NCERT Exemplar Solutions Class 12 Maths Application of Derivatives — Short Answer Q1–24

NCERT Exemplar Problems, Class 12 Mathematics — Application of Derivatives, Short Answer questions 1 to 24 (Exercise 6.3). All 24 questions solved.

Questions 1 to 11 are about rates of change: differentiate every related quantity with respect to time and eliminate what you do not need. Questions 12 to 19 use tangents and normals. Questions 20 to 22 prove that a function is increasing or decreasing by studying the sign of $f’$, and Questions 23 and 24 are maxima problems.

A note on the syllabus. The rationalised NCERT Chapter 6 keeps only rate of change, increasing and decreasing functions, and maxima and minima (Exercises 6.1 to 6.3 and the Miscellaneous Exercise). It has dropped tangents and normals, and approximations. JEE Main 2026 lists only rate of change, monotonicity and maxima and minima. JEE Advanced 2026 keeps “tangents and normals” but does not list approximations. On this page, Questions 12 to 19 (tangents, normals, angles between curves) are outside the textbook and JEE Main but remain useful for JEE Advanced, and Questions 6 and 7 (approximations) are outside all of these. The rest are in the current syllabus.

Key insight. In a related-rates question, write one equation linking the quantities, differentiate it with respect to $t$ using the chain rule, and only then substitute the values. For monotonicity, a function is increasing if $f’\ge 0$ and vanishes only at isolated points.

Question 1

A spherical ball of salt is dissolving in water in such a manner that the rate of decrease of the volume at any instant is propotional to the surface. Prove that the radius is decreasing at a constant rate.

Solution. Let the radius be $r$. Then $V=\tfrac43\pi r^3$ and the surface area is $S=4\pi r^2$. The volume is decreasing at a rate proportional to the surface, so $\dfrac{dV}{dt}=-k\cdot4\pi r^2$ for some constant $k>0$. By the chain rule, $\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}$. Equating the two expressions and cancelling $4\pi r^2$ (non-zero while the ball exists) gives $\dfrac{dr}{dt}=-k$, a constant.

Question 2

If the area of a circle increases at a uniform rate, then prove that perimeter varies inversely as the radius.

Solution. Let the area be $A=\pi r^2$. A uniform rate of increase means $\dfrac{dA}{dt}=c$, a constant. Differentiating, $2\pi r\dfrac{dr}{dt}=c$, so $\dfrac{dr}{dt}=\dfrac{c}{2\pi r}$. The perimeter is $P=2\pi r$, so $\dfrac{dP}{dt}=2\pi\dfrac{dr}{dt}=2\pi\cdot\dfrac{c}{2\pi r}=\dfrac{c}{r}$. The rate at which the perimeter grows is therefore inversely proportional to the radius.

Question 3

A kite is moving horizontally at a height of 151.5 meters. If the speed of kite is 10 m/s, how fast is the string being let out; when the kite is 250 m away from the boy who is flying the kite? The height of boy is 1.5 m.

Solution. The kite is $151.5-1.5=150$ m above the boy’s hands. Let $x$ be the horizontal distance and $s$ the length of the string, so $s^2=x^2+150^2$. When $s=250$, $x=\sqrt{250^2-150^2}=200$. Differentiating, $2s\dfrac{ds}{dt}=2x\dfrac{dx}{dt}$, so $\dfrac{ds}{dt}=\dfrac{x}{s}\dfrac{dx}{dt}=\dfrac{200}{250}\times10=8$. The height stays constant, which is why only $x$ changes.

Question 4

Two men A and B start with velocities $v$ at the same time from the junction of two roads inclined at $45^\circ$ to each other. If they travel by different roads, find the rate at which they are being seperated.

Solution. After time $t$ each man is $vt$ from the junction. By the cosine rule the separation $d$ satisfies $d^2=(vt)^2+(vt)^2-2(vt)(vt)\cos45^\circ=v^2t^2\left(2-\sqrt2\right)$. So $d=vt\sqrt{2-\sqrt2}$, which is directly proportional to $t$, and its rate of change is constant.

Question 5

Find an angle $\theta$, $0<\theta<\dfrac{\pi}{2}$, which increases twice as fast as its sine.

Solution. We need $\dfrac{d\theta}{dt}=2\dfrac{d}{dt}(\sin\theta)=2\cos\theta\dfrac{d\theta}{dt}$. Since $\dfrac{d\theta}{dt}\ne0$ we can cancel it, leaving $\cos\theta=\tfrac12$. In $\left(0,\tfrac\pi2\right)$ this gives $\theta=\dfrac\pi3$.

Question 6

Find the approximate value of $(1.999)^5$.

