NEB Class 12 · Past paper
The complete NEB Class 12 2081 exam paper for Mathematics, all 22 questions with solved model answers.
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$(r_1,\theta_1)$ and $(r_2,\theta_2)$ are the polar coordinates of $z_1$ and $z_2$. What are the polar coordinates of the image of $z_1z_2$? A) $(r_1r_2,\theta_1-\theta_2)$ B) $(r_1r_2,\theta_1+\theta_2)$ C) $(r_1+r_2,\theta_1+\theta_2)$ D) $(r_1-r_2,\theta_1+\theta_2)$
B) $(r1r2,\ \theta1+\theta2)$. When two complex numbers are multiplied the moduli multiply and the arguments add: $z1z2=r1r2,[\cos(\theta1+\theta2)+i\sin(\theta1+\theta2)].$
How many numbers are there between $999$ and $10{,}000$? A) $9\times10^3$ B) $8\times10^3$ C) $9^4$ D) $8^4$
A) $9\times10^3$. The integers from $1000$ to $9999$ number
$$ \begin{aligned} 9999-1000+1 &= 9000 \ &= 9\times10^3. \end{aligned} $$
In triangle $ABC$ with sides $a,b,c$ and semi-perimeter $s$, which equals $\cos\frac{A}{2}$? A) $\sqrt{\frac{s(s-a)}{bc}}$ B) $\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$ C) $\sqrt{\frac{s(s-a)}{(s-b)(s-c)}}$ D) $\sqrt{\frac{(s-b)(s-c)}{bc}}$
A) $\sqrt{\dfrac{s(s-a)}{bc}}$. The half-angle formula gives $\cos\dfrac A2=\sqrt{\dfrac{s(s-a)}{bc}}$ (while $\sin\dfrac A2=\sqrt{\dfrac{(s-b)(s-c)}{bc}}$).
A parabolic reflector is $20$ cm in diameter and $5$ cm deep. What are the coordinates of its focus? A) $(0,5)$ B) $(0,-5)$ C) $(5,0)$ D) $(-5,0)$
C) $(5,0)$. With vertex at the origin and axis along $x$, the rim point is $(5,10)$. For $y^2=4ax$: $10^2=4a(5)\Rightarrow a=5$, so the focus is $(a,0)=(5,0).$
$\vec a,\vec b$ are non-zero vectors. Which result is the smallest number? A) $(\vec a\times\vec b)\cdot(\vec b\times\vec a)$ B) $(\vec a\times\vec a)\cdot\vec a$ C) $(\vec b\times\vec b)\cdot\vec b$ D) $(\vec a\times\vec a)\cdot\vec b$
A) $(\vec a\times\vec b)\cdot(\vec b\times\vec a)$. Options B, C, D contain $\vec a\times\vec a=\vec0$ or $\vec b\times\vec b=\vec0$, so each is $0$. Since $\vec b\times\vec a=-(\vec a\times\vec b)$, option A $=-|\vec a\times\vec b|^2\le0$, the smallest.
