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

The word 'ALGEBRA' contains 7 letters where 'A' is repeated 2 times. The number of arrangements is $\frac{7!}{2!} = \frac{5040}{2} = 2520$.

The number of diagonals of an $n$-sided polygon is given by $\frac{n(n-3)}{2} = 44 \Rightarrow n^2 - 3n - 88 = 0 \Rightarrow (n-11)(n+8) = 0$. Since sides cannot be negative, $n = 11$.

The $(r+1)^{\text{th}}$ term from the end is equal to the $(r+1)^{\text{th}}$ term from the beginning when the terms are swapped: $\binom{12}{4} \left(-\frac{2}{x^2}\right)^{12-4} \left(\frac{x^3}{2}\right)^4 = 495 \cdot \frac{256}{x^{16}} \cdot \frac{x^{12}}{16} = 7920x^{-4}$.

Let $k = n-1$. When $n=2$, $k=1$. The series becomes $\sum_{k=1}^{\infty} \frac{1}{k!} = \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots = e - 1$.

By De Moivre's Theorem, $(\cos 60^{\circ} + i \sin 60^{\circ})^6 = \cos(6 \times 60^{\circ}) + i \sin(6 \times 60^{\circ}) = \cos 360^{\circ} + i \sin 360^{\circ} = 1 + i(0) = 1$.

Multiply the numerator by $w^3$ (since $w^3 = 1$): $\frac{w^3(a + bw + cw^2)}{aw^2 + bw + c} = \frac{w(aw^2 + bw^3 + cw^4)}{aw^2 + bw + c} = \frac{w(aw^2 + b + cw^2)}{aw^2 + bw + c}$ which simplifies based on cube root of unity properties.

A normal to a circle must pass through its center. The center of the circle $x^2 + y^2 = a^2$ is $(0,0)$. Substituting $(0,0)$ into the line gives $l(0) + m(0) + n = 0 \Rightarrow n = 0$.

The line is $y = -x + k$, so $m = -1$ and $c = k$. For a parabola $y^2 = 4ax$, $a=1$. Condition of tangency is $c = \frac{a}{m} \Rightarrow k = \frac{1}{-1} = -1$.

Using chain rule: $\frac{1}{x + \sqrt{4+x^2}} \cdot \left(1 + \frac{2x}{2\sqrt{4+x^2}}\right) = \frac{1}{x + \sqrt{4+x^2}} \cdot \frac{\sqrt{4+x^2} + x}{\sqrt{4+x^2}} = \frac{1}{\sqrt{4+x^2}}$.

$A = 4\pi r^2 \Rightarrow dA = 8\pi r \, dr$. Given $r = 2$ and $dr = 0.1$, $dA = 8\pi (2)(0.1) = 1.6\pi \text{ cm}^2$.

The equation of the tangent to $x^2 - y^2 = a^2$ at $(x_1, y_1)$ is $x x_1 - y y_1 = a^2$. Thus, $4x - 3y = 7$.

Total letters = 8 (I, N, T, E, R, V, A, L). Vowels = 3 (I, E, A), Consonants = 5 (N, T, R, V, L). a. Treating vowels as one unit, total units = 5 + 1 = 6. Arrangements = $6! \times 3! = 720 \times 6 = 4320$. b. There are 4 odd positions (1, 3, 5, 7). 3 vowels can occupy 4 odd slots in $^4P_3 = 24$ ways. Remaining 5 consonants can be arranged in $5! = 120$ ways. Total = $24 \times 120 = 2880$.

a. At least 2 ladies: (2L, 3G), (3L, 2G), (4L, 1G) $\Rightarrow \binom{4}{2}\binom{6}{3} + \binom{4}{3}\binom{6}{2} + \binom{4}{4}\binom{6}{1} = 6(20) + 4(15) + 1(6) = 126 + 60 = 186$. b. At most 2 ladies: (0L, 5G), (1L, 4G), (2L, 3G) $\Rightarrow \binom{4}{0}\binom{6}{5} + \binom{4}{1}\binom{6}{4} + \binom{4}{2}\binom{6}{3} = 1(6) + 4(15) + 6(20) = 6 + 60 + 120 = 186$.

a. Theorem statement: $(a+b)^n = \sum_{r=0}^n \binom{n}{r} a^{n-r}b^r$. Since $n=12$, middle term is $T_{6+1} = \binom{12}{6}(x^2)^6(\frac{1}{2})^6 = \frac{924}{64}x^{12} = \frac{231}{16}x^{12}$. b. Differentiating $(1+x)^n$ with respect to $x$ gives $n(1+x)^{n-1} = C_1 + 2C_2x + \dots + nC_nx^{n-1}$. Setting $x=1$ yields $n2^{n-1} = C_1 + 2C_2 + \dots + nC_n$.

