CBSE Sample Papers for Class 11 Maths Set 4 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 11 Maths with Solutions Set 4 are designed as per the revised syllabus.

CBSE Sample Papers for Class 11 Maths Set 4 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 :

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 choice is correct. Select the correct choice :

Question 1.
If x, 2y, 3z are in A.E where the distinct numbers x, y, z are in G.E then the common ratio of the G.P is
(A) 3
(B) \(\frac{1}{3}\)
(C) 2
(D) \(\frac{1}{2}\)
Answer:
Option (B) is correct.

Explanation:
Given that, x, 2y, 3z are in AP and x, y, z are in GP
So y² = xz
⇒ 4y = x + 3z
From these equations we get,

Common ratio of GP as \(\frac{y}{x}=\frac{z}{y}=\frac{1}{3}\)

Question 2.
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)

For specifying a straight line, how many geometrical parameters should be known?
(A) 1
(B) 2
(C) 3
(D) 4
Answer:
Option (C) is correct.

Explanation:
We know that,
One side of triangle is x + y- 2 = 0
and Centroid is 0rigin(0,0,0)
The required vertex of equilateral triangle is (-2, -2).

Option (B) is correct.

Explanation:
We know that,
For specifying a straight line we have to know only 2 parameters one vertex and slope.

Question 3.
If y = \(\frac{\sin (x+9)}{\cos x}\) then \(\frac{d y}{d x}\) at x = 0 is equal to :
(A) cos 9
(B) sin 9
(C) 0
(D) 1

If f(x) = x – [x], ∈ R, then f'(\(\frac{1}{2}\)) is equal to :
(A) \(\frac{3}{2}\)
(B) 1
(C) 0
(D) -1
Answer:
Option (A) is correct

Explanation:
We have
y = \(\frac{\sin (x+9)}{\cos x}\)

\(\frac{d y}{d x}\) at x = 0

\(\frac{d y}{d x}\) = cos 9

Option (B) is correct

Explanation: Given that,
f(x) = x – [x]
f'(x) at x = \(\frac{1}{2}\)
⇒ f'(\(\frac{1}{2}\)) = 1

Question 4.
L is the foot of the perpendicular drawn from a point P(3,4,5) on the XY-plane. The coordinates of point L are
(A) (3,0,0)
(B) (0,4,3)
(C) (3,0,5)
(D) None of these
OR
What is the length of foot of perpendicular drawn from the point P (3, 4, 5) on Y-axis?
(A) \(\sqrt{41}\)
(B) \(\sqrt{34}\)
(C) 5
(D) None of these
Answer:
Option (D) is correct.

Explanation:
We have,
Perpendicular drawn from point P(3, 4, 5) to XY plane have coordinates as (3,4,0)

Option (B) is correct

Explanation:
Length of foot of perpendicular drawn from point P(3,4,5) on Y-axis
⇒ Distance between points P(3,4,5) and (0,4,0)
So Distance = \(\sqrt{9+0+25}\)
= \(\sqrt{34}\)

Question 5.
The following information relate to a sample of size 60, Σx² = 18000, and Σx = 960. Then, the variance is
(A) 6.63
(B) 16
(C) 22
(D) 44
Answer:
Option (D) is correct

Explanation:
We have,
n = 60, Σx = 960, Σx² = 18000
Using the formula for variance,
Variance = 44

Question 6.
If a single letter is selected at random from the word ‘PROBABILITY’. Then find the probability if is a vowel.
(A) \(\frac{1}{3}\)
(B) \(\frac{4}{11}\)
(C) \(\frac{2}{11}\)
(D) \(\frac{3}{11}\)
Answer:
Option (B) is correct

Explanation:
We know that, letter “PROBABILITY” has vowels = 4 consonants = 7
Probability of getting vowel = \(\frac{4}{11}\)

Question 7.
If |x + 2| ≤ 9, then
(A) x ∈ 6 (-7,11)
(B) x ∈ 6 (-11,7)
(C) x ∈ (-∞, -7) ∪ (11, ∞)
(D) x ∈ (-∞,-7) ∪ (-11, ∞)
Answer:
Option (B) is correct

Explanation:
Given that,
⇒ |x + 2| ≤ 9
So we have ⇒ x ≥ = -11 and x ≤ 7
x ∈ [-11, 7]

Question 8.
The eccentricity of the hyperbola whose latus rectum between the foci is is 8 and conjugate axis is equal to half of the distance
(A) \(\frac{4}{3}\)
(B) \(\frac{4}{\sqrt{3}}\)
(C) \(\frac{2}{\sqrt{3}}\)
(D) None of these
Answer:
Option (C) is correct.

