NEB Class 11 · Past paper
The complete NEB Class 11 2074 exam paper for Mathematics, all 15 questions with solved model answers.
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(a) Prove that A - B-bar = A n B, where A and B are any two sets. (b) Let A = {a, b}, B = {b, c} and C = {c, d}. Find Ax(B U C) and Ax(B n C). (c) Test the periodicity of the function f(x) = cos(pi x) and find its period.
(a) By definition $A-\bar B=A\cap(\bar B)^c=A\cap B$ (since the complement of the complement of $B$ is $B$). Element proof: $x\in A-\bar B\iff x\in A\text{ and }x\notin\bar B\iff x\in A\text{ and }x\in B\iff x\in A\cap B$.
(b) $B\cup C={b,c,d}$, $B\cap C={c}$. $$ \begin{aligned} A\times(B\cup C) &= {(a,b),(a,c),(a,d),(b,b),(b,c),(b,d)}. \ A\times(B\cap C) &= {(a,c),(b,c)}. \end{aligned} $$
(c) $f(x)=\cos\pi x$. Since $\cos\theta$ has period $2\pi$, $\cos\pi(x+T)=\cos\pi x$ requires $\pi T=2\pi\Rightarrow T=2$. So $f$ is periodic with period $2$.
(a) Prove that sin(2 sin^-1 x) = 2x sqrt(1 - x^2). (b) Using the principle of mathematical induction, prove that 2 + 2^2 + 2^3 + ... + 2^n = 2(2^n - 1). (c) If A = (2 1; 1 -2), find A A^T.
(a) Let $\sin^{-1}x=\theta\Rightarrow\sin\theta=x,\ \cos\theta=\sqrt{1-x^2}$. Then $\sin(2\sin^{-1}x)=\sin2\theta=2\sin\theta\cos\theta=2x\sqrt{1-x^2}$.
(b) $P(n):\ 2+2^2+\cdots+2^n=2(2^n-1)$. $n=1$: LHS $=2$, RHS $=2(2-1)=2$. True. Assume $P(k)$: $\sum_{r=1}^{k}2^r=2(2^k-1)$. Then $$ \begin{aligned} \sum_{r=1}^{k+1}2^r &= 2(2^k-1)+2^{k+1} \ &= 2^{k+1}-2+2^{k+1} \ &= 2\cdot2^{k+1}-2 \ &= 2(2^{k+1}-1), \end{aligned} $$ which is $P(k+1)$. Hence by induction the result holds for all $n\in\mathbb{N}$.
(c) $A=\begin{pmatrix}2&1\1&-2\end{pmatrix}$ is symmetric, so $A^{T}=\begin{pmatrix}2&1\1&-2\end{pmatrix}$. $$ \begin{aligned} AA^{T} &= \begin{pmatrix}2&1\1&-2\end{pmatrix}\begin{pmatrix}2&1\1&-2\end{pmatrix} \ &= \begin{pmatrix}4+1&2-2\2-2&1+4\end{pmatrix} \ &= \begin{pmatrix}5&0\0&5\end{pmatrix}. \end{aligned} $$
(a) Using Cramer's rule, solve: 2x + 5y = 17, 5x - 2y = -1. (b) Find the real numbers x and y if (x - 1)i + (y + 1) = (1 + i)(4 - 3i). (c) Find the value of K so that the equation 3x^2 + 7x + 6 - K = 0 has one root equal to zero.
(a) $D=\begin{vmatrix}2&5\5&-2\end{vmatrix}=-4-25=-29,\ Dx=\begin{vmatrix}17&5\-1&-2\end{vmatrix}=-34+5=-29,\ Dy=\begin{vmatrix}2&17\5&-1\end{vmatrix}=-2-85=-87.$ $$ \begin{aligned} x &= \frac{-29}{-29} \ &= 1, \ \qquad y &= \frac{-...
(a) Find the equation of a straight line through the mid point of the line segment connecting (2, -4) and (2, 4) and parallel to the line 3x - 2y = 4. (b) Find the equation of a circle having radius 10 units and equations of any two diameters are x + 2y = 8 and x + y = 6. (c) Evaluate: lim(x -> p) (x^2 - p^2)/tan(x - p).
