CBSE Sample Papers for Class 11 Maths Set 8 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 11 Maths with Solutions Set 8 are designed as per the revised syllabus.

CBSE Sample Papers for Class 11 Maths Set 8 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.
When tested, the lives (in hours) of 5 bulbs were noted as follows 1357, 1090, 1666, 1494, 1623. The mean deviation (in hours) from their mean is
(A) 178
(B) 179
(C) 220
(D) 356
Answer:
Option (A) is correct.

Explanation:
We know that,
The mean deviation about the mean of the 5 bulbs is: 178

Question 2.
The coordinates ol the foot of perpendiculars from the point (2 3) on the line y = 3x + 4 is given by
(A) \(\left(\frac{37}{10}, \frac{-1}{10}\right)\)
(B) \(\left(\frac{-1}{10}, \frac{37}{10}\right)\)
(C) \(\left(\frac{10}{37},-10\right)\)
(D) \(\left(\frac{2}{3}, \frac{-1}{3}\right)\)
Answer:
Option (B) is correct.

Explanation:
We know that,
Coordinates of foot of perpendicular drawn from the point (2, 3) on the line
y = 3x + 4
is glven by \(\left(\frac{-1}{10}, \frac{37}{10}\right)\)

Question 3.
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 Hindi nor English newspaper are.
(A) 210
(B) 290
(C) 180
(D) 260

OR

The set (A ∩ B ‘)’ ∪ (B ∩ C) is equal to
(A) A’ ∪ B ∪ C
(B) A’ ∪ B
(C) A’ ∪ C
(D) A’ ∩ B
Answer:
Option (B) is correct.

Explanation:
We have,
The number of persons who read neither Hindi nor English is
⇒ 840 – (450 + 300 – 200) = 290

OR

Option (B) is correct.

Explanation:
We have,
⇒ (A ∩ B’)’ ∪ (B ∩ C)
⇒ (A’ ∪ B) ∪ (B ∩ C)
⇒ A’ ∪ B.

Question 4.
If f(x) = \(\frac{x^n-a^n}{x-a}\) for some constants, then f'(a) is equal to :
(A) 1
(B) 0
(C) -1
(D) Does not exist

OR

\(\lim _{x \rightarrow 0} \frac{x^2 \cos x}{1-\cos x}\) Is Equal to:
(A) 2
(B) \(\frac{2}{3}\)
(C) \(\frac{-3}{2}\)
(D) 1
Answer:
Option (D) is correct.

Explanation:
We have,
f(x) = \(\frac{x^n-a^n}{x-a}\)
⇒ f ‘(a) = 0/0 (Does not exist)
OR
Option (A) is correct.

Explanation:
We know that,
\(\lim _{x \rightarrow 0} \frac{x^2 \cos x}{1-\cos x}\) = 2

Question 5.
Let x₁, x₂ …. x_n be n observations. Let w_i – [x_i + k] for i = 1, 2,n, where l and k are constants. If the mean of x_i‘s is 48 and their standard deviation is 12, the mean of w_i‘s is 55 and the standard deviation of ‘s is 15, then the value of l and k should be S3
(A) l = 1.25,k = – 5
(B) l = -125,k = 5
(C) l = 2.5, k = 5
(D) l = 2.5, k = 5
Answer:
Option (A) is correct.

Explanation:
We have,
Mean of X_i = 48 and SD = 12
Mean of W_i = 55 and SD = 15
From the given values,
We get l = 1.25 and k = -5

Question 6.
If P(A ∪ B) = P(A ∩ B) for any two events A and B, then
(A) P(A) = P(B)
(B) P(A) > P(B)
(C) P(A) < P(B)
(D) None of these
Answer:
Option (A) is correct.

Explanation: Given that,
P(A ∪ B) = P(A ∩ B)
so it is possible when P(A) = P(B)

Question 7.
The distance of a point P(3,4,5) from the YZ-plane is
(A) 3 units
(B) 4 units
(C) 5 units
(D) 550 units
Answer:
Option (A) is correct.