Solution. Take $f(x)=x^5$ at $x=2$ with $\Delta x=-0.001$. The approximation is $f(x+\Delta x)\approx f(x)+f'(x)\Delta x$, with $f'(x)=5x^4$. So $(1.999)^5\approx 2^5+5\cdot2^4\cdot(-0.001)=32-0.08=31.92$. (The exact value is $31.92008\ldots$, so the linear approximation is very close.)

Question 7

Find the approximate volume of metal in a hollow spherical shell whose internal and external radii are 3 cm and 3.0005 cm, respectively.

Solution. The metal in the shell is the small change in volume as the radius grows from $3$ to $3.0005$. With $V=\tfrac43\pi r^3$, $dV=4\pi r^2\,dr$. Taking $r=3$ and $dr=0.0005$ gives $dV=4\pi\cdot9\cdot0.0005=0.018\pi$.

Question 8

A man, 2 m tall, walks at the rate of $1\dfrac23$ m/s towards a street light which is $5\dfrac13$ m above the ground. At what rate is the tip of his shadow moving? At what rate is the length of the shadow changing when he is $3\dfrac13$ m from the base of the light?

Solution. Let $x$ be the man’s distance from the base of the light and $s$ the length of his shadow. Similar triangles (man and light pole, both measured from the tip of the shadow) give $\dfrac{2}{16/3}=\dfrac{s}{x+s}$, that is $\dfrac38=\dfrac{s}{x+s}$, so $3x+3s=8s$ and $s=\dfrac{3x}{5}$. The man walks towards the light, so $\dfrac{dx}{dt}=-\dfrac53$. The tip of the shadow is at distance $x+s=\dfrac{8x}{5}$ from the base, so it moves at $\dfrac85\cdot\left(-\dfrac53\right)=-\dfrac83$ m/s, that is $2\dfrac23$ m/s towards the light. The shadow length changes at $\dfrac{ds}{dt}=\dfrac35\cdot\left(-\dfrac53\right)=-1$ m/s. Neither rate depends on how far the man is from the light, so the figure $3\dfrac13$ m does not matter.

Question 9

A swimming pool is to be drained for cleaning. If L represents the number of litres of water in the pool $t$ seconds after the pool has been plugged off to drain and $L=200(10-t)^2$. How fast is the water running out at the end of 5 seconds? What is the average rate at which the water flows out during the first 5 seconds?

Solution. The instantaneous rate is $\dfrac{dL}{dt}=-400(10-t)$. At $t=5$ this is $-2000$, so water runs out at $2000$ litres per second. The average rate over the first 5 seconds is the total change divided by the time: $L(0)=20000$ and $L(5)=5000$, so the average outflow is $\dfrac{20000-5000}{5}=3000$ litres per second. It is larger than the rate at $t=5$ because the water drains faster early on.

Question 10

The volume of a cube increases at a constant rate. Prove that the increase in its surface area varies inversely as the length of the side.

Solution. Let the side be $a$. Then $V=a^3$, and a constant rate of increase means $\dfrac{dV}{dt}=3a^2\dfrac{da}{dt}=c$, so $\dfrac{da}{dt}=\dfrac{c}{3a^2}$. The surface area is $S=6a^2$, so $\dfrac{dS}{dt}=12a\dfrac{da}{dt}=12a\cdot\dfrac{c}{3a^2}=\dfrac{4c}{a}$. The rate of increase of the surface area is inversely proportional to $a$.

Question 11

$x$ and $y$ are the sides of two squares such that $y=x-x^2$. Find the rate of change of the area of second square with respect to the area of first square.

Solution. The areas are $A_1=x^2$ and $A_2=y^2$. We want $\dfrac{dA_2}{dA_1}=\dfrac{d(y^2)}{d(x^2)}=\dfrac{2y\,\dfrac{dy}{dx}}{2x}=\dfrac{y}{x}\dfrac{dy}{dx}$. Here $\dfrac{dy}{dx}=1-2x$ and $\dfrac yx=1-x$, so the result is $(1-x)(1-2x)=2x^2-3x+1$. The printed answer key gives $2x^3-3x+1$; that is a misprint of the power, because the product $(1-x)(1-2x)$ has degree $2$.

Question 12

Find the condition that the curves $2x=y^2$ and $2xy=k$ intersect orthogonally.

Solution. Curves intersect orthogonally when their tangents at the common point are perpendicular, so the product of the slopes is $-1$. From $2x=y^2$, $2=2y\dfrac{dy}{dx}$, so the slope is $\dfrac1y$. From $2xy=k$, $2y+2x\dfrac{dy}{dx}=0$, so the slope is $-\dfrac yx$. The product is $-\dfrac1x$, and setting this equal to $-1$ gives $x=1$. Then $y^2=2x=2$, and $k=2xy=2y$, so $k^2=4y^2=8$.

