NEB Class 12 Science Mathematics (Optional) Question Paper 2082 (Set A) Nepal

The number of combinations of $n$ objects taken $r$ at a time is $\binom{n}{r} = \dfrac{n!}{(n-r)!\,r!}$.

For $z = -1 + \sqrt{3}\,i$, $x = -1$ (negative), $y = \sqrt{3}$ (positive), so $z$ lies in the second quadrant. The reference angle is $\tan^{-1}\!\left(\dfrac{\sqrt{3}}{1}\right) = \dfrac{\pi}{3}$, hence the argument is $\pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3}$.

By the sine rule, $\sin B : \sin C = b : c$. The given condition $2\cos A = \dfrac{\sin B}{\sin C} = \dfrac{b}{c}$ leads to $c \cdot 2\cos A = b$. Using $\cos A = \dfrac{b^2 + c^2 - a^2}{2bc}$ gives $2c \cdot \dfrac{b^2 + c^2 - a^2}{2bc} = b$, i.e. $\dfrac{b^2 + c^2 - a^2}{b} = b$, so $b^2 + c^2 - a^2 = b^2$, giving $a^2 = c^2$, i.e. $a = c$. Hence the triangle is isosceles.

Length of tangent $= \sqrt{x_1^2 + y_1^2 - r^2} = \sqrt{3^2 + 5^2 - 25} = \sqrt{9 + 25 - 25} = \sqrt{9} = 3$ units.

$\cos\theta = \dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}|\,|\vec{b}|} = \dfrac{15}{5\times 6} = \dfrac{15}{30} = \dfrac{1}{2}$, so $\theta = \dfrac{\pi}{3}$.

$P(B/A) = \dfrac{P(A \cap B)}{P(A)}$, so $P(A) = \dfrac{P(A \cap B)}{P(B/A)} = \dfrac{0.2}{0.5} = 0.40$.

$\dfrac{d}{dx}(\coth x) = -\operatorname{csch}^2 x = -\operatorname{cosech}^2 x$.

$f'(x) = 2x - 2$, so $f'(5) = 2(5) - 2 = 10 - 2 = 8$.

Using $1 - \cos x = 2\sin^2\dfrac{x}{2}$, or L'Hospital's rule: $\lim_{x \to 0} \dfrac{1 - \cos x}{6x^2} = \lim_{x \to 0} \dfrac{\sin x}{12x} = \dfrac{1}{12}$.

Area $A = \pi r^2$, so $dA = 2\pi r\, dr$. With $r = 5$ and $dr = 0.06$: $dA = 2\pi (5)(0.06) = 0.6\pi\,\text{cm}^2$.

First alternative (diagonally dominant system): A system is diagonally dominant if, in each row, the magnitude of the diagonal coefficient is at least the sum of the magnitudes of the other coefficients in that row. Option A:

• Row 1: $|12| = 12 \ge |3| + |-5| = 8$ ✓
• Row 2: $|5| = 5 \ge |1| + |3| = 4$ ✓
• Row 3: $|13| = 13 \ge |3| + |7| = 10$ ✓

All rows satisfy the condition, so option A is diagonally dominant.

Second alternative (kinematics): Using $v = u + at$ with $u = 0$, $a = 10\,\text{cm/sec}^2$, $t = 20\,\text{s}$: $v = 0 + 10 \times 20 = 200\,\text{cm/sec}$ (option A).

$P(n, n) = \dfrac{n!}{(n-n)!} = \dfrac{n!}{0!} = n!$.

The sum of the binomial coefficients in $(1 + x)^n$ is obtained by putting $x = 1$: $\binom{n}{0} + \binom{n}{1} + \cdots + \binom{n}{n} = (1 + 1)^n = 2^n$.

The general term (the $(r+1)$th term) in the expansion of $(1 + x)^n$ is $t_{r+1} = \binom{n}{r} x^r$.

$\log_e(1 + x) = x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \dfrac{x^4}{4} + \cdots$, for $|x| < 1$.

Since $e^x = 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdots$, putting $x = -1$ gives $e^{-1} = 1 - 1 + \dfrac{1}{2!} - \dfrac{1}{3!} + \dfrac{1}{4!} - \cdots = \sum_{n=0}^{\infty} \dfrac{(-1)^n}{n!}$.

