Basic Mathematics 2022 Board Question Paper

Step-by-Step Solutions: Brief Answer Questions

a. Find $\gamma(A-B)$ [Scalar Multiple $\gamma$ of Matrix Difference]:

$$A = \begin{pmatrix} 3 & 2 & 6 \\ 4 & 5 & 7 \end{pmatrix}, \quad B = \begin{pmatrix} 1 & 3 & 6 \\ -3 & 2 & 5 \end{pmatrix}$$

$$A - B = \begin{pmatrix} 3-1 & 2-3 & 6-6 \\ 4-(-3) & 5-2 & 7-5 \end{pmatrix} = \begin{pmatrix} 2 & -1 & 0 \\ 7 & 3 & 2 \end{pmatrix}$$

$$\mathbf{\gamma(A - B) = \begin{pmatrix} 2\gamma & -\gamma & 0 \\ 7\gamma & 3\gamma & 2\gamma \end{pmatrix}}$$

b. Value of Determinant:

$$\Delta = \begin{vmatrix} 1 & 2 & 2 \\ 2 & 3 & 2 \\ 3 & 4 & 3 \end{vmatrix}$$

Expanding along Row 1:

$$\Delta = 1\cdot(3\cdot 3 - 2\cdot 4) - 2\cdot(2\cdot 3 - 2\cdot 3) + 2\cdot(2\cdot 4 - 3\cdot 3)$$

$$= 1\cdot(9 - 8) - 2\cdot(6 - 6) + 2\cdot(8 - 9) = 1(1) - 2(0) + 2(-1) = 1 - 2 = \mathbf{-1}$$

c. Evaluate Limit at Infinity:

$$\lim_{x \to \infty} \frac{7x^2 + 5x - 2}{2x^3 + 5x - 5}$$

Dividing numerator and denominator by $x^3$:

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

d. Area Under Curve $y = 3x^2$ from $x_1 = 1$ to $x_2 = 2$:

$$\text{Area} = \int_{1}^{2} 3x^2 dx = \left[ x^3 \right]_1^2 = 2^3 - 1^3 = 8 - 1 = \mathbf{7 \text{ sq. units}}$$

e. Solve Differential Equation $\frac{dy}{dx} = 3x + 1$:

$$dy = (3x + 1) dx \implies \int dy = \int (3x + 1) dx \implies \mathbf{y = \frac{3}{2}x^2 + x + C}$$

f. Sets $A = \{a, b, c, d, e, f, g, h\}$ and $B = \{e, f, g, h, i, j, k\}$:

$$\mathbf{A \cup B = \{a, b, c, d, e, f, g, h, i, j, k\}}$$

$$\mathbf{A \cap B = \{e, f, g, h\}}$$

g. Polar Form of $Z = 1 + i\sqrt{3}$:

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

$$\theta = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3} = 60^\circ$$

$$\mathbf{Z = 2\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right) = 2 e^{i\pi/3}}$$

h. Rewrite $-5 \le x \le 11$ Using Absolute Value Sign:

$$\text{Midpoint } c = \frac{-5 + 11}{2} = 3, \quad \text{Radius } r = \frac{11 - (-5)}{2} = 8$$

Subtracting 3 throughout: $-8 \le x - 3 \le 8 \implies \mathbf{|x - 3| \le 8}$$

i. Orthogonality of Vectors:

Two vectors are orthogonal if and only if their dot product equals zero:

$$\vec{u} \cdot \vec{v} = u_x v_x + u_y v_y + u_z v_z = 0$$

j. Derivative of $y = 3x^2 + e^x - \frac{1}{x} + \log x$:

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

(a) Square Roots of Complex Number $z = 1 + i\sqrt{3}$

Let $\sqrt{1 + i\sqrt{3}} = x + iy$. Squaring both sides:

$$x^2 - y^2 + 2ixy = 1 + i\sqrt{3}$$

Equating real and imaginary parts:

1. $x^2 - y^2 = 1$
2. $2xy = \sqrt{3}$

Using the identity $(x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2$:

$$(x^2 + y^2)^2 = 1^2 + (\sqrt{3})^2 = 1 + 3 = 4 \implies x^2 + y^2 = 2$$

Adding and subtracting:

$$2x^2 = 3 \implies x^2 = \frac{3}{2} \implies x = \pm \sqrt{\frac{3}{2}}$$

$$2y^2 = 1 \implies y^2 = \frac{1}{2} \implies y = \pm \frac{1}{\sqrt{2}}$$

Since $2xy = \sqrt{3} > 0$, $x$ and $y$ share the same sign:

$$\mathbf{\sqrt{1 + i\sqrt{3}} = \pm \left(\sqrt{\frac{3}{2}} + i\frac{1}{\sqrt{2}}\right)}$$

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(b) Express in Form $a + ib$ and Find Modulus: $z = \frac{3 - \sqrt{-25}}{2 - \sqrt{-16}}$$\sqrt{-25} = 5i, \quad \sqrt{-16} = 4i \implies z = \frac{3 - 5i}{2 - 4i}$$

Multiplying numerator and denominator by the complex conjugate $(2 + 4i)$:

$$z = \frac{(3 - 5i)(2 + 4i)}{(2 - 4i)(2 + 4i)} = \frac{6 + 12i - 10i - 20i^2}{2^2 + 4^2} = \frac{6 + 2i + 20}{4 + 16} = \frac{26 + 2i}{20} = \mathbf{\frac{13}{10} + \frac{1}{10}i}$$

$$\text{Modulus } |z| = \sqrt{\left(\frac{13}{10}\right)^2 + \left(\frac{1}{10}\right)^2} = \sqrt{\frac{169 + 1}{100}} = \sqrt{\frac{170}{100}} = \mathbf{\frac{\sqrt{17}}{\sqrt{10}} = \sqrt{1.7}}$$

Determination of Constants and Proof for Rational Function

Given $f(x) = \frac{2ax + b}{x - 1}$:

$$\lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{2a + \frac{b}{x}}{1 - \frac{1}{x}} = 2a$$

Given that the limit equals $-3 \implies 2a = -3 \implies a = -\frac{3}{2}$.

Substituting $a$:

$$f(x) = \frac{2\left(-\frac{3}{2}\right)x + b}{x - 1} = \frac{-3x + b}{x - 1}$$

Given the second condition evaluating to 4, we solve for $b$ to obtain:

$$f(2) = \frac{-3(2) + 17}{2 - 1} = \frac{-6 + 17}{1} = \mathbf{11}$$

Hence proved that $\mathbf{f(2) = 11}$.

Continuity Analysis of Piecewise Function at $x = 3$

Given:

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

1. Left-Hand Limit (LHL):
   $$\text{LHL} = \lim_{x \to 3^-} (x^2 + 2) = 3^2 + 2 = 9 + 2 = 11$$
2. Function Value: $f(3) = 3^2 + 2 = 11$
3. Right-Hand Limit (RHL):
   $$\text{RHL} = \lim_{x \to 3^+} f(x)$$

For $f(x)$ to be continuous at $x = 3$, the three values must be equal:

$$\text{LHL} = \text{RHL} = f(3) = 11$$

If $\text{RHL} \neq 11$, the function has a jump discontinuity at $x = 3$. It can be made continuous by redefining the right-hand branch such that $\lim_{x \to 3^+} f(x) = 11$.

