BCS-012: Basic Mathematics — Worked Solutions

Term-End Examination, June 2024 (all parts carry 5 marks each)

Question 1

(a) Determinant identity

Apply R₁ → R₁ + R₂ + R₃. Since each row sums to 2(a+b+c):

Factor 2(a+b+c) from R₁, then apply C₂→C₂−C₁, C₃→C₃−C₁. Expanding along R₁ (which becomes 1,0,0) gives

since (a+b+c)(a²+b²+c²−ab−bc−ca) = a³+b³+c³−3abc.

Now expand the RHS determinant directly: $\begin{vmatrix}a&b&c\\b&c&a\\c&a&b\end{vmatrix} = a(bc-a^2)-b(b^2-ac)+c(ab-c^2) = 3abc-a^3-b^3-c^3$

So RHS = 2(3abc − a³ − b³ − c³) = −2(a³+b³+c³−3abc) = LHS. Identity proved.

(b) Matrix: (X − I₂)²

X − I₂ = \begin{bmatrix}0&-2\\2&0\end{bmatrix}

(X-I_2)^2 = \begin{bmatrix}0&-2\\2&0\end{bmatrix}\begin{bmatrix}0&-2\\2&0\end{bmatrix} = \begin{bmatrix}-4&0\\0&-4\end{bmatrix} = -4I_2

(c) Sum to n terms: 0.3 + 0.33 + 0.333 + ...

Each term: 0.\underbrace{33\ldots3}_{k} = \frac{1}{3}(1-10^{-k})

(d) A.P.: 7·T₇ = 11·T₁₁, find T₁₈

7(a+6d) = 11(a+10d) ⟹ 7a+42d = 11a+110d ⟹ 4a = −68d ⟹ a = −17d

T₁₈ = a + 17d = −17d + 17d = 0

(e) Cube roots of unity

Using 1+ω+ω² = 0:

• 2+ω+ω² = 2−1 = 1
• 3+ω+ω² = 3−1 = 2

(f) Quadratic with given roots

Roots: \frac{m-n}{m+n},\ -\frac{m+n}{m-n}

Sum = \dfrac{(m-n)^2-(m+n)^2}{(m+n)(m-n)} = \dfrac{-4mn}{m^2-n^2}, Product = −1

Equation: x^2+\dfrac{4mn}{m^2-n^2}x-1=0, i.e.

(g) ∫x√(3−2x) dx

Let u = 3−2x, x = (3−u)/2, dx = −du/2:

Substituting back:

(Verified by differentiating: gives x√(3−2x).)

(h) Rate of change of radius

V = (4/3)πr³ ⟹ dV/dt = 4πr² dr/dt

900 = 4π(25)² (dr/dt) = 2500π (dr/dt)

Question 2

(a) Solve by matrix method

Coefficient matrix determinant: $\det A = 2(-10+5)-(-1)(-15+4)+3(15-8) = -10-11+21 = 0$

Since det A = 0, the matrix (inverse) method fails — the system does not have a unique solution. Checking consistency by elimination:

• R₃ − 2R₁: 7y − 11z = −1
• 2R₂ − 3R₁: 7y − 11z = −1 (same equation)

The two reduce to one equation, so rank = 2 and the system is consistent with infinitely many solutions. Setting z = t:

(b) Induction: 1+4+7+...+(3n−2) = ½n(3n−1)

Base (n=1): LHS = 1, RHS = ½(1)(2) = 1 ✓

Inductive step: Assume true for n = k. For n = k+1: $\text{LHS} = \tfrac12 k(3k-1) + (3k+1) = \tfrac12(3k^2+5k+2) = \tfrac12(k+1)(3k+2)$ which equals RHS for n = k+1. Hence true for all n ∈ ℕ by induction.

(c) Quadratic with roots 1/(5−√72), 1/(5+6√2)

Since √72 = 6√2, the roots are \frac{1}{5-6\sqrt2} and \frac{1}{5+6\sqrt2}.

