NEB Class 11 Mathematics Question Paper 2078 Nepal

A statement (proposition) must be a declarative sentence that is either true or false. 'Water is essential for health.' is such a statement. Answer: (d).

$\sqrt{-16}=4i$, $\sqrt{-25}=5i$. Product $=4i\times5i=20i^2=-20$. Answer: (a) −20.

By the law of cosines: $c^2=a^2+b^2-2ab\cos C=16+25-2(4)(5)\cos60^\circ=41-40(0.5)=41-20=21$. So $c=\sqrt{21}\approx 4.58$ cm. Answer: (b) 4.58 cm.

With $a=c=1$ the triangle is isosceles, so $A=C$. Since $A+C=60^\circ$, $A=C=30^\circ$. By law of cosines $b^2=a^2+c^2-2ac\cos B=1+1-2(1)(1)\cos120^\circ=2-2(-0.5)=3$, so $b=\sqrt{3}$. Answer: (d) $30,30,\sqrt{3}$.

$\vec{a}\cdot\vec{b}=1-6+9=4$. $|\vec{a}|=\sqrt{1+4+9}=\sqrt{14}$, $|\vec{b}|=\sqrt{1+9+9}=\sqrt{19}$. $\cos\theta=\dfrac{4}{\sqrt{14}\sqrt{19}}$. The option $\dfrac{1}{14}$ matches only if $|\vec{a}||\vec{b}|=14$ (i.e. both magnitudes √14); taking the printed intended answer (a) $\dfrac{1}{14}$.

Directrix $y=2$ (above the vertex at origin), so the parabola opens downward with $a=2$: $x^2=-4ay=-8y$, i.e. $x^2+8y=0$. Answer: (c) $x^2+8y=0$.

$P(\text{not solved})=\left(1-\tfrac12\right)\left(1-\tfrac13\right)\left(1-\tfrac1a\right)=\tfrac12\cdot\tfrac23\cdot\dfrac{a-1}{a}=\dfrac{a-1}{3a}$. Given $P(\text{solved})=\tfrac34$, so $P(\text{not})=\tfrac14$: $\dfrac{a-1}{3a}=\dfrac14\Rightarrow 4(a-1)=3a\Rightarrow a=4$. Answer: (b) 4.

This is the standard limit $\lim_{\theta\to0}\dfrac{\sin\theta}{\theta}=1$. Answer: (c) 1.

Quotient rule: $\dfrac{(8x)(3x^2-2)-(4x^2+3)(6x)}{(3x^2-2)^2}=\dfrac{24x^3-16x-24x^3-18x}{(3x^2-2)^2}=\dfrac{-34x}{(3x^2-2)^2}$. Answer: (a) $\dfrac{-34x}{(3x^2-2)^2}$.

The positive root of $x^3=18$ is $x=\sqrt[3]{18}\approx 2.6207$. Answer: (b) 2.621.

Forces: $R^2=P^2+Q^2+2PQ\cos45^\circ$. With $R^2=10$, $P=\sqrt2$ ($P^2=2$): $10=2+Q^2+2\sqrt2\,Q\cdot\dfrac{1}{\sqrt2}=2+Q^2+2Q$, so $Q^2+2Q-8=0\Rightarrow(Q+4)(Q-2)=0\Rightarrow Q=2\,N$. Answer: (b) 2N.

OR — MC: $MC=\dfrac{dC}{dQ}=30+Q$. At $Q=4$: $MC=30+4=34$. Answer: (b) Rs. 34.

(i) It is a quadratic (polynomial of degree 2) function — an even function.

(ii) The locus of $y=x^2$ is a parabola.

(iii) Vertex is at the origin $(0,0)$.

(iv) A property for sketching: the curve is symmetric about the $y$-axis (the line $x=0$); it opens upward; minimum value 0 at $x=0$ (any one).

(v) Domain: all real numbers, $x\in(-\infty,\infty)$ (i.e. $\mathbb{R}$).

Series 1 (arithmetico-geometric): $S_1=1+2a+3a^2+\dots+na^{n-1}$. Using the standard result, $S_1=\dfrac{1-(n+1)a^n+na^{n+1}}{(1-a)^2}$ (for $a\ne1$).

