Solved papers class 11 Maths CBSE ALL INDIA 2022 – 2023

📝 SECTION B

Very Short Answer (VSA) · 5 × 2 = 10 marks

Q21.
🔢 Sets (VSA)

If P = {1, 2, 3, 4}, Q = {3, 4, 5, 6}, R = {5, 6, 7, 8} and S = {7, 8, 9, 10}. Find:

(a) \(P \cup Q\)

(b) \(P \cup Q \cup R\)

OR

Which of the given pairs of sets are disjoint?

(i) A = {1, 2, 3, 4} and B = {x: x is a natural number and 4 ≤ x ≤ 6}

(ii) A = {x: x is an even integer} and B = {x: x is an odd integer}

🔍 VIEW ANSWER

✅ STEP-BY-STEP SOLUTION:

Main Part:
(a) \(P \cup Q\) \(= \{1,2,3,4\} \cup \{3,4,5,6\}\)\( = \{1,2,3,4,5,6\}\)
(b) \(P \cup Q \cup R\)\( = \{1,2,3,4,5,6\} \cup \{5,6,7,8\}\)\( = \{1,2,3,4,5,6,7,8\}\)

OR Part (Disjoint sets):
(i) A = {1,2,3,4}, B = {4,5,6} (since natural numbers 4≤x≤6). They have common element 4 → not disjoint.
(ii) Even integers = {…,-4,-2,0,2,4,…}, Odd integers = {…,-3,-1,1,3,…}. No common element → disjoint.

✅ Final answers: (a) {1,2,3,4,5,6} (b) {1,2,3,4,5,6,7,8} | OR: (i) not disjoint, (ii) disjoint

Q22.
📐 Trigonometry (Proof)

Prove that: \(\frac{\sin x – \sin y}{\cos x + \cos y} = \tan\frac{x – y}{2}\)

🔍 VIEW PROOF

✅ STEP-BY-STEP PROOF:

Step 1: Use sum-to-product identities:

\(\sin x – \sin y\) \(= 2 \cos\frac{x+y}{2} \sin\frac{x-y}{2}\)

\(\cos x + \cos y\) \(= 2 \cos\frac{x+y}{2} \cos\frac{x-y}{2}\)

Step 2: Substitute into LHS:

\(\frac{\sin x – \sin y}{\cos x + \cos y}\) \(= \frac{2 \cos\frac{x+y}{2} \sin\frac{x-y}{2}}{2 \cos\frac{x+y}{2} \cos\frac{x-y}{2}}\)

Step 3: Cancel common factor \(2 \cos\frac{x+y}{2}\) (provided \(\cos\frac{x+y}{2} \neq 0\)):

\(= \frac{\sin\frac{x-y}{2}}{\cos\frac{x-y}{2}} = \tan\frac{x-y}{2}\)

✅ Hence proved: LHS = RHS.

Q23.
🧮 Quadratic Equations

Solve the quadratic equation: \(x^{2} + 3x + 9 = 0\)

OR

Solve: \(-x^{2} + x + 2 = 0\)

🔍 VIEW SOLUTION

✅ STEP-BY-STEP SOLUTIONS:

Main: \(x^{2} + 3x + 9 = 0\)

Using quadratic formula \(x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}\) with a=1, b=3, c=9:

Discriminant \(D = 9 – 36 = -27 = 27i^2\)

\(\sqrt{D} = \sqrt{27}i = 3\sqrt{3}i\)

\(x = \frac{-3 \pm 3\sqrt{3}i}{2}\)

Thus roots are \(\frac{-3 + 3\sqrt{3}i}{2}\) and \(\frac{-3 – 3\sqrt{3}i}{2}\)

OR: \(-x^{2} + x + 2 = 0\) (Multiply by -1)

\(x^{2} – x – 2 = 0\)

Factor: \((x – 2)(x + 1) = 0\)

Thus \(x = 2\) or \(x = -1\)

✅ Main: \(\frac{-3 \pm 3\sqrt{3}i}{2}\) | OR: x = 2, -1

Q24.
📈 Arithmetic Progression

Insert five numbers between 8 and 26 so that the resulting series is an A.P.

🔍 VIEW SOLUTION

✅ STEP-BY-STEP SOLUTION:

Let the AP be: 8, A₁, A₂, A₃, A₄, A₅, 26

Here first term a = 8, seventh term a₇ = 26.

Using formula aₙ = a + (n-1)d:

a₇ = 8 + 6d = 26

⇒ 6d = 18 ⇒ d = 3

Thus the five numbers are:

A₁ = 8 + 3 = 11

A₂ = 11 + 3 = 14

A₃ = 14 + 3 = 17

A₄ = 17 + 3 = 20

A₅ = 20 + 3 = 23

✅ The five numbers are: 11, 14, 17, 20, 23

Check: 8,11,14,17,20,23,26 forms AP with common difference 3.

Q25.
📊 Mean Deviation

Find the mean deviation about the mean for the following data:

\(x_i\)1030507090
\(f_i\)42428168
🔍 VIEW SOLUTION

✅ STEP-BY-STEP CALCULATION:

Step 1: Calculate mean (\(\bar{x}\))

\(x_i\)\(f_i\)\(f_i x_i\)
10440
3024720
50281400
70161120
908720
Total\(\sum f_i = 80\)

\(\sum f_i x_i\)

\(= 4000\)

\(\bar{x} = \frac{4000}{80} = 50\)

Step 2: Calculate \(|x_i – \bar{x}|\) and \(f_i|x_i – \bar{x}|\)

\(x_i\)\(f_i\)\(|x_i – 50|\)\(f_i|x_i – 50|\)
10440160
302420480
502800
701620320
90840320
Total  \(\sum f_i|x_i – \bar{x}|\) \(= 1280\)

Step 3: Mean deviation about mean

M.D. = \(\frac{\sum f_i|x_i – \bar{x}|}{\sum f_i} = \frac{1280}{80} = 16\)

✅ Mean deviation about mean = 16