NEB Class 12 Science Mathematics Question Paper 2081 (Set W) Nepal

$^nC_r = \dfrac{n!}{r!(n-r)!}$.

$\dfrac{1-i}{1+i} = \dfrac{(1-i)^2}{(1+i)(1-i)} = \dfrac{1-2i+i^2}{2} = \dfrac{-2i}{2} = -i = \cos 270^\circ + i\sin 270^\circ$.

$c^2 = a^2 + b^2 - 2ab\cos C = 1 + 3 - 2\sqrt3\cos30^\circ = 4 - 2\sqrt3\cdot\tfrac{\sqrt3}{2} = 4 - 3 = 1$, so $c=1=a$. Since $a=c$, the triangle is isosceles. Checking angles: with sides $1,\sqrt3,1$... $b^2 = 3 = a^2+c^2-2ac\cos B$; $3 = 1+1-2\cos B \Rightarrow \cos B = -\tfrac12 \Rightarrow B=120^\circ$ (obtuse). So it is isosceles and obtuse-angled.

Since $e > 1$ (as $\sqrt{a^2+b^2} > a$), the conic is a hyperbola: $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$.

$\cos\theta = \dfrac{(\vec i+\vec j)\cdot(\vec j+\vec k)}{|\vec i+\vec j||\vec j+\vec k|} = \dfrac{0+1+0}{\sqrt2\cdot\sqrt2} = \dfrac{1}{2} \Rightarrow \theta = 60^\circ$.

$P(A/B) = \dfrac{P(A\cap B)}{P(B)} = \dfrac{0.4}{0.75} = 0.533$. Compare with options: $P(B) = 0.75$, $P(A\cap B)=0.4$, $P(B/A) = \dfrac{0.4}{0.5}=0.8$. Since $P(A/B)=0.533$, it is less than $P(B/A)=0.8$.

$\dfrac{d}{dx}\tanh^{-1}x = \dfrac{1}{1-x^2}$, for $|x| < 1$.

$\dfrac{\tan\pi x}{\sec^2\pi x} = \tan\pi x\cos^2\pi x = \sin\pi x\cos\pi x = \tfrac12\sin 2\pi x$. At $x=\tfrac12$: $\tfrac12\sin\pi = 0$.

$2y = 2-x^2 \Rightarrow y = 1 - \tfrac{x^2}{2}$, $\dfrac{dy}{dx} = -x$. At $x=1$: slope $= -1$, so angle $= \tan^{-1}(-1) = 135^\circ = \dfrac{3\pi}{4}$.

A linear differential equation of the form $\dfrac{dy}{dx} + Py = Q$ requires an integrating factor. Option C, $\sin^2 x\dfrac{dy}{dx} + y = 2$, can be written in linear form and thus has an integrating factor.

First option: After forward elimination, the matrix becomes an upper triangular matrix.

OR (projectile): $H = \dfrac{u^2\sin^2\theta}{2g} = 40$. $u^2 = (40\sqrt2)^2 = 3200$. $\dfrac{3200\sin^2\theta}{20} = 40 \Rightarrow \sin^2\theta = \dfrac{800}{3200} = \tfrac14 \Rightarrow \sin\theta = \tfrac12 \Rightarrow \theta = 30^\circ$.

a. Circular permutations of $n$ objects $= (n-1)!$.

b. General term: $T_{r+1} = \binom{n}{r}a^{n-r}x^r$.

c. $\log_e(1-x) = -\left(x + \dfrac{x^2}{2} + \dfrac{x^3}{3} + \dots\right)$.

d. Euler's form: $\cos\theta + i\sin\theta = e^{i\theta}$.

e. $\sum_{k=1}^{n} k^3 = \left(\dfrac{n(n+1)}{2}\right)^2$.

a. Multiply numerator and denominator appropriately using $\omega^3=1$: factor $\omega$ from the denominator — $b + c\omega + a\omega^2 = \omega^{-1}(b\omega + c\omega^2 + a) = \omega^2(a + b\omega + c\omega^2)$. Hence the ratio $= \dfrac{a+b\omega+c\omega^2}{\omega^2(a+b\omega+c\omega^2)} = \omega^{-2} = \omega$.

b. Rewrite: $4x-5y+2z=1$, $3x-4z=-10$, $2y-3z=-6$. Solving by row reduction gives the values of $x, y, z$.

a. Using the sine rule and expanding $\sin(A-B)$, $\sin(B-C)$, the condition reduces to $a^2 + c^2 = 2b^2$, i.e. $a^2, b^2, c^2$ are in A.P.

b. Major axis $= 2\times$ minor axis $\Rightarrow a = 2b$. Ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$. Passing through $(1,0)$: $\dfrac{1}{a^2}=1 \Rightarrow a^2=1$, so $a=1$ and $b = \tfrac12$, $b^2 = \tfrac14$. Ellipse: $\dfrac{x^2}{1} + \dfrac{y^2}{1/4} = 1$, i.e. $x^2 + 4y^2 = 1$.

a. For $y^2 = 12x$ ($4a=12 \Rightarrow a=3$), the point $(6,9)$: $y_1^2 = 81$, $12x_1 = 72$. Since $y_1^2 > 12x_1$ (81 > 72), the point lies outside the parabola, so two real tangents can be drawn from it.

c. By induction: for $n=1$, $1+2=3$ divisible. Assume $n^3+2n$ divisible by 3. Then $(n+1)^3 + 2(n+1) = n^3+3n^2+3n+1+2n+2 = (n^3+2n) + 3(n^2+n+1)$, which is divisible by 3. Hence true for all $n$.

a. Slope $= f'(x_1) = \left.\dfrac{dy}{dx}\right|_{(x_1,y_1)}$.

b. $\dfrac{d}{dx}\operatorname{cosech} x = -\operatorname{cosech} x\coth x$.

c. It is a linear differential equation (of the first order).

d. $\int\dfrac{dx}{a^2-x^2} = \dfrac{1}{2a}\ln\left|\dfrac{a+x}{a-x}\right| + C$.

e. L'Hospital's rule applies to indeterminate forms ($\tfrac00$ or $\tfrac\infty\infty$): the limit of $\tfrac{f}{g}$ equals the limit of $\tfrac{f'}{g'}$.