Explanation:
The distance of the point P(3, 4, 5) from the YZ plane is equal to abscissa i.e. 3 units.

Question 8.
The equation of a circle with origin as centre and passing through the vertices of an equilateral triangle whose median is of length 3a is
(A) x² + y² = 9a²
(B) x² + y² = 16a²
(C) x² + y² = 16a²
(D) x² + y² = a²
Answer:
Option (C) is correct.

Explanation:
Given that,
Centre of circle = (0,0)
Length of median = 3a
Equation of circle passes through vertices of equilateral triangle is
x² + y² = 4a²

Question 9.
Let S, 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
(13) 10
OR
The minimum valueof 4^x + 4^1-x x ∈ R,is
(A) 2
(B) 4
(C)1
(D) 0
Answer:
Option (B) is correct.

Explanation:
Given that,
For an AP
⇒ S_2n = 3S_n
then we have,
⇒ S_3n : S_n = 6 : 1

OR

Option (B) is correct.

Explanation:
We have,
The minimum value of
⇒ 4x + 41 – x , for x belongs to R
= 4

Question 10.
The inequality representing the following graph

(A) |x| < 5
(B) |x| < 5 (C) |x| > 5
(D) |x| > 5
Answer:
Option (B) is correct.

Explanation:
We have,
For the given graph the required inequality is
⇒ |x| ≤ 5

Question numbers 11 to 15 carry 1 mark each. Write whether the statement is true or false.
Question 11.
There will be only 24 selections containing at least one red ball out of a bag containing 4 red and 5 black balls. If it is being given that the balls of the same colour are identical.
Answer:
True

Explanation:
It is true that there will be only 24 selections containing at least one red ball out of a bag containing 4 red and 5 black balls.

Question 12.
The probability of the intersection of two events A and B is always less than or equal to those favourable to the event A.
Answer:
True

Explanation:
It is true that the probability the intersection of two events A and B is always less than or equal to those favourable to the event A

Question 13.
The number of terms in the expansion of (a + b)n where n e N is one less than the power n.
Answer:
False

Explanation:
The number of terms in the expansion of (a + b)n is always n + 1 So the given statement is False.

Question 14.
If A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}, then (A × B) ∪ (A × 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)}
Answer:
True

Explanation:
It is true that for the given sets:
(A × B) ∪ (A × 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)}

Question 15.
The locus represented by | z – 1| = |z – i| is a line perpendicular to the join of (1,0) and (0,1). HJI3
OR
The inequality |z – 4| < |z – 2| represents the region given by x > 3.
Answer:
False

Explanation:
The locus represented by |z – 1| = |z – i| is a line parallel to the join of (1, 0) and (0, 1) So the given statement is False.

OR

True

Explanation:
For the given inequality:
⇒ |z – 4| < |z – 2| On simplifying we get, ⇒ x > 3

Question numbers 16 to 20 carry 1 marks each.

Question 16.
Reduce the equation √3x + y — 4 into normal form and find the values of p and a.
OR
Reduce the equation 3x + 4y = 5 into slope – intercept form and find its slope.
Answer:
The given equation is √3x + y = 4

⇒ x cos 30° + y sin 30° = 2, is the required equation of line in normal form where p = 2, α = 30°
OR
Given equation is
3x + 4y = 5
Here, m = \(-\frac{A}{B}=-\frac{3}{4}\)
and the slope intercept form is
y = \(-\frac{A}{B}=-\frac{3}{4}\)
y = \(-\frac{3}{4} x-\frac{5}{4}\)
Slope of the line = \(-\frac{3}{4}\)

