CBSE Sample Papers for Class 11 Maths Set 3 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 11 Maths with Solutions Set 3 are designed as per the revised syllabus.

CBSE Sample Papers for Class 11 Maths Set 3 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.
While shuffling a pack of 52 playing cards, 2 are accidentally dropped. Find the probability that the missing cards to be of different colours. [1]
(A) \(\frac{29}{52}\)
(B) \(\frac{1}{2}\)
(C) \(\frac{26}{51}\)
(D) \(\frac{27}{51}\)
Answer:
(C) \(\frac{26}{51}\)

Explanation:
The probability that the missing cards to be of different colors is
= \(\left[\frac{\mathrm{C}(26,1) \times \mathrm{C}(26,1)}{\mathrm{C}(52,2)}\right]\)
= \(\frac{26}{51}\)

Question 2.

is equal to
(B) 0
(C) 1
(D) -1
OR

is equal to
(A) \(\frac{-1}{2}\)
(B) 1
(C) \(\frac{1}{2}\)
(D) -1
Answer:
(C) 1

Explanation:
We know that on rationalizing the denominator and then applying limits using standard rules for sine function we get the required limit as 1.

Explanation:
We have

Question 3.
X-axis is the intersection of two planes
(A) XY and XZ
(B) YZ and ZX
(C) XY and YZ
(D) None of these
Answer:
(A) XY and XZ

Explanation:
We know that X-axis is the intersection of XY and XZ planes.

Question 4.
If 9 times the 9th term of an A.E is equal to 13 times the 13th term, then the 22nd terms of the A.E is
(A) 0
(B) 22
(C) 220
(D) 198
OR
If the sum of n terms of an A.E is given by S„ = 3n + 2rt, then the common difference of the A.E is
(A) 3
(B) 2
(C) 6
(D) 4
Answer:
(A) 0

Explanation:
Given that
⇒ 9 (a + 8d) = 13 (a + 12 d)
⇒ 4a = – 84 d
⇒ a = – 21 d
⇒ a + 21 d = 0
Hence 22nd term is zero.

(D) 4

Explanation:
Given that
S_n = 3n + 2n²
Common difference = Twice of coefficient of n² in standard form
so Common difference = 4

Question 5.
Consider the first 10 positive integers. If we multiply each number by – 1 and, then add 1 to each number, the variance of the numbers, so obtain is
(A) 8.25
(B) 6.5
(C) 3.87
(D) 2.87
Answer:
(A) 8.25

Explanation:
We know that the variance of numbers on multiplying each number by -1 and then adding 1 remains 8.25.

Question 6.
If x is a real number and |x| < 3, then (A) x > 3
(B) -3 < x < 3
(C) x ≤ – 3
(D) – 3 ≤ x < 3
Answer:
(B) -3 < x < 3

Explanation:
Given that |x | < 3
so – 3 < x < 3

Question 7.
If the coefficients of 2nd, 3rd and the 4th terms in the expansion of (1 + x)ⁿ are in A.E, then value of n is
(A) 2
(B) 7
(C) 11
(D) 14
OR
If A and B are coefficient of xⁿ in the expansions of (1 + x)²ⁿ and (1 + x)^{2n – 1} respectively, then \(\frac{A}{B}\)
(A) 1
(B) 2
(C) \(\frac{1}{2}\)
(D) \(\frac{1}{n}\)
[Hint: \(\frac{A}{B}=\frac{{ }^{2 n} C_n}{{ }^{2 n-1} C_n}\) = 2]
Answer:
(B) 7

Explanation:
Given
2^nd, 3^rd and 4^th terms are in AP
We have, 2C(n, 2) = C(n, 1) + C(n, 3)
On simplifying n = 7

(B) 2

Explanation:
We have
\(\frac{\mathrm{A}}{\mathrm{B}}=\frac{\mathrm{C}(2 n, n)}{\mathrm{C}(2 n-1, n)}\)
= 2

Question 8.
Equation of diagonals of the square formed by the lines x = 0, y = 0, x = 1 and y = 1 are
(A) y = x, y + x = 1
(B) y = x, x+ y = 2
(C) 2y = x, y + x = \(\frac{1}{3}\)
(D) y = 2x, y + 2x = 1
Answer:
(A) y = x, y + x = 1

Explanation:
We have
Equation of diagonals formed by square from lines x = 0, y = 0, x = 1, y = 1 is x = y and x + y = 1.

