC₁₂ = ⁿC₈, then n is equal to 
(a) 20     
(b) 12
(c) 6
(d) 30

Q27. The number of possible outcomes when a coin is tossed 6 times is
(a) 36                    
(b) 64                       
(c) 12                       
(d) 32
Sol: (b) Number of outcomes when a coin tossed = 2 (Head or Tail)
∴Total possible outcomes when a coin tossed 6 times = 2x2x2x2x2x 2 = 64

Q28. The number of different four-digit numbers that can be formed with the digits
2, 3, 4, 7 and using each digit only once is
(a) 120                  
(b) 96                       
(c) 24                       
(d) 100
Sol: (c) Given digits 2,3,4 and 7, we have to form four-digit numbers using these digits.
∴Required number of ways = ⁴P₄ = 4!=4x3x2x1 = 24

Q29. The sum of the digits in unit place of all the numbers formed with the help of 3,4, 5 and 6 taken all at a time is
(a)    432               
(b)    108                  
(c)      36                   
(c)    18
Sol: (b) If the unit place is ‘3’ then remaining three places can be filled in 3! ways.
Thus ‘3’ appears in unit place in 3! times.
Similarly each digit appear in unit place 3! times.
So,    sum of digits in unit place = 3!(3 + 4 + 5 + 6) = 18 x 6 = 108

Q30. The total number of words formed by 2 vowels and 3 consonants taken from 4 vowels and 5 consonants is
(a) 60                 
(b) 120                  
(c) 7200              
(d)  720
Sol: (c) Given number of vowels = 4 and number of consonants = 5 We have to form words by 2 vowels and 3 consonants.
So, lets first select 2 vowels and 3 consonants.
Number of ways of selection = ⁴C₂ x ⁵C₃ = 6 x 10 = 60 Now, these letters can be arranged in 5! ways.
So, total number of words = 60 x 5! = 60 x 120 = 7200

 

Q31. A five-digit number divisible by 3 is to be formed using the numbers 0, 1,2,4, and 5 without repetitions. The total number of ways this can be done is
(a)  216               
(b)  600                  
(c)  240                 
(d) 3125
[Hint: 5 digit numbers can be formed using digits 0, 1, 2, 4, 5 or by using digits 1, 2, 3, 4, 5 since sum of digits in these cases is divisible by 3.]

Sol:(a) We know that a number is divisible by 3 if the sum of its digits is divisible by 3.
Now sum of the given six digits is 15 which is divisible by 3. So to form a number of five-digit which is divisible by 3 we can remove either ‘O’ or ‘3’. If digits 1, 2, 3,4, 5 are used then number of required numbers = 5!
If digits 0, 1,2,4, 5 are used then first place from left can be filled in 4 ways and remaining 4 places can be filled in 4! ways. So in this case required numbers are 4 x 4! ways.
So, total number of numbers = 120 + 96 = 216

Q32. Everybody in a room shakes hands with everybody else. If the total number of hand shakes is 66, then the total number of persons in the room is
(a)    11                 
(b)     12                    
(c)     13                   
(d)   14
Sol: (b) Between any two person there is one hand shake.

Q33. The number of triangles that are formed by choosing the vertices from a set of 12 points, seven of which lie on the same line is
(a)    105                (b)     15                     (c)     175                  (d)  185
Sol: (d) Number of ways of selecting 3 points from given 12 points = ¹²C₃
But any three points selected from given seven collinear points does not form triangle.
Number of ways of selecting three points from seven collinear points = ⁷C₃Required number of triangles = ¹²C₃ – ⁷C₃ = 220 -35 = 185

Q34. The number of parallelograms that can be formed form a set of four parallel lines intersecting another set of three parallel lines is
(a)    6                     
(b)   18                    
(c)   12                   
(d)   9
Sol: (b) To form parallelogram we required a pair of line from a set of 4 lines and another pair of line from another set of 3 lines.
Required number of parallelograms = ⁴C₂ x ³C₂ = 6×3 = 18

Q35. The number of ways in which a team of eleven players can be selected from 22 players always including 2 of them and excluding 4 of them is
(a)   ¹⁶C₁₁               
(b)    ¹⁶C₅                    
(c)     ¹⁶C₉                  
(d)   ²⁰C₉

