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BCS-012 Mathematics Solved Notes

Uploaded June 19, 2026, 11:55 p.m.

BCS-012 Mathematics — Important Solved Notes

BCA IGNOU — Sample Questions with Step-by-Step Solutions

Based on the important topics discussed in this chat and recent BCS-012 question patterns.

  • Subject: BCS-012 Basic Mathematics
  • Format: Separated 5-mark and 10-mark answers
  • Focus: Complex numbers, matrices, induction, inequalities, calculus, vectors, LPP

Contents

  1. How to Write Answers According to Marks
  2. Set 1: Important Whiteboard Topics
    • 2.1 De Moivre's Theorem [5 marks]
    • 2.2 Cube Roots of Unity / Omega [5 marks]
    • 2.3 Matrix Inverse [5 marks]
    • 2.4 Cramer's Rule [5 marks]
    • 2.5 Linear Inequality [5 marks]
    • 2.6 Mathematical Induction [5 marks]
    • 2.7 Linear Programming: Tailors Problem [10 marks]
    • 2.8 Linear Programming: Donation Problem [10 marks]
  3. Set 2: Other Frequently Asked Questions
    • 3.1 Sum of the Series [5 marks]
    • 3.2 Quadratic Equation with Roots α², β² [5 marks]
    • 3.3 Complex Number Condition [5 marks]
    • 3.4 Limit [5 marks]
    • 3.5 Differentiation [5 marks]
    • 3.6 Local Maxima and Minima [5 marks]
    • 3.7 Integration [5 marks]
    • 3.8 Definite Integral [5 marks]
    • 3.9 Area Between Curves [5 marks]
    • 3.10 Direction Cosines [5 marks]
    • 3.11 Shortest Distance Between Two Lines [5 marks]
    • 3.12 Scalar Triple Product [5 marks]
    • 3.13 Harmonic Progression [5 marks]
    • 3.14 Cubic Equation with Integer Roots [5 marks]
    • 3.15 Rational Inequality [5 marks]
    • 3.16 Find k from a Differential Equation [5 marks]
    • 3.17 Area of Triangle by Determinants [5 marks]
    • 3.18 Logarithmic Differentiation [5 marks]
  4. Set 3: Continued Solutions for Other Questions
    • 4.1 Matrix Polynomial [5 marks]
    • 4.2 Infinite G.P. [5 marks]
    • 4.3 Quadratic Equation with Roots α²+3, β²+3 [5 marks]
    • 4.4 Differential Equation Verification [5 marks]
    • 4.5 Length of a Curve [5 marks]
    • 4.6 Integral: ∫ x³/(x+1)² dx [5 marks]
    • 4.7 Integral: ∫ x/√(x+3) dx [5 marks]
    • 4.8 Definite Log Integral [5 marks]
    • 4.9 Purely Imaginary Complex Expression [5 marks]
    • 4.10 Complex Number Condition: Real Axis [5 marks]
    • 4.11 Induction: Divisibility [5 marks]
    • 4.12 Induction: Sum of Odd Numbers [5 marks]
    • 4.13 Absolute Value Inequality [5 marks]
    • 4.14 Intersection of Two Lines in 3D [5 marks]
    • 4.15 Triangle Inequality for Vectors [5 marks]
    • 4.16 Unit Vector Perpendicular to Two Vectors [5 marks]
    • 4.17 Matrix Inverse from Product [5 marks]
    • 4.18 Linear Programming [5 marks]
  5. Quick Revision Formula Sheet

1. How to Write Answers According to Marks

For 5-mark questions

Write in this order:

Formula/definition → Substitution → Calculation → Final answer.

Do not skip important steps. In induction, always show base case, assumption, and proof for the next case.

For 10-mark Linear Programming questions

Write in this order:

Variables → Objective function → Constraints → Corner points → Value table → Final answer.

2. Set 1: Important Whiteboard Topics

2.1 De Moivre's Theorem [5 marks]

Question. Using De Moivre's theorem, find (1+i)^4.

Solution. Write 1+i in polar form:

r = \sqrt{1^2+1^2} = \sqrt{2}, \qquad \theta = \tan^{-1}\left(\frac{1}{1}\right) = 45^\circ = \frac{\pi}{4}.

So,

1+i = \sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right).

By De Moivre's theorem,

(1+i)^4 = (\sqrt{2})^4\left(\cos\left(4\cdot\frac{\pi}{4}\right) + i\sin\left(4\cdot\frac{\pi}{4}\right)\right) = 4(\cos\pi + i\sin\pi) = 4(-1+0i) = -4.

Therefore, (1+i)^4 = -4.

2.2 Cube Roots of Unity / Omega [5 marks]

Question. If 1, \omega, \omega^2 are cube roots of unity, show that

\frac{(2+3\omega+2\omega^2)^9}{(4+3\omega+3\omega^2)^9} = 1.

Solution. For cube roots of unity,

\omega^3 = 1, \qquad 1+\omega+\omega^2 = 0.

Thus,

\omega+\omega^2 = -1, \qquad 1+\omega^2 = -\omega.

Now,

2+3\omega+2\omega^2 = 2(1+\omega^2)+3\omega = 2(-\omega)+3\omega = \omega.

Also,

4+3\omega+3\omega^2 = 4+3(\omega+\omega^2) = 4+3(-1) = 1.