$A,B$ are dependent events with $P(A\cap B)=\frac1{18},\ P(B)=\frac1{18}$ and $2P(A)=5P(B)$. Find $P(B/A)$. A) $\frac1{324}$ B) $\frac1{18}$ C) $\frac5{36}$ D) $\frac25$
D) $\dfrac25$. From
$$ \begin{aligned} 2P(A) &= 5P(B) \ &= \dfrac5{18}, \end{aligned} $$
$P(A)=\dfrac5{36}.$ Then
$$ \begin{aligned} P(B/A) &= \dfrac{P(A\cap B)}{P(A)} \ &= \dfrac{1/18}{5/36} \ &= \dfrac25. \end{aligned} $$
$f'(x)=3x^2$ and $g'(x)=2x$. Using L'Hospital's rule, find $\lim_{x\to1}\frac{f(x)}{g(x)}$. A) $\frac32$ B) $\frac13$ C) $\frac12$ D) $\frac16$
A) $\dfrac32$. By L'Hospital's rule $$ \begin{aligned} \lim{x\to1}\dfrac{f(x)}{g(x)} &= \lim{x\to1}\dfrac{f'(x)}{g'(x)} \ &= \lim{x\to1}\dfrac{3x^2}{2x} \ &= \dfrac{3(1)}{2} \ &= \dfrac32. \end{aligned} $$
Which is $\int\frac{1}{x^2+4},dx$? A) $\frac12\tan^{-1}\frac2x+C$ B) $\frac12\tan^{-1}\frac x2+C$ C) $2\tan^{-1}\frac x2+C$ D) $\tan^{-1}\frac x2+C$
B) $\dfrac12\tan^{-1}\dfrac x2+C$. Using $\int\dfrac{dx}{x^2+a^2}=\dfrac1a\tan^{-1}\dfrac xa$ with $a=2.$
What is the degree of the ODE $\frac{d^3s}{dt^3}=\left(4+\left(\frac{d^2s}{dt^2}\right)^2\right)^{1/2}$? A) 1 B) 2 C) 3 D) 4
B) 2. Squaring to clear the radical, $\left(\dfrac{d^3s}{dt^3}\right)^2=4+\left(\dfrac{d^2s}{dt^2}\right)^2.$ The highest-order derivative $\dfrac{d^3s}{dt^3}$ then appears to power $2$, so the degree is $2.$
Which is an example of a homogeneous differential equation of first order? A) $3x,dy+2y,dx=4$ B) $x,dy-y,dx+1=0$ C) $x,dx+y,dy=2$ D) $(x^2+xy),dy-(xy-y^2),dx=0$
D) $(x^2+xy),dy-(xy-y^2),dx=0$. Every term is of the same (second) degree in $x,y$, so it can be written as $\dfrac{dy}{dx}=F!\left(\dfrac yx\right)$; the others contain constant terms and are not homogeneous.
Given $4x+3y=26$ (i) and $3x-2y=11$ (ii), a student eliminates $y$. Which operation is applied? A) $4\times$(ii)$+3\times$(i) B) $4\times$(ii)$-3\times$(i) C) $2\times$(i)$+3\times$(ii) D) $2\times$(i)$-3\times$(ii). OR: What is the greatest height of a projectile with initial velocity $160$ m/s and angle $30^\circ$ ($g=10$)? A) 203 m B) 230 m C) 320 m D) 640 m
C) $2\times$(i)$+3\times$(ii). $$ \begin{aligned} 2(4x+3y)+3(3x-2y) &= 8x+6y+9x-6y \ &= 17x, \end{aligned} $$ so the $y$-terms cancel. OR: C) 320 m. $$ \begin{aligned} H &= \dfrac{u^2\sin^2\theta}{2g} \ &= \dfrac{160^2\left(\tfrac12\ri...
(a) Write the sum of the series $1^2+2^2+3^2+\cdots+n^2$. (b) State the principle of mathematical induction. (c) For what condition does $\log_e\left(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots\right)$ necessarily exist? (d) Write $e^x$ in expanded form ($2<e<3$).
(a) $1^2+2^2+\cdots+n^2=\dfrac{n(n+1)(2n+1)}{6}.$ (b) Principle of mathematical induction: a statement $P(n)$ is true for all $n\in\mathbb{N}$ if (i) $P(1)$ is true, and (ii) whenever $P(k)$ is true then $P(k+1)$ is true. (c) The bracket...
Use the row-equivalent (matrix) method to solve: $2x+3y+5z-23=0,\ 3x-3y+8z-21=0,\ x+4y-z-6=0$.
Write the augmented matrix and reduce: $$\left[\begin{array}{ccc|c}2&3&5&23\3&-3&8&21\1&4&-1&6\end{array}\right].$$ Using $R_1\leftrightarrow R_3$ then eliminating $x$ and $y$ by row operations leads to $z=3$, back-substitution giving $y=2$ and $x=1.$ $$\boxed{x=1,\ y=2,\ z=3.}$$ (Check: $2+6+15=23,\ 3-6+24=21,\ 1+8-3=6.$)
(a) Find the equation of the hyperbola with vertices $(0,-4),(0,4)$ and foci $(0,6),(0,-6)$. (b) Find the dot product of $2\vec i-\vec j+2\vec k$ and $3\vec i-2\vec j+6\vec k$.
