NEB Class 12 Science Mathematics Question Paper 2082 Nepal

Using De Moivre / Euler form, $\frac{z_1}{z_2} = \frac{\cos\theta_1 + i\sin\theta_1}{\cos\theta_2 + i\sin\theta_2} = \cos(\theta_1 - \theta_2) + i\sin(\theta_1 - \theta_2)$. So option B is correct.

Discriminant $D = b^2 - 4ac = (-1)^2 - 4(1)(1) = 1 - 4 = -3 < 0$. Since $D < 0$, the roots are imaginary (complex). So option D is correct.

The maximum value of $\sin x + \cos x$ is $\sqrt{2} \approx 1.414 < 2$. Hence the equation can never be satisfied, so it has no solution. Option B is correct.

Let $\cos^{-1}\frac{1}{2} = \theta$, so $\theta = \frac{\pi}{3}$. Then $\sin(2\theta) = \sin\frac{2\pi}{3} = \frac{\sqrt{3}}{2}$. Option A is correct.

$\vec{a} \times \vec{b} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{vmatrix} = \vec{i}(2\cdot2 - 0\cdot0) - \vec{j}(0\cdot2 - 0\cdot0) + \vec{k}(0\cdot0 - 2\cdot0) = (4, 0, 0)$. Option A is correct.

The equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = -1$ (i.e. $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$) is a hyperbola with transverse axis along the $y$-axis. Its foci are at $(0, \pm be)$, where $e$ is the eccentricity. Option B is correct.

For a binomial distribution, mean $= np = 10 \times \frac{1}{2} = 5$. Option B is correct.

$f'(x) = 4x$, so $f'(1) = 4$. Option C is correct.

The order is the order of the highest derivative present, which is $\frac{d^2y}{dx^2}$ (second derivative). So the order is $2$. Option B is correct.

The last equation $0 \cdot x_2 = -5$ is impossible (it states $0 = -5$), which is a contradiction. Hence the system is inconsistent and has no solution. Option C is correct.

For the statics part: For like parallel forces, the resultant divides the line internally in the inverse ratio of the forces. If $x$ is the distance of the resultant from Q, then $5 \times (6 - x) = 15 \times x$ (taking moments about the line of the resultant), giving $30 - 5x = 15x$, so $20x = 30$, $x = 1.5$ m from Q. Option B (1.5 m) is correct.

For the OR (economics) part: At equilibrium $P_s = P_d$, so $160 + 2Q^2 = 240 - 3Q^2 \Rightarrow 5Q^2 = 80 \Rightarrow Q^2 = 16 \Rightarrow Q = 4$. Option B (4) is correct for the OR alternative.

a) $(1 + x)^n = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \binom{n}{3}x^3 + \cdots + \binom{n}{n}x^n$.

b) First four coefficients: $\binom{n}{0} = 1$, $\binom{n}{1} = n$, $\binom{n}{2} = \frac{n(n-1)}{2!}$, $\binom{n}{3} = \frac{n(n-1)(n-2)}{3!}$.

c) Putting $x = 1$: $(1+1)^n = 2^n = \sum_{r=0}^{n}\binom{n}{r}$. So the sum of all binomial coefficients is $2^n$.

d) When $n$ is even, there is a single middle term, the $\left(\frac{n}{2} + 1\right)$th term: $T_{\frac{n}{2}+1} = \binom{n}{n/2}x^{n/2}$.

e) $C(n, r_1) = C(n, r_2)$ holds when either $r_1 = r_2$ or $r_1 + r_2 = n$.

Let $P(n): 1^2 + 2^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}$.

Base step: For $n = 1$, LHS $= 1^2 = 1$, RHS $= \frac{1\cdot2\cdot3}{6} = 1$. So $P(1)$ is true.

Inductive step: Assume $P(k)$ is true: $1^2 + 2^2 + \cdots + k^2 = \frac{k(k+1)(2k+1)}{6}$.

Then for $n = k+1$:

Thus $P(k+1)$ is true whenever $P(k)$ is true. By the principle of mathematical induction, $P(n)$ holds for every natural number $n$.

Given $\tan^{-1}x + \tan^{-1}y + \tan^{-1}z = \frac{\pi}{2}$.