Sol: (c) Total number of players = 22
We have to select a team of 11 players.
We have to exclude 4 particular of them, so only 18 players are now available. Also from these 2 particular players are always included. Therefore we have to select 9 more players from the remaining 16 players.
So, required number of ways = ¹⁶C₉

Q36. The number of 5-digit telephone numbers having at least one of their digits repeated is
(a) 900000
(b) 10000                 
(c) 30240                
(d) 69760
Sol: (d) Total number of telephone numbers when there is no restriction = 10⁵ Also number of telephone numbers having all digits different = ^l0P₅ Required number of ways = 10⁵ – ^l0P₅ = 1000000 -10x9x8x7x6 = 1000000-30240 = 69760

Q37. The number of ways in which we.can choose a committee from four men and six women, so that the committee includes at least two men and exactly twice as many women as men is
(a) 94                    
(b) 126                     
(c) 128                    
(d) none of these
Sol: (a) Number of men = 4; Number of women = 6
It is given that committee includes at least two men and exactly twice as many women as men.
So, we can select either 2 men and 4 women or 3 men and 6 women.
∴ Required number of committee formed = ⁴C₂ x ⁶C₄ + ⁴C₃ x ⁶C₆
= 6×15 + 4×1=94

Q38. The total number of 9-digit numbers which have all different digits is
(a) 10!                   
(b) 9!                        
(c) 9×9!               
(d) 10×10!
Sol: (c) We have to form 9-digit number which has all different digit.
First digit from the left can be filled in 9 ways (excluding ‘0’).
Now nine digits are left including ‘O’.
So remaining eight places can be filled with these nine digits in ⁹P_S ways.
So, total number of numbers = 9 x ⁹P₈ = 9×9!

Q39. The number of words which can be formed out of the letters of the word ARTICLE, so that vowels occupy the even place is
(a) 1440               
(b) 144                     
(c) 7!                        
(d) ⁴C₄ x ³C₃
Sol: (b) We have word ARTICLE.
Vowels are A, I, E and consonants are R, T, C, L.
Now vowels occupy three even places (2^nd, 4^th and 6^th) in 3! ways.
In remaining four places four consonants can be arranged in 4! ways.
So, total number of words = 3! x4! = 6×24= 144

Q40. Given five different green dyes, four different blue dyes and three different red dyes, the number of combinations of dyes which can be chosen taking at least one green and one blue dye is
(a) 3600               
(b) 3720                   
(c) 3800                  
(d) 3600
[Hint: Possible numbers of choosing or not choosing 5 green dyes, 4 blue dyes and 3 red dyes are 2⁵, 2⁴ and 2³, respectively.]

Fill in the Blanks Type Questions

True/False Type Questions

Q51. There are 12 points in a plane of which 5 points are collinear, then the number of lines obtained by joining these points in pairs is ¹²C₂ – ⁵C₂.
Sol: False
Required number of lines = ¹²C₂ – ⁵C₂.+ 1

Q52. Three letters can be posted in fiv.e letter boxes in 3⁵
Sol: False
Each letter can be posted in any one of the five letter boxes.
So, total number of ways of posting three letters = 5x5x5 = 125

Q53. In the permutations of n things r taken together, the number of permutations in which m particular things occur together is

Q54. In a steamer there are stalls for 12 animals and there are horses, cows and calves (not less than 12 each) ready to be shipped. They can be loaded in 3¹² ways.
Sol:True
In each stall any one of the three animals can be shipped.
So total number of ways of loading = 3x3x3x…xl2 times = 3¹²

Q55. If some or all of n objects are taken at a time, then the number of combinations is 2ⁿ– 1.
Sol: True
If some or all objects taken at a time, then number of combinations would be ⁿC₁ + ⁿC₂ + ⁿC₃ + … + ⁿC_n = 2ⁿ – 1

Q56. There will be only 24 selections containing at least one red ball out of a bag containing 4 red and 5 black balls. It is being given that the balls of the same colour are identical.
Sol: False
Number of ways of selecting any number of objects from given n identical objects is 1.
Now selecting zero or more red ball from 4 identical red balls = 1 + 1 + 1 + 1 + 1=5
Selecting at least 1 black ball from 5 identical black balls =1 + T+1 + 1 + 1= 5 So, total number of ways = 5 x 5 = 25

