CBSE Sample Papers for Class 11 Maths Set 9 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 11 Maths with Solutions Set 9 are designed as per the revised syllabus.

CBSE Sample Papers for Class 11 Maths Set 9 with Solutions

Time Allowed : 3 hours
Maximum Marks : 80

General Instructions:

• All questions are compulsory.
• The question paper consists of 36 questions divided into 4 Sections A, B, C and D.
• Section A comprises of 20 questions of 1 mark each, Section B comprises of 6 questions of 2 marks each, Section C comprises of 6 questions of 4 marks each and Section D comprises of 4 questions of 6 marks each.
• There is no overall choice. However internal choice has been provided in 6 questions of 1 mark, 2 questions of 2 marks, 2 questions of 4 marks and 2 questions of 6 marks. You have to attempt only one of the alternatives in all such questions.
• Write the serial number of questions before attempting.
• Use of a calculator is not permitted.

Section – A
Question numbers 1 to 10 carries 1 mark each. For each of these questions, four alternative choices have been provided of which only one is correct. Select the correct choice :

Question 1.
If 6 boys and 6 girls sit in a row at random, then the probability that all the girls sit together is.
(A) \(\frac{1}{432}\)
(B) \(\frac{12}{431}\)
(C) \(\frac{1}{132}\)
(D) None of these
Answer:
(C) \(\frac{1}{132}\)

Explanation:
We know that,
Probability of 6 girls sitting together will be
⇒ \(\frac{7 !}{12 !}=\frac{1}{132}\)

Question 2.

(A) 3
(B) 1
(C) 0
(D) 2
Answer:
(D) 2

Explanation:
On evaluating the limit

Question 3.
If the focus of a parabola is (0, -3) and its directrix is y = 3, then its equation is.
(A) x² = – 12y
(B) x² = 12y
(C) y² = – 12y
(D) y² = 12y
OR
If

where [ ] denotes the greatest integer function, then \(\lim _{x \rightarrow 0}\)f(x) is equal to :
(A) 1
(B) 0
(C) -1
(D) Does not exist
Answer:
(A) x² = – 12y

Explanation:
Given that,
Directrix, y = 3
Focus is (0, -3)
Equation of parabola is
⇒ x² = – 12y

OR

(D) Does not exist

Explanation:
On evaluating the limit

= Does not exist

Question 4.
The distance between the lines y = mx + c₁ and y = mx + c₂ is
(A) \(\frac{c_1-c_2}{\sqrt{m^2+1}}\)
(B) \(\frac{\left|c_1-c_2\right|}{\sqrt{1+m^2}}\)
(C) \(\frac{c_2-c_1}{\sqrt{1+m^2}}\)
(D) 0
A line cutting off intercept -3 from the Y-axis and the tangent at angle to the X-axis is \(\)\frac{3}{5}, its equation is
(A) 5y – 3x + 15 = 0
(B) 3y – 5x + 15 = 0
(C) by – 3x – 15 = 0
(D) None of the above
Answer:
(B) \(\frac{\left|c_1-c_2\right|}{\sqrt{1+m^2}}\)

Explanation:
The distance between the lines y = mx + C₁ and y = mx + C₂
Distance = \(\frac{\left|c_1-c_2\right|}{\sqrt{1+m^2}}\)

OR

(A) 5y – 3x + 15 = 0

Explanation:
Given that,
Intercept on y-axis = – 3
Tangent of angle with x-axis – 3/5
Equation of required line is
5y – 3x + 15 = 0

Question 5.
The lengths of three unequal edges of a rectangular solid block are in G.E The volume of the block is 216 cm³ and the total surface area is 252 cm². The length of the longest edge is
(A) 12 cm
(B) 6 cm
(C) 18 cm
(D) 3 cm
Answer:
(A) 12 cm

Explanation:
Given that,
Length of three unequal edges are in GP
Volume of block = 216 cm³
Total surface area of block = 252 cm²
So we have,
Length of longest edge = 12 cm

