Basic Mathematics 2023 Board Question Paper

Step-by-Step Solutions: Brief Answer Questions

a. Set Intersection $(A \cap B \cap C)$:

$A = \{1, 2, 3, 4, 5\}, \quad B = \{3, 4, 5, 6, 7\}, \quad C = \{1, 3, 5, 7\}$ Elements common to all three sets:

$$\mathbf{A \cap B \cap C = \{3, 5\}}$$

b. Polar Form of $-1 + i\sqrt{3}$:

$$r = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2$$

Since $x < 0$ and $y > 0$, the complex number lies in the second quadrant:

$$\theta = \pi - \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3} = 120^\circ$$

$$\mathbf{-1 + i\sqrt{3} = 2\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right)}$$

c. Rewrite $-4 \le x \le -1$ Using Modulus Sign:

$$\text{Center } c = \frac{-4 + (-1)}{2} = -\frac{5}{2} = -2.5, \quad \text{Radius } r = \frac{-1 - (-4)}{2} = \frac{3}{2} = 1.5$$

$$\mathbf{\left|x + \frac{5}{2}\right| \le \frac{3}{2} \quad \Longleftrightarrow \quad |2x + 5| \le 3}$$

d. Evaluate $\lim_{x \to \infty} \frac{5x^3 + 3x + 7}{2x^3 + 7x + 9}$:

Dividing numerator and denominator by $x^3$:

$$\lim_{x \to \infty} \frac{5 + \frac{3}{x^2} + \frac{7}{x^3}}{2 + \frac{7}{x^2} + \frac{9}{x^3}} = \frac{5 + 0 + 0}{2 + 0 + 0} = \mathbf{\frac{5}{2}}$$

e. Unit Vector Perpendicular to $\vec{a} = \hat{i} + 3\hat{j} + 2\hat{k}$ and $\vec{b} = 2\hat{i} - 4\hat{j} + \hat{k}$:

$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 3 & 2 \\ 2 & -4 & 1 \end{vmatrix} = \hat{i}(3 - (-8)) - \hat{j}(1 - 4) + \hat{k}(-4 - 6) = 11\hat{i} + 3\hat{j} - 10\hat{k}$$

$$|\vec{a} \times \vec{b}| = \sqrt{11^2 + 3^2 + (-10)^2} = \sqrt{121 + 9 + 100} = \sqrt{230}$$

$$\mathbf{\hat{n} = \pm \frac{11\hat{i} + 3\hat{j} - 10\hat{k}}{\sqrt{230}}}$$

f. Derivative of $y = e^{2x}$:

$$\frac{dy}{dx} = \frac{d}{dx}(e^{2x}) = e^{2x} \cdot \frac{d}{dx}(2x) = \mathbf{2e^{2x}}$$

g. Calculate $5(A + B)$:

$$A = \begin{bmatrix} 2 & -4 \\ 4 & 1 \end{bmatrix}, \quad B = \begin{bmatrix} 3 & 6 \\ 5 & 2 \end{bmatrix} \implies A + B = \begin{bmatrix} 5 & 2 \\ 9 & 3 \end{bmatrix}$$

$$\mathbf{5(A + B) = \begin{bmatrix} 25 & 10 \\ 45 & 15 \end{bmatrix}}$$

h. Area Under $y = 2x + 3$ from $x = 2$ to $x = 4$:

$$\text{Area} = \int_2^4 (2x + 3) dx = \left[ x^2 + 3x \right]_2^4 = (16 + 12) - (4 + 6) = 28 - 10 = \mathbf{18 \text{ sq. units}}$$

i. Solve $\frac{dy}{dx} = 3x^2$:

$$dy = 3x^2 dx \implies \int dy = \int 3x^2 dx \implies \mathbf{y = x^3 + C}$$

j. Determinant Value:

$$\begin{vmatrix} 1 & 3 & 2 \\ 3 & 4 & 1 \\ 2 & 5 & 1 \end{vmatrix} = 1(4 - 5) - 3(3 - 2) + 2(15 - 8) = 1(-1) - 3(1) + 2(7) = -1 - 3 + 14 = \mathbf{10}$$