Therefore,

\frac{(2+3\omega+2\omega^2)^9}{(4+3\omega+3\omega^2)^9} = \frac{\omega^9}{1^9} = (\omega^3)^3 = 1^3 = 1.

Hence proved.

2.3 Matrix Inverse [5 marks]

Question. Find the inverse of

A = \begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{pmatrix}.

Solution. Using elementary row operations on [A : I],

[A : I] = \left[\begin{array}{ccc|ccc} 1 & 2 & 2 & 1 & 0 & 0 \\ 2 & 1 & 2 & 0 & 1 & 0 \\ 2 & 2 & 1 & 0 & 0 & 1 \end{array}\right].

Reducing the left side to identity gives

\left[\begin{array}{ccc|ccc} 1 & 0 & 0 & -\tfrac{3}{5} & \tfrac{2}{5} & \tfrac{2}{5} \\ 0 & 1 & 0 & \tfrac{2}{5} & -\tfrac{3}{5} & \tfrac{2}{5} \\ 0 & 0 & 1 & \tfrac{2}{5} & \tfrac{2}{5} & -\tfrac{3}{5} \end{array}\right].

Therefore,

A^{-1} = \begin{pmatrix} -\tfrac{3}{5} & \tfrac{2}{5} & \tfrac{2}{5} \\ \tfrac{2}{5} & -\tfrac{3}{5} & \tfrac{2}{5} \\ \tfrac{2}{5} & \tfrac{2}{5} & -\tfrac{3}{5} \end{pmatrix} = \frac{1}{5}\begin{pmatrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{pmatrix}.

Thus,

A^{-1} = \frac{1}{5}\begin{pmatrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{pmatrix}.

2.4 Cramer's Rule [5 marks]

Question. Use Cramer's rule to solve

2x - y + z = 4, \quad 3x - y = 5, \quad 2y - z = 1.

Solution. Write the system in standard form:

2x - y + z = 4, \quad 3x - y + 0z = 5, \quad 0x + 2y - z = 1.

The coefficient determinant is

D = \begin{vmatrix} 2 & -1 & 1 \\ 3 & -1 & 0 \\ 0 & 2 & -1 \end{vmatrix} = 5.

Now,

D_x = \begin{vmatrix} 4 & -1 & 1 \\ 5 & -1 & 0 \\ 1 & 2 & -1 \end{vmatrix} = 10, \qquad D_y = \begin{vmatrix} 2 & 4 & 1 \\ 3 & 5 & 0 \\ 0 & 1 & -1 \end{vmatrix} = 5, \qquad D_z = \begin{vmatrix} 2 & -1 & 4 \\ 3 & -1 & 5 \\ 0 & 2 & 1 \end{vmatrix} = 5.

By Cramer's rule,

x = \frac{D_x}{D} = \frac{10}{5} = 2, \qquad y = \frac{D_y}{D} = \frac{5}{5} = 1, \qquad z = \frac{D_z}{D} = \frac{5}{5} = 1.

Therefore, x=2,\ y=1,\ z=1.

2.5 Linear Inequality [5 marks]

Question. Solve \left|\dfrac{x-3}{2}\right| \le 1.

Solution.

\left|\frac{x-3}{2}\right| \le 1 \quad\text{means}\quad -1 \le \frac{x-3}{2} \le 1.

Multiplying throughout by 2,

-2 \le x-3 \le 2.

Adding 3 throughout,

1 \le x \le 5.

Hence, x \in [1,5].

2.6 Mathematical Induction [5 marks]

Question. Prove by mathematical induction:

1+2+2^2+\cdots+2^{n-1} = 2^n - 1.

Solution. Let P(n): 1+2+2^2+\cdots+2^{n-1} = 2^n-1.

For n=1: \text{LHS}=1,\ \text{RHS}=2^1-1=1. Thus P(1) is true.

Assume P(k) is true, that is,

1+2+2^2+\cdots+2^{k-1} = 2^k - 1.

Now, for n=k+1,

1+2+2^2+\cdots+2^{k-1}+2^k = (2^k-1)+2^k = 2\cdot 2^k - 1 = 2^{k+1}-1.

Therefore, P(k+1) is true. Hence, by mathematical induction,

1+2+2^2+\cdots+2^{n-1} = 2^n-1.

2.7 Linear Programming: Tailors Problem [10 marks]

Question. Tailor A earns Rs. 600 per day and Tailor B earns Rs. 800 per day. A can stitch 6 shirts and 4 pants per day. B can stitch 10 shirts and 4 pants per day. Find how many days each should work to produce at least 60 shirts and 32 pants at minimum labour cost.

Solution. Let x = number of days Tailor A works, y = number of days Tailor B works.

Cost function: C = 600x + 800y (to be minimized).

Constraints:

6x+10y \ge 60 \;\Rightarrow\; 3x+5y \ge 30,
4x+4y \ge 32 \;\Rightarrow\; x+y \ge 8,
x \ge 0,\ y \ge 0.

Solve the boundary equations:

3x+5y=30, \qquad x+y=8.

From x = 8-y:

3(8-y)+5y = 30 \;\Rightarrow\; 24-3y+5y=30 \;\Rightarrow\; 2y=6 \;\Rightarrow\; y=3.

Thus, x = 8-3 = 5.

Corner points: (10,0),\ (5,3),\ (0,8).

Corner point Cost C = 600x+800y
(10,0) 6000
(5,3) 600(5)+800(3) = 5400
(0,8) 6400

Minimum cost is Rs. 5400, at (5,3).