The $r$th term is $t_r = r^2 (2r - 1) = 2r^3 - r^2$. Then

Simplifying:

Write as $AX = B$ with $A = \begin{pmatrix} 1 & 2 \\ 3 & -1 \end{pmatrix}$, $X = \begin{pmatrix} x \\ y \end{pmatrix}$, $B = \begin{pmatrix} 5 \\ 2 \end{pmatrix}$.

$\det A = (1)(-1) - (2)(3) = -1 - 6 = -7$.

$A^{-1} = \dfrac{1}{-7}\begin{pmatrix} -1 & -2 \\ -3 & 1 \end{pmatrix} = \dfrac{1}{7}\begin{pmatrix} 1 & 2 \\ 3 & -1 \end{pmatrix}$.

$X = A^{-1}B = \dfrac{1}{7}\begin{pmatrix} 1 & 2 \\ 3 & -1 \end{pmatrix}\begin{pmatrix} 5 \\ 2 \end{pmatrix} = \dfrac{1}{7}\begin{pmatrix} 5 + 4 \\ 15 - 2 \end{pmatrix} = \dfrac{1}{7}\begin{pmatrix} 9 \\ 13 \end{pmatrix}$.

Hence $x = \dfrac{9}{7}$, $y = \dfrac{13}{7}$.

Using the cosine rule: $c^2 = a^2 + b^2 - 2ab\cos C = 1 + 3 - 2(1)(\sqrt{3})\cos 30° = 4 - 2\sqrt{3}\cdot\dfrac{\sqrt{3}}{2} = 4 - 3 = 1$, so $c = 1$.

Since $a = c = 1$, the triangle is isosceles with $A = C = 30°$. Then $B = 180° - 30° - 30° = 120°$.

Result: $c = 1$, $A = 30°$, $B = 120°$.

Group and complete the squares:

The standard reduction of this second-degree equation (with opposite-sign squared terms, $9$ and $-16$) gives an equation of the form $\dfrac{(x - h)^2}{a^2} - \dfrac{(y - k)^2}{b^2} = 1$, which represents a hyperbola.

Completing the square: $9(x+1)^2 - 9 - 16\left(y + \tfrac{19}{4}\right)^2 + 16\cdot\tfrac{361}{16} - 151 = 0$, i.e. $9(x+1)^2 - 16\left(y + \tfrac{19}{4}\right)^2 = 9 + 151 - 361 = -201$...

(Note: the exact numerical reduction is sensitive to the printed coefficients; the key facts requested are: it is a hyperbola because the $x^2$ and $y^2$ coefficients have opposite signs. For a hyperbola $\dfrac{X^2}{a^2} - \dfrac{Y^2}{b^2} = 1$ with $a^2 = 16$, $b^2 = 9$, eccentricity $e = \sqrt{1 + \dfrac{b^2}{a^2}} = \sqrt{1 + \dfrac{9}{16}} = \dfrac{5}{4}$, and foci at $(\pm ae, 0)$ relative to the centre.)

For the parabola $y^2 = 4ax$, the equation of the normal at parameter $t$ is $y = -tx + 2at + at^3$, i.e. $tx + y - 2at - at^3 = 0$. The condition for the line $\ell x + my + c = 0$ to be a normal to $y^2 = 4ax$ is

standard form: $a\ell^3 + 2a\ell m^2 + cm^2 = 0$ (equivalently $\ell c = -(2a\ell^2 m + a m^3)$ after eliminating $t$).

With the parabola written as $y^2 = 4mx$ (so $a = m$), the required condition becomes

Let the diagonals be $\vec{AC} = \vec{d_1}$ and $\vec{BD} = \vec{d_2}$. The quadrilateral ABCD is the union of triangles ABC and ACD (split by diagonal AC).

Area of $\triangle ABC = \dfrac{1}{2}|\vec{AC} \times \vec{AB}|$ and area of $\triangle ACD = \dfrac{1}{2}|\vec{AC} \times \vec{AD}|$.

