NEB Class 12 Mathematics Question Paper 2082 (Set R) Nepal

We know that $P(n, r) = C(n, r) \\times r!$, which rearranges to $C(n, r) \\times r! = P(n, r)$. So option B is correct.

By Cramer's rule, y = \\dfrac{\\begin{vmatrix} p_1 & r_1 \\\\ p_2 & r_2 \\end{vmatrix}}{\\begin{vmatrix} p_1 & q_1 \\\\ p_2 & q_2 \\end{vmatrix}}. The numerator replaces the $y$-column ($q$) with the constants ($r$), keeping the $p$-column. This matches option A.

In a triangle, the half-angle formula gives $\\cos\\dfrac{B}{2} = \\sqrt{\\dfrac{s(s-b)}{ca}}$, so $\\sec\\dfrac{B}{2} = \\dfrac{1}{\\cos\\frac{B}{2}} = \\sqrt{\\dfrac{ca}{s(s-b)}}$. This matches option D.

The line $y = mx + c$ is tangent to $x^2 + y^2 = a^2$ when the perpendicular distance from the centre equals the radius: $\\dfrac{|c|}{\\sqrt{1+m^2}} = a$, i.e. $c = \\pm a\\sqrt{1+m^2}$. This matches option C.

\\vec{a} \\times \\vec{b} = \\begin{vmatrix} \\hat{i} & \\hat{j} & \\hat{k} \\\\ 0 & 0 & 1 \\\\ 0 & 1 & 0 \\end{vmatrix} = \\hat{i}(0\\cdot0 - 1\\cdot1) - \\hat{j}(0\\cdot0 - 1\\cdot0) + \\hat{k}(0\\cdot1 - 0\\cdot0) = (-1, 0, 0). This matches option A.

$P(B/A) = \\dfrac{P(A \\cap B)}{P(A)} = \\dfrac{1/5}{1/3} = \\dfrac{3}{5}$. This matches option A.

$\\dfrac{dy}{dx} = 6x - 1$. At $x = -1$, slope of tangent $= 6(-1) - 1 = -7$. Slope of normal $= -\\dfrac{1}{-7} = \\dfrac{1}{7}$. This matches option C.

$\\int \\dfrac{dx}{9x^2 + 1} = \\int \\dfrac{dx}{(3x)^2 + 1}$. Let $u = 3x$, $du = 3\\,dx$, so $= \\dfrac{1}{3}\\int \\dfrac{du}{u^2 + 1} = \\dfrac{1}{3}\\tan^{-1}(3x) + C$. This matches option A.

A differential equation $\\dfrac{dy}{dx} = f(x, y)$ is homogeneous if $f(\\lambda x, \\lambda y) = f(x, y)$, i.e. numerator and denominator are of the same degree. Option C has $\\dfrac{x^2 + y^2}{x^2 - y^2}$ (both degree 2), so it is homogeneous. Options A and B have numerator degree 2 over denominator degree 1.

Rewrite as $\\dfrac{dy}{dx} + \\dfrac{1}{\\cos^2 x}y = \\dfrac{1}{\\cos^2 x}$, i.e. $\\dfrac{dy}{dx} + (\\sec^2 x)y = \\sec^2 x$. Here $P = \\sec^2 x$, so I.F. $= e^{\\int \\sec^2 x\\,dx} = e^{\\tan x}$. This matches option B.

First alternative: From $x - 2y = 1$, $x = 1 + 2y$. Substitute into $3x + 4y = 13$: $3(1 + 2y) + 4y = 13 \\Rightarrow 3 + 10y = 13 \\Rightarrow 10y = 10$. So option A ($10y = 10$).

Second alternative (projectile): Maximum height H = \\dfrac{u^2\\sin^2\\theta}{2g} = \\dfrac{40^2 \\times \\sin^2 30^\\circ}{2 \\times 10} = \\dfrac{1600 \\times \\frac{1}{4}}{20} = \\dfrac{400}{20} = 20\\,\\text{m}. So option A ($20\\,\\text{m}$).

In both alternatives the correct option is A.