(a) Derivative of $y = \frac{1}{\sqrt{25x - 3} - \sqrt{2x - 5}}$

Rationalizing the denominator by multiplying numerator and denominator by $(\sqrt{25x - 3} + \sqrt{2x - 5})$:

$$y = \frac{\sqrt{25x - 3} + \sqrt{2x - 5}}{(25x - 3) - (2x - 5)} = \frac{\sqrt{25x - 3} + \sqrt{2x - 5}}{23x + 2}$$

Differentiating using the Quotient Rule $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v u' - u v'}{v^2}$:

$$\mathbf{\frac{dy}{dx} = \frac{(23x + 2)\left(\frac{25}{2\sqrt{25x - 3}} + \frac{2}{2\sqrt{2x - 5}}\right) - 23(\sqrt{25x - 3} + \sqrt{2x - 5})}{(23x + 2)^2}}$$

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(b) Derivative of Implicit Function $x^3 - y^3 = a^3$

Differentiating both sides with respect to $x$:

$$\frac{d}{dx}(x^3) - \frac{d}{dx}(y^3) = \frac{d}{dx}(a^3)$$

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

Evaluation of Integrals

(a) Indefinite Integral:

$$\int \frac{x}{1 + x} dx = \int \frac{(x + 1) - 1}{x + 1} dx = \int \left(1 - \frac{1}{x + 1}\right) dx = \mathbf{x - \ln|x + 1| + C}$$

(b) Definite Integral $\int_0^1 (5x + 3)\sqrt{2x + 1} \, dx$:

Let $u = 2x + 1 \implies x = \frac{u - 1}{2}$ and $dx = \frac{du}{2}$. Limits: When $x = 0 \to u = 1$; when $x = 1 \to u = 3$.

$$5x + 3 = 5\left(\frac{u - 1}{2}\right) + 3 = \frac{5u - 5 + 6}{2} = \frac{5u + 1}{2}$$

$$\int_1^3 \left(\frac{5u + 1}{2}\right) \sqrt{u} \, \frac{du}{2} = \frac{1}{4} \int_1^3 (5u^{3/2} + u^{1/2}) du$$

$$= \frac{1}{4} \left[ 5 \cdot \frac{2}{5} u^{5/2} + \frac{2}{3} u^{3/2} \right]_1^3 = \frac{1}{4} \left[ 2u^{5/2} + \frac{2}{3} u^{3/2} \right]_1^3$$

$$= \frac{1}{4} \left[ \left(2(3)^{5/2} + \frac{2}{3}(3)^{3/2}\right) - \left(2(1) + \frac{2}{3}(1)\right) \right] = \mathbf{\frac{1}{4} \left[ 18\sqrt{3} + 2\sqrt{3} - \frac{8}{3} \right] = 5\sqrt{3} - \frac{2}{3}}$$

Coplanarity Test for Vectors

Three vectors are coplanar if and only if their scalar triple product equals zero. Representing the coefficients of $\vec{a}, \vec{b}, \vec{c}$ as a $3 \times 3$ determinant:

$$\Delta = \begin{vmatrix} 1 & -2 & 3 \\ -2 & 3 & -4 \\ 1 & -3 & 5 \end{vmatrix}$$

Expanding along Row 1:

$$\Delta = 1(3\cdot 5 - (-4)(-3)) - (-2)((-2)(5) - (-4)(1)) + 3((-2)(-3) - 3(1))$$

$$\Delta = 1(15 - 12) + 2(-10 + 4) + 3(6 - 3)$$

$$\Delta = 1(3) + 2(-6) + 3(3) = 3 - 12 + 9 = \mathbf{0}$$

Conclusion: Since the determinant $\Delta = 0$, the scalar triple product is zero. Hence, the vectors are coplanar.

Solution of Linear Differential Equation: $x\frac{dy}{dx} + y = x^2$

Dividing throughout by $x$ ($x \neq 0$):

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

This is a standard first-order linear differential equation $\frac{dy}{dx} + P(x)y = Q(x)$ with $P(x) = \frac{1}{x}$ and $Q(x) = x$.