(a) Midpoint of $(2,-4),(2,4)$ is $(2,0)$. Slope of $3x-2y=4$ is $\tfrac32$; a parallel line has slope $\tfrac32$: $$y-0=\tfrac32(x-2)\Rightarrow\boxed{3x-2y-6=0}.$$ (b) The centre is the intersection of the diameters: solving $x+2y=8$ a...
(a) Find dy/dx when x = 2a tan(theta), y = a sec(theta). (b) Evaluate: integral of (3 sin x - 4)^2 cos x dx. (c) Examine whether the function f(x) = 15x^2 - 14x + 1 is increasing or decreasing at x = 2/5 and x = 5/2.
(a) $\dfrac{dx}{d\theta}=2a\sec^2\theta,\ \dfrac{dy}{d\theta}=a\sec\theta\tan\theta$. $$ \begin{aligned} \frac{dy}{dx} &= \frac{a\sec\theta\tan\theta}{2a\sec^2\theta} \ &= \frac{\tan\theta}{2\sec\theta} \ &= \frac{\sin\theta}{2}. \end{...
(a) Define conjunction of the statements. Prepare a truth table for the compound statement (p ^ q) ^ ~(p v q). Draw the conclusion about the statement from the truth table. OR Solve the inequality |2x - 1| >= 3 and draw its graph. (b) Draw the graph of the function f(x) = x - x^2 indicating its characteristics.
(a) The conjunction $p\wedge q$ ("$p$ and $q$") is true only when both $p$ and $q$ are true. $p$ $q$ $p\wedge q$ $p\lor q$ $\sim(p\lor q)$ $(p\wedge q)\wedge\sim(p\lor q)$ ------------------ T T T T F F T F F T F F F T F T F F F F F F T ...
(a) In any triangle ABC, prove that tan((B-C)/2) = ((b-c)/(b+c)) cot(A/2). OR Solve: sec x . tan x = sqrt(2). (b) Show that determinant |1 x x^2; 1 y y^2; 1 z z^2| = (y-z)(z-x)(x-y).
(a) By the sine rule $b=k\sin B,\ c=k\sin C$: $$ \begin{aligned} \frac{b-c}{b+c} &= \frac{\sin B-\sin C}{\sin B+\sin C} \ &= \frac{2\cos\frac{B+C}{2}\sin\frac{B-C}{2}}{2\sin\frac{B+C}{2}\cos\frac{B-C}{2}} \ &= \cot\frac{B+C}{2}\tan\fra...
(a) Using row-equivalent method or inverse matrix method, solve: x + y + z = 1, x + 2y + 3z = 4, x + 3y + 7z = 13. (b) If one root of the equation is the square of the other, prove that b^3 + a^2 c + a c^2 = 3abc.
(a) Subtract the first equation from the second and third: $$ \begin{aligned} (x+2y+3z)-(x+y+z) &= 4-1\Rightarrow y+2z \ &= 3. \ (x+3y+7z)-(x+2y+3z) &= 13-4\Rightarrow y+4z \ &= 9. \end{aligned} $$ Subtract: $2z=6\Rightarrow z=3$, then $y=3-2(3)=-3$, and $x=1-(-3)-3=1$. $$\boxed{x=1,\ y=-3,\ z=3}$$ Check: $1+3(-3)+7(3)=1-9+21=13$. Correct.
(b) Let the roots of $ax^2+bx+c=0$ be $\alpha$ and $\alpha^2$. Then $\alpha+\alpha^2=-\dfrac ba$ and $\alpha^3=\dfrac ca$. Cube the sum: $(\alpha+\alpha^2)^3=\alpha^3+\alpha^6+3\alpha^3(\alpha+\alpha^2)$: $$ \begin{aligned} \left(-\frac ba\right)^3 &= \frac ca+\left(\frac ca\right)^2+3\cdot\frac ca\left(-\frac ba\right)\Rightarrow-\frac{b^3}{a^3} \ &= \frac ca+\frac{c^2}{a^2}-\frac{3bc}{a^2}. \end{aligned} $$ Multiplying by $a^3$: $-b^3=a^2c+ac^2-3abc$, hence $\boxed{b^3+a^2c+ac^2=3abc}$.
(a) Find the equations of the tangent and normal to the circle x^2 + y^2 - 2x - 4y + 3 = 0 at (2, 3). (b) Evaluate: lim(x -> 2) (sqrt(x) - sqrt(6 - x^2))/(x - 2). OR A function f(x) is defined as f(x) = 2x + 3 for x < 1, 4 for x = 1, 6x - 1 for x > 1. Is the function continuous at x = 1? If not, how can you make it continuous?