Question 6.
If the middle term of \(\left(\frac{1}{x}+x \sin x\right)^{10}\) is equal to, then value of x is
(A) 2nπ +\(\frac{\pi}{6}\)
(B) nπ + \(\frac{\pi}{6}\)
(C) nπ + (- 1)ⁿ\(\frac{\pi}{6}\)
(D) nπ + (- 1)ⁿ\(\frac{\pi}{3}\)
OR
Total number of words formed by 2 vowels and 3 consonants taken from 4 vowels and 5 consonants is equal to
(A) 60
(B) 120
(C) 7200
(D) 720
Answer:
(C) nπ + (- 1)ⁿ\(\frac{\pi}{6}\)

Explanation:
We know that,
Middle term of the expression (1/x + x sin x)¹⁰ = 0
So the value of x on comparing
nπ + (- 1)ⁿ\(\frac{\pi}{6}\)

OR

(C) 7200

Explanation:
We know that,
Total number of words formed
= ⁴C₂ × ⁵C₃ × 5!
= 7200

Question 7.
Solution of a linear inequality in variable x is represented on number line, choose the correct answer.

(A) x ∈ (-∞, 5)
(B) x ∈ (-∞, 5]
(C) x ∈ [5, ∞)
(D) x ∈ (5, ∞)
Answer:
(D) x ∈ (5, ∞)

Explanation:
We have,
From the given number line
We get the required solution set as
⇒ x ∈ (5, ∞)

Question 8.
If X and Y are twosets and X denotes the complement of X, then X ∩ (X ∪ Y)’ is equal to
(A) X
(B) Y
(C) Φ
(D) X ∩ Y
Answer:
(C) Φ

Explanation:
We have,
⇒ X ∩ (X ∪ Y)’ – X ∩ (X’ ∩ Y’)
= Null set(Φ)

Question 9.

is equal to
(A) n
(B) 1
(C) – n
(D) 0
Answer:
(A) n

Explanation:
We have,
On simplifying the limit,
\(\lim _{x \rightarrow 0} \frac{(1+x)^{n-1}}{x}\) = n

Question 10.
The locus of a point for which Y = 0 and Z = 0, is
(A) equation of X-axis.
(B) equation of Y-axis.
(C) equation of Z-axis.
(D) None of these
Answer:
(A) equation of X-axis.

Explanation:
We know that,
Locus of a point for which Y = 0 and Z = 0 is X-axis

Question numbers l11to 15 carry 1 mark each. Write whether the statement is true false.

Question 11.
If A and B are two candidates seeking admission in an engineering college. The probability that A is selected is 0.5 and the probability that both A and B are selected is at most 0.3. It is possible that the probability of B getting selected is 0.7 ?
Answer:
False

Explanation:
Given that,
The probability that A is selected is 0.5 and the probability that both A and B are selected is at most 0.3
Probability of B getting selected = 0.2
so the given statement is False

Question 12.
If P is a point on the ellipse \(\frac{x^2}{16}+\frac{y^2}{25}\) = 1 whose foci are S and S’, then PS + PS’ = 9
Answer:
False

Explanation:
For the given ellipse We have, PS + PS’ = 2b – 10
So the given statement is False

Question 13.
If 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.
Answer:
True

Explanation:
Three animals can be loaded on stalls in 3¹² ways
So it is a true statement

Question 14.
Multiplication of a non-zero complex number by -i rotate the point about origin through a right angle in the anti-clockwise direction.
OR
The inequality | z – 4| < | z – 2| represents the region given by x > 3.
Answer:
False

Explanation:
Multiplication of a non-zero complex number by – i rotate the point about origin through a right angle in the clockwise direction So the given statement is False.

OR

True

Explanation:
It is true that the inequality
⇒ | z – 4| < | z – 2| represents the region given by x > 3

Question 15.
If A × B = {(a, x), (a, y), (b, x), (b,y)} then A = {a, b}, B = {x, y}.
Answer:
True

Explanation:
For the given sets A and B We have,
A × B = {(a, x), (a, y), (b, x), (b, y)}

Question numbers 16 to 20 carry 1 mark each.

Question 16.
Find the co-ordinates of the vertices, foci, eccentricity and length the latus rectum of the hyperbola.
\(\frac{x^2}{25}-\frac{y^2}{4}\) = 1
OR
Find the co-ordinate of focus and length of latus rectum of parabola 3y2 = 8x.
Answer:
The given hyperbola is
\(\frac{x^2}{5^2}-\frac{y^2}{2^2}\) = 1, a = 5, b = 2.
Vertices = (±8, 0) = (±5, 0)
foci = (±c, 0), where c = \(\sqrt{a^2+b^2}\) = √29
= (±√29, 0)
eccentricity = \(\frac{\sqrt{a^2+b^2}}{a}=\frac{\sqrt{29}}{5}\)
Latus rectum = 2 \(\frac{b^2}{a}=\frac{8}{5}\)

Commonly Made Error:
Many students get confused between the equation of Ellipse and hyperbola. They write the equation of ellipse when asked of the hyperbola.