Therefore, Tailor A should work 5 days and Tailor B should work 3 days. The least cost is Rs. 5400.

2.8 Linear Programming: Donation Problem [10 marks]

Question. Kiaan wants to buy colour boxes and books to donate. He wants at least 4 books and 4 colour boxes. A colour box costs Rs. 200 and a book costs Rs. 400. The expenditure must not exceed Rs. 4000. Find the maximum number of items he can buy.

Solution. Let x = number of colour boxes, y = number of books.

Objective function: Z = x+y (to be maximized).

Constraints:

x \ge 4, \quad y \ge 4, \quad 200x+400y \le 4000.

Dividing by 200: x+2y \le 20.

Corner points: (4,4),\ (4,8),\ (12,4).

Corner point Z = x+y
(4,4) 8
(4,8) 12
(12,4) 16

Maximum value is 16, at (12,4).

Therefore, Kiaan should buy 12 colour boxes and 4 books.

Total cost: 200(12)+400(4) = 2400+1600 = 4000.

Maximum number of items: 16.

3. Set 2: Other Frequently Asked Questions

3.1 Sum of the Series [5 marks]

Question. Find the sum to n terms of 9+99+999+\cdots.

Solution. Write each term as a power of 10 minus 1:

9 = 10-1, \quad 99 = 10^2-1, \quad 999 = 10^3-1.

Therefore,

S_n = (10-1)+(10^2-1)+\cdots+(10^n-1) = (10+10^2+\cdots+10^n) - n.

Using the G.P. sum formula,

10+10^2+\cdots+10^n = \frac{10(10^n-1)}{10-1} = \frac{10(10^n-1)}{9}.

Thus,

S_n = \frac{10(10^n-1)}{9} - n = \frac{10^{n+1}-10-9n}{9}.

Hence, S_n = \dfrac{10^{n+1}-10-9n}{9}.

3.2 Quadratic Equation with Roots α², β² [5 marks]

Question. If \alpha and \beta are roots of 2x^2-8x-5=0, find the quadratic equation whose roots are \alpha^2 and \beta^2.

Solution. For 2x^2-8x-5=0,

\alpha+\beta = \frac{-(-8)}{2} = 4, \qquad \alpha\beta = \frac{-5}{2}.

Now,

\alpha^2+\beta^2 = (\alpha+\beta)^2-2\alpha\beta = 4^2-2\left(-\frac{5}{2}\right) = 16+5 = 21.

Also,

\alpha^2\beta^2 = (\alpha\beta)^2 = \left(-\frac{5}{2}\right)^2 = \frac{25}{4}.

The required equation is

x^2-(\alpha^2+\beta^2)x+\alpha^2\beta^2 = 0 \;\Rightarrow\; x^2-21x+\frac{25}{4}=0.

Multiplying by 4:

4x^2-84x+25 = 0.

3.3 Complex Number Condition [5 marks]

Question. If |z-1| = |z-i|, show that \text{Re}(z) = \text{Im}(z).

Solution. Let z = x+iy. Then,

|z-1| = |(x-1)+iy| = \sqrt{(x-1)^2+y^2}, \qquad |z-i| = |x+i(y-1)| = \sqrt{x^2+(y-1)^2}.

Given,

\sqrt{(x-1)^2+y^2} = \sqrt{x^2+(y-1)^2}.

Squaring both sides,

(x-1)^2+y^2 = x^2+(y-1)^2.

Expanding,

x^2-2x+1+y^2 = x^2+y^2-2y+1.

Thus, -2x = -2y \;\Rightarrow\; x=y.

Therefore, \text{Re}(z) = \text{Im}(z).

3.4 Limit [5 marks]

Question. Find \displaystyle\lim_{x\to 0}\frac{\sqrt{x+5}-\sqrt{5}}{x}.

Solution. Rationalize the numerator:

\lim_{x\to 0}\frac{\sqrt{x+5}-\sqrt{5}}{x} = \lim_{x\to 0}\frac{\sqrt{x+5}-\sqrt{5}}{x}\cdot\frac{\sqrt{x+5}+\sqrt{5}}{\sqrt{x+5}+\sqrt{5}} = \lim_{x\to 0}\frac{x+5-5}{x(\sqrt{x+5}+\sqrt{5})} = \lim_{x\to 0}\frac{1}{\sqrt{x+5}+\sqrt{5}} = \frac{1}{2\sqrt{5}}.

Hence, \dfrac{1}{2\sqrt{5}}.

3.5 Differentiation [5 marks]

Question. If y = [x(x-1)(x+2)]^{5/7}, find \dfrac{dy}{dx}.

Solution. Let

u = x(x-1)(x+2) = x(x^2+x-2) = x^3+x^2-2x.

Then y = u^{5/7}. Using the chain rule,

\frac{dy}{dx} = \frac{5}{7}u^{-2/7}\frac{du}{dx}.

Now, \dfrac{du}{dx} = 3x^2+2x-2.

Therefore,

\frac{dy}{dx} = \frac{5}{7}[x(x-1)(x+2)]^{-2/7}(3x^2+2x-2).

Thus,

\frac{dy}{dx} = \frac{5(3x^2+2x-2)}{7[x(x-1)(x+2)]^{2/7}}.

3.6 Local Maxima and Minima [5 marks]

Question. Find the local extrema of

f(x) = \frac{3}{4}x^4-8x^3+\frac{45}{2}x^2+105.