Q58. A candidate is required to answer 7 questions, out of 12 questions which are divided into two groups, each containing 6 He is not permitted to attempt more than 5 questions from either group. He can choose the seven questions in 650 ways.
Sol: False
A candidate can attempt questions in following maimer

Q59. To fill 12 vacancies there are 25 candidates of which 5 are from scheduled castes. If 3 of the vacancies are reserved for scheduled caste candidates while the rest are open to all, the number of ways in which the selection can be made is ⁵C₃ x ²²C₉.
Sol:True
We can select 3 scheduled caste candidate out of 5 in ⁵C₃ ways.
And we can select 9 other candidates out of 22 in ²²C₉ways.
.’. Total number of selections = ⁵C₃ x ²²C₉

NCERT Exemplar ProblemsMathsPhysicsChemistryBiology

We hope the NCERT Exemplar Class 11 Maths Chapter 7 Permutations and Combinations help you. If you have any query regarding NCERT Exemplar Class 11 Maths Chapter 7 Permutations and Combinations, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Class 11 Maths Chapter 8 Binomial Theorem are part of NCERT Exemplar Class 11 Maths. Here we have given NCERT Exemplar Class 11 Maths Chapter 8 Binomial Theorem.

NCERT Exemplar Class 11 Maths Chapter 8 Binomial Theorem

Short Answer Type Questions:

Q1. Find the term independent of x, where x≠0, in the expansion of \({ \left( \frac { 3{ x }^{ 2 } }{ 2 } -\quad \frac { 1 }{ 3x }  \right)  }^{ 15 }\)

Q2. If the term free from x is the expansion of  \({ \left( \sqrt { x } -\frac { k }{ { x }^{ 2 } }  \right)  }^{ 10 }\) is 405, then find the value of k.

Sol: Given expansion is \({ \left( \sqrt { x } -\frac { k }{ { x }^{ 2 } }  \right)  }^{ 10 }\)

Q3. Find the coefficient of x in the expansion of (1 – 3x + 1x²)( 1 -x)¹⁶.

Sol: (1 – 3x + 1x²)( 1 -x)¹⁶

Q4. Find the term independent of x in the expansion of \({ \left( 3x-\frac { 2 }{ { x }^{ 2 } }  \right)  }^{ 15 }\)

Sol: Given Expression \({ \left( 3x-\frac { 2 }{ { x }^{ 2 } }  \right)  }^{ 15 }\)

Q5. Find the middle term (terms) in the expansion of

Q6. Find the coefficient of x¹⁵ in the expansion of \({ \left( x-{ x }^{ 2 }\quad  \right)  }^{ 10 }\)

Sol: Given expression is   \({ \left( x-{ x }^{ 2 }\quad  \right)  }^{ 10 }\)

Q7. Find the coefficient of \(\frac { 1 }{ { x }^{ 17 } } \) in the expansion of \({ \left( { x }^{ 4 }-\frac { 1 }{ { x }^{ 3 } } \quad  \right)  }^{ 15 } \)

Q8. Find the sixth term of the expansion (y^1/2 + x^1/3)ⁿ, if the binomial coefficient of the third term from the end is 45.

>

Q9. Find the value of r, if the coefficients of (2r + 4)th and (r – 2)th terms in the expansion of (1 + x)¹⁸ are equal.

Q10. If the coefficient of second, third and fourth terms in the expansion of (1 + x)²” are in A.P., then show that 2n² – 9n + 7 = 0.

Q11. Find the coefficient of x⁴ in the expansion of (1 + x + x² + x³)¹¹.

Long Answer Type Questions

Q12. If p is a real number and the middle term in the expansion \({ \left( \frac { p }{ 2 } +2\quad \right) }^{ 8 } \) is 1120, then find the value of p.