So $\tan^{-1}x + \tan^{-1}y = \frac{\pi}{2} - \tan^{-1}z$.

Taking tangent of both sides and using $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\frac{x+y}{1-xy}$:

Thus $z(x+y) = 1 - xy$, i.e. $zx + zy = 1 - xy$, giving $xy + yz + zx = 1$. Proved.

A plane parallel to $2x + 5y - 2 = 0$ has the form $2x + 5y + d = 0$ (the $z$-coefficient is $0$). It passes through $(-3, -2, -1)$:

Hence the required plane is $2x + 5y + 16 = 0$.

Note: the stated production shares (70% and 45%) sum to 115%, which is inconsistent; the figures are taken as printed. Treating the defective contributions directly:

Probability that a selected item is from P and defective $= 0.70 \times 0.10 = 0.07$.

Probability that a selected item is from Q and defective $= 0.45 \times 0.08 = 0.036$.

So the (joint) probability of a defective item coming from P is $0.07$ and from Q is $0.036$. (If a normalised conditional probability is required, total defective $= 0.07 + 0.036 = 0.106$, giving $P(\text{from P} \mid \text{defective}) = \frac{0.07}{0.106} \approx 0.660$ and $P(\text{from Q} \mid \text{defective}) = \frac{0.036}{0.106} \approx 0.340$.)

Let $X$ = Pokhara price, $Y$ = Chitwan price. Given $\bar{X} = 160$, $\sigma_X = 8$, $\bar{Y} = 120$, $\sigma_Y = 12$, $r = 0.6$.

The regression line of $X$ on $Y$:

Regression coefficient $b_{XY} = r\frac{\sigma_X}{\sigma_Y} = 0.6 \times \frac{8}{12} = 0.4$.

For $Y = 200$:

So $X = 160 + 32 = 192$. The most likely price in Pokhara is Rs. 192.

a) (Lagrange's) Mean Value Theorem: If $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$, then there exists at least one point $c \in (a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.

b) $\frac{d}{dx}(\cosh x) = \sinh x$.

c) $\int \sqrt{a^2 + x^2}\, dx = \frac{x}{2}\sqrt{a^2 + x^2} + \frac{a^2}{2}\ln\left|x + \sqrt{a^2 + x^2}\right| + C$.

d) Geometrically, there is at least one point $c$ on the curve $y = f(x)$ between $a$ and $b$ where the tangent is parallel to the chord (secant line) joining the points $(a, f(a))$ and $(b, f(b))$.

e) When the degree of the numerator $f(x)$ is less than the degree of the denominator $g(x)$, the rational function $\frac{f(x)}{g(x)}$ is called a proper (rational) fraction.

Rewrite as $\frac{dy}{dx} + \frac{1}{x}y = x^2$, a linear ODE with $P = \frac{1}{x}$, $Q = x^2$.

Integrating factor: $\text{IF} = e^{\int \frac{1}{x}dx} = e^{\ln x} = x$.

Solution: $y \cdot x = \int x \cdot x^2\, dx + C = \int x^3\, dx + C = \frac{x^4}{4} + C$.

Hence $xy = \frac{x^4}{4} + C$, i.e. $y = \frac{x^3}{4} + \frac{C}{x}$.

Maximize $P = 50x + 60y$ subject to $3x + 4y \le 36$, $9x + 4y \le 60$, $x, y \ge 0$.

Corner (vertex) evaluation:

• Intersection of $3x + 4y = 36$ and $9x + 4y = 60$: subtracting gives $6x = 24 \Rightarrow x = 4$, then $3(4) + 4y = 36 \Rightarrow 4y = 24 \Rightarrow y = 6$. Point $(4, 6)$.
• $x$-intercepts: $9x = 60 \Rightarrow x = \frac{20}{3}$ from second; $3x=36 \Rightarrow x=12$ from first. The binding $x$-axis vertex is $\left(\frac{20}{3}, 0\right)$.
• $y$-intercept: $4y = 36 \Rightarrow y = 9$ from first; $4y = 60 \Rightarrow y = 15$ from second. Binding $y$-axis vertex is $(0, 9)$.