Explanation:
We have,
Latus rectum = 8
Length of conjugate axis is half of the distance . between the foci
⇒ Eccentricity, e = \(\frac{2}{\sqrt{3}}\)

Question 9.
Equation of Y-axis is considered as
(A) x = 0, y = 0
(B) y = 0, z = 0
(C) z = 0, x = 0
(D) None of these
Answer:
Option (C) is correct.

Explanation:
We know that,
Equation of Y-axis is z = 0 and x = 0.

Question 10.
If a, b, c, d and e be the observation of the mean m and standard deviation s, then find the standard deviation of the observations a + k, b + k, c + k, d + k and e + k is
(A) s
(B) ks
(C) s + k
(D) \(\frac{s}{k}\)
Answer:
Option (A) is correct

Explanation:
We have,
Mean = m
SD = s
For the observations a+k, b+k, c+k, d+k and c+k
The SD remains unaffected of addition of any constant value
So, Required SD = s

Question numbers 11 to 15 carry 1 mark each. Write whether the statement is true/false.

Question 11.
The last two digits of the numbers 3400 are 01.
OR
If the expansion of \(\left(x-\frac{1}{x^2}\right)^{2 \pi}\) contains a term independent of x, then n is a multiple of 2.
Answer:
True

Explanation:
The last two digits of the number (3)⁴⁰⁰ will be equal to 3⁽⁰⁰⁾ = 01

False

Explanation:
If a term in expansion of (x – 1/x²)²ⁿ is independent of x then it is not compulsory that n is a multiple of 2.
Hence given statement is False.

Question 12.
If x < -5 and x > -2, then x e (- ∞, -5).
Answer:
True

Explanation:
It is true that if x ≤ 5 then x belongs to . (-∞, -5)

Question 13.
Equation of the line passing through the point (a cos³ θ, a sin³ θ) and perpendicular to the line x sec θ + y cosec θ = a is x cos θ – y sin θ = a sin 2θ.
Answer:
False

Explanation:
Equation of the line passing through the point (a cos³θ, a sin³θ) and perpendicular to the line x sec θ + y cosec θ = a is x cos θ – y sin θ = a sin 3θ. Hence given statement is False.

Question 14.
The point (1,2) lies inside the circle x² + y² – 2x + 6y + 1 = 0
Answer:
False

Explanation:
Given Circle x² + y² – 2x + 6y + 1 = 0 For Point (1, 2)
We have,
⇒ 1 + 4 – 2 + 12 + 1 = 16 > 0
Hence given point lies outside the circle.

Question 15.
The probability that a person visiting a zoo will see the giraffe is 0.72, the probability that he will see the bears is 0.84 and the probability that he will see both is 0.52.
Answer:
False

Explanation:
The probability that the person will see both cannot be 0.52 so the given statement is FALSE.

Question 16.
If R = {x, y) : x, y ∈ z, x² + y² = 64}, then, write R in roster form.
OR
Let A = {1, 2}, B – {2, 3,4}, C = {4,5}, find A × (B ∩ C).
Answer:
Given, R = {x, y): x, y ∈ z, x² + y² = 64}
∵ x² + y² = 64 ⇒ y² = 64 – x²
R = {(0,8), (0, -8), (8,0), (-8,0)}
OR
B ∩ C = {2, 3, 4} ∩ {4, 5} = {4}
A × (B ∩ C) = {1, 2} × {4}
= {(1, 4), (2, 4)}

Question 17.
Let A, B and C be the sets such that A ∪ B = A ∩ B = A ∩ C. Show that B = C.
Answer:
Let x ∈ B …(i)
⇒ x ∈ A ∪ B
[Since B c A ∪ B, all elements of B are in A ∪ B]
⇒ x ∈ A ∪ C
(Given, A ∪ B = A ∪ Q
⇒ x ∈ A or x ∈ C ,..(ii)
Taking x ∈ A
∵ x ∈ A
Also, x ∈ B [From (i)]
∵ x ∈ A ∩ B
So, x ∈ A ∩ C
∵ (Given, A ∩ B = A ∩ C)
i.e., x ∈ A and x ∈ C
i.e., X ∈ C
∵ If x ∈ B, then X ∈ C
i.e., If an element belongs to set B, than it must belong to set C also,
∵ B ∈ C
Similarly, we can prove C ∈ B
Since, B ∈ C and C ∈ B
⇒ B = C
Hence proved