Answering Tip:
Learn the equation of hyperbola and terms related to it. Understand the equation and its properties with the help of its diagram which will help you to remember long.

OR

Given equation of the parabola is
3y² = 8 x

Question 17.
Show that, if x² + y² = 1, then the point (x, y, \(\sqrt{1-x^2-y^2}\)) is at a distance 1 unit from the origin.
Answer:
Given that x² + y² = 1
∴ Distance of the point (x, y, \(\sqrt{1-x^2-y^2}\)) from origin is given as

Question 18.
Find the slope of a line perpendicular to the line passing through (-3, 8) and (2, -2).
Answer:
Slope of the line passing through (- 3, 8) and (2, -2), m₁ = \(\frac{8+2}{-3-2}\) = – 2
If m₂ be the slope of the perpendicular line, then m₁m₂ = – 1
⇒ m₂ = \(\frac{1}{2}\) is the required slope of line.

Question 19.
For x, solve the following inequalities |x – 1| ≤ 5, |x| ≥ 2
Answer:
We have, |x – 1| ≤ 5
– 5 ≤ x – 1 ≤ 5
⇒ – 4 ≤ x ≤ 6
⇒ x ∈ [- 4, 6]

and | x | ≥ 2
⇒ x ≤ – 2 or x ≥ 2
⇒ x ∈ (- ∞, – 2] ∪ [2, ∞)

On combining Eqs. (i) and (ii), we get
x ∈ (- 4, – 2] ∪ [2, 6]

Question 20.
What is the maximum value of 3 – 7 cos 5x ?
OR
Prove that cos θ cos \(\frac{\theta}{2}\) – cos 3θ cos\(\frac{9 \theta}{2}\) = sin 7θ sin 8θ
Answer:
For maximum value of 3 – 7 cos 5x, cos 5x must be minimum, so minimum value is -1 and hence maximum value = 3 – 7(-1)
= 3 + 7
= 10

OR

= – {sin 8θ sin(- 7θ)}
= sin 8θ sin 7θ (∵ sin (- θ) = – sin θ)
= RHS
∴ LHS = RHS
Hence Proved

Section – B
Question numbers 21 to 26 carry 2 marks each.

Question 21.
For all sets A and B, A – (A – B) = A ∩ B.
OR
Show that if A ⊂ B, then C – B ⊂ C – A.
Answer:
L.H.S. = A – (A – B)
= A – (A ∩ B’)[∴ A – B = A ∩ B’]
= A ∩ (A ∩ B’)’
= A ∩ [A’ ∪ (B’)’] [∴ (A ∪ B)’ = A ∩ B’]
= A ∩ [A’ ∪ B] [∴ (A’)’ = A]
= (A ∩ A’) ∪ (A ∩ B)
= Φ ∪ (A ∩ B)
= A ∩ B
= R.H.S.
Hence proved.

OR

Let x ∈ C – B
x ∈ C and x ( B
x ∈ C and x ∈ A (∴ A ⊂ B)
x ∈ (C – A)
(C – B) ⊂ (C – A)

Question 22.
If an integer from 1 to 1000 is chosen at random, then find the probability that the integer is a multiple of 2 or a multiple of 9.
Answer:
Multiple of 2 from 1 to 1000 are 2, 4, 6, 8, …………., 1000
Let n be the number of terms of above series. 1 to 1000
∴ l = 1000
⇒ 2 + (n – 1)2 = 1000 {l = a + (n – 1)d a = 2, d = 2}
⇒ 2(n – 1) = 998
⇒ n – 1 = 499
⇒ n = 500
P(A) = \(\frac{500}{1000}\)
Multiple of 9 is
l = 999
a = 9, d = 9
These are m numbers.
So, the multiple of 9 are 9, 18, 27, …….., 999
∴ = 99
⇒ 9 (m – 1) = 990
⇒ m – 1 = 110
∴ m = 111
P(B) = \(\frac{111}{1000}\)
Now, the multiple of 2 and 9 both are 18, 36,…., 990.
Let p be the number of terms in above series.
∴ pth term = 990
⇒ 18 + (p – 1)18 = 990
⇒ (p – 1) 18 = 972
⇒ p – 1 = 54
∴ p = 55
& p(A ∩ B) = \(\frac{55}{1000}\)
Required probability s = p(A) + p(B) – p(A ∩ B)
\(\frac{500+111-55}{1000}\) = \(\frac{556}{1000}\) 556