Objective values:

• $(0, 0)$: $P = 0$.
• $\left(\frac{20}{3}, 0\right)$: $P = 50 \times \frac{20}{3} = \frac{1000}{3} \approx 333.3$.
• $(4, 6)$: $P = 50(4) + 60(6) = 200 + 360 = 560$.
• $(0, 9)$: $P = 60 \times 9 = 540$.

Maximum $P = 560$ at $x = 4$, $y = 6$.

Statics part: Let $W$ be the weight of the uniform rod (acting at its midpoint, 150 cm from each end). A weight of 5 N is placed at one end; the system balances about a pivot 25 cm from that end.

Taking moments about the pivot (25 cm from the loaded end):

• The 5 N weight is 25 cm on one side: moment $= 5 \times 25$.
• The rod's weight $W$ acts at its centre, $150 - 25 = 125$ cm on the other side: moment $= W \times 125$.

Balancing: $5 \times 25 = W \times 125 \Rightarrow 125 = 125W \Rightarrow W = 1$ N.

The weight of the rod is 1 N.

OR (economics) part a: Profit $\pi = PQ - C = (108 - 5Q)Q - (-12Q + Q^2) = 108Q - 5Q^2 + 12Q - Q^2 = 120Q - 6Q^2$. Then $\frac{d\pi}{dQ} = 120 - 12Q = 0 \Rightarrow Q = 10$, and $\frac{d^2\pi}{dQ^2} = -12 < 0$ (maximum). Price $P = 108 - 5(10) = 58$. So profit is maximum at price Rs. 58 (with $Q = 10$).

Dynamics part: Using $v^2 = u^2 + 2as$ with $u = 3$, $v = 5$, $s = 16$:

Then $F = ma \Rightarrow 4.9 = m \times 0.5 \Rightarrow m = 9.8$ kg.

The mass of the body is 9.8 kg.

OR (economics) part b: Principal $P = 100000$, rate $r = 10\%$, $n = 5$ years.

i) No extra deposits: $A = P(1 + r)^n = 100000(1.1)^5 = 100000 \times 1.61051 = \text{Rs. } 161051$.

ii) Initial Rs. 100000 grows to Rs. 161051 (as above). Additionally Rs. 20000 deposited at the end of each year for 5 years forms an ordinary annuity:

Total value $= 161051 + 122102 = \text{Rs. } 283153$.

She must choose 4 relatives from the 7 relatives and the remaining $6 - 4 = 2$ guests from the $12 - 7 = 5$ non-relatives.

Number of ways $= \binom{7}{4} \times \binom{5}{2} = 35 \times 10 = 350$.

So she can invite the guests in 350 ways.

Claim: $1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}$. Let $P(n)$ be this statement.

Base step: For $n = 1$, LHS $= 1$, RHS $= \frac{1\cdot2}{2} = 1$. True.

Inductive step: Assume $P(k)$: $1 + 2 + \cdots + k = \frac{k(k+1)}{2}$. Then

which is $P(k+1)$. By induction, $1 + 2 + \cdots + n = \frac{n(n+1)}{2}$ for all natural numbers $n$.

Write $A\mathbf{x} = \mathbf{b}$ with $A = \begin{pmatrix} 2 & -1 & 1 \\ 1 & -2 & 3 \\ 4 & 1 & 2 \end{pmatrix}$, $\mathbf{b} = \begin{pmatrix} -1 \\ 4 \\ 4 \end{pmatrix}$.

$\det A = 2(-2\cdot2 - 3\cdot1) - (-1)(1\cdot2 - 3\cdot4) + 1(1\cdot1 - (-2)\cdot4) = 2(-7) + 1(-10) + 1(9) = -14 - 10 + 9 = -15$.

Using Cramer's rule: $\det A_x = \begin{vmatrix} -1 & -1 & 1 \\ 4 & -2 & 3 \\ 4 & 1 & 2 \end{vmatrix} = -1(-4-3) +1(8-12) + 1(4+8) = 7 - 4 + 12 = 15$, so $x = \frac{15}{-15} = -1$.

$\det A_y = \begin{vmatrix} 2 & -1 & 1 \\ 1 & 4 & 3 \\ 4 & 4 & 2 \end{vmatrix} = 2(8-12) +1(2-12) + 1(4-16) = -8 -10 -12 = -30$, so $y = \frac{-30}{-15} = 2$.