(a) Transverse axis along $y$: $a=4$ (vertex), $c=6$ (focus), $$ \begin{aligned} b^2 &= c^2-a^2 \ &= 36-16 \ &= 20. \ \frac{y^2}{16}-\frac{x^2}{20} &= 1. \end{aligned} $$ (b) $$ \begin{aligned} (2)(3)+(-1)(-2)+(2)(6) &= 6+2+12 \ &= 2...
(a) Show that $2\left(a\sin^2\frac{C}{2}+c\sin^2\frac{A}{2}\right)$ is less than the perimeter of $\triangle ABC$. (b) The condition for the line $y=mx+c$ to touch the circle $x^2+y^2=a^2$ is $c^2=a^2(1+m^2)$. Justify it with an example.
(a) Using $2\sin^2\theta=1-\cos2\theta$: $2a\sin^2\tfrac C2=a(1-\cos C)$ and $2c\sin^2\tfrac A2=c(1-\cos A).$ Adding, $$a+c-(a\cos C+c\cos A)=a+c-b$$ by the projection formula $b=a\cos C+c\cos A.$ Since the perimeter is $a+b+c$ and
Supply $(x)$ and price $(y)$ of a commodity for 2017-2023: $x=80,84,86,88,92,96,97$; $y=12,14,15,16,18,20,22$. Find the likely price $y$ when the supply $x$ is $120$.
Fit the regression line of $y$ on $x$, $y=a+bx$, using $\bar x=89,\ \bar y\approx16.71$ and $$ \begin{aligned} b &= \frac{n\Sigma xy-\Sigma x\Sigma y}{n\Sigma x^2-(\Sigma x)^2}\approx0.550 \ \qquad a &= \bar y-b\bar x\approx-32.27. \end{aligned} $$ So $y\approx0.550x-32.27.$ At $x=120$: $$y\approx0.550(120)-32.27\approx33.8.$$ The likely price is about $\mathbf{33.8}$ (units).
(a) Write the derivative of $\sinh x$. (b) Under what condition is $\frac{d}{dx}(\operatorname{sech}^{-1}x)=-\frac{1}{x\sqrt{1-x^2}}$? (c) Write the meaning of $dy$ and $\Delta y$ in percentage error $\frac{\Delta y-dy}{y}\times100%$. (d) Write $\int\frac{1}{\sqrt{x^2-a^2}},dx$. (e) Write the integrating factor of $\frac{dy}{dx}+Py=Q$.
(a) $\dfrac{d}{dx}\sinh x=\cosh x.$ (b) Valid for $0<x<1$ (the domain of $\operatorname{sech}^{-1}x$ taking the positive branch). (c) $\Delta y$ is the actual change in $y$ for a change $\Delta x$; $dy=f'(x),dx$ is the approximate (diff...
(a) Find the equation of the tangent to $y=2x^3+x^2-7x-2$ at $x=2$. (b) Evaluate $\int\frac{2x^2-x+4}{x^3+4x},dx$.
(a) $\dfrac{dy}{dx}=6x^2+2x-7$; at $x=2$,
$$ \begin{aligned} \text{slope} &= 24+4-7 \ &= 21. \end{aligned} $$
The point on the curve is $\big(2,,2(8)+4-14-2\big)=(2,4).$ Tangent: $$ \begin{aligned} y-4 &= 21(x-2)\ \Rightarrow\ y \ &= 21x-38. \end{aligned} $$
(b) Partial fractions with $x^3+4x=x(x^2+4)$: $\dfrac{2x^2-x+4}{x(x^2+4)}=\dfrac1x+\dfrac{x-1}{x^2+4}.$ $$\int=\ln|x|+\tfrac12\ln(x^2+4)-\tfrac12\tan^{-1}\tfrac x2+C.$$
Using the Simplex method, maximize $Z=25x+35y$ subject to $2x+3y\le15,\ 3x+y\le12,\ x,y\ge0$. OR: (a) A body of weight $78$ N is suspended by two strings of length $4$ m and $3$ m attached to two points in a horizontal line $5$ m apart; find the tensions. (b) A rocket loses mass at $0.2$ kg/s and its velocity changes from $90$ m/s to $10$ m/s; find the retarding force.