$\\log_e(1-x) = -\\left(x + \\dfrac{x^2}{2} + \\dfrac{x^3}{3} + \\dfrac{x^4}{4} + \\cdots\\right) = -\\sum_{n=1}^{\\infty} \\dfrac{x^n}{n}$, valid for $|x| < 1$.

The total number of permutations (arrangements) of a set having $n$ distinct elements is $n!$ (i.e. $P(n, n) = n!$).

De-Moivre's theorem states that for any real number $\\theta$ and integer $n$, $(\\cos\\theta + i\\sin\\theta)^n = \\cos n\\theta + i\\sin n\\theta$.

$\\sum_{k=1}^{n} k^3 = 1^3 + 2^3 + \\cdots + n^3 = \\left[\\dfrac{n(n+1)}{2}\\right]^2 = \\dfrac{n^2(n+1)^2}{4}$.

Rewrite as $3x + 2y = 1$ and $4x + y = 3$. The augmented matrix is \\left[\\begin{array}{cc|c} 3 & 2 & 1 \\\\ 4 & 1 & 3 \\end{array}\\right].

Number of committees $= C(a, 2) \\times C(6, 3) = 120$. Now $C(6, 3) = 20$, so $C(a, 2) = \\dfrac{120}{20} = 6$. Thus $\\dfrac{a(a-1)}{2} = 6 \\Rightarrow a(a-1) = 12 \\Rightarrow a^2 - a - 12 = 0 \\Rightarrow (a-4)(a+3) = 0$. Taking the positive root, $a = 4$. So there are $4$ boys.

The complex number is the square of either root. Take $z = (\\sqrt{3} + i)^2 = 3 + 2\\sqrt{3}i + i^2 = 3 - 1 + 2\\sqrt{3}i = 2 + 2\\sqrt{3}i$.

Modulus: $r = \\sqrt{2^2 + (2\\sqrt{3})^2} = \\sqrt{4 + 12} = \\sqrt{16} = 4$.

Argument: $\\tan\\theta = \\dfrac{2\\sqrt{3}}{2} = \\sqrt{3} \\Rightarrow \\theta = \\dfrac{\\pi}{3}$ (first quadrant).

So the polar form is $z = 4\\left(\\cos\\dfrac{\\pi}{3} + i\\sin\\dfrac{\\pi}{3}\\right)$.

Using $\\sin^2\\dfrac{R}{2} = \\dfrac{1 - \\cos R}{2}$ and $\\sin^2\\dfrac{P}{2} = \\dfrac{1 - \\cos P}{2}$:

$p\\cdot\\dfrac{1-\\cos R}{2} + r\\cdot\\dfrac{1-\\cos P}{2} = \\dfrac{q}{2}$

$\\Rightarrow p(1 - \\cos R) + r(1 - \\cos P) = q$

$\\Rightarrow (p + r) - (p\\cos R + r\\cos P) = q$.

By the projection formula, $p\\cos R + r\\cos P = q$. Therefore:

$(p + r) - q = q \\Rightarrow p + r = 2q$.

Hence $p, q, r$ are in A.P. (since $q$ is the arithmetic mean of $p$ and $r$).

Projection of $\\vec{b}$ on $\\vec{a} = \\dfrac{\\vec{a}\\cdot\\vec{b}}{|\\vec{a}|}$.

$\\vec{a}\\cdot\\vec{b} = (4)(3) + (-3)(-2) + (2)(4) = 12 + 6 + 8 = 26$.

$|\\vec{a}| = \\sqrt{4^2 + (-3)^2 + 2^2} = \\sqrt{16 + 9 + 4} = \\sqrt{29}$.

Projection $= \\dfrac{26}{\\sqrt{29}}$.

$3x^2 - 6x - 4y^2 = 0 \\Rightarrow 3(x^2 - 2x) - 4y^2 = 0 \\Rightarrow 3(x-1)^2 - 3 - 4y^2 = 0 \\Rightarrow 3(x-1)^2 - 4y^2 = 3$.