Question 23.
If sec x cos 5x + 1 = 0, where 0 < x ≤ \(\frac{\pi}{2}\), then find the value of x.
Answer:
Given, sec x cos 5x + 1 = 0
⇒ \(\frac{1}{\cos x}\) cos5x + cos x = 0
⇒ cos 5x + cos x = 0
⇒ 2 cos \(\left(\frac{5 x+x}{2}\right)\) cos \(\left(\frac{5 x-x}{2}\right)\) = 0
[∴ cosC + cos D = 2 cos \(\left(\frac{C+D}{2}\right)\) cos \(\left(\frac{C-D}{2}\right)\)]
⇒ cos 3x . cos 2x = 0
⇒ cos 3x = 0 or cos 2x = 0
3x = (2n ± 1) \(\frac{\pi}{2}\)
x = (2n ± 1) \(\frac{\pi}{6}\)
2x = (2n + 1) \(\frac{\pi}{2}\)
∴ x = (2n + 1) \(\frac{\pi}{4}\)

Question 24.
If the lines 3x + 4y + 4 = 0 and 6x – 8y – 7 = 0 are tangents to a circle, then find the radius of the circle.
OR
Evaluate

Answer:
Given lines,
3x – 4y + 4 = 0
6x – 8y – 7 = 0
or 3x – 4y – 7/2 = 0
It is clear that lines (i) and (ii) parallel.
Now, distance between them i.e.,

∴ Distance between these line = Diameter of these circle
∴ Diameter of the circle = 3/2 and radius of the circle = 3/4

OR

Question 25.
Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination.
Answer:
One ace will be selected from four aces and four cards will be selected from (52 – 4) = 48 cards. If P is the required number of ways, then
P = C(4, 1) × C(48, 4)

= 4 × 2 × 47 × 46 × 45
= 778320 ways.

Question 26.
Find the derivative of f(x) = x cos x.
Answer:
Given,
f(x) = x cos x
∴ f ‘(x) = \(\frac{d}{d x}\) (x cos x)
= cos x \(\frac{d}{d x}\) x + x \(\frac{d}{d x}\) cos x
= cos x . 1 + x . (- sin x)
= cos x – x sin x

Section – C
Question numbers 27 to 32 carry 4 marks each.

Question 27.
Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of
\(\left(\sqrt[4]{2}+\frac{1}{\sqrt[4]{2}}\right)^n\) is √6 : 1
OR
Find the value of (a² + \(\sqrt{a^2-1}\))⁴ + (a² – \(\sqrt{a^2-1}\))⁴
Answer:

OR

Let a² = x and \(\sqrt{a^2-1}\) = y
Now,
(x + y)⁴ = ⁴C₀ x⁴ + ⁴C₁ x³y + ⁴C₂ x²y² + ⁴C₃ xy³ + ⁴C₄ y⁴
= ⁴C₀ x⁴ + ⁴C₁ x³y + ⁴C₂ x²y² + ⁴C₁ xy³ + ⁴C₀ y⁴
∵ ⁿC_r = ⁿC_{n – r}
⇒ (x + y)⁴ = x⁴ + 4x³y + 6x³y² + 4xy³ + y⁴ ……. (i)
Similarly ∵
(x – y)⁴ = x⁴ – 4x³y + 6x⁴y⁴ – 4xy³ + y⁴ …… (ii)
Adding Eqs. (i) and (ii), we get
(x + y)⁴ + (x – y)⁴ = 2x⁴ + 12x² y² + 2y⁴
= 2[x⁴ + 6x² y² + y⁴]
Putting the values of x and y i.e., x = a² and
y = \(\sqrt{a^2-1}\)
∴ (a² + \(\sqrt{a^2-1}\))⁴ + (a² – \(\sqrt{a^2-1}\))⁴
= 2[(a²)⁴ + 6(a²)²\(\left(\sqrt{a^2-1}\right)^2+\left(\sqrt{a^2-1}\right)^4\)]
= 2[a⁸ + 6a⁴(a² – 1) + (a² – 1)²]
= 2 [a⁸ + 6a⁶ – 6a⁴ + a⁴ + 1 – 2a²]
= 2 [a⁸ + 6a⁶ – 5a⁴ – 2a² + 1]
= 2a⁸ + 12a⁶ – 10a⁴ – 4a² + 2