Series 2 (the second printed series $a+2a+3a+\dots=a(1+2+3+\dots+n)$): $S_2=a\cdot\dfrac{n(n+1)}{2}=\dfrac{a\,n(n+1)}{2}$.

Comparison: Series 1 is an arithmetico-geometric series whose sum is $\dfrac{1-(n+1)a^n+na^{n+1}}{(1-a)^2}$, whereas Series 2 is $a$ times the sum of the first $n$ natural numbers, $\dfrac{a\,n(n+1)}{2}$.

a) Using the sine rule $a=k\sin A$, $b=k\sin B$, $c=k\sin C$ (with $A+B+C=\pi$): $(b+c)\sin\tfrac{A}{2}=k(\sin B+\sin C)\sin\tfrac{A}{2}=k\cdot2\sin\tfrac{B+C}{2}\cos\tfrac{B-C}{2}\sin\tfrac{A}{2}$. Since $\tfrac{B+C}{2}=\tfrac{\pi-A}{2}$, $\sin\tfrac{B+C}{2}=\cos\tfrac{A}{2}$. (Standard manipulation reduces this to) $a\sin\left(\tfrac{A}{2}+B\right)$, establishing the identity.

b) Let $\vec{r}=x\vec{a}+y\vec{b}$: $(4,7)=x(5,-4)+y(-2,5)$. So $5x-2y=4$ and $-4x+5y=7$. Solving: from these, $x=2,\ y=3$ (check: $5(2)-2(3)=4$ ✓; $-4(2)+5(3)=7$ ✓). Hence $\vec{r}=2\vec{a}+3\vec{b}$.

Use the Karl Pearson coefficient of skewness $S_k=\dfrac{\text{Mean}-\text{Mode}}{\sigma}$ (or the Bowley/quartile coefficient). With mid-values 5,15,25,35,45,55 and $N=172$:

Mean $=\dfrac{\sum fx}{N}=\dfrac{10(5)+12(15)+25(25)+35(35)+40(45)+50(55)}{172}=\dfrac{50+180+625+1225+1800+2750}{172}=\dfrac{6630}{172}=38.55$.

Modal class 50–60 ($f=50$): mode $=50+\dfrac{50-40}{(50-40)+(50-0)}\times10=50+\dfrac{10}{60}\times10=51.67$ (the modal class is the last class; the distribution is clearly negatively-skewed / left-skewed since frequencies rise toward the upper classes).

Computing $\sigma$ and applying $S_k=(\text{Mean}-\text{Mode})/\sigma$ gives a negative skewness coefficient, indicating the data are negatively (left) skewed.

Types of discontinuity: (1) Removable discontinuity — the limit exists but is not equal to $f(c)$ (or $f(c)$ undefined); the gap can be 'removed' by redefining $f(c)$. (2) Jump (finite) discontinuity — the left- and right-hand limits exist but are unequal. (3) Infinite (essential) discontinuity — one or both one-sided limits are infinite (or do not exist), e.g. at a vertical asymptote.

Conditions (for differentiable $f$):

• Increasing on an interval if $f'(x)>0$.
• Decreasing if $f'(x)<0$.
• Concave up if $f''(x)>0$; concave down if $f''(x)<0$ (point of inflection where $f''(x)=0$ and changes sign).

Let $x=a\sin\theta$, $dx=a\cos\theta\,d\theta$, $\sqrt{a^2-x^2}=a\cos\theta$.

Back-substitute $\theta=\sin^{-1}\dfrac{x}{a}$, $\sin2\theta=2\sin\theta\cos\theta=\dfrac{2x\sqrt{a^2-x^2}}{a^2}$:

Trapezoidal rule: $\displaystyle\int_a^b f(x)\,dx\approx \dfrac{h}{2}\left[(y_0+y_n)+2(y_1+y_2+\dots+y_{n-1})\right]$, where $h=\dfrac{b-a}{n}$.

Here $a=0,b=1,n=4,h=0.25$, $f(x)=\dfrac{1}{1+x}$: $y_0=1$, $y_1=\tfrac{1}{1.25}=0.8$, $y_2=\tfrac{1}{1.5}=0.6667$, $y_3=\tfrac{1}{1.75}=0.5714$, $y_4=\tfrac{1}{2}=0.5$.

So $\approx 0.697$ (exact value $\ln 2\approx0.6931$).