Solution. Differentiate:

f'(x) = 3x^3-24x^2+45x = 3x(x^2-8x+15) = 3x(x-3)(x-5).

For extrema, f'(x)=0 \Rightarrow x = 0, 3, 5.

Second derivative: f''(x) = 9x^2-48x+45.

  • At x=0: f''(0)=45>0 → local minimum, f(0)=105.
  • At x=3: f''(3)=81-144+45=-18<0 → local maximum, f(3)=\dfrac{609}{4}.
  • At x=5: f''(5)=225-240+45=30>0 → local minimum, f(5)=\dfrac{545}{4}.

Therefore,

\text{Local minima: } (0,105),\ \left(5,\frac{545}{4}\right) \qquad \text{Local maximum: } \left(3,\frac{609}{4}\right).

3.7 Integration [5 marks]

Question. Evaluate \displaystyle\int x\sqrt{x+2}\,dx.

Solution. Let u = x+2 \Rightarrow x = u-2,\ dx = du. Then,

\int x\sqrt{x+2}\,dx = \int (u-2)\sqrt{u}\,du = \int (u^{3/2}-2u^{1/2})\,du = \frac{2}{5}u^{5/2}-\frac{4}{3}u^{3/2}+C.

Putting u=x+2:

\int x\sqrt{x+2}\,dx = \frac{2}{5}(x+2)^{5/2}-\frac{4}{3}(x+2)^{3/2}+C.

3.8 Definite Integral [5 marks]

Question. Evaluate \displaystyle\int_0^1 \frac{x}{\sqrt{1-x^2}}\,dx.

Solution. Let u = 1-x^2 \Rightarrow du = -2x\,dx, so x\,dx = -\tfrac{1}{2}du. Therefore,

\int \frac{x}{\sqrt{1-x^2}}\,dx = -\frac{1}{2}\int u^{-1/2}\,du = -\sqrt{u} = -\sqrt{1-x^2}.

Applying limits,

\int_0^1 \frac{x}{\sqrt{1-x^2}}\,dx = \left[-\sqrt{1-x^2}\right]_0^1 = 0-(-1) = 1.

Hence, 1.

3.9 Area Between Curves [5 marks]

Question. Find the area bounded by y=\sqrt{x} and y=x.

Solution. Intersection points from \sqrt{x}=x. Squaring: x=x^2 \Rightarrow x(x-1)=0 \Rightarrow x=0,1.

On [0,1], \sqrt{x} > x. Therefore,

A = \int_0^1 (\sqrt{x}-x)\,dx = \int_0^1 (x^{1/2}-x)\,dx = \left[\frac{2}{3}x^{3/2}-\frac{x^2}{2}\right]_0^1 = \frac{2}{3}-\frac{1}{2} = \frac{1}{6}.

Hence, A = \dfrac{1}{6} square units.

3.10 Direction Cosines [5 marks]

Question. Find the direction cosines of the line joining (0,1,-1) and (3,2,1).

Solution. Direction ratios are (3-0,\ 2-1,\ 1-(-1)) = (3,1,2).

Magnitude is \sqrt{3^2+1^2+2^2} = \sqrt{14}.

Therefore, direction cosines are \left(\dfrac{3}{\sqrt{14}}, \dfrac{1}{\sqrt{14}}, \dfrac{2}{\sqrt{14}}\right).

3.11 Shortest Distance Between Two Lines [5 marks]

Question. Find the shortest distance between the lines

\frac{x-1}{2}=\frac{y+1}{3}=z \qquad\text{and}\qquad \frac{x+1}{5}=\frac{y-2}{1}=\frac{z-2}{0}.

Solution. For the first line: A=(1,-1,0), \vec{b_1}=(2,3,1).

For the second line: B=(-1,2,2), \vec{b_2}=(5,1,0).

Formula for shortest distance:

d = \frac{|(\vec{B}-\vec{A})\cdot(\vec{b_1}\times\vec{b_2})|}{|\vec{b_1}\times\vec{b_2}|}.

Now, \vec{B}-\vec{A} = (-2,3,2).

Also,

\vec{b_1}\times\vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 1 \\ 5 & 1 & 0 \end{vmatrix} = (-1, 5, -13).

Thus,

(\vec{B}-\vec{A})\cdot(\vec{b_1}\times\vec{b_2}) = (-2)(-1)+3(5)+2(-13) = -9.

So, |(\vec{B}-\vec{A})\cdot(\vec{b_1}\times\vec{b_2})| = 9.

Also, |\vec{b_1}\times\vec{b_2}| = \sqrt{(-1)^2+5^2+(-13)^2} = \sqrt{195}.

Therefore, d = \dfrac{9}{\sqrt{195}}.

3.12 Scalar Triple Product [5 marks]

Question. If \vec{a}=-\hat{i}+\hat{j}+2\hat{k}, \vec{b}=2\hat{i}-\hat{j}+\hat{k}, \vec{c}=3\hat{i}-\hat{j}+2\hat{k}, find (\vec{a}\times\vec{b})\cdot\vec{c}.

Solution. \vec{a}=(-1,1,2), \vec{b}=(2,-1,1), \vec{c}=(3,-1,2).

Now,

\vec{a}\times\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 2 \\ 2 & -1 & 1 \end{vmatrix} = 3\hat{i}+5\hat{j}-\hat{k} = (3,5,-1).