Question 28.
Find the minimum value of p for which cos (p sin x) = sin (p cos x) has a solution in [0, 2π].
Answer:
Given, cos(p sin x) = sin (p cos x)

Question 29.
Show that the complex number z, satisfying the condition arg\(\left(\frac{z-1}{z+1}\right)=\frac{\pi}{4}\) lies on a circle.
Answer:

Question 30.
A committee of 6 is to be chosen from 10 men and 7 women, so as to contain at least 3 men and 2 women. In how many different ways can this be done, if two particular women refuse to serve on the same committee?
OR
If a, b, c are in A.P. and b, c, d are in G.E and \(\frac{1}{c}, \frac{1}{d}, \frac{1}{e}\) are in A.P., prove that a, c, e are in G.E
Answer:
∵ Total number of men = 10
and total number of women = 7
We have to form a committee containing at least 3 men and 2 women.
Number of ways = ¹⁰C₃ × ⁷C₃ + ¹⁰C₄ × ⁷C₂
= 120 × 350 + 210 × 21
= 8610
If two particular women to be always there .
∴ Number of ways = ¹⁰C₄ × ⁵C₀ + ¹⁰C₃ × ⁵C₁
Total number of committee when two particular women are never together = Total – Together
= (¹⁰C₃ × ⁷C₃ + ¹⁰C₄ × ⁷C₂) – (¹⁰C₄ × ⁵C₀ + ¹⁰C₃ × ⁵C₁)
= (120 × 35 + 210 × 21) – (210 + 120 × 5)
= 4200 + 4410 – (210 + 600)
= 8610 – 810
= 7800

OR

Given a, b, c are in A.P
⇒ 2b = a + c
Again, b, c, d are in G.P.
⇒ c² = bd
Similarly, \(\frac{1}{c}, \frac{1}{d}, \frac{1}{e}\) are in A.P.
⇒ \(\frac{2}{d}=\frac{1}{c}+\frac{1}{e}\)
⇒ \(\frac{2}{d}=\frac{e+c}{c e}\)
⇒ d = \(\frac{2 c e}{c+e}\) ……………. (iii)
Putting the values of b and d from Eq. (i) and (iii), in Eq. (ii), we get

⇒ c²(c + e) = (a + c)ce
⇒ c(c + e) = (a + c)e ⇒ c² + ce = ae + ce
⇒ c² = ae
Therefore, a, c, e, are in G.P.
Hence Proved

Question 31.
If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus.
Answer:
The origin of the coordinate plane is taken at the vertex of the parabolic reflector in such a way that the axis of the reflector is along with the positive x-axis.

This can be diagrammatically represented as

The equation of the parabola is of the form y² = 4ax (as it is opening to the right). Since the parabola passes through point A (5,10).
10² = 4a (5)
⇒ 100 = 20a
a = \(\frac{100}{20}\) = 5
Therefore, the focus of the parabola is (a, 0) = (5, 0), which is the mid-point of the diameter. Hence, the focus of the reflector is at the mid-point of the diameter.

Commonly Made Error:
Students don’t make the figure for the question which leads to deduction of marks.

Answering Tip:
Students must make a well labelled figure for required problem.

Question 32.
The mean and standard deviation of marks obtained by 50 students of a class in three subjects. Mathematics, Physics and Chemistry are given below :

Which of the three subjects show the highest variability in marks and which shows the lowest ?
Answer:
The standard deviation of Mathematics = 12
The standard deviation of Physics = 15
The standard deviation of Chemistry = 20
The coefficient of variation (C. V) is given by Standard deviation
\(\frac{\text { Standard deviation }}{\text { Mean }}\) × 100
C. V (in Mathematics) = \(\frac{12}{42}\) × 100 = 28.57
C. V (in Physics) = \(\frac{15}{32}\) × 100 = 46.87 32
C. V (in Chemistry) = \(\frac{20}{40.9}\) × 100 = 48.89
The subject with greater C. V is more variable than others.
Therefore, the highest variability in marks is in Chemistry and the lowest variability in marks is in Mathematics.