Question 13

Prove that the curves $xy=4$ and $x^2+y^2=8$ touch each other.

Solution. Curves touch when they meet at a point and share a tangent there. From $xy=4$ we get $2xy=8=x^2+y^2$, so $(x-y)^2=0$ and $y=x$. Then $x^2=4$, so the curves meet at $(2,2)$ and $(-2,-2)$. For $xy=4$ the slope is $-\dfrac yx$, and for the circle it is $-\dfrac xy$. At both points $x=y$, so both slopes equal $-1$.

Question 14

Find the co-ordinates of the point on the curve $\sqrt x+\sqrt y=4$ at which tangent is equally inclined to the axes.

Solution. A tangent equally inclined to the axes has slope $1$ or $-1$. Differentiating, $\dfrac1{2\sqrt x}+\dfrac1{2\sqrt y}\dfrac{dy}{dx}=0$, so $\dfrac{dy}{dx}=-\dfrac{\sqrt y}{\sqrt x}$, which is always negative. Hence the slope must be $-1$, which gives $\sqrt y=\sqrt x$ and $x=y$. Then $2\sqrt x=4$, so $x=y=4$.

Question 15

Find the angle of intersection of the curves $y=4-x^2$ and $y=x^2$.

Solution. Setting $4-x^2=x^2$ gives $x=\pm\sqrt2$ and $y=2$. The slopes are $-2x$ for the first curve and $2x$ for the second. At $x=\sqrt2$ they are $m_1=-2\sqrt2$ and $m_2=2\sqrt2$. Then $\tan\theta=\left|\dfrac{m_1-m_2}{1+m_1m_2}\right|=\left|\dfrac{-4\sqrt2}{1-8}\right|=\dfrac{4\sqrt2}{7}$. By symmetry the angle at $x=-\sqrt2$ is the same.

Question 16

Prove that the curves $y^2=4x$ and $x^2+y^2-6x+1=0$ touch each other at the point $(1,2)$.

Solution. First, $(1,2)$ lies on both curves: $2^2=4\cdot1$, and $1+4-6+1=0$. Now compare the slopes. For the parabola, $2y\dfrac{dy}{dx}=4$, so the slope is $\dfrac2y=1$ at $(1,2)$. For the circle, $2x+2y\dfrac{dy}{dx}-6=0$, so the slope is $\dfrac{3-x}{y}=\dfrac{2}{2}=1$ at $(1,2)$. The slopes are equal, so the curves share a tangent there.

Question 17

Find the equation of the normal lines to the curve $3x^2-y^2=8$ which are parallel to the line $x+3y=4$.

Solution. The line $x+3y=4$ has slope $-\tfrac13$, so the normals have slope $-\tfrac13$ and the tangents at those points have slope $3$. Differentiating, $6x-2y\dfrac{dy}{dx}=0$, so $\dfrac{dy}{dx}=\dfrac{3x}{y}$. Setting this equal to $3$ gives $y=x$, and then $3x^2-x^2=8$ gives $x=\pm2$. The points are $(2,2)$ and $(-2,-2)$. The normal at $(2,2)$ is $y-2=-\tfrac13(x-2)$, that is $x+3y=8$. The normal at $(-2,-2)$ is $y+2=-\tfrac13(x+2)$, that is $x+3y=-8$.

Question 18

At what points on the curve $x^2+y^2-2x-4y+1=0$, the tangents are parallel to the $y$-axis?

Solution. Completing the squares, $(x-1)^2+(y-2)^2=4$, a circle with centre $(1,2)$ and radius $2$. A tangent is parallel to the $y$-axis (vertical) exactly where the radius is horizontal, that is at the points with the same $y$-coordinate as the centre: $(1\pm2,\,2)$. Using calculus, $\dfrac{dy}{dx}=\dfrac{1-x}{y-2}$ is infinite when $y=2$, giving $(x-1)^2=4$ again.

Question 19

Show that the line $\dfrac xa+\dfrac yb=1$, touches the curve $y=b\cdot e^{-x/a}$ at the point where the curve intersects the axis of $y$.

Solution. The curve meets the $y$-axis where $x=0$, at $y=b\,e^0=b$, so at $(0,b)$. The slope of the curve is $\dfrac{dy}{dx}=-\dfrac ba e^{-x/a}$, which at $x=0$ is $-\dfrac ba$. The line $\dfrac xa+\dfrac yb=1$ has slope $-\dfrac ba$ and passes through $(0,b)$, since $0+1=1$. So the line passes through the point $(0,b)$ with the same slope as the curve there, which means it is the tangent.

Question 20

Show that $f(x)=2x+\cot^{-1}x+\log\left(\sqrt{1+x^2}-x\right)$ is increasing in $\mathbf R$.