Divide by 3: $\\dfrac{(x-1)^2}{1} - \\dfrac{y^2}{3/4} = 1$. This is a hyperbola with $a^2 = 1$, $b^2 = \\dfrac{3}{4}$.

Eccentricity: $e = \\sqrt{1 + \\dfrac{b^2}{a^2}} = \\sqrt{1 + \\dfrac{3/4}{1}} = \\sqrt{\\dfrac{7}{4}} = \\dfrac{\\sqrt{7}}{2}$.

Let the ellipse be $\\dfrac{x^2}{a^2} + \\dfrac{y^2}{b^2} = 1$ with major axis $2a$ and minor axis $2b$. Given major axis $= 4 \\times$ minor axis: $2a = 4(2b) \\Rightarrow a = 4b$.

Passes through $(4, 2)$: $\\dfrac{16}{a^2} + \\dfrac{4}{b^2} = 1$. With $a^2 = 16b^2$: $\\dfrac{16}{16b^2} + \\dfrac{4}{b^2} = 1 \\Rightarrow \\dfrac{1}{b^2} + \\dfrac{4}{b^2} = 1 \\Rightarrow \\dfrac{5}{b^2} = 1 \\Rightarrow b^2 = 5$. Then $a^2 = 80$.

Eccentricity: $e = \\sqrt{1 - \\dfrac{b^2}{a^2}} = \\sqrt{1 - \\dfrac{5}{80}} = \\sqrt{1 - \\dfrac{1}{16}} = \\sqrt{\\dfrac{15}{16}} = \\dfrac{\\sqrt{15}}{4}$.

With $n = 6$: $\\sum X = 348$, $\\bar{X} = 58$; $\\sum Y = 480$, $\\bar{Y} = 80$.

Deviations $x = X - 58$: $-13, -8, -2, 4, 7, 12$; $y = Y - 80$: $-15, -10, -5, 0, 10, 20$.

$\\sum xy = 195 + 80 + 10 + 0 + 70 + 240 = 595$.

$\\sum x^2 = 169 + 64 + 4 + 16 + 49 + 144 = 446$.

$\\sum y^2 = 225 + 100 + 25 + 0 + 100 + 400 = 850$.

$r = \\dfrac{\\sum xy}{\\sqrt{\\sum x^2 \\cdot \\sum y^2}} = \\dfrac{595}{\\sqrt{446 \\times 850}} = \\dfrac{595}{\\sqrt{379100}} \\approx \\dfrac{595}{615.7} \\approx 0.97$.

So there is a strong positive correlation, $r \\approx 0.97$.

We need the regression of $X$ on $Y$. Using the deviations from Q16(a): $\\sum xy = 595$, $\\sum y^2 = 850$, $\\bar{X} = 58$, $\\bar{Y} = 80$.

Regression coefficient $b_{XY} = \\dfrac{\\sum xy}{\\sum y^2} = \\dfrac{595}{850} = 0.7$.

Regression line: $X - \\bar{X} = b_{XY}(Y - \\bar{Y}) \\Rightarrow X - 58 = 0.7(Y - 80)$.

At $Y = 150$: $X = 58 + 0.7(150 - 80) = 58 + 0.7 \\times 70 = 58 + 49 = 107$.

So the supply is $107$ units when the price is Rs. 150.

$\\dfrac{d}{dx}\\text{cosech}^{-1}(x) = -\\dfrac{1}{|x|\\sqrt{x^2 + 1}}$.

L'Hospital's rule states that if $\\lim_{x\\to a} f(x) = \\lim_{x\\to a} g(x) = 0$ (or both $\\to \\infty$), giving an indeterminate form $\\dfrac{0}{0}$ or $\\dfrac{\\infty}{\\infty}$, then $\\lim_{x\\to a}\\dfrac{f(x)}{g(x)} = \\lim_{x\\to a}\\dfrac{f'(x)}{g'(x)}$, provided the latter limit exists.

A tangent parallel to the y-axis is vertical, so its slope is undefined. The condition is $\\dfrac{dy}{dx} = \\infty$, equivalently $\\dfrac{dx}{dy} = 0$.