Question 18.
Write the maximum value of cos(cos x).
OR
Find the coordinates of a point on the parabola y² = 8x, whose focal distance is 4.
Answer:
The maximum and minimum value of cos x is 1 and -1.
Hence, cos (cos x) has maximum value at x = 90°
i.e. cos(cos 90°) = cos (0°) = 1
OR
Given parabola is y² = 8x (i)
8x = 4ax ⇒ a = 2
Focal distance = |x + a| = 4
⇒ |x + 2| = 4
⇒ x + 2 = ±4 ⇒ x = 2,-6
But x ≠ -6 For x = 2, y² = 8 × 2
y² = 16 ⇒ y = ±4
So, the points are (2,4) and (2,-4).

Question 19.
Find the solution of equation x² + x + 1 = 0.
Answer:
x² + x + 1 = 0

Question 20.
In how many ways 7 pictures can be hanged on 9 pegs?
Answer:
7 pictures have options of 9 pegs. So they can be selected in ⁹C₇ ways.
Now, these pictures can arrange themselves in 7! ways.
∴ Number of ways by which 7 pictures can be hanged on 9 pegs = ⁹C₇ × 7!
= \(\frac{9 !}{7 ! 2 !}\) × 7! = \(\frac{9 !}{2 !}\)

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

Question 21.
Find the angles between the lines √3x + y = 1 and x + √3y = 1
OR
Evaluate \(\lim _{x \rightarrow 0} \frac{\sin x-2 \sin 3 x+\sin 5 x}{x}\)
Answer:
The given lines are
√3x + y = 1 ⇒ y = -√3x +1, m₁ = -√3
and x + √3y = 1 ⇒ y = \(-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}\), m₂ = \(-\frac{1}{\sqrt{3}}\)
Let θ be the angle between the lines,
tan θ = \(\left|\frac{m_1-m_2}{1+m_1 m_2}\right|\)
= \(\left|\frac{-\sqrt{3}+\frac{1}{\sqrt{3}}}{1+(-\sqrt{3})\left(-\frac{1}{\sqrt{3}}\right)}\right|\)

Question 22.
If tan x = \(\frac{b}{a}\), then find the value of \(\sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}\)
Answer:

Question 23.
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)
Answer:
Given,
L = {1,2, 3, 4}, M = {3, 4, 5, 6} and N = {1,3,5}
∴ M ∪ N = {l, 3,4,5,6}
L – (M ∪ N)={ 2}
Now, L – M = {1, 2}, L – N = {2,4}
(L – M) ∩ (L – N) = {2}
Hence, L – (M ∪ N) = (L – M) ∪ (L – N)

Question 24.
Evaluate \(\lim _{x \rightarrow 0} \frac{\sqrt{1+x^3}-\sqrt{1-x^3}}{x^2}\)
Answer:

Question 25.
Find the angle in radians between the hands of a clock at 7 : 20 p.m.
Answer:
We know that the hour hand completes one rotation in 12 hours while the minute hand completes one rotation in 60 minutes.
Angle traced by the hour hand in 12 hours = 360r

∴ Angle traced by the hour hand in 7 hours 20 minutes i.e., \(\frac{22}{3}\) hours
= \(\left(\frac{360}{12} \times \frac{22}{3}\right)^0\) = 220°

Also, the angle traced by the minute hand in 60 minutes = 360Y
The angle traced by the minute hand in 20 minutes
= (\(\frac{360}{60}\) × 20) = 120°

Hence, the required angle between two hands
= 220°-120°
= 100°

Question 26.
From a group of 2 boys and 3 girls, two children are selected. Find the sample space of this experiment.
OR
If A and B are mutually exclusive events, P(A) = 0.35 and P(B) = 0.45, then find P(A’ ∩ B’)
Answer:
Let B₁, B₂ be the boys and G₁, G₂ G₃ be the girls then, the sample space is
S = {B₁B₂, B₁G₁, B₁G₂, B₁G₃, B₂G₁, B₂G₂, B₂G₃, G₁G₂ G₁G₃, G₂G₃}
OR
A and B are mutually exclusive events.
∴ Then P(A ∩ B) = 0
and P(A) = 0.35, P(B) = 0.45
P(A’ ∩ B’) = P(A ∪ B)’ = 1 – P (A ∪ B) = 1 – 0.8 = 0.2