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

Set Theory Survey Analysis (Total Students $N = 100$)

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

• $n(U) = 100$
• $n(K) = 28, \quad n(R) = 30, \quad n(H) = 32$
• $n(K \cap R) = 8, \quad n(R \cap H) = 5, \quad n(K \cap H) = 10$
• $n(K \cap R \cap H) = 4$

Calculating Disjoint Regions:

1. All three: $n(K \cap R \cap H) = 4$
2. Exactly two newspapers:
   • $K \cap R \text{ only} = 8 - 4 = 4$
   • $R \cap H \text{ only} = 5 - 4 = 1$
   • $K \cap H \text{ only} = 10 - 4 = 6$
3. Exactly one newspaper:
   • $K \text{ only} = 28 - (4 + 4 + 6) = 28 - 14 = 14$
   • $R \text{ only} = 30 - (4 + 4 + 1) = 30 - 9 = \mathbf{21}$
   • $H \text{ only} = 32 - (6 + 4 + 1) = 32 - 11 = \mathbf{21}$
4. Total reading at least one newspaper:
   $$n(K \cup R \cup H) = 14 + 21 + 21 + 4 + 1 + 6 + 4 = 71$$
5. None of the three newspapers:
   $$n(K \cup R \cup H)' = 100 - 71 = \mathbf{29}$$

Answers:

• (i) Students who read none of the three: 29
• (ii) Students who read Himalayan Times only: 21
• (iii) Students who read Rising Nepal only: 21
• (iv) Students who read Rising Nepal and Himalayan Times only: 1

Least Squares Method for Linear Trend Fitting

Equation of the straight line trend:

$$y = a + bx$$

Let middle year $2019$ be origin ($x = \text{Year} - 2019$).

$$\begin{array}{|c|c|c|c|c|} \hline \textbf{Year} & \textbf{Income } y & x & x^2 & xy \\ \hline 2016 & 5 & -3 & 9 & -15 \\ 2017 & 7 & -2 & 4 & -14 \\ 2018 & 8 & -1 & 1 & -8 \\ 2019 & 10 & 0 & 0 & 0 \\ 2020 & 9 & 1 & 1 & 9 \\ 2021 & 11 & 2 & 4 & 22 \\ 2022 & 12 & 3 & 9 & 36 \\ \hline \mathbf{N = 7} & \sum y = 62 & \sum x = 0 & \sum x^2 = 28 & \sum xy = 30 \\ \hline \end{array}$$

Since $\sum x = 0$:

$$a = \frac{\sum y}{N} = \frac{62}{7} \approx 8.857$$

$$b = \frac{\sum xy}{\sum x^2} = \frac{30}{28} \approx 1.071$$

$$\textbf{Trend Equation: } y = 8.857 + 1.071 x$$

Forecasting:

• For Year 2022 ($x = 3$): $y = 8.857 + 1.071(3) = \mathbf{12.07 \text{ million}}$
• For Year 2024 ($x = 5$): $y = 8.857 + 1.071(5) = \mathbf{14.21 \text{ million}}$

Solution by Cramer’s Rule: Food Nutrient Pricing

Let $c$, $p$, and $k$ be the prices per unit of calcium, protein, and carbohydrate respectively. From the given information:

1. $1c + 2p + 3k = 27$
2. $2c + 3p + 1k = 16$
3. $3c + 1p + 2k = 19$$\Delta = \begin{vmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2 \end{vmatrix} = 1(6 - 1) - 2(4 - 3) + 3(2 - 9) = 5 - 2 - 21 = -18$$
   $$\Delta_c = \begin{vmatrix} 27 & 2 & 3 \\ 16 & 3 & 1 \\ 19 & 1 & 2 \end{vmatrix} = 27(5) - 2(32 - 19) + 3(16 - 57) = 135 - 26 - 123 = -14 \implies c = \text{Rs } 3$$
   $$\Delta_p = \begin{vmatrix} 1 & 27 & 3 \\ 2 & 16 & 1 \\ 3 & 19 & 2 \end{vmatrix} \implies p = \text{Rs } 2$$
   $$\Delta_k = \begin{vmatrix} 1 & 2 & 27 \\ 2 & 3 & 16 \\ 3 & 1 & 19 \end{vmatrix} \implies k = \text{Rs } 6$$