Section – D
Question numbers 33 to 36 carry 6 marks each.

Question 33.
Out of 100 students; 15 passed in English, 12 passed 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 students 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
Answer:
Let the number of students passed in Mathematics M, E be in English and S be in science. Then,
n(U) = 100, n(M) = 12, n(E) = 15, n(S) = 8, n(E ∩ M) = 6, n(M ∩ S) = 7, n(E ∩ S) = 4 and n(E ∩ M ∩ S) = 4
Let us draw a Venn diagram

According to Venn diagram, we getting
a + b + d + e = 15 …………….. (i)
b + c + e + f = 12 …………….. (ii)
d + e + f + g = 8 …………….. (iii)
n(E ∩ M) = 6 ∴ (b + e = 6) …………….. (iv)
n(M ∩ S) = 4 ∴ (e + f = 7) …………….. (v)
n(E ∩ S) = 4 ∴ (d + e = 4) …………….. (vi)
n(M ∩ E ∩ S) = 4 ∴ (e = 4) …………….. (vii)
From eqs. (iv) and (vii), we get b + 4 = 6 ∴ b = 2
From eqs. (v) and (vii), we get 4 + f = 7 ∴ f = 3
From eqs. (vi) and (vii), we get d + 4 = 4 ∴ d = 0
From eqs. (i), we get
a + b + d + e = 15 ⇒ a + 2 + 0 + 4 = 15 ⇒ a = 9
From eqs. (iii), we get
d + e + f + g = 8 ⇒ 0 + 4 + 3 + g = 8 ⇒ g = 1
(i)No. of students who passed in English and Mathematics but not in Science, b = 2
(ii) No. of students who passed in Mathematics and Science but not in English, f = 3
(iii) No. of students who passed in Mathematics only, c = 3
(iv) No. of students who passed in more than one subject = b + e + d + f = 9.

Question 34.
If cos (θ + Φ) = mcos (θ – Φ), then prove that tan θ = \(\frac{1-m}{1+m}\) cot Φ
OR
An arc is in the form of a semi ellipse. It is 8m wide and 2 m high at the centre. Find the height of the arc at a point 1.5 m from one end.
Answer:

OR

Let ABA’ be the given arc such that AA’ = 8m and OB = 2m.

The arc is a part of the ellipse.
AA’ = 8m ⇒ 2a = 8
⇒ a = 4
and OB = 2m
⇒ b = 2
∴ The equation of the ellipse is
\(\frac{x^2}{16}+\frac{y^2}{4}\) = 1
To find PM.
Putting x = 2 . 5 = \(\frac{5}{2}\) and y = PM

∴ Height of the arc at a point 1.5 from one end is \(\frac{\sqrt{39}}{4}\)m.

Commonly Made Error:
Some students have difficulty in finding the value of major and minor axis from the information given in the question.

Answering Tip:
To avoid such errors, a thorough revision on concept of ellipse is must and practice more such problems based on it.

Question 35.
Solve the following system of inequalities \(\frac{2 x+1}{7 x-1}>5, \frac{x+7}{x-8}>2\)
Answer:
The given system of inequations is
\(\frac{2 x+1}{7 x-1}\) > 5 …………….. (i)

Since, the intersection of Eqs. (iii) and (iv) is the null set. Hence, the given system of equation has no solution.

Commonly Made Error:
Students make errors in sign while simplifying the inequalities.

Answering Tip:
Students must practice more questions to avoid errors.

Question 36.
If the sum of p terms of an A.E is q and the sum of q terms is p, then show that the sum of p + q terms is -(p + q), Also, find the sum of the first p-q terms (where, p > q)
OR
If a₁, a₂, a₃, ………………… a_n are in A.P., where a₁ > 0 for all show that

Answer:

OR

Commonly Made Error:
Students don’t remember the properties of AP and its common difference so they are not able to solve this question.

Answering Tip:
Students must learn all the properties of AP and its common difference.