(a) The circle $x^2+y^2-2x-4y+3=0$ has centre $(1,2)$. The tangent at $(x1,y1)=(2,3)$ is $$ \begin{aligned} xx1+yy1-(x+x1)-2(y+y1)+3 &= 0\Rightarrow2x+3y-(x+2)-2(y+3)+3 \ &= 0\Rightarrow x+y-5 \ &= 0. \end{aligned} $$ The normal passes...
(a) Find from first principles, the derivative of sqrt(2 - 3x). (b) Find the area of the region between the curve y^2 = 4x and the line x = y.
(a) $f(x)=\sqrt{2-3x}$. $$ \begin{aligned} f'(x) &= \lim_{h\to0}\frac{\sqrt{2-3(x+h)}-\sqrt{2-3x}}{h} \ &= \lim_{h\to0}\frac{(2-3x-3h)-(2-3x)}{h\big(\sqrt{2-3x-3h}+\sqrt{2-3x}\big)} \ &= \lim_{h\to0}\frac{-3h}{h\big(\sqrt{2-3x-3h}+\sqrt{2-3x}\big)} \ &= \frac{-3}{2\sqrt{2-3x}}. \end{aligned} $$
(b) The line $x=y$ and $y^2=4x$ meet where $y^2=4y\Rightarrow y=0,4$ (points $(0,0),(4,4)$). Integrating with respect to $x$ (parabola $y=2\sqrt x$ above line $y=x$ on $[0,4]$): $$ \begin{aligned} A &= \int_0^4\big(2\sqrt x-x\big),dx \ &= \left[\frac43x^{3/2}-\frac{x^2}{2}\right]_0^4 \ &= \frac43(8)-8 \ &= \frac{32}{3}-8 \ &= \frac{8}{3}\ \text{sq. units}. \end{aligned} $$
Show that f: R -> R defined by f(x) = cx + d where c (!= 0) and d are real numbers, is one to one and onto. Find f^-1(x).
One-to-one: Suppose $f(x1)=f(x2)$. Then $cx1+d=cx2+d\Rightarrow cx1=cx2\Rightarrow x1=x2$ (since $c\neq0$). So $f$ is injective. Onto: Let $y\in\mathbb{R}$. Solve $y=cx+d\Rightarrow x=\dfrac{y-d}{c}\in\mathbb{R}$, and
The sum of three numbers in A.P. is 36. When the numbers are increased by 1, 4, 43 respectively, the resulting numbers are in G.P. Find the numbers.
Let the A.P. be $12-d,\ 12,\ 12+d$ (sum $=36\Rightarrow$ middle term $12$). After adding $1,4,43$: $$ \begin{aligned} 13-d,\quad16,\quad55+d\quad\text{form a G.P.}\Rightarrow16^2 &= (13-d)(55+d). \ 256 &= 715-42d-d^2\Rightarrow d^2+42d-459 \ &= 0\Rightarrow d \ &= \frac{-42\pm60}{2} \ &= 9\ \text{or}\ -51. \end{aligned} $$
Find the equations of the bisectors of the angles between the lines 4x - 3y + 1 = 0 and 12x - 5y + 7 = 0. Also prove that the bisectors are at right angles to each other. OR Find the angle between the two lines represented by ax^2 + 2hxy + by^2 = 0. Also find the condition under which the two lines are a) perpendicular b) coincident.
Main. Bisectors: $\dfrac{4x-3y+1}{5}=\pm\dfrac{12x-5y+7}{13}\Rightarrow13(4x-3y+1)=\pm5(12x-5y+7)$. ($+$): $4x+7y+11=0$. ($-$): $7x-4y+3=0$. Slopes $-\tfrac47$ and $\tfrac74$; product $=-1$, so the bisectors are perpendicular. OR For
State De Moivre's theorem. Using De Moivre's theorem, find the cube roots of unity. If w = (-1 + sqrt(3) i)/2 be a complex cube root of unity, show that w^2 = (-1 - sqrt(3) i)/2.
De Moivre's theorem. For any integer $n$, $(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta$. Cube roots of unity. Solve $z^3=1=\cos0+i\sin0$. The roots are $$zk=\cos\frac{2k\pi}{3}+i\sin\frac{2k\pi}{3},\quad k=0,1,2.$$ $k=0$: $1$. ...