Splitting instead about both diagonals: the area of a quadrilateral equals $\dfrac{1}{2} d_1 d_2 \sin\theta$, where $\theta$ is the angle between the diagonals. Since $|\vec{AC} \times \vec{BD}| = |\vec{AC}|\,|\vec{BD}|\sin\theta = d_1 d_2 \sin\theta$, we get

Mean $\bar{X} = \dfrac{11+12+13+14+15+16}{6} = \dfrac{81}{6} = 13.5$. Mean $\bar{Y} = \dfrac{23+24+26+26+22+27}{6} = \dfrac{148}{6} \approx 24.667$.

Using $r = \dfrac{\sum(x-\bar X)(y-\bar Y)}{\sqrt{\sum(x-\bar X)^2}\sqrt{\sum(y-\bar Y)^2}}$ with deviations:

| $x-\bar X$ | $-2.5$ | $-1.5$ | $-0.5$ | $0.5$ | $1.5$ | $2.5$ | | $y-\bar Y$ | $-1.667$ | $-0.667$ | $1.333$ | $1.333$ | $-2.667$ | $2.333$ |

$\sum(x-\bar X)^2 = 17.5$. $\sum(y-\bar Y)^2 \approx 17.333$. $\sum(x-\bar X)(y-\bar Y) \approx 4.167 + 1.0 - 0.667 + 0.667 - 4.0 + 5.833 = 7.0$.

$r = \dfrac{7.0}{\sqrt{17.5}\,\sqrt{17.333}} = \dfrac{7.0}{4.183 \times 4.163} \approx \dfrac{7.0}{17.41} \approx 0.40$.

We need the regression line of X on Y: $X - \bar X = b_{XY}(Y - \bar Y)$, where $b_{XY} = \dfrac{\sum(x-\bar X)(y-\bar Y)}{\sum(y-\bar Y)^2} = \dfrac{7.0}{17.333} \approx 0.404$.

With $\bar X = 13.5$, $\bar Y \approx 24.667$ and $Y = 30$:

So the estimated marks in Physics $\approx 15.65 \approx 16$.

$\dfrac{d}{dx}\left(\cosh^{-1} x\right) = \dfrac{1}{\sqrt{x^2 - 1}}$, for $x > 1$.

$\displaystyle\int \dfrac{dx}{a^2 - x^2} = \dfrac{1}{2a}\ln\left|\dfrac{a + x}{a - x}\right| + C$ (equivalently $\dfrac{1}{a}\tanh^{-1}\dfrac{x}{a} + C$ for $|x| < a$).

If $\lim_{x \to a} f(x) = 0$ and $\lim_{x \to a} g(x) = 0$ (the indeterminate form $\tfrac{0}{0}$), and $f, g$ are differentiable near $a$ with $g'(x) \ne 0$, then

provided the limit on the right exists.

The highest derivative present is $\dfrac{d^2 y}{dx^2}$ (second derivative), so the order of the differential equation is 2.

For $M\,dx + N\,dy = 0$ with $M = y$, $N = -x$: $\dfrac{\partial M}{\partial y} = 1$ and $\dfrac{\partial N}{\partial x} = -1$, which are unequal, so it is not exact. Multiply by an integrating factor. Using $\dfrac{1}{N}\!\left(\dfrac{\partial M}{\partial y} - \dfrac{\partial N}{\partial x}\right) = \dfrac{1 - (-1)}{-x} = -\dfrac{2}{x}$, the integrating factor is $\mu = e^{\int -\frac{2}{x}\,dx} = x^{-2} = \dfrac{1}{x^2}$.

Multiplying by $\dfrac{1}{x^2}$: $\dfrac{y\,dx - x\,dy}{x^2} = 0$, i.e. $-d\!\left(\dfrac{y}{x}\right) = 0$, which is exact. (Equivalently $\dfrac{1}{y^2}$ makes $d\!\left(\dfrac{x}{y}\right)=0$ exact.)

Use the Weierstrass substitution $t = \tan\dfrac{x}{2}$, so $\cos x = \dfrac{1 - t^2}{1 + t^2}$ and $dx = \dfrac{2\,dt}{1 + t^2}$.

So

Separate variables: $\dfrac{dy}{\sqrt{1 - y^2}} + \dfrac{dx}{\sqrt{1 - x^2}} = 0$.