Question 9.
Equation of a circle which passes through (3, 6) and touches the axes is .
(A) x² + y² + 6x + 6y + 3 = 0
(B) x² + y² – 6x – 6y – 9 = 0
(C) x² + y² – 6x – 6y + 9 = 0
(D) None of these
Answer:
(C) x² + y² – 6x – 6y + 9 = 0

Explanation:
Equation of circle which passes through (3, 6) and touches the axis is
x² + y² – 6x – 6y + 9 = 0

Question 10.
Following are the marks obtained by 9 students in a mathematics test 50, 69, 20, 33, 53, 39, 40, 65, 59. The mean deviation from the median is
(A) 9
(B) 10.5
(C) 12.67
(D) 14.76
Answer:
(C) 12.67

Explanation:
We have Mean deviation from the median of observations 50, 69, 20, 33, 53, 39, 40, 65, 59 is 12.67

Question numbers 11 to 15 carry 1 mark each. Write whether the statement is True or False.

Question 11.
The lines ax + 2y + 1 = 0, bx + 3y + 1 = 0 and cx + 4y + 1 = 0 are concurrent, if a, b and c are in GP.
Answer:
False.

Explanation:
The given lines are concurrent if a, b, c are in AP so given statement is False.

Question 12.
Eighteen guests are to be seated, half on each side of a long. Four particular guests desire to sit on one particular side and three other on other side of the table. The number of ways in which the seating arrangements can be made is \(\frac{11 !}{5 ! 6 !}\) (9 !)(9!)
516!
OR
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.
Answer:
True.

Explanation:
The number of ways in which the seating arrangements can be made is (11!)(9!)(9!)/(5!)(6!)

False.

Explanation:
The number of lines obtained is
= C(12, 2) – C(5,2) + 1
Hence given statement is False

Question 13.
The line lx + my + n = 0 will touch the parabola y² = 4ax, if ln = am²
Answer:
True.

Explanation:
The line lx + my + n = 0 will touch the parabola if ln = am².

Question 14.
Let sets R and T be defined as
R = {x ∈ Z \ x is divisible by 2}
T = {x ∈ Z | x is divisible by 6}. Then T ⊂ R
Answer:
True.

Explanation:
It is true that T is a subset of R.
As numbers divisible by 6 are also divisible by 2.

Question 15.
Let z, and z2 be two complex number such that |Z₁ + Z₂| = |Z₁| + |Z₂|,then | Z₁ – Z₂| = 0.
OR
For any complex number z, the minimum value of | z | + | z – 1| is 1.
Answer:
False.

Explanation:
We know that
|Z₁ + Z₂| is not equal to |Z₁| + |Z₂|
Hence given statement is incorrect.

True.

Explanation:
It is true that the minimum value of | z | + | z – 1| is 1 for any complex number z.

Question numbers 16 to 20 carry 1 mark each.

Question 16.
Let/be the subset of Z x Z defined by
f = {(ab, a + b): a, be Z}. Is f a function from Z to Z? Justify your answer.
Answer:
Given, f = {(ab, a + b): a, b ∈ Z}
Taking a = b = 1, we have (ab, a + b) = (1, 2) ∈ f
Taking a = b = -1, we have (ab, a + b) = (1, -2) ∈ f
⇒ f – image of 1 is not unique
Hence, f is not a function

Question 17.
Solve 5x < 24, when x e N
Answer:
We have, 5x < 24
\(\frac{5 x}{5}<\frac{24}{5}\) or x < \(\frac{24}{5}\)
This is true when
x = 1, 2, 3, 4 (x being a natural number).

Question 18.
If A = {-1,1), find A × A × A.
Answer:
Given, A = {-1, 1}
A × A = {-1, 1} × {-1,1}
= {(-1, -1), (-1, 1), (1, -1), (1, 1)}
Again, A × A × A = {(- 1, – 1), (- 1,1), (1, – 1), (1, 1)} × {-1,1}
= {(- 1, – 1, – 1), (- 1, – 1, 1), (-1, 1, -1), (- 1, 1, 1), (1, -1, -1), (1, -1, 1), (1, 1, -1), (1, 1, 1)}.