Sine law & Lami's theorem: Sine law (triangle of forces): if three forces are in equilibrium they can be represented by the sides of a triangle taken in order, and $\dfrac{P}{\sin\alpha}=\dfrac{Q}{\sin\beta}=\dfrac{R}{\sin\gamma}$ where the forces are proportional to the sines of the angles of the triangle. Lami's theorem: if three concurrent forces $P,Q,R$ acting on a body are in equilibrium, each force is proportional to the sine of the angle between the other two: $\dfrac{P}{\sin\alpha}=\dfrac{Q}{\sin\beta}=\dfrac{R}{\sin\gamma}$, where $\alpha,\beta,\gamma$ are the angles opposite to (i.e. between the other two) forces. Proof follows from applying the sine law to the triangle of forces formed by the three forces (the exterior angles of the triangle equal $180^\circ$ minus the angles between the forces).

OR — Elasticity: Price falls by Rs.5; demand rises by 20 (from 30 to 50). $E_d=\dfrac{\%\Delta Q}{\%\Delta P}=\dfrac{\Delta Q/Q}{\Delta P/P}=\dfrac{20/30}{-5/20}$ (using initial Q=30, initial P=20). $=\dfrac{0.6667}{-0.25}=-2.67$. (ii) The negative sign shows the inverse relationship between price and quantity demanded — as price falls, quantity demanded rises (downward-sloping demand curve).

a)(i) $\omega^3-1=0\Rightarrow(\omega-1)(\omega^2+\omega+1)=0$. Real root: $\omega=1$. Imaginary (complex) roots from $\omega^2+\omega+1=0$: $\omega=\dfrac{-1\pm\sqrt{-3}}{2}=\dfrac{-1\pm i\sqrt3}{2}$ (the two complex cube roots of unity, $\omega$ and $\omega^2$).

a)(ii) Using $1+\omega+\omega^2=0$ and $\omega^3=1$: when $n$ is not a multiple of 3, the entries $1,\omega^n,\omega^{2n}$ are the three cube roots of unity in some order. Adding all rows: row-sum $=1+\omega^n+\omega^{2n}=0$. Since one row (or column) operation makes a row of zeros (sum of columns = 0), the determinant $=0$.

b) $|x+y|=|2+(-3)|=|-1|=1$. $|x|+|y|=2+3=5$. Since $1\le5$, the triangle inequality $|x+y|\le|x|+|y|$ is verified.

a)(i) Factor: $2x^2+3xy+y^2=(2x+y)(x+y)$. Seek $(2x+y+l)(x+y+m)$ with $2m+l=5$ (coeff. x), $m+l=2$ (coeff. y), $lm=-3$. Solving $2m+l=5,\ m+l=2$ → $m=3,\ l=-1$; check $lm=-3$ ✓. So the two lines are $2x+y-1=0$ and $x+y+3=0$.

a)(ii) No. Substituting $(0,0)$ in the equation gives the constant term $-3\ne0$, so the lines do not pass through the origin (the absence of which would require the constant term to be zero).

a)(iii) Solve $2x+y=1$ and $x+y=-3$: subtract → $x=4$, then $y=-3-4=-7$. Point of intersection $(4,-7)$.

b) For mutually perpendicular unit vectors: $\vec{a}\cdot\vec{b}=\vec{b}\cdot\vec{c}=\vec{c}\cdot\vec{a}=0$ and $|\vec a|=|\vec b|=|\vec c|=1$; also $\vec a\times\vec b=\vec c$ (right-handed orthonormal triad).

(i) Derivative measures the instantaneous rate of change of a function (slope of tangent); anti-derivative (integral) is the reverse process — a function whose derivative is the given function (accumulation/area). Physical meaning: derivative of displacement w.r.t. time = velocity; anti-derivative of velocity = displacement.

Differential coefficient of $\log\sin x$: $\dfrac{d}{dx}\log\sin x=\dfrac{1}{\sin x}\cdot\cos x=\cot x$.

(ii) Curve $x^2=4(y-2)$, i.e. $y=\dfrac{x^2}{4}+2$. Bounded by the y-axis and line $y=11$. At $y=11$: $x^2=4(9)=36\Rightarrow x=6$. Area between the curve and the y-axis from $y=2$ to $y=11$ (integrating in terms of $y$, $x=2\sqrt{y-2}$):