CBSE Sample Papers for Class 11 Maths Set 6 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 11 Maths with Solutions Set 6 are designed as per the revised syllabus.

CBSE Sample Papers for Class 11 Maths Set 6 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.
The tangent of the angle between the line that intercepts on the axes are a, – b and b, – a respectively, is
(A) \(\frac{a^2-b^2}{a b}\)
(B) \(\frac{b^2-a^2}{2}\)
(C) \(\frac{b^2-a^2}{2ab}\)
(D) None of these
Answer:
Option (C) is correct.

Explanation:
We have, Equations of lines as:
⇒ \(\frac{x}{a}-\frac{y}{b}\) = 1
⇒ \(\frac{x}{a}-\frac{y}{b}\) = 1
Slope of first line = \(\frac{b}{a}\)
Slope of second line = \(\frac{a}{b}\)
Tangent of angle between them,
tan x = \(\frac{b^2-a^2}{2 a b}\)

Question 2.
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

OR

If A = {1,3,5,7, 9,11,13,15,17}, B = {2,4, 6,…, 18} and N the set of natural numbers is the universal set, then A’ ∪ (A ∪ B) ∩ B’ is
(A) Φ
(B) N
(C) A
(D) B
Answer:
Option (B) is correct.

Explanation:
We have,
⇒ 2m – 2n = 112
The values of m = 7 and n = 4 satisfy the following equation.
So m = 7 and n = 4.

OR

Option (D) is correct.

Explanation:
We know that,
⇒ A’ ∪ (A ∪ B) ∩ B’
⇒ B ∪ (A ∪ B) ∩ A
⇒ (A ∪ B) ∩ A
⇒ B

Question 3.
\(\lim _{x \rightarrow 0} \frac{|\sin x|}{x}\) is equal to:
(A) 1
(B) -1
(C) Does not exist
(D) None of these
OR

then the quadratic equation whose roots are \(\lim _{x \rightarrow 2^{-}}\) f(x) and \(\lim _{x \rightarrow 0} \frac{\tan 2 x-x}{3 x-\sin x}\) is:
(A) x² – 6x + 9 = 0
(B) x² – 7x + 8 = 0
(C) x² – 14x + 49 = 0
(D) x² – \(\frac{7}{2}\)x + \(\frac{3}{2}\)
Answer:
Option (C) is correct.

Explanation:
We know that,
\(\lim _{x \rightarrow 0} \frac{|\sin x|}{x}\) = Does not exist
As both left hand limit and right hand limit are different.

OR

Option (D) is correct.

Explanation:
We have,
The roots of both the limits as 3 and 1/2
So the quadratic equation satisfying both the limits
x² – \(\frac{7}{2}\)x + \(\frac{3}{2}\)

Question 4.
Three numbers are chosen from 1 to 20. Find the probability that they are not consecutive.
(A) \(\frac{186}{190}\)
(B) \(\frac{187}{190}\)
(C) \(\frac{188}{190}\)
(D) \(\frac{18}{{ }^{20} C_3}\)
Answer:
Option (B) is correct.

Explanation:
We know that,
Probability of choosing three consecutive numbers from 1 to 20 is \(\frac{187}{190}\)

Question 5.
If the mean of 100 observation is 50 and their standard deviation is 5, then the sum of all squares of all the observation is
(A) 50000
(B) 250000
(C) 252500
(D) 255000
Answer:
Option (C) is correct.

Explanation:
We have,
Number of Observations = 100
Mean = 50 SD = 5
Sum of squares of all observations = 252500

Question 6.
The number of words which can be formed out of the letters of the word ARTICLES, so that vowels occupy the
(A) 6.63
(B) 144
(C) 22
(D) ⁴C₄× ³C₃
Answer:
Option (B) is correct.

Explanation:
We have,
Number of words that can be formed from word ARTICLES so that vowels occupy even places = C(4, 3) × 3! × 3! = 144

Question 7.
IfA and B are mutually exclusive events, then
(A) P(A) ” P(B̅)
(B) P(A) ≥ P(B̅)
(C) P(A) < P(B̅)
(D) None of these
Answer:
Option (A) is correct.