(a) Proof: $\sqrt{a - ib} = x - iy \implies \sqrt{a + ib} = x + iy$

Given $\sqrt{a - ib} = x - iy$. Squaring both sides:

$$a - ib = (x - iy)^2 = x^2 - y^2 - 2ixy$$

Equating real and imaginary components:

$$a = x^2 - y^2 \quad \text{and} \quad b = 2xy$$

Now consider $x + iy$:

$$(x + iy)^2 = x^2 - y^2 + 2ixy = a + ib$$

Taking the square root on both sides:

$$\mathbf{\sqrt{a + ib} = x + iy} \quad \blacksquare$$

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(b) Square Roots of $7 - 24i$

Let $\sqrt{7 - 24i} = x - iy$ with $x, y > 0$. Squaring:

$$x^2 - y^2 - 2ixy = 7 - 24i \implies x^2 - y^2 = 7 \quad \text{and} \quad 2xy = 24$$

$$(x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2 = 7^2 + (-24)^2 = 49 + 576 = 625 \implies x^2 + y^2 = 25$$

Adding and subtracting:

$$2x^2 = 32 \implies x^2 = 16 \implies x = \pm 4$$

$$2y^2 = 18 \implies y^2 = 9 \implies y = \pm 3$$

Since the imaginary part is negative ($-24i$), $x$ and $y$ have opposite signs in the sum:

$$\mathbf{\sqrt{7 - 24i} = \pm (4 - 3i)}$$

Limit and Continuity of $f(x)$ at $x = 3$

Given:

$$f(x) = \begin{cases} 2x + 5 & \text{for } x \le 3 \\ x^2 + 2 & \text{for } x > 3 \end{cases}$$

1. Left-Hand Limit (LHL):
   $$\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (2x + 5) = 2(3) + 5 = \mathbf{11}$$
2. Right-Hand Limit (RHL):
   $$\lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (x^2 + 2) = 3^2 + 2 = \mathbf{11}$$
3. Function Value:
   $$f(3) = 2(3) + 5 = \mathbf{11}$$

Since $\text{LHL} = \text{RHL} = f(3) = 11$, the limit $\lim_{x \to 3} f(x)$ exists and equals 11. Therefore, $f(x)$ is continuous at $x = 3$.

Evaluation of Limit by Rationalization

$$\lim_{x \to 2} \frac{x - \sqrt{8 - x^2}}{\sqrt{x^2 + 12} - 4}$$

At $x = 2$, this yields the indeterminate form $\frac{2 - \sqrt{4}}{\sqrt{16} - 4} = \frac{0}{0}$. Multiplying numerator and denominator by the conjugate factors $(x + \sqrt{8 - x^2})$ and $(\sqrt{x^2 + 12} + 4)$:

$$= \lim_{x \to 2} \frac{(x^2 - (8 - x^2))(\sqrt{x^2 + 12} + 4)}{((x^2 + 12) - 16)(x + \sqrt{8 - x^2})} = \lim_{x \to 2} \frac{(2x^2 - 8)(\sqrt{x^2 + 12} + 4)}{(x^2 - 4)(x + \sqrt{8 - x^2})}$$

$$= \lim_{x \to 2} \frac{2(x^2 - 4)(\sqrt{x^2 + 12} + 4)}{(x^2 - 4)(x + \sqrt{8 - x^2})} = \lim_{x \to 2} \frac{2(\sqrt{x^2 + 12} + 4)}{x + \sqrt{8 - x^2}}$$

Substituting $x = 2$:

$$= \frac{2(\sqrt{16} + 4)}{2 + \sqrt{8 - 4}} = \frac{2(4 + 4)}{2 + 2} = \frac{2(8)}{4} = \mathbf{4}$$

(a) Derivative of $y = \frac{1}{\sqrt{2x + 3} - \sqrt{2x - 3}}$

Rationalizing:

$$y = \frac{\sqrt{2x + 3} + \sqrt{2x - 3}}{(2x + 3) - (2x - 3)} = \frac{\sqrt{2x + 3} + \sqrt{2x - 3}}{6}$$

Differentiating with respect to $x$:

$$\frac{dy}{dx} = \frac{1}{6} \left[ \frac{1}{2\sqrt{2x + 3}}(2) + \frac{1}{2\sqrt{2x - 3}}(2) \right] = \mathbf{\frac{1}{6} \left( \frac{1}{\sqrt{2x + 3}} + \frac{1}{\sqrt{2x - 3}} \right)}$$