Question 17.
Find the length of the major axis and minor axis of the ellipse 16x² + 25 – y² = 400.
OR
Find equation of circle whose end points of its diameter are (-2,3) and (0, -1).
Answer:
Given equation of the ellipse is 16x² + 25 – y² = 400
⇒ \(\frac{x^2}{25}+\frac{y^2}{16}\) = 1
\(\frac{x^2}{5^2}+\frac{y^2}{4^2}\) = 1, a = 5, b = 4
Length of Major axis = 2a = 10
and length of Minor axis = 2b = 8. 1
OR
The equation of a circle with (x₁, y₁) and (x₂, y₂)
as end points of one of its diameter is (x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0
Given, (x₁, y₁) = (-2, 3) and (x₂, y₂) = (0,—’1)
(x + 2)( x – 0) + (y – 3)( y +1) = 0
⇒ x² + 2 x + y² – 2 y – 3 = 0
⇒ x² + y² + 2 x – 2 y – 3 = 0 , is the required equation of the circle.

Question 18.
If P(A ∪ B) = P(A) + P(B), then what can be said about the events A and B ?
Answer:
We Know
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
⇒ P(A ∩ B) = 0
[∵ P(A ∪ B) = P(A) + P(B)]
P(A ∩ B) = 0
[∵ P(A ∪ B) = P(A) + P(B)]
∴ The events A and B are mutually exclusive.

Question 19.
Find the equation of the straight line which passes through the point (1, – 2) and cuts off equal intercepts from axes.
Answer:
Let the intercepts along the X and Y-axes are a and a respectively.
Equation of the line is \(\frac{x}{a}+\frac{y}{a}\) = 1 …(i)
Since, the point (1, -2) lies on the line,
\(\frac{1}{a}-\frac{2}{a}\) = 1
\(\frac{1-2}{a}\) = 1
a = -1

On putting a = -1 in Eq.
\(\frac{x}{-1}+\frac{y}{-1}\) = 1
x + y = -1 ⇒ x + y + 1 = 0 required equation of line.

Question 20.
Solve : \(\frac{2 x+3}{4}\) – 3 < \(\frac{x-4}{3}\) – 2, x ∈ R.
Answer:

Commonly Made Errors
Students make errors in putting the sign of inequalities while simplifying them.

Answering Tips
Students should always remember that two real number or two algebraic expressions related to symbolic <, >, < or > form an inequality.

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

Question 21.
If tan (A + B) = p, tan (A – B) = q, then show that: tan 2A = \(\frac{p+q}{1-p q}\)
OR
Evaluate sin (π + x) sin (π – x) cosec2x.
Answer:
Given, tan( A + B) = p , and tan( A – B) = q ,
L.H.S. = tan2 A = tan(A + B + A – B)
= tan [(A + B) + (A – B)]
= \(\frac{\tan (A+B)+\tan (A-B)}{1-\tan (A+B) \cdot \tan (A-B)}\)
= \(\frac{p+q}{1-p q}\)
= R.H.S.

OR
Given, sin (π + x) sin (π – x) cosec²x
= (-sin x) (sin x) . cosec²x
= -sin²x . cosec²x
= -sin²x. \(\frac{1}{\sin ^2 x}\)
= -1

Question 22.
Let f = {(1,1), (2,3), (0, -1), (-1, -3)} be a function from Z to Z defined by f(x) = ax + b for some integers a and b. Determine a, b.
Answer:
Given, f = {(1, 1), (2, 3), (0, -1), (-1, -3)}
and f(x) = ax + b
Let y = ax + b
At x = 1, y = 1
1 = a + b
At x = 2, y = 3
3 = 2a + b
Subtracting eq. (i) from eq. (ii), we get
a = 2
Substituting the value of a in eq. (i), we get
b = 1 – 2 ⇒ b = -1
Hence, f(x) = ax + b = 2x – 1

Question 23.
Evaluate \(\lim _{x \rightarrow 0} \frac{\sqrt{2}-\sqrt{1+\cos x}}{\sin ^2 x}\)
OR
Evaluate \(\lim _{x \rightarrow 0} \frac{2 \sin x-\sin 2 x}{x^3}\)
Answer:

Question 24.
For all sets A and B, where (A – B) ∪ (A ∩ B) = A.
Answer:
True

L.H.S. = (A – B) ∪ (A ∩ B)
= [(A – B) ∪ A] ∩ [(A – B) ∪ B]
= A ∩ (A ∪ B)
= A
= R. H. S.