Solution. Differentiate each term. $\dfrac{d}{dx}\cot^{-1}x=-\dfrac1{1+x^2}$, and $\dfrac{d}{dx}\log\left(\sqrt{1+x^2}-x\right)=\dfrac{\dfrac{x}{\sqrt{1+x^2}}-1}{\sqrt{1+x^2}-x}=-\dfrac1{\sqrt{1+x^2}}$. So $f'(x)=2-\dfrac1{1+x^2}-\dfrac1{\sqrt{1+x^2}}$. Since $1+x^2\ge1$, each of $\dfrac1{1+x^2}$ and $\dfrac1{\sqrt{1+x^2}}$ is at most $1$, so $f'(x)\ge0$. Equality needs both to equal $1$, which happens only at $x=0$, an isolated point. A function with $f’\ge0$ and $f’=0$ only at an isolated point is increasing on $\mathbf R$.

Question 21

Show that for $a\ge1$, $f(x)=\sqrt3\sin x-\cos x-2ax+b$ is decreasing in $\mathbf R$.

Solution. $f'(x)=\sqrt3\cos x+\sin x-2a$. Writing $\sqrt3\cos x+\sin x=2\sin\left(x+\dfrac\pi3\right)$, we get $f'(x)=2\sin\left(x+\dfrac\pi3\right)-2a\le2-2a$. Since $a\ge1$, $2-2a\le0$, so $f'(x)\le0$ for every $x$. For $a>1$ it is strictly negative, and for $a=1$ it is zero only at isolated points where the sine equals $1$. Either way $f$ is decreasing on $\mathbf R$.

Question 22

Show that $f(x)=\tan^{-1}(\sin x+\cos x)$ is an increasing function in $\left(0,\dfrac\pi4\right)$.

Solution. By the chain rule, $f'(x)=\dfrac{\cos x-\sin x}{1+(\sin x+\cos x)^2}$. The denominator is always positive. On $\left(0,\tfrac\pi4\right)$ we have $\cos x>\sin x$, so the numerator is positive. Hence $f'(x)>0$ throughout the interval.

Question 23

At what point, the slope of the curve $y=-x^3+3x^2+9x-27$ is maximum? Also find the maximum slope.

Solution. The slope is $m=\dfrac{dy}{dx}=-3x^2+6x+9$, and we maximise this quadratic in $x$. $m’=-6x+6=0$ gives $x=1$, and $m”=-6<0$, so this is a maximum. The maximum slope is $m(1)=-3+6+9=12$. The point on the curve has $y=-1+3+9-27=-16$.

Question 24

Prove that $f(x)=\sin x+\sqrt3\cos x$ has maximum value at $x=\dfrac\pi6$.

Solution. $f'(x)=\cos x-\sqrt3\sin x$, which is zero when $\tan x=\dfrac1{\sqrt3}$, so $x=\dfrac\pi6$ in the first quadrant. The second derivative is $f”(x)=-\sin x-\sqrt3\cos x$, and $f”\left(\dfrac\pi6\right)=-\dfrac12-\dfrac32=-2<0$, so $x=\dfrac\pi6$ is a point of local maximum. The value there is $\dfrac12+\dfrac32=2$, which agrees with writing $f(x)=2\sin\left(x+\dfrac\pi3\right)$, whose greatest value is $2$.

Common mistakes

• Substituting before differentiating. In Questions 3 and 8 the value of a changing quantity (the string length, the man’s position) must be put in only after differentiating; substituting first treats it as a constant.
• Forgetting the direction of motion. In Question 8 the man is walking towards the light, so $\dfrac{dx}{dt}$ is negative, and the shadow length decreases. Reporting $+1$ m/s for the shadow is wrong.
• Taking the slope of the wrong curve. In Questions 12 to 17, the angle or condition is about the tangents at the common point, so first find the points of intersection and only then substitute into the two derivatives.
• Dropping the $a=1$ case. In Question 21, $f’\le0$ is enough for a decreasing function as long as $f’$ is zero only at isolated points, so $a=1$ is included.
• Copying the printed answer to Question 11. The area ratio $\dfrac{d(y^2)}{d(x^2)}$ is $(1-x)(1-2x)=2x^2-3x+1$; the cubic printed in the key does not arise.

Practise next

• Exemplar Long Answer Q25–34, the next questions, with optimisation problems like Questions 23 and 24.
• Exemplar Objective and Fill in the Blanks Q35–64, more rates of change, tangents and monotonicity in objective form.
• NCERT textbook Exercise 6.1 for rate of change, as in Questions 1 to 11.
• NCERT textbook Exercise 6.2 for increasing and decreasing functions, as in Questions 20 to 22.
• NCERT textbook Exercise 6.3 and the Miscellaneous Exercise for maxima and minima, as in Questions 23 and 24.