Simplex. Corner points of the feasible region and objective values: $$(0,0)\to0,\quad(4,0)\to100,\quad(0,5)\to175,\quad(3,3)\to180.$$ The binding constraints $2x+3y=15,\ 3x+y=12$ meet at $(3,3)$; the simplex iterations terminate there. $$\boxed{Z_{\max}=25(3)+35(3)=180\text{ at }(3,3).}$$
OR (a) $3^2+4^2=5^2$, so the strings are perpendicular at the body; tension $=W\times\dfrac{\text{opposite side}}{\text{hypotenuse}}$:
$$ \begin{aligned} T_4 &= 78\cdot\tfrac35 \ &= 46.8\text{ N}, \ T_3 &= 78\cdot\tfrac45 \ &= 62.4\text{ N}. \end{aligned} $$
(b)
$$ \begin{aligned} \text{Retarding force} &= \left|\dfrac{dm}{dt}\right|\Delta v \ &= 0.2\times(90-10) \ &= 16\text{ N}. \end{aligned} $$
(a) Prove that the coefficient of $x^n$ in $(1+x)^{2n}$ is twice the coefficient of $x^n$ in $(1+x)^{2n-1}$. (b) Use De Moivre's theorem to find the square root of $-(2+2\sqrt3,i)$. (c) By mathematical induction, prove that $n!<2^n$.
(a) The coefficients are $\binom{2n}{n}$ and $\binom{2n-1}{n}.$ Now
$$ \begin{aligned} \frac{\binom{2n}{n}}{\binom{2n-1}{n}} &= \frac{(2n)!/(n!,n!)}{(2n-1)!/(n!,(n-1)!)} \ &= \frac{(2n)!}{(2n-1)!}\cdot\frac{(n-1)!}{n!} \ &= 2n\cdot\frac1n \ &= 2, \end{aligned} $$
so $\binom{2n}{n}=2\binom{2n-1}{n}.\ \blacksquare$
(b) $-(2+2\sqrt3,i)=-2-2\sqrt3,i$ has modulus $4$ and argument $240^\circ$ (third quadrant): $4(\cos240^\circ+i\sin240^\circ).$ By De Moivre the square roots are $2\left(\cos\tfrac{240^\circ+360^\circ k}{2}+i\sin\cdots\right)$ for $k=0,1$: $2(\cos120^\circ+i\sin120^\circ)=-1+\sqrt3,i$ and $2(\cos300^\circ+i\sin300^\circ)=1-\sqrt3,i.$ So the square roots are $\pm(-1+\sqrt3,i).$ (Check $(-1+\sqrt3 i)^2=-2-2\sqrt3 i.$)
(c) Note: $n!<2^n$ actually holds only for $n=1,2,3$ ($1<2,\ 2<4,\ 6<8$) and reverses from $n=4$ ($24>16$). The result provable by induction is $2^n<n!$ for $n\ge4$: base $2^4=16<24=4!$; assuming $2^k<k!$, then $2^{k+1}=2\cdot2^k<2\cdot k!<(k+1)k!=(k+1)!$ since $k+1>2.$ Hence $2^n<n!$ for all $n\ge4.\ \blacksquare$
(a) If $a=3+\sqrt3,\ b=2,\ c=\sqrt3$, solve the triangle. (b) Find the foci and length of the latus rectum of $\frac{x^2}{25}+\frac{y^2}{100}=1$. (c) Find the area of the parallelogram whose diagonals are $\vec a=\vec i-2\vec k$ and $\vec b=4\vec i+3\vec j+\vec k$.