Explanation:
We have,
For mutually exclusive events,
⇒ P(A) ≤ P(B̅)

Question 8.
The following information relates to a sample of size 60, 2 18000, and Z 960. Then, the vaiiance is
(A) 6.63
(8) 16
(C) 22
(D)44
Answer:
Option (D) is correct.

Explanation:
Given that, n = 60
Sigma x = 960
Sigma x² = 18000
Using the formula for variance, we get
Variance = 44

Question 9.
The value of cos 1° cos 2° cos 3°… cos 179° is
(A) \(\frac{1}{\sqrt{2}}\)
(B) 0
(C) 1
(D) -1

OR

If tan α = \(\frac{1}{7}\), tan β = \(\frac{1}{3}\), then cos2α is equal to
(A) sin 2β
(B) sin 4β
(C) sin 3β
(D) cos2β
Answer:
Option (B) is correct.

Explanation:
We have,
⇒ cos1 cos2 cos 3 ………….. cos179
As cos 90 = 0
So cos1 cos2 cos3….cos179 = 0

OR

Option (B) is correct.

Explanation:
Given that,
tan α = \(\frac{1}{7}\)
tan β = \(\frac{1}{3}\)
On simplification, we get
⇒ cos 2α = sin 4β

Question 10.
The length of the latus rectum of ellipse 3x² + y² = 12 is
(A) 4
(B) 3
(C) 8
(D) \(\frac{4}{\sqrt{3}}\)
Answer:
Option (D) is correct.

Explanation:
We have,
For ellipse,
⇒ 3x² + y² = 12
Length of latus rectum = \(\frac{4}{\sqrt{3}}\)

Question numbers 11 to 15 carry 1 mark each. Write True or False.

Question 11.
If xy > 0, then x < 0 and y < 0 Answer: True Explanation: We know that, For xy > 0,
we have x < 0 and y < 0
So given statement is True

Question 12.
2 is not a complex number.
OR
The order relation is defined on the set of complex numbers.
Answer:
False

Explanation:
2 is a purely real complex number So given statement is False.

OR

False
Explanation:
The order relation is not defined on the set of complex numbers So the given statement is False.

Question 13.
Two sequence cannot be in both A.E and G.E together.
Answer:
False

Explanation:
Two sequences can be in A.P and G.P together
So the given statement is False.

Question 14.
The points A (-2,1), B (0, 5) and C (-1,2) are collinear.
Answer:
False

Explanation:
The points A(-2, 1) , B(0, 5) and C(-1, 2) are not collinear as they don’t lie on a straight line. So the given statement is False.

Question 15.
If the line lx + my = -1 is a tangent to the circle x² + y² = a², then the point (l, m) lies on a circle.
Answer:
False

Explanation:
The line lx + my = – 1 is a tangent to the circle x² + y² = a², then the point (l, m) do not lie on a circle
Hence the given statement is False.

Question numbers 16 to 20 carry 1 mark each.

Question 16.
If P = {1,3}, Q = {2, 3, 5}, find the number of relations from P to Q.
Answer:
Given, P = {1, 3} and Q = {2, 3, 5}
∵ n(P) = 2 and n(Q) = 3
∴ Number of relations = 2^{n(P) × n(Q)}
= 2^2×3
= 2⁶
= 64

Question 17.
Find a and b if (a -1, b + 5) = (2,3)
OR
Find the number of elements from set X to set Y if X = {1, 2, 3} and Y = {a, b}.
Answer:
Given, (a – 1, b + 5) = (2, 3)
By the equality of sets, we get
a – 1 = 2
⇒ a = 3
and b +5 = 3
b = 3 – 5 = -2
∴ a = 3 and b = – 2

OR

Given, X = {1, 2, 3} and Y = {a, b}
X × Y = {(1, a), (1, b), (2, a), (2, b), (3, a), (3, b)}
∴ Number of elements from set X to set Y = 6

Question 18.
A die is thrown and then a coin is tossed if an odd number comes up describe the sample space.
OR
A coin is tossed. If it shows head, we draw a ball from a bag consisting of 2 red and 3 black balls. If it shows tail, coin is tossed again. Find the sample space for the experiment.
Answer:
Sample space, S = {1H, 1T, 2, 3H, 3T, 4, 5H, 5T, 6}
OR
Let R₁, R₂ are the red balls and B₁, B₂, B₃ are the black balls in the bag.
∴ The sample space associated with the experiment is
S = {HR₁, HR₂, HB₁, HB₂, HB₃, TH, TT}

Question 19.
L is the foot of the perpendicular drawn from a point P (6, 7, 8) on the XY plane. What are the co-ordinates of point L.
Answer:
The co-ordinates of point L are (6, 7, 0).