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

Question 27.
If sin (θ + α) = a and sin (θ + β) = b, then prove that cos2(α – β) – 4ab cos (α – β) = 1 – 2a² – 2b².
OR
If a cos 2θ + b sin 2θ = c has α and β as its roots, then prove that tan α + tan β = \(\frac{2 b}{a+c}\)
Answer:
Given, sin(θ + α) = a and sin(θ + β) = b
cos(α – β) = cos[θ + α – θ – β]
= cos[(θ + α) – (θ + β)]
= cos(θ + α) cos (θ + β) + sin(θ + α) sin (θ + β)

⇒ a(1 – tan²θ) + b(2tanθ) = (1+ tan²θ)c
⇒ a – atan²θ + 2btanθ – ctan²θ – c = 0
⇒ -(a + c)tan²θ + 2btanθ + (a – c) = 0
⇒ (a + c)tan²θ – 2btanθ + (c – a) = 0
Since, α and P are the roots of this equation
⇒ tan α + tan β = \(\)
⇒ tan α + tan β = \(\frac{2 b}{a+c}\)
Hence proved.

Question 28.
If \(\left|\frac{z-5 i}{z+5 i}\right|\) = 1 show that z is a real number:
Answer:
Given, \(\left|\frac{z-5 i}{z+5 i}\right|\) = 1
\(\left|\frac{z-5 i}{z+5 i}\right|\) = 1
|z – 5i| = |z + 5i|
|x + iy – 5i| = |x + iy + 5i| where z = x + iy
|x + i(y – 5)| = |x + i(y + 5)|
Now, \(\)
x² + (y – 5)² = x² + (y + 5)²
(y – 5)² = (y + 5)²
y² + 25 – 10y = y² + 25 + 10y
20y = 0
y = o
z = x + i0 = x:, which is a real number.

Commonly Made Error
Students make simplification errors while simplifying the modulus of complex numbers. ;

Answering Tip
Zero is purely real as well as purely imaginary but not imaginary.

Question 29.
Find the mean deviation about the median for the following data :

Answer:
Let us find cumulative frequencies and Σf_i|x_i – M| as follows:

Now, f = 14, class corresponding to cumulative frequency 28 is 20-30.
∴ I = 20, f = 14, h = 10
Median = l + \(\frac{\frac{\mathrm{N}}{2}-\mathrm{C}_f}{f}\) × h
= 20 + \(\frac{25-14}{14}\) × 10
= 20 + \(\frac{11}{14}\) × 10 = 27.86
Σf_i|x_i – M| = 517.16

∴ Mean deviation about median (M) i.e.,
M.D (M) = \(\frac{\sum f_i\left|x_i-\mathrm{M}\right|}{\sum f_i}\)
= \(\frac{517.16}{50}\) = 10.34

Question 30.
Prove that: cos² x+cos² (x + \(\frac{\pi}{3}\)) + cos²(x – \(\frac{\pi}{3}\)) = \(\frac{3}{2}\)
OR
Find the equation of the circle passing through the points (4, 1) and (6, 5) and whose centre is on the line 4 x + y = 16.
Answer:

OR
Let the equation of the circle be
(x – h)² + (y – k)² = r² ………….(i)
(1), with centre (h, k) and radius V.
Since (1) passes through (4, 1) and (6, 5).
∴ (4 – h)² + (1 – k)² = r² …(2)
and (6 – h)² + (5 – k)² = r …(3)
(2) and (3) => (4 – h)² + (1 – k)² = (6 – h)² + (5 – k)²
⇒ 16 – 8h + h² + 1 – 2k + k² = 36 – 12h + h² + 25 – 10k + k²
⇒ -8h-2k + 17 + 12h + 10k – 61 = 0
⇒ 4h – 8k – 44 =0
⇒ h + 2k -11 = 0
Centre lies on the line 4x + y = 16
∴ 4h + k – 16 = 0
from equation (4) & (5) h = 3 and k = 4 from equ. (2)
(4 – h)² + (1 – k)² = r²
(4 -3)² +(1 -4)² = r²
1 + 9 = 10 = r²
from equ. (1)
(x – h)² + (y – k)² = r²
(x – 3)² +(y – 4)² =10
x² + y² – 6x – 8y + 15 = 0

Question 31.
Show that the set of all points such that the difference of their distance from (4,0) and (-4,0) is always equal to 2 and it represent a hyperbola.
Answer:
According to the problems
\(\sqrt{(x+4)^2+y^2}-\sqrt{(x-4)^2+y^2}\) = 2
\(\sqrt{(x+4)^2+y^2}=2+\sqrt{(x-4)^2+y^2}\)