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

The standard form of a first order linear differential equation is $\\dfrac{dy}{dx} + Py = Q$, where $P$ and $Q$ are functions of $x$ (or constants).

Let $y = \\coth^{-1}(\\sin 2x)$. We know $\\dfrac{d}{du}\\coth^{-1}(u) = \\dfrac{1}{1 - u^2}$ (for $|u| > 1$; here it gives $-\\dfrac{1}{u^2-1}$). Using chain rule with $u = \\sin 2x$:

$\\dfrac{dy}{dx} = \\dfrac{1}{1 - \\sin^2 2x}\\cdot\\dfrac{d}{dx}(\\sin 2x) = \\dfrac{2\\cos 2x}{1 - \\sin^2 2x} = \\dfrac{2\\cos 2x}{\\cos^2 2x} = \\dfrac{2}{\\cos 2x} = 2\\sec 2x$.

Factor: $x^3 - x^2 - 2x = x(x^2 - x - 2) = x(x-2)(x+1)$.

Partial fractions: $\\dfrac{1}{x(x-2)(x+1)} = \\dfrac{A}{x} + \\dfrac{B}{x-2} + \\dfrac{C}{x+1}$.

$1 = A(x-2)(x+1) + Bx(x+1) + Cx(x-2)$.

• $x = 0$: $1 = A(-2)(1) = -2A \\Rightarrow A = -\\dfrac{1}{2}$.
• $x = 2$: $1 = B(2)(3) = 6B \\Rightarrow B = \\dfrac{1}{6}$.
• $x = -1$: $1 = C(-1)(-3) = 3C \\Rightarrow C = \\dfrac{1}{3}$.

So $\\int = -\\dfrac{1}{2}\\ln|x| + \\dfrac{1}{6}\\ln|x-2| + \\dfrac{1}{3}\\ln|x+1| + C$.

First alternative (simplex): Maximize $P = 10x + 3y$ subject to $6x + 7y \\le 42$, $x + 3y \\le 9$ (this is the binding one of the two $x+3y$ constraints), $x, y \\ge 0$.

Vertices of the feasible region:

• $(0,0)$: $P = 0$.
• $(7, 0)$ from $6x = 42$: check $x + 3y = 7 \\le 9$ ✓; $P = 70$.
• $(0, 3)$ from $x + 3y = 9$: $P = 9$.
• Intersection of $6x + 7y = 42$ and $x + 3y = 9$: from the second, $x = 9 - 3y$; sub: $6(9-3y) + 7y = 42 \\Rightarrow 54 - 18y + 7y = 42 \\Rightarrow -11y = -12 \\Rightarrow y = \\dfrac{12}{11}$, $x = 9 - \\dfrac{36}{11} = \\dfrac{63}{11}$; $P = \\dfrac{630}{11} + \\dfrac{36}{11} = \\dfrac{666}{11} \\approx 60.5$.

Maximum $P = 70$ at $(x, y) = (7, 0)$.

Second alternative (statics): On the inclined plane (angle $\\theta$), to support weight $R$:

• Force $A$ parallel to the plane (length) balances the component $R\\sin\\theta$: $A = R\\sin\\theta$, so $\\sin\\theta = \\dfrac{A}{R}$.
• Force $B$ acting horizontally (parallel to base): resolving along the plane, $B\\cos\\theta = R\\sin\\theta$, giving $\\tan\\theta = \\dfrac{R\\sin\\theta}{?}$. Using $B$ that singly supports $R$: $B = R\\tan\\theta$, so $\\tan\\theta = \\dfrac{B}{R}$...

From $\\sin\\theta = \\dfrac{A}{R}$ and $\\tan\\theta = \\dfrac{A}{B}$ (since the horizontal force $B$ satisfies $B = A/\\tan...$), use $\\dfrac{1}{\\sin^2\\theta} = 1 + \\dfrac{1}{\\tan^2\\theta}$. This yields $\\dfrac{R^2}{A^2} = 1 + \\dfrac{B^2}{...}$ leading to $\\dfrac{1}{A^2} = \\dfrac{1}{B^2} + \\dfrac{1}{R^2}$.