Sum = \dfrac{10}{25-72}=-\dfrac{10}{47}, Product = \dfrac{1}{25-72}=-\dfrac1{47}

(d) Complex number power

So \left(\dfrac{1-i}{1+i}\right)^{10} = -1, i.e. a = −1, b = 0 (purely real, b = 0 confirmed; note the magnitude works out to a real integer as required — the imaginary part vanishes exactly as the question asks to verify).

Question 3

(a) Row-reduce A to I₃

R₃→R₃−2R₁: (0,−3,1). Matrix: [[1,1,3],[0,5,2],[0,−3,1]]

R₃→5R₃+3R₂: (0,0,11) → R₃→R₃/11: (0,0,1)

R₂→R₂−2R₃: (0,5,0)→/5: (0,1,0)

R₁→R₁−3R₃: (1,1,0); R₁→R₁−R₂: (1,0,0)

Result: \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=I_3. Hence A is row equivalent to I₃ (and rank(A) = 3).

(b) Solve (2x−5)/(x+2) < 5

True when both factors share sign:

Graph: open circles at −5 and −2; shade the ray left of −5 and the ray right of −2 (point x = −2 excluded, as it's undefined there).

(c) Cubic with roots in A.P.

Let roots be a−d, a, a+d. Sum = 3a = 48/32 = 3/2 ⟹ a = 1/2.

Product = a(a²−d²) = 3/32. With a=1/2: ¼−d² = 3/16 ⟹ d² = 1/16 ⟹ d = 1/4.

(d) Local maxima/minima of f(x) = x³−6x²+9x+100

f′(x) = 3(x−1)(x−3); critical points x = 1, 3. f″(x) = 6x−12.

• x=1: f″<0 → local maximum, f(1) = 104
• x=3: f″>0 → local minimum, f(3) = 100

Question 4

(a) Continuity of f(x) = 2|x|/x (x≠0), f(0)=0

• For x→0⁺: f(x) = 2x/x = 2
• For x→0⁻: f(x) = −2x/x = −2

Left limit (−2) ≠ Right limit (2), so \lim_{x\to0} f(x) does not exist.

f is discontinuous at x = 0.

(b) Line through (1,−1,−2) parallel to 3î−2ĵ+5k̂

Vector form: \vec r = (\hat i-\hat j-2\hat k)+\lambda(3\hat i-2\hat j+5\hat k)

Cartesian form: $\frac{x-1}{3}=\frac{y+1}{-2}=\frac{z+2}{5}$

(c) Arc length of y = 2x^(3/2) from (1,2) to (4,16)

dy/dx = 3√x ⟹ Length = \int_1^4\sqrt{1+9x}\,dx

Let u = 1+9x: =\dfrac{2}{27}\left[u^{3/2}\right]_{10}^{37}=\dfrac{2}{27}\left(37\sqrt{37}-10\sqrt{10}\right)\approx 14.33 units

(d) Sum of integers between 100 and 1000 divisible by 9

First term 108 (9×12), last term 999 (9×111); number of terms = 111−12+1 = 100

Question 5

(a) Maximize 5x+2y subject to the constraints

Rewriting: 2x+3y≥6, x−2y≤2, 3x+2y≤12, −3x+2y≤3, x,y≥0.

Testing intersections of boundary lines for feasibility gives the corner points of the feasible region:

Maximum value = 19, attained at (x,y) = (7/2, 3/4).

(b) Area bounded by y = x² and y² = x

Intersections: x⁴ = x ⟹ x = 0, 1 → points (0,0) and (1,1).

(c) Normal form and rank of A

R₁↔R₂: [[1,2,3],[0,1,2],[3,1,1]]

R₃→R₃−3R₁: (0,−5,−8)

R₃→R₃+5R₂: (0,0,2) → /2: (0,0,1)

Back-substitute (R₂→R₂−2R₃, R₁→R₁−3R₃−2R₂... ) reduces fully to:

Rank(A) = 3.

(d) Vector proof: if a+b+c=0, show a×b = b×c = c×a

Since c = −(a+b): $\vec b\times\vec c = \vec b\times(-(\vec a+\vec b)) = -\vec b\times\vec a-\vec b\times\vec b = \vec a\times\vec b$

Since c×a = −(a+b)×a = −b×a = a×b