List the criteria for a function y = f(x), to have the local maxima and local minima at a point. Find the local maxima and local minima of f(x) = 4x^3 - 15x^2 + 12x + 7. OR A spherical ball of salt dissolving in water decreases its volume at the rate of 0.75 cm^3/min. Find the rate at which the radius of the salt is decreasing when its radius is 6 cm.
Criteria. At $x=c$ with $f'(c)=0$: local maximum if $f''(c)<0$; local minimum if $f''(c)>0$.
Main. $f(x)=4x^3-15x^2+12x+7$. $$ \begin{aligned} f'(x) &= 12x^2-30x+12 \ &= 6(2x-1)(x-2) \ &= 0\Rightarrow x \ &= \tfrac12 \ \ 2.\qquad f''(x) &= 24x-30. \end{aligned} $$ At $x=\tfrac12$: $f''=-18<0$ (maximum),
$$ \begin{aligned} f!\left(\tfrac12\right) &= 4\cdot\tfrac18-15\cdot\tfrac14+6+7 \ &= \tfrac12-\tfrac{15}{4}+13 \ &= \tfrac{39}{4}. \end{aligned} $$ At $x=2$: $f''=18>0$ (minimum), $f(2)=32-60+24+7=3$. $$ \begin{aligned} & \text{Local maximum }=\tfrac{39}{4}\ \text{at }x=\tfrac12 \ & \text{Local minimum }=3\ \text{at }x=2. \end{aligned} $$
OR $V=\tfrac43\pi r^3$, $\dfrac{dV}{dt}=-0.75$. Then $\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}\Rightarrow-0.75=4\pi(36)\dfrac{dr}{dt}=144\pi\dfrac{dr}{dt}$, so $$ \begin{aligned} \frac{dr}{dt} &= -\frac{0.75}{144\pi} \ &= -\frac{1}{192\pi}\ \text{cm/min}. \end{aligned} $$ The radius decreases at $\dfrac{1}{192\pi}\approx1.66\times10^{-3}$ cm/min.
math-maxima-minima
(a) By definition (since the complement of the complement of is ). Element proof: .
(b) , .
(c) . Since has period , requires . So is periodic with period .
(a) Let . Then .
(b) . : LHS , RHS . True. Assume : . Then
which is . Hence by induction the result holds for all .
(c) is symmetric, so .
(a) $$ \begin{aligned} x &= \frac{-29}{-29} \ &= 1, \ \qquad y &= \frac{-...
(a) Midpoint of is . Slope of is ; a parallel line has slope : (b) The centre is the intersection of the diameters: solving a...
(a) . $$ \begin{aligned} \frac{dy}{dx} &= \frac{a\sec\theta\tan\theta}{2a\sec^2\theta} \ &= \frac{\tan\theta}{2\sec\theta} \ &= \frac{\sin\theta}{2}. \end{...
(a) The conjunction (" and ") is true only when both and are true. ------------------ T T T T F F T F F T F F F T F T F F F F F F T ...
(a) By the sine rule : $$ \begin{aligned} \frac{b-c}{b+c} &= \frac{\sin B-\sin C}{\sin B+\sin C} \ &= \frac{2\cos\frac{B+C}{2}\sin\frac{B-C}{2}}{2\sin\frac{B+C}{2}\cos\frac{B-C}{2}} \ &= \cot\frac{B+C}{2}\tan\fra...
(a) Subtract the first equation from the second and third:
Subtract: , then , and . Check: . Correct.
(b) Let the roots of be and . Then and . Cube the sum: :
Multiplying by : , hence .
(a) The circle has centre . The tangent at is The normal passes...
(a) .
(b) The line and meet where (points ). Integrating with respect to (parabola above line on ):
One-to-one: Suppose . Then (since ). So is injective. Onto: Let . Solve , and
Let the A.P. be (sum middle term ). After adding :
Main. Bisectors: . (): . (): . Slopes and ; product , so the bisectors are perpendicular. OR For
De Moivre's theorem. For any integer , . Cube roots of unity. Solve . The roots are : . ...
Criteria. At with : local maximum if ; local minimum if .
Main. .
At : (maximum),
At : (minimum), .
OR , . Then , so
The radius decreases at cm/min.