(a) By the cosine rule $\cos A=\dfrac{b^2+c^2-a^2}{2bc}.$ Here
$$ \begin{aligned} a^2 &= (3+\sqrt3)^2 \ &= 12+6\sqrt3\approx22.39, \end{aligned} $$
so $\cos A=\dfrac{4+3-22.39}{2(2)(\sqrt3)}\approx-2.22.$ Since $|\cos A|>1$ (equivalently $b+c=2+\sqrt3<a$), the given lengths violate the triangle inequality and no real triangle exists; the data as printed cannot form a triangle. (The cosine rule shown is the correct method once valid sides are given.)
(b) $a^2=100$ (major axis along $y$), $b^2=25.$
$$ \begin{aligned} c &= \sqrt{a^2-b^2} \ &= \sqrt{75} \ &= 5\sqrt3. \end{aligned} $$
Foci $(0,\pm5\sqrt3).$
$$ \begin{aligned} \text{Latus rectum} &= \dfrac{2b^2}{a} \ &= \dfrac{2(25)}{10} \ &= 5. \end{aligned} $$
(c) Area $=\tfrac12|\vec a\times\vec b|.$
$$ \begin{aligned} \vec a\times\vec b &= \begin{vmatrix}\vec i&\vec j&\vec k\1&0&-2\4&3&1\end{vmatrix} \ &= (6,-9,3), \ |\vec a\times\vec b| &= \sqrt{36+81+9} \ &= \sqrt{126} \ &= 3\sqrt{14}. \end{aligned} $$
Area $=\dfrac{3\sqrt{14}}{2}.$
(a) A curve $y^2=8x$ changes its abscissa and ordinate at the same rate; the ordinate is double the abscissa. Justify with calculation. (b) $\int\frac{dx}{a+b\cos x}$ depends on the values of $a$ and $b$. Explain. (c) An exact differential equation has the form $M(x,y),dx+N(x,y),dy=0$. Give a simple example and its characteristic.
(a) Differentiate $y^2=8x$: $2y\dfrac{dy}{dt}=8\dfrac{dx}{dt}.$ Given $\dfrac{dy}{dt}=\dfrac{dx}{dt}$: $2y=8\Rightarrow y=4$, and then
$$ \begin{aligned} x &= \dfrac{y^2}{8} \ &= 2. \end{aligned} $$
So
$$ \begin{aligned} y &= 4 \ &= 2x \end{aligned} $$
(ordinate is double the abscissa) at the point $(2,4)$, as required.
(b) Substituting $t=\tan\tfrac x2$ gives $\int\dfrac{2,dt}{(a+b)+(a-b)t^2}.$ The form of the answer depends on the sign of $\dfrac{a-b}{a+b}$: if $a>b$ the integral is a $\tan^{-1}$ (inverse-tangent) form; if $a<b$ it is a logarithmic form; if $a=b$ it reduces to a simple rational integral. Hence the result depends on the relative values of $a$ and $b.$
(c) $M,dx+N,dy=0$ is exact iff $\dfrac{\partial M}{\partial y}=\dfrac{\partial N}{\partial x}.$ Example: $2xy,dx+x^2,dy=0$ has $M=2xy,\ N=x^2$ with
$$ \begin{aligned} M_y &= 2x \ &= N_x, \end{aligned} $$
so it is exact and its solution is $x^2y=C.$
and are the polar coordinates of and . What are the polar coordinates of the image of ? A) B) C) D)
B) . When two complex numbers are multiplied the moduli multiply and the arguments add:
How many numbers are there between and ? A) B) C) D)
A) . The integers from to number
In triangle with sides and semi-perimeter , which equals ? A) B) C) D)
A) . The half-angle formula gives (while ).
A parabolic reflector is cm in diameter and cm deep. What are the coordinates of its focus? A) B) C) D)
C) . With vertex at the origin and axis along , the rim point is . For : , so the focus is
are non-zero vectors. Which result is the smallest number? A) B) C) D)
A) . Options B, C, D contain or , so each is . Since , option A , the smallest.
are dependent events with and . Find . A) B) C) D)
D) . From
Then
and . Using L'Hospital's rule, find . A) B) C) D)
A) . By L'Hospital's rule
Which is ? A) B) C) D)
B) . Using with
What is the degree of the ODE ? A) 1 B) 2 C) 3 D) 4
B) 2. Squaring to clear the radical, The highest-order derivative then appears to power , so the degree is
Which is an example of a homogeneous differential equation of first order? A) B) C) D)
D) . Every term is of the same (second) degree in , so it can be written as ; the others contain constant terms and are not homogeneous.
