NEB Class 12 Science Mathematics (Optional) Question Paper 2080 (Set A) Nepal

Each of the $r$ letters has $n$ choices of letter boxes. Therefore, the total number of ways is $n \times n \times \dots \times n$ ($r$ times) = $n^r$.

Given the operation $a o b = 2a + 3b$, substitute $a = 2$ and $b = 3$: $2 o 3 = 2(2) + 3(3) = 4 + 9 = 13$.

The general solution for $\tan^2 x = \tan^2 \theta$ is $x = n\pi \pm \theta$, where $n \in \mathbb{Z}$.

The principal value branch (range) of $\sin^{-1}x$ is $-\frac{\pi}{2} \le y \le \frac{\pi}{2}$.

Since $\vec{j} \times \vec{k} = \vec{i}$, $\vec{k} \times \vec{i} = \vec{j}$, and $\vec{i} \times \vec{j} = \vec{k}$: $\vec{i} \cdot \vec{i} + \vec{j} \cdot \vec{j} - \vec{k} \cdot \vec{k} = 1 + 1 - 1 = 1$.

Probability of first card being an ace = $\frac{4}{52} = \frac{1}{13}$. Probability of second card being an ace without replacement = $\frac{3}{51} = \frac{1}{17}$. Total probability = $\frac{1}{13} \times \frac{1}{17} = \frac{1}{221}$.

For two planes $A_1x+B_1y+C_1z+D_1=0$ and $A_2x+B_2y+C_2z+D_2=0$ to be parallel, their direction ratios must be proportional: $\frac{2}{4} = \frac{3}{6} = \frac{5}{c} \implies \frac{1}{2} = \frac{5}{c} \implies c = 10$.

The limit is in $\frac{0}{0}$ form. Applying L'Hospital's rule by differentiating numerator and denominator: $\lim_{x\rightarrow0}\frac{\frac{d}{dx}[2(e^{2x}-1)]}{\frac{d}{dx}[\log(1+x)]} = \lim_{x\rightarrow0}\frac{4e^{2x}}{\frac{1}{1+x}} = \frac{4(1)}{1} = 4$.

Using standard formula $\int\frac{dx}{x^2-a^2} = \frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right| + C$ where $a=6$: $\int\frac{dx}{x^2-36} = \frac{1}{12}\ln\left|\frac{x-6}{x+6}\right| + C$.

Since the system is in upper-triangular form and the diagonal elements $a_{11}$ and $a_{22}$ are non-zero, unique values can be determined via back substitution. Thus it has a unique solution.

Mass $m = 250\text{ gm} = 0.25\text{ kg}$. Force $F = 3\text{ N}$, time $t = 1\text{ s}$, initial velocity $u = 0$. From Impulse = Change in momentum: $F \times t = m(v - u) \implies 3 \times 1 = 0.25 \times v \implies v = \frac{3}{0.25} = 12\text{ ms}^{-1}$.

Given $Q_0 = 6$, then equilibrium price $P_0 = 60 - 2(6) - 6^2 = 60 - 12 - 36 = 12$. Consumer Surplus $\text{CS} = \int_{0}^{Q_0} P\, dQ - P_0Q_0 = \int_{0}^{6} (60 - 2Q - Q^2)\, dQ - (12 \times 6)$ $= \left[ 60Q - Q^2 - \frac{Q^3}{3} \right]_0^6 - 72 = \left(360 - 36 - \frac{216}{3}\right) - 72 = (360 - 36 - 72) - 72 = 252 - 72 = 180$.

a) The sum of coefficients of even terms (i.e., $C_1 + C_3 + C_5 + \dots$) in $(1+x)^n$ is $2^{n-1}$. b) Total terms in $(x + \frac{1}{x})^6$ is $6 + 1 = 7$. c) $e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!} - \dots = \sum_{n=0}^{\infty} \frac{(-1)^n x^n}{n!}$. d) The condition for equal roots is that the discriminant is zero: $b^2 - 4ac = 0$. e) In the polar form of a complex number, $r$ represents the modulus (or absolute value) of $z$, which is $|z| = \sqrt{x^2+y^2}$.

a) Let $z = 1 - \sqrt{3}i$. Here modulus $r = \sqrt{1^2 + (-\sqrt{3})^2} = 2$. Argument $\theta = \tan^{-1}\left(\frac{-\sqrt{3}}{1}\right) = -\frac{\pi}{3}$. So, $z = 2\left[\cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right)\right]$. By De-Moivre's theorem: $z^6 = 2^6 \left[\cos\left(6 \times -\frac{\pi}{3}\right) + i\sin\left(6 \times -\frac{\pi}{3}\right)\right] = 64 [\cos(-2\pi) + i\sin(-2\pi)] = 64 [1 + 0] = 64$.

