BCS-012 — Group 3 Solutions (Questions 4 & 5)

Question 4

(a) Find intervals where f(x) = (1/3)x³ − 3x² + 5x + 11 is (i) increasing (ii) decreasing

f′(x) = x² − 6x + 5 = (x−1)(x−5)

Critical points: x = 1, x = 5

Interval	x < 1	1 < x < 5	x > 5
Sign of f′(x)	+	−	+

Answer:

• (i) Increasing on (−∞, 1) ∪ (5, ∞)
• (ii) Decreasing on (1, 5)

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(b) Find the inverse of the matrix A = [[2,−1,5],[3,0,2],[0,1,1]]

Step 1: |A| (expand along Row 1)

|A| = 2(0·1−2·1) − (−1)(3·1−2·0) + 5(3·1−0·0) = 2(−2) + 1(3) + 5(3) = −4+3+15 = 14

Step 2: Cofactors

• C₁₁ = (0−2) = −2, C₁₂ = −(3−0) = −3, C₁₃ = (3−0) = 3
• C₂₁ = −(−1−5) = 6, C₂₂ = (2−0) = 2, C₂₃ = −(2−0) = −2
• C₃₁ = (−2−0) = −2, C₃₂ = −(4−15) = 11, C₃₃ = (0+3) = 3

Step 3: adj(A) = transpose of cofactor matrix =

[[−2, 6, −2], [−3, 2, 11], [3, −2, 3]]

A⁻¹ = (1/|A|) · adj(A) = (1/14)·[[−2,6,−2],[−3,2,11],[3,−2,3]]

Answer: A⁻¹ = (1/14) [[−2, 6, −2], [−3, 2, 11], [3, −2, 3]]

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(c) Find a quadratic equation with real coefficients, one of whose roots is −3+4i

Since the equation has real coefficients, complex roots occur in conjugate pairs.

So the other root is −3−4i.

Sum of roots = (−3+4i)+(−3−4i) = −6

Product of roots = (−3+4i)(−3−4i) = (−3)²−(4i)² = 9−(−16) = 25

Required equation: x² − (sum)x + (product) = 0

x² − (−6)x + 25 = 0

Answer: x² + 6x + 25 = 0

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(d) Show that f(x) = x² ln(1/x), x > 0, has a local maximum at x = 1/√e

f(x) = x² ln(1/x) = −x² ln x

f′(x) = −[2x·ln x + x²·(1/x)] = −[2x ln x + x] = −x(2 ln x + 1)

Setting f′(x) = 0, and since x > 0: 2 ln x + 1 = 0 ⟹ ln x = −1/2 ⟹ x = e^(−1/2) = 1/√e

Second derivative test:

f′(x) = −2x ln x − x

f″(x) = −2(ln x + 1) − 1 = −2 ln x − 3

At x = 1/√e: ln x = −1/2

f″(1/√e) = −2(−1/2) − 3 = 1 − 3 = −2 < 0

Answer: Since f″(1/√e) < 0, f(x) has a local maximum at x = 1/√e. ✓

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Question 5

(a) If a = −î+ĵ, b = ĵ+k̂, c = î+k̂, find (a×b)×c

a = (−1,1,0), b = (0,1,1), c = (1,0,1)

Step 1: a × b = | î ĵ k̂ ; −1 1 0 ; 0 1 1 |

= î(1·1−0·1) − ĵ(−1·1−0·0) + k̂(−1·1−1·0) = î(1) − ĵ(−1) + k̂(−1) = (1, 1, −1)

Step 2: (a×b) × c, where a×b = (1,1,−1), c=(1,0,1):

= | î ĵ k̂ ; 1 1 −1 ; 1 0 1 | = î(1·1−(−1)·0) − ĵ(1·1−(−1)·1) + k̂(1·0−1·1) = î(1) − ĵ(2) + k̂(−1)

Answer: (a×b)×c = î − 2ĵ − k̂

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(b) Find the shortest distance between the lines

r = î+2ĵ+k̂+λ(î−ĵ+k̂) and r = 2î+ĵ−k̂+μ(2î+ĵ+2k̂)

a₁ = (1,2,1), d₁ = (1,−1,1) a₂ = (2,1,−1), d₂ = (2,1,2)

a₂ − a₁ = (1, −1, −2)

d₁ × d₂ = | î ĵ k̂ ; 1 −1 1 ; 2 1 2 | = î(−1·2−1·1) − ĵ(1·2−1·2) + k̂(1·1−(−1)·2) = î(−3) − ĵ(0) + k̂(3) = (−3, 0, 3)

|d₁×d₂| = √(9+0+9) = √18 = 3√2

(a₂−a₁)·(d₁×d₂) = (1)(−3)+(−1)(0)+(−2)(3) = −3+0−6 = −9

Shortest distance = |(a₂−a₁)·(d₁×d₂)| / |d₁×d₂| = |−9| / 3√2 = 9/(3√2) = 3/√2

Answer: Shortest distance = 3/√2 = 3√2/2 units

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(c) Use induction to show that 1 + 1/3 + 1/3² + .... + 1/3ⁿ < 3/2, ∀n ∈ N

Let Sₙ = 1 + 1/3 + 1/3² + ... + 1/3ⁿ. We first show by induction that:

Sₙ = 3/2 − (1/2)(1/3)ⁿ ...(★)

Base case (n=1): S₁ = 1+1/3 = 4/3.

RHS of (★) = 3/2 − (1/2)(1/3) = 3/2 − 1/6 = 4/3. ✓ matches.

Inductive step: Assume Sₖ = 3/2 − (1/2)(1/3)ᵏ holds.

Sₖ₊₁ = Sₖ + (1/3)^(k+1) = 3/2 − (1/2)(1/3)ᵏ + (1/3)^(k+1) = 3/2 + (1/3)ᵏ[−1/2 + 1/3] = 3/2 + (1/3)ᵏ(−1/6) = 3/2 − (1/2)(1/3)^(k+1)

This matches (★) for n = k+1. So by induction, (★) holds for all n ∈ N.

Since (1/2)(1/3)ⁿ > 0 for every n ∈ N, it follows that Sₙ = 3/2 − (1/2)(1/3)ⁿ < 3/2.

Answer: Hence 1 + 1/3 + 1/3² + ... + 1/3ⁿ < 3/2 for all n ∈ N. ✓ (proved)