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(b) Implicit Derivative of $x^3 + y^3 = 27$

Differentiating both sides with respect to $x$:

$$3x^2 + 3y^2 \frac{dy}{dx} = 0 \implies 3y^2 \frac{dy}{dx} = -3x^2 \implies \mathbf{\frac{dy}{dx} = -\frac{x^2}{y^2}}$$

Integration by Parts

(a) $\int x^2 e^x dx$:

Using $\int u v' dx = u v - \int u' v dx$ with $u = x^2, v' = e^x$:

$$\int x^2 e^x dx = x^2 e^x - \int 2x e^x dx$$

Integrating $\int x e^x dx$ again by parts:

$$\int x e^x dx = x e^x - \int e^x dx = x e^x - e^x$$

Substituting back:

$$\int x^2 e^x dx = x^2 e^x - 2(x e^x - e^x) + C = \mathbf{e^x (x^2 - 2x + 2) + C}$$

(b) $\int x^2 \ln x \, dx$:

Let $u = \ln x \implies u' = \frac{1}{x}$, and $v' = x^2 \implies v = \frac{x^3}{3}$:

$$\int x^2 \ln x \, dx = \frac{x^3}{3} \ln x - \int \frac{x^3}{3} \cdot \frac{1}{x} dx = \frac{x^3}{3} \ln x - \frac{1}{3} \int x^2 dx = \mathbf{\frac{x^3}{3} \ln x - \frac{x^3}{9} + C}$$

Vector Coplanarity Analysis

Vectors are coplanar if the determinant formed by their linear expansion coefficients vanishes ($\Delta = 0$). Expanding the scalar triple product determinant:

$$\begin{vmatrix} 1 & -2 & 3 \\ -2 & 3 & -4 \\ 1 & -3 & 5 \end{vmatrix} = 1(15 - 12) - (-2)(-10 + 4) + 3(6 - 3) = 3 - 12 + 9 = \mathbf{0}$$

Since the determinant is 0, the vectors are coplanar.

Linear Differential Equation: $(1 + x^2)\frac{dy}{dx} + 2xy = 4x^2$

Dividing by $(1 + x^2)$:

$$\frac{dy}{dx} + \frac{2x}{1 + x^2} y = \frac{4x^2}{1 + x^2}$$

1. Integrating Factor:
   $$\text{I.F.} = e^{\int \frac{2x}{1 + x^2} dx} = e^{\ln(1 + x^2)} = 1 + x^2$$
2. General Solution:
   $$y \cdot (1 + x^2) = \int \frac{4x^2}{1 + x^2} (1 + x^2) dx + C = \int 4x^2 dx + C = \frac{4}{3}x^3 + C$$
   $$\mathbf{y = \frac{\frac{4}{3}x^3 + C}{1 + x^2}}$$

3-Set Survey Problem ($N = 500$)

Let $K$ = Kathmandu Post, $R$ = Rising Nepal, $H$ = Himalayan Times.

• $n(U) = 500$
• $n(K) = 280, \quad n(R) = 190, \quad n(H) = 110$
• $n(K \cap R) = 75, \quad n(R \cap H) = 50, \quad n(K \cap H) = 45$
• None read $= 55 \implies n(K \cup R \cup H) = 500 - 55 = 445$

a. Read All Three Newspapers:

$$n(K \cup R \cup H) = n(K) + n(R) + n(H) - n(K \cap R) - n(R \cap H) - n(K \cap H) + n(K \cap R \cap H)$$

$$445 = 280 + 190 + 110 - 75 - 50 - 45 + x$$

$$445 = 410 + x \implies \mathbf{x = 35 \text{ students}}$$

b. Read Exactly Two Newspapers Only:

$$(75 - 35) + (50 - 35) + (45 - 35) = 40 + 15 + 10 = \mathbf{65 \text{ students}}$$

c. Read Exactly One Newspaper Only:

• $K \text{ only} = 280 - (40 + 35 + 10) = 195$
• $R \text{ only} = 190 - (40 + 35 + 15) = 100$
• $H \text{ only} = 110 - (10 + 35 + 15) = 50$$\text{Total One Only} = 195 + 100 + 50 = \mathbf{345 \text{ students}}$$

Least Squares Trend Line Fitting for Industrial Profits

Trend Equation: $y = a + bx$. Setting the base origin at the middle year ($x = \text{Year} - \text{Year}_0$). From normal equations:

$$\sum y = N a + b \sum x$$

$$\sum xy = a \sum x + b \sum x^2$$

Solving yields the slope $b$ and intercept $a$, enabling extrapolation for future financial years (2023 and 2024).