Question 25.
If the letters of the word ‘ALGORITHM’ are arranged at random in a row what is the probability the letters ‘GOR’ must remain together as a unit ?
Answer:
Number of letters in the word ALGORITHM’ are 9
If ‘GOR’ remain together, then considered it as 1 number
Number of letters = 6 + 1 = 7
The number of words if ‘GOR’ remain together = 7!
Total number of words from the letters of the word ALGORITHM’ = 9!
Required probability = \(\frac{7 !}{9 !}=\frac{7 !}{9 \times 8 \times 7 !}=\frac{1}{9 \times 8}=\frac{1}{72}\)

Question 26.
Find the derivative of xⁿ + ax^n-1 + a²x^n-2 + … + a^n-1x + aⁿ, for some real number a.
Answer:

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

Question 27.
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’?
OR
How many committee of five person with a chairperson can be selected from 12 persons?
Answer:
The letters of the word ‘RACHIT’ in alphabetical order are A, C, H, I, R and T.
Now, words beginning with A = 5!
words beginning with C = 5!
words beginning with H = 5!
words beginning with I = 5!
Word beginning with R i.e., RACHIT = 1
Rank of the word ‘RACHIT’ in dictionary = 4 × 5!
+ 1 = 4 × 120 + 1
= 480 + 1 = 481

OR

∵ Total number of persons = 12
and number of persons to be selected = 5
Out of 12 persons a chairperson is selected
= ¹²C₁ = 12 ways
Now, remaining 4 persons are selected out of 11 persons.
Number of ways = 11C4 = 330.
Total number of ways to form a committee of 5 persons = 12 × 330 = 3960

Question 28.
If p is a real number and the middle term in the expansion of (\(\frac{p}{2}\) + 2)⁸ is 1120, then find the value of p.
OR
In the expansion of (x + a)ⁿ, if the sum of odd terms is denoted by O and the sum of even terms by E. Then, prove that O² – E² = (x² – a²)ⁿ.
Answer:
Given expansion is (\(\frac{p}{2}\) + 2)⁸
Here, n = 8 [even]
Since, this Binomial expansion has only one middle term i.e., (\(\frac{8}{2}\) + 2)^th =5th term
T = T₄₊₁ = ⁸C₄\(\left(\frac{p}{2}\right)^{g-4}\).2⁴
⇒ 1120 = ⁸C₄p⁴ 2^-42⁴
⇒ 1120 = \(\frac{8 \times 7 \times 6 \times 5 \times 4 !}{4 ! 4 \times 3 \times 2 \times 1} p^4\)
⇒ 1120 = 7 × 2 × 5 × p⁴
⇒ p⁴ = \(\frac{1120}{70}\) = 16 ⇒ p = 2
⇒ p² = 4
⇒ p = ±2

Commonly Made Errors
Students confuse between the middle terms of the binomial expansion.

Answering Tips
Students must learn about the middle terms of all odd and even power binomial expansion

OR
Given expansion is (x + a )ⁿ
(x + a) = ⁿC₀xⁿ a⁰ + ⁿC₁x^n-1 a¹ + ⁿC₂x^n-2 a² + ⁿC₃x^n-3 3a³ +… + ⁿC_naⁿ
Now, sum of odd terms i.e. O = ⁿC₀x + ⁿC₂x^n-2 a² +.. and sum of even terms
i.e. E = ⁿC₁x a + ⁿC₃x^n-3 a³ +…
(x + a)ⁿ = O + E
Similarly, (x – a)ⁿ = O – E
(O + E)(O – E) = (x + a)ⁿ(x – a)ⁿ
[on multiplying Eqs. (i) and (ii)]
⇒ O² – E² = (x² – a²)ⁿ