(Dot product positive: $\vec a\cdot\vec b = |\vec a||\vec b|\cos\theta > 0$ when $\theta < 90^\circ$; e.g. $(\vec i)\cdot(\vec i+\vec j) = 1 > 0$.)

a. Compute Karl Pearson's correlation coefficient $r$ from the six pairs (strong positive correlation expected).

b. Find the regression line of weight (Y) on age (X), $Y = a + bX$, then substitute $X=12$ to estimate the weight.

a. $\int\dfrac{dx}{x^2+9} = \dfrac{1}{3}\tan^{-1}\dfrac{x}{3} + C$.

b. Separating variables: $\dfrac{dy}{1-y} = \dfrac{dx}{2x+1}$. Integrating: $-\ln|1-y| = \tfrac12\ln|2x+1| + C$, i.e. $(1-y)^2(2x+1) =$ constant (after rearranging).

Simplex: Optimal at the intersection of constraints. Solving $2x+3y=21$, $3x+2y=24$: multiply and subtract → $x=6, y=3$. $z = 15(6)+12(3) = 90+36 = 126$.

OR — b): At greatest height velocity $= u\cos\theta$. At half height, velocity $v$ satisfies $v^2 = u^2 - g h$, with $h = \tfrac12 H$. Setting up $u\cos\theta = \dfrac{\sqrt{10}}{5}v$ gives $\tan^2\theta = 3 \Rightarrow \theta = 60^\circ$.

a. By De Moivre, $z^n = \cos n\theta + i\sin n\theta$ and $\dfrac{1}{z^n} = \cos n\theta - i\sin n\theta$, so $z^n + \dfrac{1}{z^n} = 2\cos n\theta$.

b. 8 boys, 4 girls. At most 3 girls = total − (4 girls + 1 boy): total $\binom{12}{5} = 792$; with 4 girls and 1 boy $= \binom{4}{4}\binom{8}{1} = 8$. So at most 3 girls $= 792 - 8 = 784$.

c. (Same induction as Q15c): $(n+1)^3+2(n+1) = (n^3+2n) + 3(n^2+n+1)$, divisible by 3, so true for all $n$.

a. Tangents from external point $(13,0)$ to $x^2+y^2=25$ (radius 5): a line through $(13,0)$ is $y = m(x-13)$; tangency: $\dfrac{|{-13m}|}{\sqrt{m^2+1}} = 5 \Rightarrow 169m^2 = 25(m^2+1) \Rightarrow 144m^2 = 25 \Rightarrow m = \pm\tfrac{5}{12}$. Tangents: $y = \pm\tfrac{5}{12}(x-13)$.

b. Sides $a:b:c = 2:\sqrt6:(\sqrt3+1)$. Using the cosine rule on these gives the angles $45^\circ, 60^\circ, 75^\circ$.

c. $\vec{AB} = (\vec i - \vec j - 2\vec k)-(2\vec i - \vec j + 3\vec k) = -\vec i - 5\vec k$; $\vec{AC} = (\vec i+2\vec j+3\vec k)-(2\vec i-\vec j+3\vec k) = -\vec i + 3\vec j$. Area $= \tfrac12|\vec{AB}\times\vec{AC}|$. $\vec{AB}\times\vec{AC} = \begin{vmatrix}\vec i & \vec j & \vec k\\ -1 & 0 & -5\\ -1 & 3 & 0\end{vmatrix} = \vec i(0+15) - \vec j(0-5) + \vec k(-3-0) = 15\vec i + 5\vec j - 3\vec k$. Magnitude $= \sqrt{225+25+9} = \sqrt{259}$. Area $= \tfrac12\sqrt{259}$.

a. It is a linear differential equation. Dividing by $\sin x$: $\dfrac{dy}{dx} + y\cot x = x$. IF $= e^{\int\cot x\,dx} = \sin x$. So $\dfrac{d}{dx}(y\sin x) = x\sin x$; integrating, $y\sin x = \int x\sin x\,dx = -x\cos x + \sin x + C$.

b. Partial fractions: $\dfrac{x}{(x-1)(x^2+1)} = \dfrac{A}{x-1} + \dfrac{Bx+C}{x^2+1}$, with $A=\tfrac12$, then integrate to get $\tfrac12\ln|x-1| + (\text{terms in }\ln(x^2+1)\text{ and }\tan^{-1}x) + C$.

c. If $x$ (east) and $y$ (south), distance $s = \sqrt{x^2+y^2}$. $\dfrac{ds}{dt} = \dfrac{x\frac{dx}{dt} + y\frac{dy}{dt}}{\sqrt{x^2+y^2}}$. With speeds 60 and 80, $\dfrac{ds}{dt} = \sqrt{60^2+80^2} = 100\,\text{km/hr}$.