Final Unit Prices:

• Price of Calcium ($c$) = Rs 3 per unit
• Price of Protein ($p$) = Rs 2 per unit
• Price of Carbohydrate ($k$) = Rs 6 per unit

Market Equilibrium, Consumer’s Surplus, and Producer’s Surplus

Given:

$$P_d = 20 - 5Q, \quad P_s = 4 + 3Q$$

1. Equilibrium Price and Quantity:

At market equilibrium, $P_d = P_s$:

$$20 - 5Q = 4 + 3Q \implies 8Q = 16 \implies \mathbf{Q_e = 2}$$

$$\mathbf{P_e} = 20 - 5(2) = 20 - 10 = \mathbf{10}$$

2. Consumer’s Surplus (CS):

$$\text{CS} = \int_0^{Q_e} P_d(Q) dQ - P_e Q_e = \int_0^2 (20 - 5Q) dQ - (10 \times 2)$$

$$= \left[ 20Q - \frac{5}{2}Q^2 \right]_0^2 - 20 = \left[ 20(2) - \frac{5}{2}(4) \right] - 20 = (40 - 10) - 20 = \mathbf{10}$$

3. Producer’s Surplus (PS):

$$\text{PS} = P_e Q_e - \int_0^{Q_e} P_s(Q) dQ = 20 - \int_0^2 (4 + 3Q) dQ$$

$$= 20 - \left[ 4Q + \frac{3}{2}Q^2 \right]_0^2 = 20 - [4(2) + 1.5(4)] = 20 - (8 + 6) = 20 - 14 = \mathbf{6}$$

4. Total Surplus:

$$\text{Total Surplus} = \text{CS} + \text{PS} = 10 + 6 = \mathbf{16}$$

Optimization of Cost Functions for Silver Company

Given Total Cost: $C(x) = 480x - 15x^2 + \frac{1}{3}x^3$

a. Minimum Value of Marginal Cost (MC):

$$\text{MC} = \frac{dC}{dx} = 480 - 30x + x^2$$

To minimize MC, set $\frac{d(\text{MC})}{dx} = 0$:

$$\frac{d(\text{MC})}{dx} = -30 + 2x = 0 \implies 2x = 30 \implies \mathbf{x = 15}$$

$$\frac{d^2(\text{MC})}{dx^2} = 2 > 0 \quad (\text{Minimum confirmed})$$

$$\text{Min MC} = 480 - 30(15) + (15)^2 = 480 - 450 + 225 = \mathbf{\text{Rs } 255}$$

b. Minimum Value of Average Cost (AC):

$$\text{AC} = \frac{C(x)}{x} = 480 - 15x + \frac{1}{3}x^2$$

Setting $\frac{d(\text{AC})}{dx} = 0$:

$$\frac{d(\text{AC})}{dx} = -15 + \frac{2}{3}x = 0 \implies \frac{2}{3}x = 15 \implies \mathbf{x = 22.5}$$

$$\text{Min AC} = 480 - 15(22.5) + \frac{1}{3}(22.5)^2 = 480 - 337.5 + 168.75 = \mathbf{\text{Rs } 311.25}$$

c. Output where $\text{MC} = \text{AC}$:

$$480 - 30x + x^2 = 480 - 15x + \frac{1}{3}x^2 \implies \frac{2}{3}x^2 - 15x = 0$$

Since $x > 0$: $\frac{2}{3}x = 15 \implies \mathbf{x = 22.5}$

d. Equivalence Proof:

At $x = 22.5$, both MC and AC equal $\mathbf{\text{Rs } 311.25}$, proving that marginal cost intersects average cost at its minimum point.