Solution by Cramer’s Rule

System of equations:

1. $2x + 5y + 7z = 12$
2. $x + 2y - z = 0$
3. $x + y + z = 9$$\Delta = \begin{vmatrix} 2 & 5 & 7 \\ 1 & 2 & -1 \\ 1 & 1 & 1 \end{vmatrix} = 2(2 + 1) - 5(1 + 1) + 7(1 - 2) = 6 - 10 - 7 = -11$$
   $$\Delta_x = \begin{vmatrix} 12 & 5 & 7 \\ 0 & 2 & -1 \\ 9 & 1 & 1 \end{vmatrix} = 12(3) - 5(9) + 7(-18) = 36 - 45 - 126 = -135 \implies x = \frac{\Delta_x}{\Delta}$$
   $$\Delta_y = \begin{vmatrix} 2 & 12 & 7 \\ 1 & 0 & -1 \\ 1 & 9 & 1 \end{vmatrix} \implies y = \frac{\Delta_y}{\Delta}$$
   $$\Delta_z = \begin{vmatrix} 2 & 5 & 12 \\ 1 & 2 & 0 \\ 1 & 1 & 9 \end{vmatrix} \implies z = \frac{\Delta_z}{\Delta}$$

Market Equilibrium & Economic Surplus Analysis

Given:

$$P_d = 50 - 2Q_d, \quad P_s = 14 + 4Q_s$$

a. Equilibrium Price and Quantity:

Equating $P_d = P_s$:

$$50 - 2Q = 14 + 4Q \implies 6Q = 36 \implies \mathbf{Q_e = 6}$$

$$\mathbf{P_e} = 50 - 2(6) = 50 - 12 = \mathbf{38}$$

b. Consumer’s Surplus (CS) and Producer’s Surplus (PS):

$$\text{CS} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 6 \times (50 - 38) = 3 \times 12 = \mathbf{36}$$

$$\text{PS} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 6 \times (38 - 14) = 3 \times 24 = \mathbf{72}$$

c. Total Surplus:

$$\text{Total Surplus} = \text{CS} + \text{PS} = 36 + 72 = \mathbf{108}$$

Profit Maximization and Revenue Optimization

Given:

$$TC = \frac{1}{3}Q^3 - 15Q^2 + 480Q + 750, \quad P = 536 - 2Q$$

a. Revenue Function and Profit Function:

$$\mathbf{TR} = P \times Q = (536 - 2Q)Q = \mathbf{536Q - 2Q^2}$$

$$\mathbf{\Pi(Q)} = TR - TC = (536Q - 2Q^2) - \left(\frac{1}{3}Q^3 - 15Q^2 + 480Q + 750\right)$$

$$\mathbf{\Pi(Q) = -\frac{1}{3}Q^3 + 13Q^2 + 56Q - 750}$$

b. Profit Maximization Output $Q$:

$$\frac{d\Pi}{dQ} = -Q^2 + 26Q + 56 = 0 \implies Q^2 - 26Q - 56 = 0$$

$$(Q - 28)(Q + 2) = 0 \implies \mathbf{Q = 28 \text{ units}} \quad (Q > 0)$$

$$\frac{d^2\Pi}{dQ^2} = -2Q + 26 \Big|_{Q=28} = -56 + 26 = -30 < 0 \quad (\text{Maximum confirmed})$$

c. Maximum Marginal Profit:

Marginal Profit $M\Pi = -Q^2 + 26Q + 56$. Setting $\frac{d(M\Pi)}{dQ} = -2Q + 26 = 0 \implies Q = 13$.

$$\text{Max } M\Pi = -(13)^2 + 26(13) + 56 = -169 + 338 + 56 = \mathbf{225}$$

d. Maximum Revenue:

$$\frac{d(TR)}{dQ} = 536 - 4Q = 0 \implies 4Q = 536 \implies Q = 134$$

$$\mathbf{\text{Max } TR} = 536(134) - 2(134)^2 = 71,824 - 35,912 = \mathbf{\text{Rs } 35,912}$$