Q15. In the expansion of (x + a)ⁿ, if the sum of odd term is denoted by 0 and the sum of even term by Then, prove that

Q17. Find the term independent ofx in the expansion of (1 +x + 2x³)\({ \left( \frac { 3 }{ 2 } { x }^{ 2 }-\frac { 1 }{ 3x } \quad \quad  \right)  }^{ 9 } \)

Objective Type Questions

Q18. The total number of terms in the expansion of (x + a)¹⁰⁰ + (x – a)¹⁰⁰ after simplification is

(a) 50
(b) 202
(c) 51
(d) none of these

Q19. If the integers r > 1, n > 2 and coefficients of (3r)th and (r + 2)nd terms in the binomial expansion of (1 + x)²ⁿ are equal, then
(a) n = 2r             
(b) n = 3r            
(c) n = 2r + 1       
(d) none of these

Q20. The two successive terms in the expansion of (1 + x)²⁴ whose coefficients are in the ratio 1 : 4 are
(a) 3^rd and 4^th
(b) 4^th and 5^th
(c) 5^th and 6^th
(d) 6^th and 7^th

Q21. The coefficients of xⁿ in the expansion of (1 + x)²ⁿ and (1 + x)²ⁿ ~¹ are in the ratio
(a) 1 : 2                   
(b) 1 : 3                  
(c) 3 : 1
(d) 2:1

Q22. If the coefficients of 2^nd, 3^rd and the 4^th terms in the expansion of (1 + x)ⁿ are in A.P., then the value of n is
(a) 2           
(b) 7 
(c) 11               
(d) 14

Q23. If A and B are coefficients of  xⁿ   in the expansions of (1 + x)²ⁿ and (1 + x)²ⁿ–¹  respectively, then A/B  equals to

NCERT Exemplar ProblemsMathsPhysicsChemistryBiology

We hope the NCERT Exemplar Class 11 Maths Chapter 8 Binomial Theorem help you. If you have any query regarding NCERT Exemplar Class 11 Maths Chapter 8 Binomial Theorem, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Class 11 Maths Chapter 9 Sequence and Series are part of NCERT Exemplar Class 11 Maths. Here we have given NCERT Exemplar Class 11 Maths Chapter 9 Sequence and Series.

NCERT Exemplar Class 11 Maths Chapter 9 Sequence and Series

Short Answer Type Questions:

Q1. The first term of an A.P. is terms is zero, show that the sum of its next q terms
\(\frac { -a(p+q)q }{ p-1 }  \)

Sol: Let the common differeence of the given A.P be d

Q2. A man saved Rs. 66000 in 20 years. In each succeeding year after the first year, he saved Rs. 200 more than what he saved in the previous year. How much did he save in the first year?
Sol: Let us assume that the man saved Rs.a in the first year.
In each succeeding year, an increment of Rs. 200 is made. So, it forms an A.P. whose
First term = a, Common difference, d = 200 and n=20

=> 6600 = 2a + 19 x 200 => 2a = 2800
∴a = 1400

 Q3. A man accepts a position with an initial salary of Rs. 5200 per month. It is understood that he will receive an automatic increase of Rs. 320 in the very next month and each month thereafter.
(i) Find his salary for the tenth month.
(ii) What is his total earnings during the first year?

Q5. A carpenter was hired to build 192 window frames. The first day he made five frames and each day, thereafter he made two more frames than he made the day before. How many days did it take him to finish the job?
Sol: Here, a = 5 and d = 2

Q6. The sum of interior angles of a triangle is 180°. Show that the sum of the interior angles of polygons with 3, 4, 5, 6, … sides form an arithmetic progression. Find the sum of the interior angles for a 21 sided polygon.
Sol: We know that, sum of interior angles of a polygon of side n is (n – 2) x 180°.

Q7. A side of an equilateral triangle is 20 cm long. A second equilateral triangle is inscribed in it by joining the mid-points of the sides of the first triangle. This process is continued for third, fourth, fifth, triangles. Find the perimeter of the sixth inscribed equilateral triangle.
Sol: Let the given equilateral triangle be ∆ ABC with each side of 20 cm.
By joining the mid-points of this triangle, we get another equilateral triangle of side equal to half of the length of side of ∆ABC.
Continuing in this way, we get a set of equilateral triangles with side equal to half of the side of the previous triangle.
Now,
Perimeter of first triangle = 20 x 3 = 60 cm;
Perimeter of second triangle = 10 x 3 = 30 cm;
Perimeter of third triangle = 5×3 = 15 cm;