Question 19.
What is the distance of the point (3, 4, 5) from the YZ plane ?
Answer:
Any point on YZ plane is (0, 1, 1)

Question 20.
When a coin is tossed, write two events which are mutually exclusive and exhaustive.
OR
One number is chosen at random from the number 1 to 21. What is the probability that it is prime.
Answer:
When we toss a coin, sample space, S = {H, T}
Let A = event of getting head = {H}
and B = event of getting tail = {T}
Then, A ∩ B = Φ and A ∪ B = S
∴ A, B are mutually exclusive and exhaustive.

Commonly Made Error:
Students get confused between the mutually ! exclusive and exhaustive events.

Answering Tip:
Always remember if A ∩ B = Φ, then A and B are mutually exclusive events, and if A ∪ B = S, then A and B are exhaustive events.

Sample space n(s) = 21
Prime numbers from 1 to 21 are 2, 3, 5, 7, 11, 13, 17, 19.
It ‘E’ be the event of getting a prime number, then
n(E) = 8
∴ P(E) = \(\frac{n(\mathrm{~F})}{n(\mathrm{~S})}\) = \(\frac{8}{21}\)
The probability that the number is prime = \(\frac{8}{21}\)

Commonly Made Error:
Some students do not know the correct prime 1 numbers between 1 and 21, they consider 1 as prime number.

Answering Tip:
Probability of an event,
P(A) = \(\frac{\text { Number of observed frequencies }}{\text { Total frequency }}\)

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

Question 21.
1 boy and 2 girls are in a room A and 3 boys and 1 girl are in room B. Write the sample space for the experiment in which room is selected and then a person.
Answer:
Let B₁, G₁, G₂ are in room A and B₂, B₃, B₄, G₃ are in room B. Then, sample space is
S = {AB₁, AG₁, AG₂, BB₂, BB₃, BB₄, BG₃}.

Question 22.
Differentiate \(\frac{x}{\sin x}\) with respect to x.
Answer:

= cosec x – x cot x cosec x
f'(x) = (1 – x cot x) . cosec x

Commonly Made Error:
Some students make arithmetic errors while applying quotient rule which leads to incorrect answer.

Answering Tip:
Differentiation rules for different functions and terms need attention.

Question 23.
Solve \(\frac{x+3}{x-1}\) 0, x ∈ R .
Answer:
We have, \(\frac{x+3}{x-1}\) > 0 …………. (i)
Equation x – 1 and x + 3 to zero, we obtain x = 1, -3 as critical points. Plot these points on real line as shown in figure. The real line is divided into three regions. In the right most region the expression on L.H.S. of (i) is positive and in the remaining two regions it is alternatively negative + – +

and positive. Since the expression in (i) is positive, so the solution set is given by
\(\frac{x+3}{x-1}\) > 0 ⇒ (- ∞, – 3) ∪ (1, ∞)

Question 24.
Find the symmetric difference of sets A = {1, 3, 5, 6, 7} and B = {3, 7, 8, 9}.
Answer:
Given, sets are A = {1, 3, 5, 6, 7} and B = {3, 7, 8, 9}
Now, A-B = {1, 3, 5, 6, 7} – {3, 7,8, 9}
= {1, 5, 6}
and B – A = {3, 7, 8,9} – {1, 3,5,6, 7}
= {8, 9}
∴ Required symmetric difference,
A ∆ B = (A – B) ∪ (B – A)
= {1, 5, 6} ∪ {8, 9}
= {1, 5, 6, 8, 9}

Question 25.
Evaluate the left hand and right hand limits of the following function at x = 2. Does \(\lim _{x \rightarrow 2}\) f(x) exists ?
f(x) = 2x + 3 if x ≤ 2
f(x) = x + 5 if x > 2
OR
Show that \(\lim _{x \rightarrow 4} \frac{|x-4|}{x-4}\) does not exist.
Answer:

Commonly Made Error:
Several students make errors because of confusion between the left-hand limit and the right-hand limit.

Answering Tip:
Understand the difference between the left hand limit and right hand limit and its method to solve with appropriate sign convention.

∴ LHL ≠ RHL
So, limit does not exist.

