Tools

Markdown + LaTeX Viewer

Paste GitHub-flavored Markdown below. Inline math uses $...$ (e.g. $x^2+1$ ) and block math uses $$...$$ (e.g. $$\int_0^1 x\,dx$$). Everything renders live, client-side.

Upload a .md file

Stored in the database and viewable by anyone with the link. Max 1 MB, .md/.markdown only.

Markdown source

# BCS-012 — Group 1 Solutions (Question 1)

## (b) If A = [[2,3],[4,6]], show that A² = 8A

A² = A × A

- Row1: (2×2+3×4, 2×3+3×6) = (4+12, 6+18) = (16, 24)
- Row2: (4×2+6×4, 4×3+6×6) = (8+24, 12+36) = (32, 48)

So A² = [[16,24],[32,48]]

Also 8A = 8×[[2,3],[4,6]] = [[16,24],[32,48]]

**Answer:** Since A² = [[16,24],[32,48]] = 8A, the result is verified. ✓

---

## (c) Prove by induction: 2 + 4 + 6 + ... + 2n = n(n+1), ∀n ∈ N

**Step 1 (Base case, n = 1):** LHS = 2, RHS = 1(1+1) = 2. True.

**Step 2 (Assume true for n = k):** 2+4+6+...+2k = k(k+1)

**Step 3 (Prove for n = k+1):**

2+4+...+2k+2(k+1) = k(k+1) + 2(k+1)  [using assumption]
= (k+1)(k+2) = (k+1)[(k+1)+1]

This is the statement for n = k+1.

**Answer:** By the Principle of Mathematical Induction, 2+4+...+2n = n(n+1) holds for all n ∈ N. ✓

---

## (d) If pth term of an A.P. is q and qth term is p, show that rth term = p+q−r

Let first term = a, common difference = d.

- Tₚ = a+(p−1)d = q   ...(1)
- T_q = a+(q−1)d = p   ...(2)

Subtracting (2) from (1): (p−q)d = q−p ⟹ d = −1

From (1): a = q−(p−1)(−1) = q+p−1

rth term: T_r = a+(r−1)d = (p+q−1)+(r−1)(−1) = p+q−1−r+1 = p+q−r

**Answer:** rth term of the A.P. = p + q − r. ✓

---

## (e) If a+ib = (cosθ+i sinθ)², show that a²+b² = 1

(cosθ+i sinθ)² = cos²θ + 2i sinθ cosθ + i² sin²θ
= (cos²θ − sin²θ) + i(2 sinθ cosθ)
= cos 2θ + i sin 2θ  (double-angle identities)

So a+ib = cos2θ + i sin2θ ⟹ a = cos2θ, b = sin2θ

a² + b² = cos²2θ + sin²2θ = 1

**Answer:** a² + b² = 1. ✓

---

## (f) Find the sum to n terms of the series 1² + 3² + 5² + ... (up to n terms)

The kth term is the kth odd number squared: Tₖ = (2k−1)² = 4k² − 4k + 1

Sₙ = Σ(4k²−4k+1) = 4Σk² − 4Σk + Σ1, for k = 1 to n

= 4·[n(n+1)(2n+1)/6] − 4·[n(n+1)/2] + n
= (2/3)n(n+1)(2n+1) − 2n(n+1) + n

Taking n/3 common after simplifying numerator:

2n(n+1)(2n+1) − 6n(n+1) + 3n = n[2(n+1)(2n+1) − 6(n+1) + 3]
= n[(4n²+6n+2) − (6n+6) + 3] = n(4n² − 1)

**Answer:** Sₙ = n(4n² − 1)/3 = n(2n−1)(2n+1)/3

---

## (g) Find the roots of 8x³ − 14x² + 7x − 1 = 0, given the roots are in G.P.

Let the roots be a/r, a, ar.

**Product of roots** = (a/r)(a)(ar) = a³ = −(−1)/8 = 1/8 ⟹ a = 1/2

**Sum of roots** = a/r + a + ar = 14/8 = 7/4

⟹ a(1/r + 1 + r) = 7/4 ⟹ (1/2)(1+r+1/r) = 7/4 ⟹ r + 1/r = 5/2

⟹ 2r² − 5r + 2 = 0 ⟹ (2r−1)(r−2)=0 ⟹ r = 2 or r = 1/2

Taking r = 2: roots are a/r = 1/4, a = 1/2, ar = 1

**Check:**
- Sum = 1/4+1/2+1 = 7/4 ✓
- Product = 1/4×1/2×1 = 1/8 ✓
- Sum of products taken two at a time = (1/4)(1/2)+(1/2)(1)+(1/4)(1) = 1/8+1/2+1/4 = 7/8 ✓ (matches coefficient 7/8)

**Answer:** The roots are 1/4, 1/2, and 1.

---

## (h) Show that ∫₀⁵ √(5−x) / (√x + √(5−x)) dx = 5/2

Let I = ∫₀⁵ √(5−x)/(√x+√(5−x)) dx   ...(1)

Using the property ∫₀ᵃ f(x)dx = ∫₀ᵃ f(a−x)dx, replace x by (5−x):

I = ∫₀⁵ √x / (√(5−x)+√x) dx   ...(2)

Adding (1) and (2):

2I = ∫₀⁵ [√(5−x)+√x] / [√x+√(5−x)] dx = ∫₀⁵ 1 dx = 5

⟹ I = 5/2

**Answer:** I = 5/2. ✓ (proved)

Preview