Q8. In a potato race 20 potatoes are placed in a line at intervals of 4 m with the first potato 24 m from the starting point. A contestant is required to bring the potatoes back to the starting place one at a time. How far would he run in bringing back all the potatoes?
Sol: Distance travelled to bring first potato = 24 + 24 = 2 x 24 = 48 m
Distance travelled to bring second potato = 2(24 + 4) = 2 x 28 = 56 m
Distance travelled to bring third potato = 2(24 + 4 + 4) = 2 X 32 = 64 m; and so on…
Clearly, 48, 56, 64,… is an A.P. with first term 48 and common difference 8. Also, number of terms is 20.
Total distance run in bringing back all the potatoes,

Q9. In a cricket tournament 16 school teams participated. A sum of Rs. 8000 is to be awarded among themselves as prize money. If the last placed team is awarded Rs. 275 in prize money and the award increases by the same amount for successive finishing places, how much amount will the first place team receive?
Sol: Let the first place team get Rs. a as the prize money.
Since award money increases by the same amount for successive finishing places, we get an A.P.

Q11. Find the sum of the series (3³ – 2³) + (5³ – 4³) + (7³ – 6³) + … to
(i) n terms                                               
(ii) 10 terms
Sol: Given series is: (3³ — 2³) + (5³ – 4³) + (7³ – 6³) + … n terms

Q12. Find the rth term of an A.P. sum of whose first n terms is 2n +3n²
Sol: Sum of k terms of A.P., S_n = 2n + 3n²

Long Answer Type Questions

Q13. If A is the arithmetic mean and G₁ , G₂ be two geometric means between any two numbers, then prove that 2A = \(\frac { { G }_{ 1 }^{ 2 } }{ { G }_{ 2 } } +\frac { { G }_{ 2 }^{ 2 } }{ { G }_{ 2 } } \quad   \)
Sol: Let the numbers be a and b.

Q15. 1f the sum of p terms of an AP. is q and the sum of q terms isp, then show that the sum ofp + q terms is —(p + q). Also, find the sum of first p — q terms (where, p > q).
Sol: Let first term and common difference of the A.P. be a and d, respectively. Given, S_p = q

Q16. If pth, qth and rth terms of an A.P. and G.P. are both a, b and c, respectively, then show that a^b-c b^c-a-c^a-b = 1.
Sol: Let A and d be the first term and common difference of A.P., respectively. Also, let B and R be the first term and common ratio of G.P., respectively.
It is given that,

Objective Type Questions:

Q17. If the sum of n terms of an A.P. is given by S_n = 3n + 2n², then the common difference of the A.P. is
(a) 3                       
(b) 2                         
(c) 6                        
(d) 4
Sol: (d) Given, S_n = 3n + 2n²

Q18. If the third term of G.P. is 4, then the product of its first 5 terms is
(a) 4³
(b) 4⁴
(c) 4⁵
(d) none of these
Sol: (c) Let a and r the first term and common ratio, respectively.

Q19. If 9 times the 9th term of an A.P. is equal to 13 times the 13th term, then the 22nd term of the A.P. is
(a) 0
(b) 22
(c) 198
(d) 220
Sol: (a) Let the first term and coiqmon difference of given A.P. be a and d, respectively.

Q20. If x, 2y and 3z are in A.P. where the distinct numbers x, y and z are in G.P., then the common ratio of the G.P.is
Sol: Since x, 2y and 3z are in A.P., we get

Q21. If in an A.P., S_n = qn² and S_m = qm², where S_r denotes the sum of r terms of the AP, then S_q equals

Q22. Let S_n denote the sum of the first n terms of an A.P. If S_2n = 3S_n, then S_3n : S_n  is equal to
(a) 4
(b) 6
(c) 8
(d) 10
Sol: (b) Let first term be a and common difference be d.
Then, S_2n = 3S_n

Q25. If t_n denotes the nth term of the series 2 + 3+ 6+11 + 18+…, then t₅₀ is
(a) 49² – 1     
(b) 49²   
(c) 50²+l
(d) 49² +2

Q26. The lengths of three unequal edges of a rectangular solid block are in G.P. If the volume of the block is 216 cm³ and the total surface area is 252 cm², then the length of the longest edge is
(a) 12 cm               
(b) 6 cm                   
(c) 18 cm                
(d) 3 cm
Sol: (a) Let the length, breadth and height of rectangular solid block be a/r, a and ar, respectively.