Question 20.
Solve \(\frac{1}{x-2}\) < 0, x ∈ R
Answer:
We have, \(\frac{1}{x-2}\) < 0
⇒ x – 2 < 0 [∵ \(\frac{a}{b}\) < 0 and a > 0 ⇒ b < 0]
⇒ x < 2 ⇒ x ∈ (-∞, 2)

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

Question 21.
Find the domain of the real function f(x) = \(\sqrt{x^2-4}\)
Answer:
Given, f (x) = \(\sqrt{x^2-4}\)
f is real only when x² – 4 ≥ 0,
i.e., (x – 2)(x + 2) ≥ 0
∴ Domain off = ( -∞, – 2] ∪ [2, ∞)

Question 22.
Find the coefficient of x4 in (1 – x) (2 + x)5 using binomial theorem.
OR
Find the 4th term from the end in the expansion of \(\left(\frac{x^3}{2}-\frac{2}{x^3}\right)^9\)
Answer:
Given, (1 – x)² (2 + x)⁵
= (1 + x² – 2x)(2⁵ + ⁵C₁ 2^5-1 x¹ + ⁵C₂ 2^5-2 x² + ⁵C₃ 2^5-3 x³ + ⁵C₄ 2^5-4 x⁴ + ⁵C₅ 2^5-5 x⁵)
= (1 + x² – 2x)(2⁵ + 5 . 2⁴ . x + 10 . 2³ x² +10 . 2² x³ + 5 . 2 . x⁴ + x⁵)
= (1 + x² – 2x)(2⁵ + 80x + 80x² + 40x³ + 10x⁴ + x⁵)
∴ Coefficient of x⁴ = 80 – 80 + 10
= 10

Commonly Made Error
Students had difficulty in using the binomial theorem formulae to solve for power and often substituted the coefficients when equating to the power of a given term.

Answering Tip
Revise the concept of the binomial theorem thoroughly and practice more such problems based on it.

OR

Question 23.
Evaluate \(\lim _{x \rightarrow 0} \frac{\log (6+x)-\log (6-x)}{x}\)
Answer:

Question 24.
If U = {1, 2, 3, 4,5,6, 7, 8, 9}, A = {2,4,6, 8} and B = {2, 3,5, 7} verify that (i) (A ∪ B)’ =-A’ ∩ B
Answer:
A ∪ B = {2, 4. 6, 8) ∪ {2, 3, 5, 7}
= {2, 3, 4, 5, 6, 7, 8}
L. H. S. = (A ∪ B)’ = U – (A ∪ B)
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 3, 4, 5, 6, 7, 8}
= {1, 9} ………….(i)

Now, Complement of A = A’ = U – A = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 4, 6, 8}
= {1, 3, 5, 7, 9}
and Complement of B = B’ = U – B = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 3, 5, 7}
= {1, 4, 6, 8, 9}
R. H. S. = A’ ∩ B’ = {1, 3, 5, 7, 9} ∩ {1, 4, 6, 8, 9}
= {1, 9} …(ii)
From eqs. (i) and (ii), we get L.H.S. = R. H. S.
Hence, (A ∪ B)’ = A’ ∩ B’ is verified.

Question 25.
Find the value of 5^1/2 5^1/4 5^1/8… up to infinity.
Answer:

Question 26.
Find the domain and range of f(x) = |2x – 3| – 3.
OR
Evaluate \(\lim _{x \rightarrow 0} \frac{(x+2)^{1 / 3}-2^{1 / 3}}{x}\)
Answer:
Given, f (x) = |2x – 3| – 3
The domain of the expression is all real number except where the expression is undefined. In this case, there is not real number that makes the expression undefined
Domain off = (- ∞, ∞) = R
The absolute value of expression has a ‘V shape. The range of a positive absolute value expression starts at its vertex and extends to infinity.
Range of f = [-3, ∞) or {y : y ≥ – 3}

