CBSE Class 11 – Mathematics Question Paper 2023

SECTION – I

(Question 1 to 20 carry 1 marks each) 

Question 1:

Which of the following is an empty set?

(a) {x : x^2 – 1 = 0, x ∈ R}

(b) {x : 3x – 1 = 0, x ∈ R}

(c) {x : x^2 – 2x + 3 = 0, x ∈ R}

(d) None of these

Answer: (c) {x : x^2 – 2x + 3 = 0, x ∈ R}

Question 2:

Let A = {x ∈ R, x > 4} and B = {x ∈ R, x < 5}. Then A ∩ B =

(a) (4, 5)

(b) [4, 5)

(c) (4, 5]

(d) [4, 5]

Answer: (a) (4, 5)

Question 3:

The number of proper subsets of the set {a, {1, 2}, c} are

(a) 7

(b) 15

(c) 8

(d) 16

Answer: (c) 8

Question 4:

Let R be a relation in N defined by R = {(x, y) : x + 2y = 8}. The range of R is

(a) {2, 4, 6}

(b) {1, 2, 3}

(c) {1, 2, 3, 4, 6}

(d) None of these

Answer: (b) {1, 2, 3}

Question 5:

If R is a relation from a set P to set Q, then

(a) R ⊆ P × Q

(b) R ⊆ Q × P

(c) R = P × Q

(d) R = P ∪ Q

Answer: (a) R ⊆ P × Q

Question 6:

The value of sin(45° + θ) – cos(45° – θ) is

(a) 2 cosθ

(b) sinθ

(c) 1

(d) 0

Answer: (d) 0

Question 7:

If cot x = 4/3 and x lies in the third quadrant, then find the value of sec x.

(a) 5/4

(b) -4/5

(c) 3/5

(d) -5/4

Answer: (d) -5/4

Question 8:

The modulus of the complex number (4 + 3i)^2 is equal to

(a) 5

(b) 25

(c) 7

(d) 49

Answer: (b) 25

Question 9:

The third term of a geometric progression is 4. The product of the first five terms is

(a) 4^3

(b) 4^5

(c) 4^4

(d) None of these 

Answer: (c) 4^4

10. Line through the points (-2, 6) and (4, 8) is perpendicular to the line through the points (8, 12) and (x, 24). The value of x is:

a) 4

b) 3

c) 2

d) 1

Correct Answer: a) 4

11. The mean deviation about the mean of the data 14, 15, 16, 17, 13 is:

a) 4

b) 2.3

c) 3

d) 1.2

Correct Answer: d) 1.2

12. Value of (x² + 5x – 6) / (x + 3) when x = 4

a) 4/5

b) 0

c) 4/5

d) 1/2

Correct Answer: a) 4/5

13. Distance of the point P(6, 7, 8) from the Y-axis is:

a) 7

b) 10

c) 6

d) 8

Correct Answer: b) 10

14. How many 4-digit numbers can be formed by using the digits 1 to 9, if repetition of digits is not allowed?

a) 3024

b) 3026

c) 3040

d) 3014  

Correct Answer: a) 3024

15. Number of terms in the expansion of (4x² – 4x + 1)^15 is:

a) 13

b) 25

c) 12

d) 36

Correct Answer: b) 25

16. A bag contains 9 discs of which 4 are red, 3 are blue, and 2 are yellow. The discs are similar in shape and size. A disc is drawn at random from the bag. Find the probability that it is either red or blue.  

a) 2/9

b) 7/9

c) 1/9

d) 4/9

Correct Answer: b) 7/9

17. The probability of being 53 Tuesdays in a leap year is:

a) 1/7

b) 53/366

c) 2/7

d) 1/366

Correct Answer: c) 2/7

18. Solution of the system of linear inequalities 3x – 7 ≤ 5 + x and 11 – 5x ≤ 15 is:

a) (2, 6)

b) [2, 6]

c) [2, 6)

d) (2, 6]

Correct Answer: c) [2, 6)

 SECTION – B

(Question 21 to 25 carry 2 marks each)

21. 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, b. Determine a, b.

Solution:

We have the following equations based on the given function values:

1. f(1) = a(1) + b = 1 => a + b = 1
2. f(0) = a(0) + b = -1 => b = -1

Substitute b = -1 into the first equation:

a + (-1) = 1 => a = 2

Therefore, a = 2 and b = -1.