Fill in the Blanks

Q27. If a, b and c are in G.P., then the value of \(\frac { a-b }{ b-c }   \) is equal to ________

Q28. The sum of terms equidistant from the beginning and end in an A.P. is equal to _____.
Sol: Let a be the first term and d be ihe common difference of the A.P.

Q29. The third term of a G.P. is 4. The product of the first five terms is  
Sol: Let a and r the first term and common ratio, respectively.

True/False Type Questions

Q30. Two sequences cannot be in both A.P. and G.P. together.
Sol: False
Consider the sequence 3,3,3; which is A.P. and G.P. both.

Q31. Every progression is a sequence but the converse, i.e., every sequence is also a progression need not necessarily be true.
Sol: True                                                            –
Consider the progression a, a + d, a + 2d, … and sequence of prime number 2, 3, 5, 7, 11,…
Clearly, progression is a sequence but sequence is not progression because it does not follow a specific pattern.

Q32. Any term of an A.P. (except first) is equal to half the sum of terms which are equidistant from it.
Sol: True

Q33. The sum or difference of two G.P.s, is again a G.P.

Q34. If the sum of n terms of a sequence is quadratic expression, then it always represents an A.P.
Sol: False

Match the questions given under Column I with their appropriate answers given under the column II.

NCERT Exemplar ProblemsMathsPhysicsChemistryBiology

We hope the NCERT Exemplar Class 11 Maths Chapter 9 Sequence and Series help you. If you have any query regarding NCERT Exemplar Class 11 Maths Chapter 9 Sequence and Series, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Class 11 Maths Chapter 10 Straight Lines are part of NCERT Exemplar Class 11 Maths. Here we have given NCERT Exemplar Class 11 Maths Chapter 10 Straight Lines.

NCERT Exemplar Class 11 Maths Chapter 10 Straight Lines

Short Answer Type Questions

Q1. Find the equation of the straight line which passes through the point (1, -2)

Q2. Find the equation of the line passing through the point (5,2) and perpendicular to the line joining the points (2, 3) and (3, -1)

Q3. Find the angle between the lines y = (2 -√3) (x + 5) and y = (2 + -√3) {x – 7)
Sol: Slope of the line = (2 -√3)(x + 5) is: m_l = (2 -√3 )

Q4. Find the equation of the lines which passes through the point (3, 4) and cuts off intercepts from the coordinate axes such that their sum is 14.

Q5. Find the points on the line x+y = 4 which lie at a unit distance from the line 4x + 3y= 10
Sol. Let the required point be (h, k) lies on the line x + y = 4

Q7. Find the equation of lines passing through (1,2) and making angle 30° with y-axis.

Q8. Find the equation of the line passing through the point of intersection of lx + y = 5 and x + 3 y +8 = 0 and parallel to the line 3x + 4y = 1.

Q9. For what values of a and b the intercepts cut off on the coordinate axes by the line ax + by + 8 = 0 are equal in length but opposite in signs to those cut off by the line 2x – 3y + 6 = 0 on the axes.
Sol: Given line is:

Q10. If the intercept of a line between the coordinate axes is divided by the point (-5,4) in the ratio 1 : 2, then find the equation of the line.
Sol: Let the line through the point P(-5, 4) meets axis at A(h, 0) and B(0, k)

According to the question, we have AP: BP =1:2


Q11. Find the equation of a straight line on which length of perpendicular from the origin is four units and the line makes an angle of 120° with the positive direction of x-axis.
Sol: Given that the line makes and angle 120° with positive direction of x-axis.

Q12. Find the equation of one of the sides of an isosceles right angled triangle whose hypotenuse is given by 3x + Ay = 4 and the opposite vertex of the hypotenuse is (2, 2).

Long Answer Type Questions

Q13. If the equation of the base of an equilateral triangle is x + y – 2 and the vertex is (2, -1), then find the length of the side of the triangle.