On squaring both sides, we get
x² + 8x + 16 + y² = 4 + x² – 8x + 16 + y² + 4\(\sqrt{(x-4)^2+y^2}\)
⇒ 16x – 4 = 4\(\sqrt{(x-4)^2+y^2}\)
⇒ 4(4x -1) = 4\(\sqrt{(x-4)^2+y^2}\)
⇒ 16x² – 8x + 1 = x² + 16 – 8x + y²
⇒ 15x² – y² = 15 which is a Hyperbola.

Question 32.
Find the term independent ofxin the expansion of (1 + x + 2x³)\(\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9\)
Answer:

For term independent of x, putting 18 – 3r = 0,19 – 3r = 0 and 21 – 3r = 0, we get
r = 6,r = \(\frac{19}{3}\),r = 7
Since, the possible value of r are 6 and 7.
Hence, second term is not independent of x.
The term independent of x is

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

Question 33.
In a group of students, 225 students know French, 100 know Spanish and 45 know both. Each student know either French or Spanish. How many students are there in the groups.
Answer:
Let F and S denote the no. of students who know French and Spanish, respectively.
Given, n(F) – 225, w(S) = 100, n(F ∩ S) = 45
Using identity,
n(F ∪ S) = n(F) + n(S) – n(F ∩ S)
= 225 + 100-45 = 325-45
= 280.

Question 34.
Evaluate \(\lim _{x \rightarrow 0} \frac{\sqrt{2}-\sqrt{1+\cos x}}{\sin ^2 x}\)
OR
\(\lim _{y \rightarrow 0} \frac{(x+y) \sec (x+y)-x \sec x}{y}\)
Answer:

OR

Question 35.
Solve the following system of inequalities graphically
3x + 2y ≤ 150, x + 4y ≤ 80, x ≤ 15, y ≥ 0.
Answer:
The inequalities are
3x + 2y ≤ 150, x + 4y ≤ 80, x ≤ 15, y ≥ 0.
(a) The line 3x + 2y ≤ 150 passes through the points (50,0), and (0, 75). AB represented this line.
Putting x = 0, y = 0 in 3x + 2y ≤ 150 ⇒ 0 ≤ 150 This shows that origin lies in this region.

The region lying below AB and the points lying on AB represent the inequality 3x – 2y ≤ 150.
(b) The equation x + 4y = 80 passes through the points (80, 0), (0,20).
This is represented by CD.
Since 0 ≤ 80, a region lying below CD and the points of CD, represents the inequality x + 4y ≤ 80.
(c) x ≤ 15 is the region lying on the left of EF whose equation is x = 15 and the points lying on EF.
(d) y ≥ 0 is the region above x-axis and points lying on y = 0.
Thus, the multi-shaded region is the solution to the given inequalities.

Question 36.
If a is the A.M of b and c and a is one AM and G₁, and G₂ are two geometric means of two numbers b and c means are G₁, and G₂ then prove that (G₁)³ + (G₂)³ = 2abc.
OR
The diameter of circles (in mm) drawn in a design are given below :

Calculate the standard deviation and mean diameter of the circles.
[Hint: First make the data continuous by making the classes as 32.5 – 36.5, 36.5 – 40.5, 40.5 – 44.5, 44.5 – 48.5, 48.5- 52.5 and then proceed].
Answer:
It is given that a is the A.M. of b and c.
∴ a = \(\frac{b+c}{2}\) ⇒ b + c = 2a ……………… (i)
Since G₁ and G₂ are two geometric means between b and c. Therefore, b, G₁ G₂, c is a G.P with common
ratio r = \(\left(\frac{c}{b}\right)^{\frac{1}{3}}\)
G₁ = br = b\(\left(\frac{c}{b}\right)^{\frac{1}{3}}=c^{\frac{1}{3}} b^{\frac{2}{3}}\) and
G₂ = br² = b\(\left(\frac{c}{b}\right)^{\frac{2}{3}}=b^{\frac{1}{3}} c^{\frac{2}{3}}\)
⇒ G₁³ = b²c and G₂³ = bc²
⇒ G₁³ + G₂³ = b²c + bc² => G₁³ + G₂³ = bc(b + c)
⇒ G₁ + G₂ = 2 abc [using (i)]
Hence proved

After making the classes continuous, we have :