a. Let $x = 1/3$. The series is $x + \frac{x^3}{3} + \frac{x^5}{5} + \dots = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right) = \frac{1}{2}\ln\left(\frac{4/3}{2/3}\right) = \frac{1}{2}\ln 2$. b. Given $y = \ln(1+x) \Rightarrow e^y = 1+x \Rightarrow x = e^y - 1 = y + \frac{y^2}{2!} + \frac{y^3}{3!} + \dots$

a. $\sqrt{3}-i = 2(\cos(-\pi/6) + i\sin(-\pi/6)) \Rightarrow (\sqrt{3}-i)^4 = 2^4(\cos(-2\pi/3) + i\sin(-2\pi/3)) = 16(-\frac{1}{2} - i\frac{\sqrt{3}}{2}) = -8 - 8\sqrt{3}i$. b. $1 = \cos(2k\pi) + i\sin(2k\pi) \Rightarrow 1^{1/3} = \cos\frac{2k\pi}{3} + i\sin\frac{2k\pi}{3}$ for $k=0,1,2$, giving $1, -\frac{1}{2}+i\frac{\sqrt{3}}{2}, -\frac{1}{2}-i\frac{\sqrt{3}}{2}$.

a. Parallel line form: $3x + 4y + c = 0$. Distance from center $(0,0)$ equals radius $2$: $\frac{|c|}{\sqrt{3^2+4^2}} = 2 \Rightarrow |c| = 10 \Rightarrow 3x + 4y \pm 10 = 0$. b. Distance from $(0,0)$ to line $lx+my-1=0$ is $\frac{|-1|}{\sqrt{l^2+m^2}} = a \Rightarrow l^2+m^2 = \frac{1}{a^2}$, representing a circle of radius $\frac{1}{a}$.

a. $a=3, b=1$. Center $= (-3,0)$. Vertices $= (-3 \pm 3, 0) = (0,0)$ and $(-6,0)$. $e = \sqrt{1 - 1/9} = \frac{\sqrt{8}}{3}$. Foci $= (-3 \pm \sqrt{8}, 0)$. Directrix: $x + 3 = \pm \frac{9}{\sqrt{8}}$. b. Vertical ellipse since foci are on y-axis: $be = 4 \Rightarrow b(2/3) = 4 \Rightarrow b = 6$. $a^2 = b^2(1-e^2) = 36(1 - 4/9) = 20$. Equation: $\frac{x^2}{20} + \frac{y^2}{36} = 1$.

a. $v = \frac{dS}{dt} = 3t^2 - 4t$. At $t=2$, $v = 3(4)-8 = 4\text{ m/s}$. $a = \frac{dv}{dt} = 6t - 4$. At $t=2$, $a = 6(2)-4 = 8\text{ m/s}^2$. b. $A = \pi r^2 \Rightarrow \frac{dA}{dt} = 2\pi r \frac{dr}{dt} = 2\pi (25)(0.20) = 10\pi \text{ cm}^2\text{/sec}$.

a. $(10-1)! = 9! = 362880$. b. Treat 2 boys as 1 unit. Total units = $6$. Circular arrangement = $(6-1)! = 5!$. The 2 boys can swap in $2!$ ways. Total = $120 \times 2 = 240$. c. Fix 4 boys first at the table in $(4-1)! = 3! = 6$ ways. This creates 4 spaces between them where 4 girls can sit in $4! = 24$ ways. Total = $6 \times 24 = 144$.

a. Tangent to $y^2=4ax$ is $y = mx + a/m$. Substituting in $x^2 = 4by$ yields $x^2 - 4bmx - 4ba/m = 0$. For tangency, Discriminant = $16b^2m^2 + 16ba/m = 0 \Rightarrow m = -(a/b)^{1/3}$. Common tangent equation: $a^{1/3}x + b^{1/3}y + a^{2/3}b^{2/3} = 0$. b. $y = mx + 2/m$. Passes through $(2,5) \Rightarrow 5 = 2m + 2/m \Rightarrow 2m^2 - 5m + 2 = 0 \Rightarrow m=2$ or $m=1/2$. Tangents: $y = 2x + 1$ and $y = \frac{1}{2}x + 4$. Points of contact are $(a/m^2, 2a/m) \Rightarrow (1/2, 2)$ and $(8, 8)$.

a. $\frac{1}{1 + \sinh^2 x} \cdot \cosh x = \frac{\cosh x}{\cosh^2 x} = \text{sech } x$. b. The limit expression standard form is $\frac{x - \frac{1}{2}\sin 2x}{x^3}$. Applying L'Hospital rule repeatedly leads to $\lim_{x \rightarrow 0} \frac{1 - \cos 2x}{3x^2} = \lim_{x \rightarrow 0} \frac{2\sin 2x}{6x} = \frac{4}{6} = \frac{2}{3}$. c. $V = \frac{4}{3}\pi r^3 \Rightarrow dV = 4\pi r^2 dr = 4\pi (3^2)(0.01) = 0.36\pi \text{ cm}^3$.