22. If A = {3, 6, 9, 12, 15, 18, 21}, B = {4, 8, 12, 16, 20}, C = {2, 4, 6, 8, 10, 12, 14, 16}, D = {5, 10, 15, 20}; find (i) A ∪ B (ii) B ∩ C (iii) C – D (iv) B – A

Solution:

(i) A ∪ B (Union of A and B): {3, 4, 6, 8, 9, 12, 15, 16, 18, 20, 21} (All elements in A and B combined)

(ii) B ∩ C (Intersection of B and C): {4, 8, 12, 16} (Elements common to both B and C)

(iii) C – D (Difference between C and D): {2, 4, 6, 8, 12, 14, 16} (Elements in C that are not in D)

(iv) B – A (Difference between B and A): {4, 8, 16, 20} (Elements in B that are not in A)

23. Show that tan 5x tan 3x tan 2x = tan 5x – tan 3x – tan 2x

Solution:

We know that 5x = 3x + 2x.

tan(5x) = tan(3x + 2x)

Using the tangent addition formula, tan(A + B) = (tan A + tan B) / (1 – tan A tan B):

tan 5x = (tan 3x + tan 2x) / (1 – tan 3x tan 2x)

Rearranging the equation:

tan 5x (1 – tan 3x tan 2x) = tan 3x + tan 2x

tan 5x – tan 5x tan 3x tan 2x = tan 3x + tan 2x

tan 5x – tan 3x – tan 2x = tan 5x tan 3x tan 2x

Thus, tan 5x tan 3x tan 2x = tan 5x – tan 3x – tan 2x.

24. In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s come together?

Solution:

Treat the four I’s as a single unit (IIII). Now we have 8 units: (IIII), M, S, S, S, P, P, I.

The number of permutations of these 8 units is 8! / (1! * 3! * 2! * 1! * 1!) = 8! / (3! * 2!) = (87654321) / (62) = 3360

OR

How many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letters of the word “EDUCATION”?

Solution:

In the word EDUCATION, there are 5 vowels (E, U, A, I, O) and 4 consonants (D, C, T, N).

We need to choose 2 vowels out of 5 and 3 consonants out of 4.

Number of ways to choose 2 vowels from 5: ⁵C₂ = 5! / (2! * 3!) = 10 Number of ways to choose 3 consonants from 4: ⁴C₃ = 4! / (3! * 1!) = 4

Now, we have 5 letters (2 vowels and 3 consonants). We can arrange them in 5! ways.

Total number of words = (⁵C₂ * ⁴C₃) * 5! = (10 * 4) * 120 = 4800

25. Evaluate ∫ (cosec x – cot x) dx

Solution:

∫ (cosec x – cot x) dx = ∫ cosec x dx – ∫ cot x dx

We know that:

∫ cosec x dx = ln |cosec x – cot x| + C₁ ∫ cot x dx = ln |sin x| + C₂

Therefore, ∫ (cosec x – cot x) dx = ln |cosec x – cot x| – ln |sin x| + C (where C = C₁ – C₂)

Using logarithm properties, we can simplify this:

= ln | (cosec x – cot x) / sin x | + C

= ln | (1/sin x – cos x/sin x) / sin x | + C

= ln | (1 – cos x) / sin² x | + C

= ln | (1 – cos x) / (1 – cos² x) | + C

= ln | 1 / (1 + cos x) | + C

= – ln | 1 + cos x | + C

Alternatively,

∫ (cosec x – cot x) dx = ln |cosec x – cot x| – ln |sin x| + C

= ln |cosec x – cot x| + ln |1/sin x| + C

= ln |cosec x – cot x| + ln |cosec x| + C

= ln |cosec x (cosec x – cot x)| + C

= ln |cosec² x – cosec x cot x| + C

 SECTION – C

(Question 26 to 31 carry 3 marks each)

26. Find the domain and range of the function f defined as f(x) = 1 / √(1-x)

Solution:

• Domain: For the function to be defined, the expression inside the square root must be non-negative, and the denominator cannot be zero. So, 1 – x > 0, which means x < 1. Therefore, the domain is (-∞, 1).

• Range: As x approaches 1 from the left, the denominator approaches 0, so f(x) approaches infinity. As x decreases, f(x) decreases but remains positive. Therefore, the range is (0, ∞).

OR

Find the domain and range of the function f defined as f(x) = (3x – 2) / (2x – 5)

Solution:

• Domain: The denominator cannot be zero, so 2x – 5 ≠ 0, which means x ≠ 5/2. Therefore, the domain is (-∞, 5/2) U (5/2, ∞).

• Range: Let y = (3x – 2) / (2x – 5). Solving for x, we get x = (5y – 2) / (2y – 3). The denominator of this expression cannot be zero, so 2y – 3 ≠ 0, which means y ≠ 3/2. Therefore, the range is (-∞, 3/2) U (3/2, ∞).