Given (i) and (ii), a student eliminates . Which operation is applied? A) (ii)(i) B) (ii)(i) C) (i)(ii) D) (i)(ii). OR: What is the greatest height of a projectile with initial velocity m/s and angle ()? A) 203 m B) 230 m C) 320 m D) 640 m
C) (i)(ii). so the -terms cancel. OR: C) 320 m. $$ \begin{aligned} H &= \dfrac{u^2\sin^2\theta}{2g} \ &= \dfrac{160^2\left(\tfrac12\ri...
(a) Write the sum of the series . (b) State the principle of mathematical induction. (c) For what condition does necessarily exist? (d) Write in expanded form ().
(a) (b) Principle of mathematical induction: a statement is true for all if (i) is true, and (ii) whenever is true then is true. (c) The bracket...
Use the row-equivalent (matrix) method to solve: .
Write the augmented matrix and reduce: Using then eliminating and by row operations leads to , back-substitution giving and (Check: )
(a) Find the equation of the hyperbola with vertices and foci . (b) Find the dot product of and .
(a) Transverse axis along : (vertex), (focus), (b) $$ \begin{aligned} (2)(3)+(-1)(-2)+(2)(6) &= 6+2+12 \ &= 2...
(a) Show that is less than the perimeter of . (b) The condition for the line to touch the circle is . Justify it with an example.
(a) Using : and Adding, by the projection formula Since the perimeter is and
Supply and price of a commodity for 2017-2023: ; . Find the likely price when the supply is .
Fit the regression line of on , , using and
So At : The likely price is about (units).
(a) Write the derivative of . (b) Under what condition is ? (c) Write the meaning of and in percentage error . (d) Write . (e) Write the integrating factor of .
(a) (b) Valid for (the domain of taking the positive branch). (c) is the actual change in for a change ; is the approximate (diff...
(a) Find the equation of the tangent to at . (b) Evaluate .
(a) ; at ,
The point on the curve is Tangent:
(b) Partial fractions with :
Using the Simplex method, maximize subject to . OR: (a) A body of weight N is suspended by two strings of length m and m attached to two points in a horizontal line m apart; find the tensions. (b) A rocket loses mass at kg/s and its velocity changes from m/s to m/s; find the retarding force.
Simplex. Corner points of the feasible region and objective values: The binding constraints meet at ; the simplex iterations terminate there.
OR (a) , so the strings are perpendicular at the body; tension :
(b)
(a) Prove that the coefficient of in is twice the coefficient of in . (b) Use De Moivre's theorem to find the square root of . (c) By mathematical induction, prove that .
(a) The coefficients are and Now
so
(b) has modulus and argument (third quadrant): By De Moivre the square roots are for : and So the square roots are (Check )
(c) Note: actually holds only for () and reverses from (). The result provable by induction is for : base ; assuming , then since Hence for all
(a) If , solve the triangle. (b) Find the foci and length of the latus rectum of . (c) Find the area of the parallelogram whose diagonals are and .
(a) By the cosine rule Here
so Since (equivalently ), the given lengths violate the triangle inequality and no real triangle exists; the data as printed cannot form a triangle. (The cosine rule shown is the correct method once valid sides are given.)
(b) (major axis along ),
Foci
(c) Area
Area
(a) A curve changes its abscissa and ordinate at the same rate; the ordinate is double the abscissa. Justify with calculation. (b) depends on the values of and . Explain. (c) An exact differential equation has the form . Give a simple example and its characteristic.
(a) Differentiate : Given : , and then
So
(ordinate is double the abscissa) at the point , as required.
(b) Substituting gives The form of the answer depends on the sign of : if the integral is a (inverse-tangent) form; if it is a logarithmic form; if it reduces to a simple rational integral. Hence the result depends on the relative values of and
(c) is exact iff Example: has with
so it is exact and its solution is