NEB Class 12 Science Mathematics Question Paper 2077 Nepal

Write the general term: $\dfrac{n}{(n+1)!}=\dfrac{(n+1)-1}{(n+1)!}=\dfrac{1}{n!}-\dfrac{1}{(n+1)!}$.

Summing from $n=1$ to $\infty$ gives a telescoping series:

Hence the sum is $1$.

The ZX-plane has equation $y=0$. Let the plane divide $PQ$ in ratio $k:1$. The $y$-coordinate of the dividing point is $\dfrac{k(-5)+1(4)}{k+1}=0$, so $-5k+4=0$, giving $k=\dfrac{4}{5}$.

Hence the ZX-plane divides $PQ$ in the ratio $4:5$ (internally).

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

$\vec{a}\cdot\vec{b}=(1)(1)+(2)(-1)+(-1)(1)=1-2-1=-2$.

$|\vec{a}|=\sqrt{1^2+2^2+(-1)^2}=\sqrt{6}$.

Projection $=\dfrac{-2}{\sqrt{6}}=-\dfrac{2}{\sqrt{6}}=-\dfrac{\sqrt{6}}{3}$.

Using $1+\cos 2y=2\cos^2 y$ and $1-\cos 2x=2\sin^2 x$:

Separating variables: $\dfrac{dy}{\cos^2 y}=-\dfrac{dx}{\sin^2 x}$, i.e. $\sec^2 y\,dy=-\csc^2 x\,dx$.

Integrating: $\tan y = \cot x + C$.

Mean $\bar{x}=\dfrac{3+5+9+11+7+6}{6}=\dfrac{41}{6}\approx 6.833$.

Absolute deviations from mean: $|3-6.833|=3.833$, $|5-6.833|=1.833$, $|9-6.833|=2.167$, $|11-6.833|=4.167$, $|7-6.833|=0.167$, $|6-6.833|=0.833$.

Sum $=13$. Mean deviation $=\dfrac{13}{6}\approx 2.167$.

Abelian group: A group $(G,*)$ in which the operation $*$ is commutative, i.e. $a*b=b*a$ for all $a,b\in G$.

Proof: In any group $(a*b)^{-1}=b^{-1}*a^{-1}$. Since $G$ is abelian, $b^{-1}*a^{-1}=a^{-1}*b^{-1}$. Hence $(a*b)^{-1}=a^{-1}*b^{-1}$.

Verification: $(a*b)*(a^{-1}*b^{-1})=a*(b*a^{-1})*b^{-1}=a*(a^{-1}*b)*b^{-1}=(a*a^{-1})*(b*b^{-1})=e*e=e$, confirming $a^{-1}*b^{-1}$ is the inverse of $a*b$.

Normal to parabola: The normal to $y^2=4mx$ at parameter $t$ is $y+tx=2mt+mt^3$. Comparing the given line $ax+by+c=0$ i.e. $y=-\dfrac{a}{b}x-\dfrac{c}{b}$ with the normal $y=-tx+(2mt+mt^3)$, we get $t=\dfrac{a}{b}$ and $-\dfrac{c}{b}=2mt+mt^3$. Eliminating $t$ gives the condition $am b^2 \cdot$ (substitute $t=a/b$): $-\dfrac{c}{b}=2m\dfrac{a}{b}+m\dfrac{a^3}{b^3}$, i.e. $-cb^2 = 2mab^2+ma^3$, so $ma^3+2mab^2+cb^2=0$, i.e. $\boxed{m a^{3}+2 m a b^{2}+c b^{2}=0}$.

Or — Ellipse: $\dfrac{(x+2)^2}{16}+\dfrac{(y-5)^2}{9}=1$ has centre $(-2,5)$, $a^2=16,\ b^2=9$ with major axis along $x$. So $a=4,\ b=3$, $c=\sqrt{a^2-b^2}=\sqrt{7}$.

Vertices: $(-2\pm 4,\,5)=(2,5)$ and $(-6,5)$.

Foci: $(-2\pm\sqrt{7},\,5)$.

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

Denominator $1+\sin x+\cos x=\dfrac{(1+t^2)+2t+(1-t^2)}{1+t^2}=\dfrac{2+2t}{1+t^2}=\dfrac{2(1+t)}{1+t^2}$.

Integral $=\displaystyle\int\dfrac{2\,dt/(1+t^2)}{2(1+t)/(1+t^2)}=\int\dfrac{dt}{1+t}=\ln|1+t|+C=\ln\left|1+\tan\dfrac{x}{2}\right|+C$.

From first principles: $\dfrac{d}{dx}e^{\tan x}=\lim_{h\to0}\dfrac{e^{\tan(x+h)}-e^{\tan x}}{h}=e^{\tan x}\lim_{h\to0}\dfrac{e^{\tan(x+h)-\tan x}-1}{h}$. Since $\tan(x+h)-\tan x\to 0$, using $\dfrac{e^{u}-1}{u}\to1$ and $\lim_{h\to0}\dfrac{\tan(x+h)-\tan x}{h}=\sec^2 x$, we get $\dfrac{d}{dx}e^{\tan x}=e^{\tan x}\sec^2 x$.