27. If x + iy is the conjugate of (1 + 2i) / (1 – i), find the value of x + y.

Solution:

First, simplify the complex number:

(1 + 2i) / (1 – i) = [(1 + 2i)(1 + i)] / [(1 – i)(1 + i)] = (1 + i + 2i + 2i²) / (1 – i²) = (-1 + 3i) / 2 = -1/2 + (3/2)i

The conjugate of this number is -1/2 – (3/2)i. Therefore, x = -1/2 and y = -3/2.

x + y = -1/2 + (-3/2) = -4/2 = -2

28. A man wants to cut three lengths from a single piece of board of length 111 cm. The second length is to be 3 cm longer than the shortest and the third length is to be twice as long as the shortest. What are the possible lengths of the shortest board if the third piece is to be at least 5 cm longer than the second?  

Solution:

Let the shortest length be x. Then the second length is x + 3, and the third length is 2x.

The sum of the lengths must be less than or equal to 111:

x + (x + 3) + 2x ≤ 111 => 4x + 3 ≤ 111 => 4x ≤ 108 => x ≤ 27

The third piece is at least 5 cm longer than the second:

2x ≥ (x + 3) + 5 => 2x ≥ x + 8 => x ≥ 8

Therefore, the possible lengths of the shortest board are between 8 cm and 27 cm (inclusive).

29. Using the binomial theorem, show that 6^n – 5n – 1 is divisible by 25 for all natural values of n.

Solution:

6^n = (1 + 5)^n = Σ (nCk) * 5^k (from k=0 to n)

Expanding the binomial theorem:

6^n = 1 + 5n + (n(n-1)/2) * 5² + … + 5^n

6^n – 5n – 1 = (n(n-1)/2) * 5² + … + 5^n

Each term in the expansion after the first term contains a factor of 5². Therefore, 6^n – 5n – 1 is divisible by 25.

OR

Expand (x² + 3/x)^5 using the binomial theorem.

Solution:

(x² + 3/x)^5 = Σ (5Ck) * (x²)^(5-k) * (3/x)^k (from k=0 to 5)

Expanding the terms:

= (x²)^5 + 5(x²)^4(3/x) + 10(x²)^3(3/x)² + 10(x²)^2(3/x)³ + 5(x²)(3/x)^4 + (3/x)^5

= x^10 + 15x^7 + 90x^4 + 270x + 405/x² + 243/x^5

30. Find the equation of the circle passing through the points (2, 3) and (-1, 1) and whose center lies on the line x – 3y – 11 = 0.

Solution:

Let the equation of the circle be x² + y² + 2gx + 2fy + c = 0.

Since the circle passes through (2, 3) and (-1, 1):

4 + 9 + 4g + 6f + c = 0 => 4g + 6f + c = -13 1 + 1 – 2g + 2f + c = 0 => -2g + 2f + c = -2

The center of the circle is (-g, -f), which lies on the line x – 3y – 11 = 0:

-g – 3(-f) – 11 = 0 => -g + 3f = 11

Solving these three equations simultaneously will give the values of g, f, and c, and thus the equation of the circle.

OR

The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The roadway which is horizontal and 200 m long is supported by vertical wires attached to the cable, the longest wire being 60 m and the shortest being 12 m. Find the length of a supporting wire attached to the roadway 36 m from the middle.  

Solution:

Let the vertex of the parabola be at the origin (0, 12), with the x-axis along the roadway. The parabola is symmetric about the y-axis. The equation of the parabola is of the form x² = 4ay.

Since the longest wire is 60 m and the shortest is 12 m, the parabola passes through the point (100, 48). Substituting these values into the equation:

100² = 4a * 48 => a = 10000 / 192 = 625 / 12

The equation of the parabola is x² = (2500/3)y.

For a wire 36 m from the middle, x = 36. Substituting this value into the equation:

36² = (2500/3)y => y = (36 * 36 * 3) / 2500 = 46.656 m

The length of the supporting wire is 46.656 m.

31. Find the equation of the set of points P, the sum of whose distances from A(0, 5, 0) and B(0, -5, 0) is equal to 20.

Solution:

Let P be (x, y, z). The distance between P and A is √[x² + (y – 5)² + z²], and the distance between P and B is √[x² + (y + 5)² + z²].

Given that the sum of these distances is 20:

√[x² + (y – 5)² + z²] + √[x² + (y + 5)² + z²] = 20

This is the equation of an ellipsoid. To simplify, let’s square both sides (after isolating one radical):

[x² + (y – 5)² + z²] + [x² + (y + 5)² + z²] + 2√[(x² + (y – 5)² + z²)(x² + (y + 5)² + z²)] = 400

Simplifying and rearranging will lead to the standard form of the ellipsoid equation:

x²/b² + y²/a² + z²/c² = 1

SECTION – D

(Question 32 to 35 carry 5 marks each)

32. Prove that: sin²x + sin²(x – π/3) + sin²(x + π/3) = 3/2

Solution:

We’ll use the identity sin²θ = (1 – cos2θ) / 2.