Q14. A variable line passes through a fixed point P. The algebraic sum of the perpendiculars drawn from the points (2, 0), (0, 2) and (1, 1) on the line is zero. Find the coordinates of the point P

Q15. In what direction should a line be drawn through the point (1, 2) so that its point of intersection with the line x+y = 4 is at a distance √6/3 from the given point.

Q16. A straight line moves so that the sum of the reciprocals of its intercepts made on axes is constant. Show that the line passes through a fixed point.

Q17. Find the equation of the line which passes through the point (-4, 3) and the portion of the line intercepted between the axes is divided internally in the ratio 5 : 3 by this point.
Sol: Let the line through the point P(-A, 3) meets axis at A(h, 0) and 0(0, k)
Now according to the question AP : BP =5:3

Q18. Find the equations of the lines through the point of intersection of the lines x-y+ 1=0 and 2x – 3y + 5 = 0 and whose distance from the point (3, 2) is 7/5

Q19. If the sum of the distances of a moving point in a plane from the axes is 1, then find the locus of the point.

Q20. P_l, P₂ are points on either of the two lines y — √3 |x| = 2 at a distance of 5 units from their point of intersection. Find the coordinates of the foot of perpendiculars drawn from P₁, P₂ on the bisector of the angle between the given lines.


Q21. If p is the length of perpendicular from the origin on the line \(\frac { x }{ a } +\quad \frac { y }{ b }    \)  and a²,p² and are in the A.P , then show that a⁴+b⁴ = 0

Objective Type Questions

Q22. A line cutting off intercept -3 from the y-axis and the tangent of angle to the x-axis is 3/5, its equation is

Q23. Slope of a line which cuts off intercepts of equal lengths on the axes is
(a) -1
(b) 0       
(c) 2       
(d) √3
Sol: (a) Equation of the according to the question is \(\frac { x }{ a } +\quad \frac { y }{ a }     \)
=> x+y = a
Required slope = -1

Q24. The equation of the straight line passing through the point (3, 2) and perpendicular to the line y = x is
(a) x-y = 5           
(b) x+y = 5
(c)x+y=l
(d)x-y=1
Sol:(b) Slope of the given line y = x is 1.
Thus, slope of line perpendicular to y = x is -1.
Line passes through the point (3, 2).
So, equation of the required line is:y-2=-l (x – 3) => x + y = 5

Q25. The equation of the line passing through the point (1,2) and perpendicular to the line x +y + 1 = 0 is
(a) y-x +1=0           
(b) y — x—1=0
(c) y-x + 2 = 0                                      
(d) y — x — 2=0
Sol: (b) Slope of the given line +1=0 is-1.
So, slope of line perpendicular to above line is 1.
Line passes through the point (1,2).
Therefore, equation of the required linens:
y-2 = 1(x- 1) => y-x-1=0.

Q26. The tangent of angle between the lines whose intercepts on the axes are a, -b and b, -a, respectively, is

Q27. If the line \(\frac { x }{ a } +\frac { y }{ b }  \)  passes through the points (2, -3) and (4, -5), then (a, b) a b is
(a) (1,1)                  
(b) (-1,1)                  
(c) (1,-1)                  
(d) (-1,-1)

Q28. The distance of the point of intersection of the lines 2x – 3y + 5 = 0 and 3x + 4y = 0 from the line 5x – 2y = 0 is

Q29. The equations of the lines which pass through the point (3, -2) and are inclined at 60° to the line √3 x + y = 1 is

Q30. The equations of the lines passing through the point (1,0) and at a distance √3/2 from the origin, are

Q33. If the coordinates of the middle point of the portion of a line intercepted between the coordinate axes is (3, 2), then the equation of the line will be
(a) 2x + 3y = 12                                     
(b) 3x + 2y=l2
(c) 4x-3y = 6                                        
(d) 5x- 2y=10
Sol:

Q34. Equation of the line passing through (1,2) and parallel to the line y = 3x – 1 is
(a)y + 2=x+l                                    
(b) y + 2 = 3(x + 1)
(c) y -2 =  3(x — 1)                             
(d) y-2=x-l
Sol: (c) Line is parallel to the line y = 3x – 1.
So, slope of the line is‘3’.
Also, line passes through the point (1,2).
So, equation of the line is: y – 2 = 3(x – 1)