b) Let $P(n): n(n+1)(2n+1)$ is divisible by 6. Base Case: For $n=1$, $1(1+1)(2(1)+1) = 1 \times 2 \times 3 = 6$, which is divisible by 6. True. Inductive Step: Assume $P(k)$ is true, so $k(k+1)(2k+1) = 6m$ for some integer $m$. Now for $n=k+1$: $(k+1)(k+2)(2(k+1)+1) = (k+1)(k+2)(2k+3) = (k+1)(k+2)(2k+1 + 2)$ $= (k+1)(2k+1)(k+2) + 2(k+1)(k+2)$ $= k(k+1)(2k+1) + 2(k+1)(2k+1) + 2(k+1)(k+2)$ $= 6m + 2(k+1)[2k+1 + k+2] = 6m + 2(k+1)(3k+3) = 6m + 6(k+1)^2 = 6[m + (k+1)^2]$, which is divisible by 6. Thus proved.

a) Let $\alpha = \cos^{-1}\frac{1}{2} = \frac{\pi}{3}$, so $\sin\alpha = \frac{\sqrt{3}}{2}$. Let $\beta = \sin^{-1}\frac{3}{5}$, so $\sin\beta = \frac{3}{5}$ and $\cos\beta = \sqrt{1 - (3/5)^2} = \frac{4}{5}$. Now, $\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta = \left(\frac{\sqrt{3}}{2}\right)\left(\frac{4}{5}\right) + \left(\frac{1}{2}\right)\left(\frac{3}{5}\right) = \frac{4\sqrt{3} + 3}{10}$.

b) Let $\vec{a} = -\vec{i}-\vec{j}+\vec{k}$ and \vec{b} = 3\vec{i}-\vec{j}+2\vec{k}. $\vec{a} \times \vec{b} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ -1 & -1 & 1 \\ 3 & -1 & 2 \end{vmatrix} = \vec{i}(-2+1) - \vec{j}(-2-3) + \vec{k}(1+3) = -\vec{i} + 5\vec{j} + 4\vec{k}$. Area of parallelogram = $|\vec{a} \times \vec{b}| = \sqrt{(-1)^2 + 5^2 + 4^2} = \sqrt{1 + 25 + 16} = \sqrt{42}$ sq. units.

a) Both X and Y are already in perfectly increasing order. Let's assign ranks: Rank of X ($R_1$): 1, 2, 3, 4, 5 Rank of Y ($R_2$): 1, 2, 3, 4, 5 Difference $d = R_1 - R_2$: 0, 0, 0, 0, 0 $\implies \sum d^2 = 0$. Rank correlation coefficient $R = 1 - \frac{6\sum d^2}{n(n^2-1)} = 1 - 0 = 1$.

b) Binomial distribution with $n = 4$, $p = 20\% = 0.2$, $q = 0.8$. We want $P(X = 2) = \binom{4}{2} p^2 q^{4-2} = 6 \times (0.2)^2 \times (0.8)^2 = 6 \times 0.04 \times 0.64 = 0.1536$.

a) It represents the slope of the tangent line to the curve at that particular point. b) $\int\frac{1}{a^{2}+x^{2}}dx = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) + C$. c) $\Delta y$ represents the actual exact change in the dependent variable $y$ as $x$ changes by $\Delta x$, whereas $dy$ represents the approximate change in $y$ estimated using the tangent line derivative ($dy = f'(x)dx$). d) $\frac{d}{dx}(\sinh x) = \cosh x$. e) A first-order linear differential equation has the form: $\frac{dy}{dx} + P(x)y = Q(x)$.

a) Separating variables: $\frac{1}{1+y^2} dy = \frac{1}{1+x^2} dx$. Integrating both sides: $\int \frac{1}{1+y^2} dy = \int \frac{1}{1+x^2} dx \implies \tan^{-1}y = \tan^{-1}x + C$.

b) $f(x) = x^2 - x - 2$.