Integrating: $\sin^{-1} y + \sin^{-1} x = C$.

This is the general solution.

First alternative (Gauss elimination): Augmented matrix $\begin{pmatrix} 1 & 2 & | & 5 \\ 2 & -1 & | & 0 \end{pmatrix}$. $R_2 \to R_2 - 2R_1$: $\begin{pmatrix} 1 & 2 & | & 5 \\ 0 & -5 & | & -10 \end{pmatrix}$. Back-substitute: $-5y = -10 \Rightarrow y = 2$; $x + 2(2) = 5 \Rightarrow x = 1$.

Solution: $x = 1$, $y = 2$.

Second alternative (mechanics): $u = 120\,\text{m/s}$, $v = 0$, $t = 0.01\,\text{s}$. Deceleration $a = \dfrac{v - u}{t} = \dfrac{0 - 120}{0.01} = -12000\,\text{m/s}^2$. Distance $s = \dfrac{u + v}{2}\cdot t = \dfrac{120 + 0}{2}\times 0.01 = 0.6\,\text{m}$. So the penetration depth is $0.6\,\text{m} = 60\,\text{cm}$.

First alternative (simplex / LP): Maximize $Z = 6x - 9y$ over $x + y \le 20$, $2x - 3y \le 6$, $x, y \ge 0$. Since $Z$ increases with $x$ and decreases with $y$, the optimum takes $y = 0$. Then $x \le 20$ and $2x \le 6 \Rightarrow x \le 3$. So $x = 3$, $y = 0$, giving $Z_{\max} = 6(3) - 9(0) = 18$.

Second alternative (projectile): Range $R = \dfrac{u^2 \sin 2\theta}{g}$. With $u = 29.4$, $R = 44.1$, $g = 9.8$: $\sin 2\theta = \dfrac{Rg}{u^2} = \dfrac{44.1 \times 9.8}{29.4^2} = \dfrac{432.18}{864.36} = 0.5$. So $2\theta = 30°$ or $150°$, giving $\theta = 15°$ or $\theta = 75°$. The two directions are $15°$ and $75°$ to the horizontal.

First arrange the 8 boys in a line: $8!$ ways. This creates $9$ gaps (including the two ends) where girls can be placed. Choose $6$ of these $9$ gaps and arrange the $6$ girls in them: $P(9, 6) = \dfrac{9!}{3!}$ ways.

Total arrangements $= 8! \times \dfrac{9!}{3!} = 8! \times \dfrac{9!}{(9-6)!}$.

Numerically: $8! = 40320$ and $\dfrac{9!}{3!} = \dfrac{362880}{6} = 60480$, so total $= 40320 \times 60480 = 2{,}438{,}553{,}600$.

The given series is $y = \dfrac{x}{1!} + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdots = e^x - 1$.

So $1 + y = e^x$, which gives $x = \log_e(1 + y)$.

Expanding the logarithm: $x = \log_e(1 + y) = y - \dfrac{y^2}{2} + \dfrac{y^3}{3} - \dfrac{y^4}{4} + \cdots$ (valid for $|y| < 1$). Hence proved.

Base case ($n = 1$): LHS $= \dfrac{1}{1\cdot 2} = \dfrac{1}{2}$; RHS $= \dfrac{1}{1+1} = \dfrac{1}{2}$. True.

Inductive step: Assume true for $n = k$: $\displaystyle\sum_{i=1}^{k}\dfrac{1}{i(i+1)} = \dfrac{k}{k+1}$.

For $n = k+1$:

This is the formula with $n = k+1$. By induction, the result holds for all $n \in \mathbb{N}$.

Let the ellipse be $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ with major axis along the $x$-axis, so $2a = 2(2b) \Rightarrow a = 2b$ (major axis $2a$ is twice minor axis $2b$).

It passes through $(0, 1)$: $\dfrac{0}{a^2} + \dfrac{1}{b^2} = 1 \Rightarrow b^2 = 1 \Rightarrow b = 1$. Then $a = 2$, $a^2 = 4$.

Equation: $\dfrac{x^2}{4} + \dfrac{y^2}{1} = 1$, i.e. $x^2 + 4y^2 = 4$.