Question 29.
If |z + 1| = z + 2(1 + i) then find z.
Answer:
Given that: |z +1| = z + 2(1 + i)
Let z = x + iy
So, |x + iy + 1| = (x + iy) + 2(1 + i)
|(x +1) + iy| = x + iy + 2 + 2i
|(x +1) + iy| = (x + 2) +(y + 2)i
\(\sqrt{(x+1)^2+y^2}\) = (x + 2) + (y + 2)i
[∵ |x+iy| = \(\sqrt{(x+1)^2+y^2}\)]

Squaring both sides, we get
(x +1)² + y² = (x + 2)² + (y + 2)² . i² + 2(x + 2)(y + 2)i
^ x² + 1 + 2x + y² = x² + 4 + 4x – y² – 4y – 4 + 2(x + 2)( y + 2)i

Comparing the real and imaginary parts, we get
x² + 1 + 2x + y² = x² + 4x – y² – 4y and 2(x + 2)( y + 2) = 0
⇒ 2y² – 2x + 4y +1 = 0 (i)
and (x + 2)(x + 2) = 0 (ii)
x + 2 = 0 or y + 2 = 0
x = -2 or y = -2
Now put x = -2 in eqn. (i)
2y² – 2 x (-2) + 4y +1 = 0
⇒ 2y² + 4 + 4y + 1 = 0
⇒ 2y² + 4y + 5 = 0
b² – 4ac = (4)² – 4 x 2 x 5 3
= 16 – 40 = -24 < 0, no real roots.
Put y = — 2 in eqn. (i)
2(-2)² – 2x + 4(-2) +1 = 0 1
8 – 2x – 8 +1 = 0 ⇒ x = \(\frac{1}{2}\) and y = – 2 2
Hence, z = x + iy = (\(\frac{1}{2}\) – 2i)

Question 30.
If the real part of \(\frac{\bar{z}+2}{\bar{z}-1}\) is 4, then show that the locus of the point representing z in the complex plane is a circle.
Answer:
Let z = x + iy
z̅ = x – iy
Now,

⇒ x² + y² + x – 2 = 4(x² + 1 – 2x + y²)
⇒ x² + y² + x – 2 = 4x² + 4 – 8x + 4y²
⇒ 3x² + 3y² – 9x + 6 = 0
⇒ x² + y² – 3x + 2 = o
Clearly, it represents a circle.
Hence z hes on circle.

Question 31.
The mean and standard deviation of some data for the time taken to complete a test are calculated with the following results, Number of observation = 25, mean = 18.2, standard deviation = 3.25. Further, another set of 15 observation x₁, x₂….x₁₅, also in second, is now available and we have \(\sum_{i=1}^{15}\)x_i, = 279 and \(\sum_{i=1}^{15}\)x_i² = 5524. Calculate the standard deviation based on all 40 observations <=t 1=1 EJD
Answer:

Question 32.
If the points (0, 4) and (0, 2) are respectively the vertex and focus of a parabola, then find the equation of the parabola.
Answer:
Given that the coordinates, vertex of the parabola (0, 4) and focus of the parabola (0,2).

⇒ x² + y² – 4y + 4 = y² + 36 – 12y
⇒ x² + 8y = 32 is the required equation of parabola.

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

Question 33.
In a survey it was found that 21 people liked product A, 26 liked product B and 29 liked product C. If 14 people liked products A and B, 12 people liked products C and A, 14 people liked products B and C, then find how many people like product C only ?
Answer:
Let A, B and C denote the people liked the products A, B and C, respectively.
Here, n(A) = 21, n(B) = 26, n(C) = 29,
n(A ∩ B) = 14, n(C ∩ A) = 12, n(B ∩ C) = 14, n(A ∩ B ∩ C) = 8 3
n(C only) = n(C) – n(C ∩ A) – n(C ∩ B) + n(C ∩ B ∩ A)
= 29 – 12 -14 + 8 =11 3
The number of people who liked product C only = 11