Or — MVT: If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then $\exists\,c\in(a,b)$ with $f'(c)=\dfrac{f(b)-f(a)}{b-a}$.

For $f(x)=2x^2-10x+29$ on $[2,7]$: $f(2)=8-20+29=17$, $f(7)=98-70+29=57$. $\dfrac{f(7)-f(2)}{7-2}=\dfrac{40}{5}=8$. $f'(x)=4x-10=8\Rightarrow x=4.5\in(2,7)$. MVT verified with $c=4.5$.

At maximum height the final velocity $v=0$. Using $v=u-gt$: $0=40-10t$, so $t=4\,\text{s}$.

Part 1 (inclined plane): Length of plane $L=\dfrac{h}{\sin 30^\circ}=\dfrac{44.1}{0.5}=88.2\,\text{m}$. Acceleration along plane $a=g\sin30^\circ=4.9\,\text{ms}^{-2}$. Total time $T$: $L=\tfrac12 aT^2\Rightarrow 88.2=\tfrac12(4.9)T^2\Rightarrow T^2=36\Rightarrow T=6\,\text{s}$. Equal time intervals are at $t=2,4,6$ s. Distances from top: $s\propto t^2$, so at $t=2,4,6$: $s=\tfrac12(4.9)(4)=9.8$, $\tfrac12(4.9)(16)=39.2$, $88.2$ m. The three parts (in order from top) are $9.8\,\text{m},\ 39.2-9.8=29.4\,\text{m},\ 88.2-39.2=49.0\,\text{m}$, i.e. ratio $1:3:5$.

Or — depth of well: Let depth $=d$, fall time $t_1$ with $d=\tfrac12 g t_1^2$, sound time $t_2=d/352.8$, and $t_1+t_2=4\tfrac{2}{9}=\dfrac{38}{9}\,\text{s}$. Solving gives $d\approx 78.4\,\text{m}$ (fall time $\approx4\,\text{s}$, sound time $\approx0.222\,\text{s}$).

Resultant of two like parallel forces: Two like parallel forces $P$ and $Q$ have a resultant $R=P+Q$ acting parallel to them, in the same direction, whose line of action divides the line joining their points of application internally in the ratio inverse to the forces: if the resultant acts at a point dividing $AB$ such that $P\cdot AC=Q\cdot CB$, then $\dfrac{AC}{CB}=\dfrac{Q}{P}$. (For two unlike parallel forces, $R=P-Q$ acting nearer the larger force, dividing externally.)

Or — Varignon's theorem: Geometrically, the moment of a force about a point equals twice the area of the triangle whose base is the force vector (its line segment of representation) and whose apex is the point. Varignon's theorem: the algebraic sum of the moments of two intersecting forces about any point in their plane equals the moment of their resultant about the same point. Proof uses the geometric (twice-triangle-area) interpretation: representing each force and the resultant from the common point, the areas of the triangles add, establishing $\text{moment}(R)=\text{moment}(P)+\text{moment}(Q)$.

A system is diagonally dominant if in each equation $|a_{ii}|\ge\sum_{j\ne i}|a_{ij}|$. As written:

Eq.1: $|3|<|12|+|-1|=13$ — fails.

However, rearranging equations to bring the largest coefficient onto the diagonal: take Eq.3 ($10x+4y-2z=20$) for $x$: $|10|\ge|4|+|-2|=6$ ✓; Eq.1 ($3x+12y-z=28$) for $y$: $|12|\ge|3|+|-1|=4$ ✓; Eq.2 ($x+4y+7z=2$) for $z$: $|7|\ge|1|+|4|=5$ ✓.

So after suitable reordering the system is diagonally dominant; in the order originally given it is not.

Let $f(x)=x^3-7x^2+14x-6$. $f(0)=-6<0$, $f(1)=1-7+14-6=2>0$, so a root lies in $(0,1)$.

Applying bisection repeatedly (midpoint, check sign of $f$):

• $f(0.5)=0.125-1.75+7-6=-0.625<0$ → root in $(0.5,1)$
• $f(0.75)\approx 0.422-3.94+10.5-6\approx 0.984>0$ → $(0.5,0.75)$
• $f(0.625)\approx 0.244-2.73+8.75-6\approx 0.260>0$ → $(0.5,0.625)$
• $f(0.5625)\approx -0.20$ → $(0.5625,0.625)$
• continuing until interval width $<10^{-2}$ gives root $\approx 0.586$.

Root $\approx 0.585$–$0.586$.

Corner points of the feasible region (intersection of $2x_1+x_2=10$ and $x_1+3x_2=10$): solving gives $x_1=4,\ x_2=2$. Vertices: $(0,0)$, $(5,0)$, $(4,2)$, $(0,10/3)$.

Evaluate $F=15x_1+10x_2$: at $(0,0)=0$; at $(5,0)=75$; at $(4,2)=60+20=80$; at $(0,10/3)=33.3$.

Maximum $F=80$ at $x_1=4,\ x_2=2$ (confirmed by simplex iterations).