Therefore,

(\vec{a}\times\vec{b})\cdot\vec{c} = (3,5,-1)\cdot(3,-1,2) = 9-5-2 = 2.

Hence, 2.

3.13 Harmonic Progression [5 marks]

Question. Find the 10th term of the H.P. \dfrac{1}{5},\dfrac{1}{11},\dfrac{1}{17},\dfrac{1}{23},\ldots

Solution. For H.P., take reciprocals. The corresponding A.P. is 5,11,17,23,\ldots

Here, a=5, d=11-5=6.

The 10th term of the A.P. is a_{10} = a+9d = 5+9(6) = 59.

Therefore, the 10th term of the H.P. is \dfrac{1}{59}.

3.14 Cubic Equation with Integer Roots [5 marks]

Question. Find the roots of x^3-13x^2+15x+189=0, given that the roots are integers and one root exceeds another by 2.

Solution. Try integer factors of 189. Put x=7:

7^3-13(7)^2+15(7)+189 = 343-637+105+189 = 0.

Thus, x=7 is a root, so (x-7) is a factor.

Dividing x^3-13x^2+15x+189 by (x-7), we get x^2-6x-27.

Now, x^2-6x-27 = (x+3)(x-9).

Therefore, x^3-13x^2+15x+189 = (x-7)(x+3)(x-9).

Hence, the roots are 7,\ -3,\ 9.

3.15 Rational Inequality [5 marks]

Question. Solve \dfrac{9}{x-3} < 5.

Solution. Bring all terms to one side:

\frac{9}{x-3}-5 < 0.

Taking L.C.M.,

\frac{9-5(x-3)}{x-3} < 0 \;\Rightarrow\; \frac{24-5x}{x-3} < 0.

Critical points: 24-5x=0 \Rightarrow x=\dfrac{24}{5}, and x-3=0 \Rightarrow x=3.

Check the sign on intervals (-\infty,3), \left(3,\dfrac{24}{5}\right), \left(\dfrac{24}{5},\infty\right).

The expression is negative on (-\infty,3) \cup \left(\dfrac{24}{5},\infty\right).

Therefore, x \in (-\infty,3) \cup \left(\dfrac{24}{5},\infty\right).

3.16 Find k from a Differential Equation [5 marks]

Question. If y = 2e^x+e^{-x} and \dfrac{d^2y}{dx^2} = ky, find k.

Solution. Given y = 2e^x+e^{-x}.

Differentiate: \dfrac{dy}{dx} = 2e^x-e^{-x}.

Again differentiate: \dfrac{d^2y}{dx^2} = 2e^x+e^{-x}.

But y = 2e^x+e^{-x}. Therefore, \dfrac{d^2y}{dx^2} = y.

Given \dfrac{d^2y}{dx^2}=ky, so ky=y \Rightarrow k=1.

Hence, k=1.

3.17 Area of Triangle by Determinants [5 marks]

Question. Using determinants, find the area of the triangle whose vertices are (1,2),\ (-2,3),\ (-3,-4).

Solution. Area of triangle:

A = \frac{1}{2}\left|\begin{matrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{matrix}\right|.

So,

A = \frac{1}{2}\left|\begin{matrix} 1 & 2 & 1 \\ -2 & 3 & 1 \\ -3 & -4 & 1 \end{matrix}\right|.

Using the determinant formula,

A = \frac{1}{2}|1(3-(-4))+(-2)(-4-2)+(-3)(2-3)| = \frac{1}{2}|7+12+3| = \frac{1}{2}(22) = 11.

Therefore, A = 11 square units.

3.18 Logarithmic Differentiation [5 marks]

Question. If y = \log_e\left[e^x\left(\dfrac{x-2}{x+2}\right)^{3/4}\right], find \dfrac{dy}{dx}.

Solution. Using log rules,

y = \log_e(e^x)+\log_e\left(\frac{x-2}{x+2}\right)^{3/4} = x+\frac{3}{4}\log_e\left(\frac{x-2}{x+2}\right) = x+\frac{3}{4}\{\log_e(x-2)-\log_e(x+2)\}.

Differentiate:

\frac{dy}{dx} = 1+\frac{3}{4}\left(\frac{1}{x-2}-\frac{1}{x+2}\right) = 1+\frac{3}{4}\left(\frac{(x+2)-(x-2)}{(x-2)(x+2)}\right) = 1+\frac{3}{4}\left(\frac{4}{x^2-4}\right) = 1+\frac{3}{x^2-4} = \frac{x^2-1}{x^2-4}.

Thus,

\frac{dy}{dx} = \frac{x^2-1}{x^2-4}.

4. Set 3: Continued Solutions for Other Questions

4.1 Matrix Polynomial [5 marks]

Question. If A = \begin{pmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{pmatrix}, find f(A), where f(x) = x^2-5x+6.

Solution. For a matrix, f(A) = A^2-5A+6I.

First calculate

A^2 = \begin{pmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{pmatrix}\begin{pmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{pmatrix} = \begin{pmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{pmatrix}.

Now,

-5A = \begin{pmatrix} -10 & 0 & -5 \\ -10 & -5 & -15 \\ -5 & 5 & 0 \end{pmatrix}, \qquad 6I = \begin{pmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{pmatrix}.

Therefore,

f(A) = A^2-5A+6I = \begin{pmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{pmatrix}+\begin{pmatrix} -10 & 0 & -5 \\ -10 & -5 & -15 \\ -5 & 5 & 0 \end{pmatrix}+\begin{pmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{pmatrix} = \begin{pmatrix} 1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4 \end{pmatrix}.