Q35. Equations of diagonals of the square formed by the lines x = 0, y = 0, x = 1 and y = 1 are

Q37. The point (4, 1) undergoes the following two successive transformations:
(i) Reflection about the line y = x
(ii) Translation through a distance 2 units along the positive x-axis Then the final coordinates of the point are
(a)  (4,3)               
(b) (3,4)                
(c)  (1,4)              
(d) (7/2,7/2)
Sol: (b) Reflection of A (4, 1) in y = x is 5(1,4).
Now translation of point B through a distance ‘2’ units along the positive x-axis shifts B to C( 1 + 2,4) or C(3,4).

Q38. A point equidistant from the lines 4x + 3y+ 10 = 0, 5x – 12y + 26 = 0 and lx + 24y – 50 = 0 is
(a)    (1,-1)              
(b)    (1, 1)               
(c)    (0,0)              
(d)   (0, 1)
Sol: (c) Clearly distance of each of three lines from (0, 0) is 2 units.

Q39. A line passes through (2, 2) and is perpendicular to the line 3x + y = 3. Its y-intercept is

Q40. The ratio in which the line 3x + 4y + 2 = 0 divides the distance between the lines 3x + 4y + 5 = 0 and 3x + 4y – 5 = 0 is
(a) 1:2 (b) 3:7 (c) 2:3 (d) 2:5

Q41. One vertex of the equilateral triangle with centroid at the origin and one side asx + y- 2 = 0is
(a) (-1,-1) (b) (2,2) (c) (-2,-2) (d) (2,-2)

Fill in the Blanks Type Questions

Q42. If a, b, c are in A.P., then the straight lines ax + by + c = 0 will always pass through __________    

Q43.The line which cuts off equal intercept from the axes and pass through the point (1, -2) is

Q44. Equations of the lines through the point (3, 2) and making an angle of 45° with the line x – 2y = 3 are _____

Q45. The points (3,4) and (2, -6) are situated on the_____ of the line 3x – 4y – 8= 0.

Sol: Given line is 3x – 4y – 8 = 0
For point (3, 4), 3(3) – 4(4) – 8 = -15 < 0
For point (2, -6), 3(2) – 4(—6) – 8 = 22 > 0
Hence, the points (3,4) and (2, -6) lies on opposite side of the line.

Q46. A point moves so that square of its distance from the point (3, -2) is numerically equal to its distance from the line 5x – 12y = 3. The equation of its locus is______ .
Sol:

Q47. Locus of the mid-points of the portion of the line x sin θ + y cos θ = p intercepted between the axes is ______

True/False Type Questions

Q48. If the vertices of a triangle have integral coordinates, then the triangle can not be equilateral.

This is a contradiction to the fact that the area is a rational number. Hence, the triangle cannot be equilateral.

Q49. The points A(-2, 1), B(0, 5), C(-l, 2) are collinear.
Sol: False

Q50. Equation of the line passing through the point (a cos³ , a sin³ ) and perpendicular to the line
x sec + y cosec = a isx cos -y sin = a sin 2

Q51. The straight line 5x + 4y = 0 passes through the point of intersection of the straight lines x + 2y— 10 = 0 and 2x +y + 5 = 0.

Q53. The equation of the line joining the point (3, 5) to the point of intersection of the lines 4x +y – 1 = 0 and lx – 3v – 35 = 0 is equidistant from the points (0, 0) and (8, 34).

Sol: True

Q55. The line ax + 2y + 1 = 0, bx + 2y + 1 = 0 and cx + 4y + 1 = 0 are concurrent, if a, b and c are in GP.
Sol: False

Q56. Line joining the points (3, -4) and (-2, 6) is perpendicular to the line joining the points (-3, 6) and (9, -18).

NCERT Exemplar ProblemsMathsPhysicsChemistryBiology

We hope the NCERT Exemplar Class 11 Maths Chapter 10 Straight Lines help you. If you have any query regarding NCERT Exemplar Class 11 Maths Chapter 10 Straight Lines, drop a comment below and we will get back to you at the earliest.