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

Question 27.
Evaluate :
(1 + cos \(\frac{\pi}{8}\))(1 + cos \(\frac{3 \pi}{8}\))(1 + cos \(\frac{5 \pi}{8}\))(1 + cos \(\frac{7 \pi}{8}\))
OR
Prove that:
2 sin²β + 4 cos (α + β) sin a sin β + cos ²(α + β) = cos ²α
Answer:
We have,

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

Question 28.
Write the complex number z = \(\frac{1-i}{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}}\) in polar form.
Answer:

Question 29.
Prove that there is no term involving x6 in the expansion of \(\left(2 x^2-\frac{3}{x}\right)^{11}\)
Answer:

Question 30.
If a, b, c are in G.E, then the following are also in G.R
(i) a², b², c²
(ii) a³ b³ c³
OR
If the ellipse with equation 9x² + 25y² = 225, then find the eccentricity and foci.
Answer:

Question 31.
The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The Roadway which is horizontal and 100 m long is supported by vertical wires attached to the Cable, the longest wire being 30 m and the shortest is 6 m. Find the length of a supporting wire attached to the roadway 18 m from the middle.
Answer:
The vertex is at the lowest point of the cable. The origin of the coordinate plane is taken as the vertex of the parabola, while its vertical axis is taken along the position y-axis. This can be diagrammatically represented as

Here, AB and OC are the longest and the shortest wires, respectively, attached to the cable
DF is the supporting wire attached to the roadway, 18 m from the middle.
Here, AB = 30 m, OC = 6 m, and BC = \(\frac{100}{2}\) = 50
The equation of the parabola is of the form x² = 4ay (as it is opening upwards).
The coordinate of a point A are (50, 30 – 6) = (50, 24). Since A (50, 24) is a point on the parabola.
(50 )² = 4a(24)
⇒ a = \(\frac{50 \times 50}{4 \times 24}=\frac{625}{24}\)

Equation of the parabola, x² = 4 × \(\frac{625}{24}\) × y
6x² = 625y
The x-coordinate of point D is 18.
Hence, at X = 18,
6 (18 )² = 625y
⇒ y = \(\frac{6 \times 18 \times 18}{625}\)
⇒ y = 3.11 (approx.)
∴ DE = 3.11 m
DF = DE + EF = 3.11m + 6 m = 9.11 m
Thus, the length of the supporting wire attached to the roadway 18 m from the middle is
Approximately 9.11 m.

Question 32.
Find the mean, variance and standard deviation using short cut method.

Answer:

Here, y_i = \(\frac{x_i-92.5}{5}\) ∵ h = 5 and a = 92.5

Mean = 93,
Variance = 105.58
and Standard deviation = \(\sqrt{105.58}\) = 10.27

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

Question 33.
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 ?
Answer:
Let C and T denote the students who play cricket and tennis, respectively.
Given, n(C) = 25, n(T) = 20, n(C ∩ T) = 10 and n(U) = 60
Using identity,
n(C ∪ T) = n(C) + n(T) – n(C ∩ T)
= 25 + 20 – 10
= 35 3
Number of students who play neither = n(U) – n(C ∪ T)
= 60 – 35
= 25

Question 34.
There are 15 points in a plane out of which only 6 are in a straight line, then
(a) How many different straight lines can be made?
(b) How many triangles can be made?
OR
From 6 different novels and 3 different dictionaries, 4 novels and a dictionary is to be selected and arranged in a row on the shelf so that the dictionary is always in the middle. Then find the number of such arrangements.
Answer:
(a) Straight line can be formed by joining 2 points, so number of ways in which we select 2 points from 15 points is = ¹⁵C₂
But it also include that in which 6 points are in straight line. From those 6 points only one line can be formed.
So, we have to subtract ⁶C₂ – 1 from the above result.