Hence,

f(A) = \begin{pmatrix} 1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4 \end{pmatrix}.

4.2 Infinite G.P. [5 marks]

Question. Find the sum of an infinite G.P. whose first term is 15 and fourth term is \dfrac{3}{25}.

Solution. Let the first term be a and common ratio be r. Given a = 15.

The fourth term is ar^3 = \dfrac{3}{25}.

So, 15r^3 = \dfrac{3}{25} \Rightarrow r^3 = \dfrac{3}{25\cdot 15} = \dfrac{1}{125}.

Therefore, r = \dfrac{1}{5}.

The sum of an infinite G.P. is S_\infty = \dfrac{a}{1-r}, |r|<1.

Thus,

S_\infty = \frac{15}{1-\tfrac{1}{5}} = \frac{15}{\tfrac{4}{5}} = \frac{75}{4}.

Hence, S_\infty = \dfrac{75}{4}.

4.3 Quadratic Equation with Roots α²+3, β²+3 [5 marks]

Question. If \alpha and \beta are roots of x^2-4x+5=0, find the quadratic equation whose roots are \alpha^2+3 and \beta^2+3.

Solution. For x^2-4x+5=0: \alpha+\beta=4, \alpha\beta=5.

Now,

\alpha^2+\beta^2 = (\alpha+\beta)^2-2\alpha\beta = 16-10 = 6.

Sum of new roots:

(\alpha^2+3)+(\beta^2+3) = \alpha^2+\beta^2+6 = 6+6 = 12.

Product of new roots:

(\alpha^2+3)(\beta^2+3) = \alpha^2\beta^2+3(\alpha^2+\beta^2)+9 = (\alpha\beta)^2+3(6)+9 = 25+18+9 = 52.

Required equation: x^2-(\text{sum})x+\text{product}=0.

Thus, x^2-12x+52=0.

4.4 Differential Equation Verification [5 marks]

Question. If y = ax+\dfrac{b}{x}, show that x^2\dfrac{d^2y}{dx^2}+x\dfrac{dy}{dx}-y=0.

Solution. Given y = ax+bx^{-1}.

Differentiate: \dfrac{dy}{dx} = a-bx^{-2} = a-\dfrac{b}{x^2}.

Again differentiate: \dfrac{d^2y}{dx^2} = 2bx^{-3} = \dfrac{2b}{x^3}.

Now,

x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}-y = x^2\left(\frac{2b}{x^3}\right)+x\left(a-\frac{b}{x^2}\right)-\left(ax+\frac{b}{x}\right) = \frac{2b}{x}+ax-\frac{b}{x}-ax-\frac{b}{x} = 0.

Hence proved.

4.5 Length of a Curve [5 marks]

Question. Find the length of the curve y = 4+3x from (0,4) to (2,10).

Solution. Formula for length of curve:

L = \int_a^b \sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx.

Here, y=4+3x \Rightarrow \dfrac{dy}{dx}=3.

The limits are x=0 to x=2. Therefore,

L = \int_0^2 \sqrt{1+3^2}\,dx = \int_0^2 \sqrt{10}\,dx = \sqrt{10}[x]_0^2 = 2\sqrt{10}.

Hence, L = 2\sqrt{10}.

4.6 Integral: \int \dfrac{x^3}{(x+1)^2}\,dx [5 marks]

Question. Evaluate I = \displaystyle\int \dfrac{x^3}{(x+1)^2}\,dx.

Solution. Let u=x+1 \Rightarrow x=u-1,\ dx=du. Then,

I = \int \frac{(u-1)^3}{u^2}\,du.

Expand: (u-1)^3 = u^3-3u^2+3u-1.

So,

\frac{(u-1)^3}{u^2} = u-3+\frac{3}{u}-\frac{1}{u^2}.

Thus,

I = \int\left(u-3+\frac{3}{u}-\frac{1}{u^2}\right)du = \frac{u^2}{2}-3u+3\ln|u|+\frac{1}{u}+C.

Putting u=x+1:

I = \frac{(x+1)^2}{2}-3(x+1)+3\ln|x+1|+\frac{1}{x+1}+C.

4.7 Integral: \int \dfrac{x}{\sqrt{x+3}}\,dx [5 marks]

Question. Evaluate \displaystyle\int \dfrac{x}{\sqrt{x+3}}\,dx.

Solution. Let u=x+3 \Rightarrow x=u-3,\ dx=du. Then,

\int \frac{x}{\sqrt{x+3}}\,dx = \int \frac{u-3}{\sqrt{u}}\,du = \int (u^{1/2}-3u^{-1/2})\,du = \frac{2}{3}u^{3/2}-6u^{1/2}+C.

Putting u=x+3:

\int \frac{x}{\sqrt{x+3}}\,dx = \frac{2}{3}(x+3)^{3/2}-6(x+3)^{1/2}+C.

4.8 Definite Log Integral [5 marks]

Question. Evaluate \displaystyle\int_a^b \frac{\log x}{x}\,dx, a>b>0.

Solution. Let u=\log x \Rightarrow du = \dfrac{1}{x}dx. Therefore,

\int \frac{\log x}{x}\,dx = \int u\,du = \frac{u^2}{2} = \frac{(\log x)^2}{2}.