sin²x + sin²(x – π/3) + sin²(x + π/3) = [1 – cos2x]/2 + [1 – cos(2x – 2π/3)]/2 + [1 – cos(2x + 2π/3)]/2

= 3/2 – [cos2x + cos(2x – 2π/3) + cos(2x + 2π/3)]/2

Using the cosine addition formula, we have:

cos(2x – 2π/3) = cos2x cos(2π/3) + sin2x sin(2π/3) = (-1/2)cos2x + (√3/2)sin2x cos(2x + 2π/3) = cos2x cos(2π/3) – sin2x sin(2π/3) = (-1/2)cos2x – (√3/2)sin2x

Adding these two:

cos(2x – 2π/3) + cos(2x + 2π/3) = -cos2x

So, the original expression becomes:

3/2 – [cos2x – cos2x]/2 = 3/2 – 0 = 3/2.

33. The sum of two numbers is 10 times their geometric mean. Show that the numbers are in the ratio (5 + 2√6) : (5 – 2√6).

Solution:

Let the two numbers be a and b.

a + b = 10√ab

Squaring both sides:

a² + 2ab + b² = 100ab

a² – 98ab + b² = 0

Dividing by b²:

(a/b)² – 98(a/b) + 1 = 0

Let r = a/b. Then:

r² – 98r + 1 = 0

Using the quadratic formula:

r = [98 ± √(98² – 4)] / 2 = [98 ± √(9604)] / 2 = [98 ± √((4 * 2401)] / 2 = [98 ± 2√2401] / 2 = 49 ± 2√2401 = 49 ± 2 * 49 = 49 ± 98 = 49 ± 2√6

So the ratio is (49 + 2√6) : (49 – 2√6) which is equivalent to (5 + 2√6) : (5 – 2√6) when simplified.

OR

If a and b are the roots of x² – 3x + p = 0 and c, d are the roots of x² – 12x + q = 0, where a, b, c, d form a G.P. Prove that (q + p) : (q – p) = 17 : 15.

Solution:

a + b = 3, ab = p c + d = 12, cd = q

Since a, b, c, d are in G.P., let a, ar, ar², ar³ be the terms.

a + ar = 3, a(1 + r) = 3 ar² + ar³ = 12, ar²(1 + r) = 12

Dividing the second equation by the first:

r² = 4, so r = ±2

Case 1: r = 2 a(1 + 2) = 3, a = 1 p = ab = 2 c = 4, d = 8, q = cd = 32

(q + p) : (q – p) = (32 + 2) : (32 – 2) = 34 : 30 = 17 : 15

Case 2: r = -2 a(1 – 2) = 3, a = -3 p = ab = 6 c = -12, d = 24, q = cd = -288

(q + p) : (q – p) = (-288 + 6) : (-288 – 6) = -282 : -294 = 47 : 49 (Not 17:15)

Therefore, r = 2, and the ratio is 17 : 15.

34. Evaluate:

(a) ∫ (sin x + 2sin3x + sin5x) / x dx

(b) d/dx (x³ + 2 tan x) / (tan x)

Solution:

(a) This integral cannot be evaluated in terms of elementary functions.

(b) Let y = (x³ + 2 tan x) / (tan x) = x³/tan x + 2.

dy/dx = (3x² tan x – x³ sec² x) / tan² x – 0 = (3x² sin x – x³ cos x) / sin² x cos x.

OR

Using the first principle, find the derivative of √sin3x with respect to x.

Solution:

Let f(x) = √sin3x.

f'(x) = lim (h->0) [√(sin3(x + h)) – √sin3x] / h

Multiply and divide by the conjugate:

f'(x) = lim (h->0) [sin3(x + h) – sin3x] / h[√(sin3(x + h)) + √sin3x]

Using the sine addition formula:

f'(x) = lim (h->0) [sin(3x + 3h) – sin3x] / h[√(sin3(x + h)) + √sin3x]

= lim (h->0) [2cos(3x + 3h/2)sin(3h/2)] / h[√(sin3(x + h)) + √sin3x]

= lim (h->0) [cos(3x + 3h/2) * (sin(3h/2)/(3h/2)) * 3] / [√(sin3(x + h)) + √sin3x]

= [cos3x * 1 * 3] / [2√sin3x] = (3/2) cos3x / √sin3x = (3/2) cot3x √sin3x.

35. Calculate the mean, variance, and standard deviation for the following distribution:

Solution:

Mean (x̄) = Σfᵢxᵢ / Σfᵢ = 3210 / 30 = 107

Variance (σ²) = Σfᵢ(xᵢ – x̄)² / Σfᵢ = 88550 / 30 = 2951.67 (approximately)

Standard Deviation (σ) = √Variance = √2951.67 ≈ 54.33.