BCS-012 — Group 2 Solutions (Questions 2 & 3)

Question 2

(a) Find the angle between a = −î+2ĵ+3k̂ and b = î+4ĵ+k̂

a·b = (−1)(1)+(2)(4)+(3)(1) = −1+8+3 = 10

|a| = √(1+4+9) = √14

|b| = √(1+16+1) = √18 = 3√2

cosθ = (a·b)/(|a||b|) = 10/(√14·3√2) = 10/(3√28) = 10/(6√7) = 5/(3√7) = 5√7/21

Answer: θ = cos⁻¹(5√7/21) ≈ 38.2°

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(b) Use first derivative test to find local maxima/minima of f(x) = x³ − 27x + 5

f′(x) = 3x² − 27 = 3(x²−9) = 3(x−3)(x+3) ⟹ critical points x = 3, −3

Interval	x < −3	−3 < x < 3	x > 3
Sign of f′(x)	+	−	+

At x = −3: f′ changes (+) → (−) ⟹ local maximum. f(−3) = −27+81+5 = 59

At x = 3: f′ changes (−) → (+) ⟹ local minimum. f(3) = 27−81+5 = −49

Answer: Local maximum = 59 at x = −3; Local minimum = −49 at x = 3.

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(c) Find the nth term of the H.P.: 1/7, 2/15, 1/8, 2/17, ....

Re-write every term with numerator 2: 1/7 = 2/14, 2/15 = 2/15, 1/8 = 2/16, 2/17 = 2/17

So the H.P. is 2/14, 2/15, 2/16, 2/17, ...

The denominators 14, 15, 16, 17, ... form an A.P. with first term 14 and common difference 1.

nth denominator = 14 + (n−1)(1) = n + 13

Answer: nth term of the H.P. = 2/(n+13)

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(d) Find the area bounded by y = x² and y = x

Points of intersection: x² = x ⟹ x(x−1) = 0 ⟹ x = 0, 1

For 0 ≤ x ≤ 1, the line y = x lies above the parabola y = x².

Area = ∫₀¹ (x − x²) dx = [x²/2 − x³/3]₀¹ = (1/2 − 1/3) − 0 = 1/6

Answer: Required area = 1/6 square units.

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Question 3

(a) Use Cramer's rule to solve: 3x+y+z=6, x+y=3, 2y+z=1

Writing in standard form:

• 3x + y + z = 6
• x + y + 0z = 3
• 0x + 2y + z = 1

D = | 3 1 1 ; 1 1 0 ; 0 2 1 | = 3(1−0) − 1(1−0) + 1(2−0) = 3 − 1 + 2 = 4

Dx = | 6 1 1 ; 3 1 0 ; 1 2 1 | = 6(1−0) − 1(3−0) + 1(6−1) = 6 − 3 + 5 = 8

Dy = | 3 6 1 ; 1 3 0 ; 0 1 1 | = 3(3−0) − 6(1−0) + 1(1−0) = 9 − 6 + 1 = 4

Dz = | 3 1 6 ; 1 1 3 ; 0 2 1 | = 3(1−6) − 1(1−0) + 6(2−0) = −15 − 1 + 12 = −4

x = Dx/D = 8/4 = 2 y = Dy/D = 4/4 = 1 z = Dz/D = −4/4 = −1

Check: 3(2)+1+(−1)=6 ✓; 2+1=3 ✓; 2(1)+(−1)=1 ✓

Answer: x = 2, y = 1, z = −1

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(b) Solve the inequality 3/(7−x) < 2, x ∈ R

Case 1: 7 − x > 0, i.e. x < 7

3 < 2(7−x) ⟹ 3 < 14−2x ⟹ 2x < 11 ⟹ x < 11/2

Combined with x < 7: x < 11/2

Case 2: 7 − x < 0, i.e. x > 7 (inequality reverses on multiplying by negative)

3 > 2(7−x) ⟹ 3 > 14−2x ⟹ 2x > 11 ⟹ x > 11/2

Combined with x > 7: x > 7

(x = 7 is excluded as the expression is undefined there.)

Answer: Solution: x ∈ (−∞, 11/2) ∪ (7, ∞)

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(c) Evaluate ∫(2ˣ + 2⁻ˣ)² dx

(2ˣ+2⁻ˣ)² = 2²ˣ + 2·2ˣ·2⁻ˣ + 2⁻²ˣ = 4ˣ + 2 + 4⁻ˣ

∫(4ˣ + 2 + 4⁻ˣ) dx = 4ˣ/ln4 + 2x − 4⁻ˣ/ln4 + C

Answer: = (4ˣ − 4⁻ˣ)/ln4 + 2x + C

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(d) If a = 2î−ĵ+5k̂ and b = 3î+ĵ+λk̂, find λ so that a and b are at right angles

For a ⊥ b, a·b = 0

(2)(3) + (−1)(1) + (5)(λ) = 0

6 − 1 + 5λ = 0 ⟹ 5λ = −5 ⟹ λ = −1

Answer: λ = −1