Applying limits,

\int_a^b \frac{\log x}{x}\,dx = \left[\frac{(\log x)^2}{2}\right]_a^b = \frac{1}{2}\{(\log b)^2-(\log a)^2\}.

Hence,

\int_a^b \frac{\log x}{x}\,dx = \frac{1}{2}\left[(\log b)^2-(\log a)^2\right].

4.9 Purely Imaginary Complex Expression [5 marks]

Question. If z\in\mathbb{C}, |z|=1, and z\neq -1, show that \dfrac{z-1}{z+1} is purely imaginary.

Solution. Let z=x+iy. Since |z|=1, x^2+y^2=1.

Now,

\frac{z-1}{z+1} = \frac{x-1+iy}{x+1+iy}.

Multiply numerator and denominator by the conjugate of the denominator:

\frac{x-1+iy}{x+1+iy}\cdot\frac{x+1-iy}{x+1-iy}.

The denominator is (x+1)^2+y^2.

The numerator is (x-1+iy)(x+1-iy). Expanding:

(x^2-1+y^2)+2iy.

Since x^2+y^2=1, the real part becomes x^2+y^2-1 = 0.

Therefore,

\frac{z-1}{z+1} = \frac{2iy}{(x+1)^2+y^2},

which is purely imaginary. Hence proved.

4.10 Complex Number Condition: Real Axis [5 marks]

Question. If |z-i|=|z+i|, show that \text{Im}(z)=0.

Solution. Let z=x+iy. Then,

|z-i| = |x+i(y-1)| = \sqrt{x^2+(y-1)^2}, \qquad |z+i| = |x+i(y+1)| = \sqrt{x^2+(y+1)^2}.

Given,

\sqrt{x^2+(y-1)^2} = \sqrt{x^2+(y+1)^2}.

Squaring both sides,

x^2+(y-1)^2 = x^2+(y+1)^2.

Cancel x^2:

y^2-2y+1 = y^2+2y+1.

Thus, -2y = 2y \Rightarrow 4y=0 \Rightarrow y=0.

Therefore, \text{Im}(z) = 0.

4.11 Induction: Divisibility [5 marks]

Question. Use mathematical induction to show that 2^{3n}-1 is divisible by 8 for every natural number n.

Solution. Let P(n): 2^{3n}-1 is divisible by 8.

For n=1: 2^{3(1)}-1 = 2^3-1 = 8-1 = 7.

This is not divisible by 8, so the usual correct form of this standard question is that 2^{3n}-1 is divisible by 7, not 8.

If the intended question is 2^{3n}-1 divisible by 7, then:

2^3 \equiv 1 \pmod{7},

so

2^{3n} = (2^3)^n \equiv 1^n \equiv 1 \pmod{7}.

Hence, 2^{3n}-1 \equiv 0 \pmod{7}.

Therefore, 2^{3n}-1 is divisible by 7.

Exam note. If your paper says "divisible by 8," recheck the printed expression. The statement "2^{3n}-1 divisible by 8" is false for n=1.

4.12 Induction: Sum of Odd Numbers [5 marks]

Question. Prove by mathematical induction:

1+3+5+\cdots+(2n-1) = n^2.

Solution. Let P(n): 1+3+5+\cdots+(2n-1) = n^2.

For n=1: \text{LHS}=1,\ \text{RHS}=1^2=1. So P(1) is true.

Assume P(k) is true: 1+3+5+\cdots+(2k-1) = k^2.

Now add the next odd number 2(k+1)-1 = 2k+1:

1+3+5+\cdots+(2k-1)+(2k+1) = k^2+(2k+1) = k^2+2k+1 = (k+1)^2.

Thus, P(k+1) is true. Hence,

1+3+5+\cdots+(2n-1) = n^2.

4.13 Absolute Value Inequality [5 marks]

Question. Solve \left|\dfrac{2x-1}{3}\right| \le 2.

Solution.

\left|\frac{2x-1}{3}\right| \le 2 \quad\text{means}\quad -2 \le \frac{2x-1}{3} \le 2.

Multiplying by 3: -6 \le 2x-1 \le 6.

Adding 1: -5 \le 2x \le 7.

Dividing by 2: -\dfrac{5}{2} \le x \le \dfrac{7}{2}.

Hence, x \in \left[-\dfrac{5}{2}, \dfrac{7}{2}\right].

4.14 Intersection of Two Lines in 3D [5 marks]

Question. Show that the lines

\frac{x-5}{4}=\frac{y-7}{4}=\frac{z+3}{-5} \qquad\text{and}\qquad \frac{x-8}{7}=\frac{y-4}{1}=\frac{z-5}{3}

intersect.

Solution. Let the first line be equal to \lambda:

x=5+4\lambda, \quad y=7+4\lambda, \quad z=-3-5\lambda.

Let the second line be equal to \mu:

x=8+7\mu, \quad y=4+\mu, \quad z=5+3\mu.

For intersection,

5+4\lambda = 8+7\mu \tag{1}
7+4\lambda = 4+\mu \tag{2}

From (2): \mu = 4\lambda+3.

Substitute in (1):

5+4\lambda = 8+7(4\lambda+3) \Rightarrow 5+4\lambda = 29+28\lambda \Rightarrow -24 = 24\lambda \Rightarrow \lambda = -1.

Then, \mu = 4(-1)+3 = -1.

Now check z: -3-5(-1) = 2, and 5+3(-1) = 2. Both give the same point.

Hence, the lines intersect at (1,3,2).