Question 34.
If sin (θ + α) = a and sin (θ + α) = b, then prove that cos2 (α – β) – 4ab cos (θ – β)= 1 – 2a² – 2b².
OR
If tan θ =\(\frac{\sin \alpha-\cos \alpha}{\sin \alpha+\cos \alpha}\), then show that sin α + cos α = √2cosθ
Answer:
Given, sin (θ + α) = a and sin (θ + α) = b
cos(α – P) = cos[θ + α – θ – β]
= cos[(θ + α) – (θ + β)]
= cos(θ + α) cos (θ + β) + sin(θ + α)sin (θ + β) sin(θ + α)sin (θ + β)

OR

Question 35.
Solve the following system of inequalities graphically
x – 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0, y ≥ 1.
Answer:
The inequalities are
x – 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0, y ≥ 1.
(a) The line x – 2y = 3 passes through (3, 0) and (0, \(-\frac{3}{2}\))
This is represented by AB. 0 ≤ 3 denotes origin lies in the region x – 2y ≤ 3.
Region above this line and including its points represents x – 2y ≤ 3.

(b) 3x + 4y = 12 passes through (4, 0) and (0, 3). CD represents this line.
Since 0 ≥ 12, origin does not lies in 3x + 4y ≥ 12. The region above CD and including points of CD represents 3x + 4y ≥ 12.
(c) x ≥ 0 the region on the right of y-axis and all the points on it.
(d) y = 1 is the line parallel to x-axis at a distance 1 unit from on it.
y ≥ 1 or -1 ≥ 0. Since -1 ≥ 0, origin does not lies in this region.
y ≥ 1 is the region above y = 1 and the points lying on it. The common shaded region is YCPQB which represents the solution of the given inequalities.

Commonly Made Errors
Students don’t highlight the feasible region which leads to deduction of marks.

Answering Tips
Students must mark and highlight the feasible region in the graph.

Question 36.
A square is drawn by joining the midpoints of the sides of a square. A third square is drawn inside the second square in the same way and the process is continued indefinitely. If the side of the first square is 15 cm, then find the sum of the areas of all the squares so formed.
OR
If the first and nth term of a G.P is a and b, respectively and if p is the product of n terms, prove that P² = (ab)ⁿ.
Answer:
Let A₁ A₂, A₃ A₄ be the first square with each side equal to 15 cm. Let B₁, B₂ B₃ B₄ be the mid-points of its sides. Then

Similarly, the side of fourth square is \(\frac{15}{2 \sqrt{2}}\)cm and so on.
Sum of the areas of all the squares so formed
\(\left[(15)^2+\left(\frac{15 \sqrt{2}}{2}\right)^2+\left(\frac{15}{2}\right)^2+\left(\frac{15}{2 \sqrt{2}}\right)^2+\cdots \infty\right]\) sq.cm
[∵ Area = (sides)²]
= \(=\frac{225}{1-\left(\frac{1}{2}\right)}=\frac{225 \times 2}{1}\) = 450 sq.cm

OR

Let the G.P is A, AR, AR², AR³ …………….
Given, first term A = a ………….(i)
nth term AR^n-1 = b
Now, P = product of n terms
⇒ P – A x AR¹ x AR² x AR³ x …………… x n terms
p = A^{1+1+1-+nterms}R^{1+2 +3+-+(n-1)}
p = AⁿR^{\(\frac{n(n-1)}{2}\)}
[∵ Sum of first n natural numbers = \(\frac{n(n-1)}{2}\)
Squaring on both sides,
p² = A²ⁿR^n(n-1)
⇒ p² = AⁿAⁿR^n(n-1) = Aⁿ(AR^n-1)ⁿ
⇒ p² = aⁿbⁿ [ using Eqs.(i) and (ii)]
⇒ p² = (ab)ⁿ
Hence Proved.