1. $f(x)$ is a polynomial, so it is continuous on $[-1, 2]$.
2. $f'(x) = 2x - 1$, which exists on $(-1, 2)$, so it is differentiable. Thus LMVT applies. There exists $c \in (-1, 2)$ such that $f'(c) = \frac{f(2)-f(-1)}{2 - (-1)}$. $f(2) = (3)(0) = 0$, $f(-1) = (0)(-3) = 0$. $2c - 1 = \frac{0 - 0}{3} = 0 \implies c = \frac{1}{2}$. Since $\frac{1}{2} \in (-1, 2)$, LMVT is verified.

Introduce slack variables $s_1, s_2 \ge 0$:

1. $x + 3y + s_1 = 21$
2. $2x + 3y + s_2 = 24$ Objective function: $-10x - 15y + Z = 0$

Initial Simplex Tableau:

Most negative indicator in Z row is -15 (y column). Pivot element is 3. Divide row 1 by 3:

Next pivot column is x column (indicator -5). Pivot row is s2 row (ratio 3/(1/3)=21 vs 3/1=3). Pivot is 1. Performing row operations to clear x column:

Since all indicators in the Z-row are $\ge 0$, maximum value is reached. Maximum $Z = 120$ at $x = 3, y = 6$.

Initially, let the resultant act at point C at a distance $x$ from M. Then: $A \cdot x = B \cdot (l - x) \implies (A+B)x = Bl \implies x = \frac{Bl}{A+B}$. When opposite forces $T$ are introduced (suppose $+T$ added to $A$ and $-T$ added to $B$ since they are opposite in direction), the new forces are $A+T$ and $B-T$. Let the new resultant act at distance $x'$ from M: $(A+T)x' = (B-T)(l-x') \implies (A+B)x' = (B-T)l \implies x' = \frac{(B-T)l}{A+B}$. The displacement is $\Delta x = x - x' = \frac{Bl - (B-T)l}{A+B} = \frac{Tl}{A+B}$. Proved.

According to Leontief model, Total output $X = (I - P)^{-1}Q$. $I - P = \begin{bmatrix}1&0\\ 0&1\end{bmatrix} - \begin{bmatrix}0.1&0.4\\ 0.2&0.2\end{bmatrix} = \begin{bmatrix}0.9&-0.4\\ -0.2&0.8\end{bmatrix}$. Determinant $|I-P| = (0.9)(0.8) - (-0.4)(-0.2) = 0.72 - 0.08 = 0.64$. $(I-P)^{-1} = \frac{1}{0.64} \begin{bmatrix}0.8&0.4\\ 0.2&0.9\end{bmatrix}$. Now, $X = \frac{1}{0.64} \begin{bmatrix}0.8&0.4\\ 0.2&0.9\end{bmatrix} \begin{bmatrix}3200\\ 1800\end{bmatrix} = \frac{1}{0.64} \begin{bmatrix} 0.8(3200) + 0.4(1800) \\ 0.2(3200) + 0.9(1800) \end{bmatrix} = \frac{1}{0.64} \begin{bmatrix} 2560 + 720 \\ 640 + 1620 \end{bmatrix} = \frac{1}{0.64} \begin{bmatrix} 3280 \\ 2260 \end{bmatrix} = \begin{bmatrix} 5125 \\ 3531.25 \end{bmatrix}$.

a) Augmented matrix: $\begin{bmatrix} 2 & 1 & -1 & : & 9 \\ 3 & -1 & 2 & : & -1 \\ 4 & 1 & -3 & : & 17 \end{bmatrix}$. Perform row operations to reach reduced row echelon form: $R_2 \leftarrow 2R_2 - 3R_1 \implies \begin{bmatrix} 2 & 1 & -1 & : & 9 \\ 0 & -5 & 7 & : & -29 \\ 4 & 1 & -3 & : & 17 \end{bmatrix}$ $R_3 \leftarrow R_3 - 2R_1 \implies \begin{bmatrix} 2 & 1 & -1 & : & 9 \\ 0 & -5 & 7 & : & -29 \\ 0 & -1 & -1 & : & -1 \end{bmatrix}$ Swap $R_2$ and $-1 \times R_3$: $\begin{bmatrix} 2 & 1 & -1 & : & 9 \\ 0 & 1 & 1 & : & 1 \\ 0 & -5 & 7 & : & -29 \end{bmatrix}$ $R_3 \leftarrow R_3 + 5R_2 \implies \begin{bmatrix} 2 & 1 & -1 & : & 9 \\ 0 & 1 & 1 & : & 1 \\ 0 & 0 & 12 & : & -24 \end{bmatrix} \implies z = -2$. By back substitution, $y + (-2) = 1 \implies y = 3$. $2x + 3 - (-2) = 9 \implies 2x + 5 = 9 \implies x = 2$. Solution: $(2, 3, -2)$.