4.15 Triangle Inequality for Vectors [5 marks]

Question. If \vec{a} and \vec{b} are two vectors, show that |\vec{a}+\vec{b}| \le |\vec{a}|+|\vec{b}|.

Solution. Start with the square of the left side:

|\vec{a}+\vec{b}|^2 = (\vec{a}+\vec{b})\cdot(\vec{a}+\vec{b}).

Expanding,

|\vec{a}+\vec{b}|^2 = |\vec{a}|^2+2\vec{a}\cdot\vec{b}+|\vec{b}|^2.

Using \vec{a}\cdot\vec{b} \le |\vec{a}||\vec{b}|, we get

|\vec{a}+\vec{b}|^2 \le |\vec{a}|^2+2|\vec{a}||\vec{b}|+|\vec{b}|^2.

So,

|\vec{a}+\vec{b}|^2 \le (|\vec{a}|+|\vec{b}|)^2.

Taking the square root on both sides,

|\vec{a}+\vec{b}| \le |\vec{a}|+|\vec{b}|.

4.16 Unit Vector Perpendicular to Two Vectors [5 marks]

Question. If \vec{a}=4\hat{i}+\hat{j}+3\hat{k}, \vec{b}=-2\hat{i}+\hat{j}-2\hat{k}, find a unit vector perpendicular to both \vec{a} and \vec{b}.

Solution. A vector perpendicular to both \vec{a} and \vec{b} is \vec{a}\times\vec{b}.

\vec{a}=(4,1,3), \vec{b}=(-2,1,-2).

Now,

\vec{a}\times\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 1 & 3 \\ -2 & 1 & -2 \end{vmatrix}.

So,

\vec{a}\times\vec{b} = \hat{i}(1(-2)-3(1)) - \hat{j}(4(-2)-3(-2)) + \hat{k}(4(1)-1(-2)).

Thus,

\vec{a}\times\vec{b} = -5\hat{i}+2\hat{j}+6\hat{k}.

Magnitude:

|\vec{a}\times\vec{b}| = \sqrt{(-5)^2+2^2+6^2} = \sqrt{25+4+36} = \sqrt{65}.

Therefore, a unit vector perpendicular to both is

\frac{-5\hat{i}+2\hat{j}+6\hat{k}}{\sqrt{65}}.

The negative of this vector is also a valid answer.

4.17 Matrix Inverse from Product [5 marks]

Question. If matrices A and B satisfy AB = BA = 6I_3, find A^{-1}.

Solution. We know that if AB=I, then B=A^{-1}. Here,

AB = BA = 6I_3.

Divide both sides by 6:

A\left(\frac{1}{6}B\right) = I_3, \qquad \left(\frac{1}{6}B\right)A = I_3.

Therefore,

A^{-1} = \frac{1}{6}B.

If

B = \begin{pmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{pmatrix},

then

A^{-1} = \frac{1}{6}\begin{pmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{pmatrix}.

4.18 Linear Programming [5 marks]

Question. Maximize P = 5x+2y subject to

10x+2y \ge 2100, \qquad x+\frac{1}{2}y \le 600, \qquad y \le 800, \qquad x\ge 0,\ y\ge 0.

Solution. Simplify the first two inequalities:

10x+2y \ge 2100 \;\Rightarrow\; 5x+y \ge 1050,
x+\frac{1}{2}y \le 600 \;\Rightarrow\; 2x+y \le 1200.

So the constraints are

5x+y \ge 1050, \quad 2x+y \le 1200, \quad y \le 800, \quad x,y \ge 0.

Feasible corner points: (50,800),\ (200,800),\ (600,0),\ (210,0).

Corner point P = 5x+2y
(50,800) 5(50)+2(800) = 1850
(200,800) 5(200)+2(800) = 2600
(600,0) 5(600)+2(0) = 3000
(210,0) 5(210)+2(0) = 1050

Maximum value is 3000, at (600,0).

Therefore, P_{\max} = 3000 at (x,y) = (600,0).

5. Quick Revision Formula Sheet

Complex Numbers

\omega^3=1, \qquad 1+\omega+\omega^2=0, \qquad \omega+\omega^2=-1.

If z = r(\cos\theta+i\sin\theta), then z^n = r^n(\cos n\theta+i\sin n\theta).

Matrices

A^{-1} = \frac{\text{adj}\,A}{|A|}, \quad |A|\neq 0.

For a matrix polynomial, f(A) = A^2-5A+6I if f(x)=x^2-5x+6.

Cramer's Rule

x = \frac{D_x}{D}, \qquad y = \frac{D_y}{D}, \qquad z = \frac{D_z}{D}.

Mathematical Induction

Base case → Assumption for n=k → Proof for n=k+1.

Calculus

\int x^n\,dx = \frac{x^{n+1}}{n+1}+C, \quad n\neq -1.
\frac{d}{dx}e^x = e^x, \qquad \frac{d}{dx}e^{-x} = -e^{-x}, \qquad \frac{d}{dx}\log x = \frac{1}{x}.

Vectors

|\vec{a}+\vec{b}| \le |\vec{a}|+|\vec{b}|.

Unit vector perpendicular to \vec{a},\vec{b}:

\hat{n} = \frac{\vec{a}\times\vec{b}}{|\vec{a}\times\vec{b}|}.

Linear Programming

Maximum or minimum occurs at a corner point of the feasible region.

End of Notes — BCS-012 Basic Mathematics (BCA, IGNOU)