$\det A_z = \begin{vmatrix} 2 & -1 & -1 \\ 1 & -2 & 4 \\ 4 & 1 & 4 \end{vmatrix} = 2(-8-4) +1(4-16) -1(1+8) = -24 -12 -9 = -45$, so $z = \frac{-45}{-15} = 3$.

Check in eq. 1: $2(-1) - 2 + 3 = -1$ ✓; eq. 2: $-1 - 4 + 9 = 4$ ✓; eq. 3: $-4 + 2 + 6 = 4$ ✓.

Hence $x = -1$, $y = 2$, $z = 3$.

From $2l + 2m - n = 0$ we get $n = 2l + 2m$. Substitute into $lm + mn + nl = 0$:

More directly: $lm + n(m + l) = 0 \Rightarrow lm + (2l + 2m)(l + m) = 0$.

Expand: $lm + 2(l + m)^2 = 0 \Rightarrow lm + 2(l^2 + 2lm + m^2) = 0 \Rightarrow 2l^2 + 5lm + 2m^2 = 0$.

Divide by $m^2$ and let $t = l/m$: $2t^2 + 5t + 2 = 0 \Rightarrow t = \frac{-5 \pm \sqrt{25 - 16}}{4} = \frac{-5 \pm 3}{4}$, so $t = -\frac{1}{2}$ or $t = -2$.

Line 1 ($l : m = -1 : 2$): then $n = 2l + 2m = 2(-1) + 2(2) = 2$. So $(l, m, n) \propto (-1, 2, 2)$, magnitude $\sqrt{1 + 4 + 4} = 3$. Direction cosines $\left(-\frac{1}{3}, \frac{2}{3}, \frac{2}{3}\right)$.

Line 2 ($l : m = -2 : 1$): then $n = 2(-2) + 2(1) = -2$. So $(l, m, n) \propto (-2, 1, -2)$, magnitude $3$. Direction cosines $\left(-\frac{2}{3}, \frac{1}{3}, -\frac{2}{3}\right)$.

Angle: $\cos\theta = l_1 l_2 + m_1 m_2 + n_1 n_2 = \left(-\frac{1}{3}\right)\left(-\frac{2}{3}\right) + \frac{2}{3}\cdot\frac{1}{3} + \frac{2}{3}\cdot\left(-\frac{2}{3}\right) = \frac{2}{9} + \frac{2}{9} - \frac{4}{9} = 0$.

Since $\cos\theta = 0$, the angle between the two lines is $\theta = 90°$ (the lines are perpendicular).

Take unit vectors in the $xy$-plane making angles $A$ and $B$ with the positive $x$-axis:

The angle between $\hat{a}$ and $\hat{b}$ is $(A - B)$. The cross product (in the $z$-direction):

Also $|\hat{b} \times \hat{a}| = |\hat{a}||\hat{b}|\sin(A - B) = \sin(A - B)$ (taking $A > B$, with $\hat{k}$ direction).

Equating the $\hat{k}$ components:

Proved.

Statement: If $f$ is continuous on $[a, b]$, differentiable on $(a, b)$, and $f(a) = f(b)$, then there exists at least one $c \in (a, b)$ such that $f'(c) = 0$.

Geometric interpretation: If the curve $y = f(x)$ has equal ordinates at the endpoints, then there is at least one point between them where the tangent to the curve is horizontal (parallel to the $x$-axis).

Verification for $f(x) = x(x - 3)^2$ on $[0, 3]$:

• $f$ is a polynomial, so continuous on $[0, 3]$ and differentiable on $(0, 3)$.
• $f(0) = 0\cdot(−3)^2 = 0$ and $f(3) = 3\cdot0^2 = 0$, so $f(0) = f(3)$.

All conditions hold. Now $f(x) = x(x-3)^2 = x(x^2 - 6x + 9) = x^3 - 6x^2 + 9x$, so

Setting $f'(x) = 0$: $x = 1$ or $x = 3$. The value $x = 1 \in (0, 3)$, so $c = 1$ satisfies Rolle's theorem. Hence Rolle's theorem is verified.

At $x = 5$: numerator $= 25 - 25 = 0$ and denominator $= 5 + 20 - 25 = 0$, a $\frac{0}{0}$ form.

Applying L'Hospital's rule (differentiate numerator and denominator):