Question 26.
Prove that: (cos x + cos y)² + (sin x – sin y)² = 4 cos² \(\frac{x+y}{2}\)
OR
Proved that: \(\frac{(\sin 7 x+\sin 5 x)+(\sin 9 x+\sin 3 x)}{(\cos 7 x+\cos 5 x)+(\cos 9 x+\cos 3 x)}\) = tan 6x
(cos 7x + cos 5x) + (cos 9x + cos 3x)
Answer:
L.H.S = (cos x + cos y)² + (sin x – sin y)²

Hence Proved.

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

Question 27.
Prove that:
cos A cos 2A cos 4A cos 8A = \(\frac{\sin 16 A}{16 \sin A}\)
OR
If tan A – tan B = x, cot B – cot A = y, prove that cot (A – B) = \(\frac{1}{x}+\frac{1}{y}\)
Answer:

= cos A cos 2A cos 4A cos 8A
= L.H.S.
Hence Proved.

Given tanA – tan B = x ………… (i)
and cot B – cot A = y ………… (ii)
From (ii), cot B – cot A = y

Question 28.
Show that the middle term in the expansion of (1 + x)²ⁿ is \(\frac{\{1 \cdot 3 \cdot 5 \cdot 7 \ldots(2 n-1)\} 2^n x^n}{n !}\) where n ∈ N.
OR
If the first three terms in the expansion of (a + b)ⁿ are 27, 54 and 36 respectively, then find a, b and n.
Answer:
The total number of terms in (1 + x)²ⁿ is 2n + 1
Hence, middle term is \(\frac{2 n+1+1}{2}\) = n + 1
∴ T_{n + 1} = ²ⁿC_n (1)ⁿ (x)ⁿ

Hence Proved.

Since
(a + b)ⁿ = ⁿC₀ aⁿ + ⁿC₁ a^{n – 1} b + ⁿC₂ a^{n – 2} b² + ……………. + ⁿC_n bⁿ
Given that
T₁ = ^xC₀ a^x = 27
⇒ aⁿ = 27 …………. (i)
T₂ = ^xC₂ a^{x – 1}b = 54
⇒ na^{n – 1} = 54 …………….. (ii)
T₃ = ²C₂ a^{n – 2} = 36
⇒ \(\frac{n(n-1)}{2}\) a^{n – 1}b² = 36 …………….. (iii)

From (iv) and (v), we get n = 3 ……………. (vi)
From (i) and (vi), we get a = 3 ……………. (vii)
From (v) and (viii), we get b = 2

Commonly Made Error:
Students get confused between the ratio of the terms and their coefficients which leads to incorrect result.

Answering Tip:
In the successive terms of the expansion, powers of first quantity a go on decreases by 1, whereas the powers of the second quantity b increases by 1.

Question 29.
Find the co-ordinates of the foci, the vertices, the eccentricity and the length of the latus rectum of hyperbola 9y² – 4x² = 36.
Answer:
Given hyperbola is
9y² – 4x² = 36
⇒ \(\frac{y^2}{2^2}-\frac{x^2}{3^2}\) = 1 (∵ Transverse axis is along X-axis.)
⇒ a =2, b =3
c = \(\sqrt{a^2+b^2}\) =\(\sqrt{4+9}\) = √13.
e = \(\frac{c}{a}=\frac{\sqrt{13}}{2}\)
⇒ foci = (0, ± c) = (0, ± √l3)
vertices = (0, ± a) = (0, ± 2)
eccentricity, e = \(\frac{\sqrt{13}}{2}\)
Length of latus rectum = \(\frac{2 b^2}{a}=\frac{2 \cdot 9}{2}\) = 9 units

Question 30.
In how many ways 7 positive and 5 negative signs can be arranged in a row so that no two negative signs occur together?
Answer:
Since there is no condition for positive (+) sign, fix them in a row.
+ + + + + + +
There are 6 places in between each plus and one before and one after these positive (+) sign.
i.e., there are 8 places for negative (-) sign and 5(-) negative signs are there.
∴ These negative (-) signs can be placed in ⁸C₅ ways.
Hence, required number of arrangements are = ⁸C₅
= \(\frac{8 !}{3 ! 5 !}\)
= \(\frac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1}\) = 56

Question 31.
Find real valued 0 such that \(\frac{3+2 i \sin \theta}{1-2 i \sin \theta}\) is purely real.
Answer:

Question 32.
Find the mean and variance for first n natural numbers.
Answer:
The first n natural numbers are 1,2, 3,…, n.
∴ Mean, x̄ = \(\frac{1+2+3+\ldots .+n}{n}\)

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

Question 33.
Using the properties of sets and their complements prove that
(A ∪ B) – C = (A – C) ∪ (B – C)
Answer:
(A ∪ B) – C = (A – C) ∪ (B – C)
Let x ∈ [(A ∪ B) – C]
x ∈ (A ∪ B) and x ∉ C
(x ∈ A or x ∈ B) and x ∉ C
(x ∈ A and x ∉ C) or (x ∈ B and x ∉ C)
x ∈ {(A – C) or x ∈ (B – C)}
x ∈ {(A – C) ∪ (B – C)}
(A ∪ B) – C ∈ (A – C) ∪ (B – C) …. (i)
Again, let y ∈ [(A – C) ∪ (B – C)]
y ∈ (A – C) or y ∈ (B – C)
(y ∈ A and y ∉ C) or (y ∈ B and y ∈ C)
(y ∈ A or y ∈ B) and y ∉ C
y ∈ {(A ∪ B) and y ∉ C
y ∈ {(A∪B) – C)
(A – C) ∪(B – C) ∈ (A ∪ B) – C ….. (ii)
From eqs. (i) and (ii),
(A ∪ B) – C = (A – C) ∪ (B – C)
Hence proved

Commonly Made Error:
Some students are unable to apply appropriate properties which result in proving the answer wrong.

Answering Tip:
Learn the properties of sets and their complements properly.

Question 34.
If the A.M. between rth and sth terms of an A.P be equal to A.M. between rth and sth terms of the A.P, then show that p + q = r + s.
OR
\(\cos ^4 \frac{\pi}{8}+\cos ^4 \frac{3 \pi}{8}+\cos ^4 \frac{5 \pi}{8}+\cos ^4 \frac{7 \pi}{8}\)
Answer:
Let a be the first term and d be the common difference of the given A.R Then
a_p = p^th term = a + (p – 1)d;
a_q = q^th term = a + (q – 1)d;
a_r = r^th term =a + (r – 1)d;
and a_s = s^th term =a + (s – 1)d
It is given that
A.M. between a_p and a_q – A.M. between a_r and a_s
⇒ \(\frac{1}{2}\)(a_p + a_q) = \(\frac{1}{2}\)(a_r + a_s)
⇒ a_p + a_q = a_r + a_s
⇒ a + (p – 1)d + a + (q – 1)d = a + (r – 1)d + a + (s – 1)d
⇒ (p + q – 2 )d = (r + s – 2 )d
⇒ p + q = r + s
Hence proved

Question 35.
Solve the following system of linear inequalities if
3x + 2y ≥ 24, 3x + y ≤ 15, x ≥ 4.
Answer:
Consider the inequation 3x + 2y ≥ 24 as an equation,
we have 3x + y = 24

Hence, line 3x + y = 24 intersect coordinate axes at points (8,0) and (0,12).
Now, (0, 0) does not satisfy the inequation 3x + 2y ≥ 24.
Therefore, the half-plane of the solution set does not contains (0,0).
Consider the inequation 3x + y < 15 as an equation, we have
⇒ 3x + y = 15
y = 15 – 3x

Line 3x + y = 15 intersects coordinate axes at points (5, 0) and (0,15).
Now, point (0, 0) satisfy the inequation 3x + y ≤ 15.
Therefore, the half plane of the solution contain origin.
Consider the inequality x ≥ 4 as an equation, we have x = 4
It represents a straight line parallel to Y-axis passing through (4, 0). Now, point (0, 0) does not satisfy the inequation x ≥ 4.
Therefore, half plane does not contains (0, 0),
The graph of the above inequations is given below.

It is clear from the graph that there is no common region corresponding to this inequality.
Hence, the given system of inequalities has no solution.

Question 36.
Let S be the sum, P be the product and R be the sum of reciprocals of n terms in a G.P. Prove that P²Rⁿ = Sⁿ.
OR

Answer:
Let the G.P is a ,ar, ar², ar³ ……………. ar^{n – 1}
Given, S = Sum of n terms = a, ar, ar², ar³ ……………. ar^{n – 1}