b) Let roots be $\alpha, \beta$. Given $\alpha + \beta = 5$ and $\alpha^2 + \beta^2 = 13$. Since $(\alpha+\beta)^2 - 2\alpha\beta = 13 \implies 25 - 2\alpha\beta = 13 \implies 2\alpha\beta = 12 \implies \alpha\beta = 6$. The quadratic equation is $x^2 - (\alpha+\beta)x + \alpha\beta = 0 \implies x^2 - 5x + 6 = 0$.

c) General term $T_n = \frac{1+2+\dots+n}{(n+1)!} = \frac{n(n+1)}{2(n+1)!} = \frac{n}{2(n!)} = \frac{1}{2(n-1)!}$. Sum $= \sum_{n=1}^{\infty} \frac{1}{2(n-1)!} = \frac{1}{2} \left( 1 + \frac{1}{1!} + \frac{1}{2!} + \dots \right) = \frac{e}{2}$. Proved.

a) Dividing the plane equation by 24: $\frac{3x}{24} + \frac{4y}{24} + \frac{6z}{24} = 1 \implies \frac{x}{8} + \frac{y}{6} + \frac{z}{4} = 1$. Intercepts on the x, y, and z axes are 8, 6, and 4 respectively.

b) Length of major axis $2a = 12 \implies a = 6$. Eccentricity $e = \frac{2}{3}$. Since $b^2 = a^2(1-e^2) = 36(1 - \frac{4}{9}) = 36 \times \frac{5}{9} = 20$. Equation of ellipse: $\frac{x^2}{36} + \frac{y^2}{20} = 1$.

c) From first equation, $2n = 4l + 3m \implies n = 2l + 1.5m$. Substitute into $lm - mn + nl = 0 \implies lm - (m-l)n = 0 \implies lm - (m-l)(2l + 1.5m) = 0 \implies lm - (2lm + 1.5m^2 - 2l^2 - 1.5lm) = 0 \implies 2l^2 + 0.5lm - 1.5m^2 = 0 \implies 4l^2 + lm - 3m^2 = 0$. Factoring: $(4l - 3m)(l + m) = 0$. Case 1: $l = -m \implies n = 2(-m) + 1.5m = -0.5m$. DRs: $(-1, 1, -0.5) \propto (2, -2, 1)$. DCs: $(\frac{2}{3}, -\frac{2}{3}, \frac{1}{3})$. Case 2: $4l = 3m \implies m = \frac{4}{3}l \implies n = 2l + 2l = 4l$. DRs: $(1, \frac{4}{3}, 4) \propto (3, 4, 12)$. DCs: $(\frac{3}{13}, \frac{4}{13}, \frac{12}{13})$. Angle $\cos\theta = |l_1l_2 + m_1m_2 + n_1n_2| = |\frac{6}{39} - \frac{8}{39} + \frac{12}{39}| = \frac{10}{39} \implies \theta = \cos^{-1}\left(\frac{10}{39}\right)$.

a) The concept used is Partial Fractions. Let $\frac{x^2}{(x+2)^2(x+3)} = \frac{A}{x+2} + \frac{B}{(x+2)^2} + \frac{C}{x+3}$. $x^2 = A(x+2)(x+3) + B(x+3) + C(x+2)^2$. For $x = -2: 4 = B(1) \implies B = 4$. For $x = -3: 9 = C(-1)^2 \implies C = 9$. Comparing coefficients of $x^2$: $1 = A + C \implies A = 1 - 9 = -8$. Thus, $\int \left( \frac{-8}{x+2} + \frac{4}{(x+2)^2} + \frac{9}{x+3} \right) dx = -8\ln|x+2| - \frac{4}{x+2} + 9\ln|x+3| + C$.

b) No, Rolle's theorem does not hold (or exist) because one of its essential conditions is violated. Rolle's theorem explicitly requires that $f(a) = f(b)$ so that the slope of the secant line is zero, guaranteeing a point where the derivative is zero.

c) Example: $\frac{dy}{dx} = \frac{x+y}{x}$. Put $y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}$. $v + x\frac{dv}{dx} = \frac{x+vx}{x} = 1 + v \implies x\frac{dv}{dx} = 1 \implies dv = \frac{dx}{x}$. Integrating both sides gives $v = \ln|x| + C \implies \frac{y}{x} = \ln|x| + C \implies y = x\ln|x| + Cx$.