(b) Non collinear point 9 and collinear point 6
Case I
We choose 1 Non collinear and 2 collinear point to make a triangle
⁹C₁ × ⁶C₂ = \(\frac{9 !}{8 ! \cdot 1 !} \times \frac{6 !}{4 ! \cdot 2 !}\)
= \(\frac{9 \times 6 \times 5}{2}\) = 135

Case II
We choose 2 Non collinear point and 1 collinear point to make a triangle
⁹C₂ × ⁶C₁
= \(\frac{9 !}{2 ! \cdot 7 !} \times \frac{6 !}{1 ! \cdot 5 !}\)
= \(\frac{9 \times 8 \times 6}{2}\) = 216

Case III
We choose 3 non collinear points to make a triangle
⁹C₃ = \(\frac{9 !}{3 ! 6 !}\)
= \(\frac{9 \cdot 8 \cdot 7}{3 \cdot 2}\) = 84
Total number of triangles = 135 + 216 + 84
= 435

OR

The number of ways in which 4 novels can be selected from 6 different novels = ⁶C₄ = 15 1
The number of ways in which 1 dictionary can be selected out of 3 different dictionaries = ³C₁ = 3
Now, it is given that dictionary is always in the middle.
Hence, the arrangement look like M M D M M
Dictionary is always in the middle as the novels are different, hence they can be arranged in 4! ways. 1
Hence, total number of such arrangement
= 15 × 3 × 4!
= 15 × 3 × 4 × 3 × 2 × 1
= 1080

Question 35.
Show that the following system of linear inequalities has no solution.
x + 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0, y ≥ 1.
Answer:
Consider the inequation x + 2y < 3 as an equation, we have
x + 2 y = 3
⇒ x = 3 – 2 y
⇒ 2y = 3 – x
Now (0, 0) satisfy the inequation x + 2y ≤ 3.
So, half plane contains (0, 0) as the solution and the line x + 2y = 3 intersect the coordinate axis at (3, 0) and (0, \(\frac{3}{2}\)).

Consider the inequation 3x + 4y ≥ 12 as an equation, we have 3x + 4y = 12
⇒ 4y = 12 – 3x
Thus, coordinate axis intersected by the line 3x + 4y = 12 at points (4, 0) and (0, 3).
Now, (0, 0) does not satisfy the inequation 3x + 4y = 12.
Therefore, half plane of the solution does not contained (0, 0).
Consider the inequation y ≥ 1 as an equation, we have y = 1.
It represents a straight line parallel to X-axis passing through the point (0, 1).
Now, (0, 0) does not satisfy the inequation y ≥ 1. Therefore, the half-plane of the solution does not contains (0, 0).
Clearly x ≥ 0 represents the region lying on the right side of Y-axis.
The solution set of the given linear constraints will be the intersection of the above region.

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

Commonly Made Error
Students do not shade the feasible region which leads to deduction of marks.

Answering Tip
Students must shade the feasible region and highlight the required region.

Question 36.
The ratio of the A.M. and G.M. of two positive numbers a and b is m : n. Show that a : b = ©L!
a : b = (m + \(\left.\sqrt{m^2-n^2}\right)\)): (m – \(\left.\sqrt{m^2-n^2}\right)\))
OR
For the function f(r) = \(\frac{x^{100}}{100}+\frac{x^{99}}{99}+\ldots+\frac{x^2}{2}\) + x + 1, prove that/'(r) = 100/'(0).
Answer:
Let the A.M. of the number a and b is A and G.M. of a and b is G
then, A = \(\frac{a+b}{2}\) and G = \(\sqrt{ab}\)
Given, A : G = m : n
i.e \(\frac{A}{G}=\frac{m}{n} \Rightarrow \frac{a+b}{2 \sqrt{a b}}=\frac{m}{n}\)

Applying componendo and dividendo rule,

OR
Given,
f(x) = \(\frac{x^{100}}{100}+\frac{x^{99}}{99}+\ldots .+\frac{x^2}{2}\) + x + 1
f'(x) = \(\frac{100 x^{99}}{100}+\frac{99 x^{98}}{99}+\ldots .+\frac{2 x}{2}\) + 1
[∵ \(\)xⁿ = nx^n-1]
⇒ f ‘(x) = x⁹⁹ + x⁹⁸ + …. + x + 1
⇒ f ‘(1) = (1)⁹⁹ + (1)⁹⁸ + … + 1 + 1
= 1 + 1 + 1 + …………. + 1 (100 terms)
= 100
and f ‘(0) = 1
f ‘(1) = 100 x 1
∴ f ‘(1) = 100 f ‘(0)
Hence Proved