NCERT Exemplar Solutions Class 12 Maths Application of Derivatives — Objective and Fill in the Blanks Q35–64

NCERT Exemplar Problems, Class 12 Mathematics — Application of Derivatives, Objective Type and Fill in the Blanks questions 35 to 64. All 30 questions solved.

These are the quick-fire questions of the unit. Almost every one reduces to a single derivative and a sign check: set $f'(x)=0$ for stationary points, read the sign of $f’$ for increasing or decreasing, and read the slope $\dfrac{dy}{dx}$ for tangents and normals. Rate questions need only the chain rule applied to a relation between the changing quantities.

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 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 37 to 39, 41 to 45, 60 and 61 are tangent and normal questions, so they matter for JEE Advanced but not for the rationalised NCERT chapter or JEE Main. Question 40 is an approximation question and falls outside all three. Questions 35, 36, 46 to 59 and 62 to 64 are fully in syllabus.

Key insight. Every question here comes down to the sign or the zeros of one derivative. For a tangent or normal, $f’$ gives the slope; for monotonicity, the sign of $f’$ on an interval; for extrema, the zeros of $f’$ and a sign change or second-derivative test.

Question 35

The sides of an equilateral triangle are increasing at the rate of $2$ cm/sec. The rate at which the area increases, when side is $10$ cm is:

• (A) $10$ cm²/s
• (B) $\sqrt3$ cm²/s
• (C) $10\sqrt3$ cm²/s
• (D) $\dfrac{10}{3}$ cm²/s

Solution. The area of an equilateral triangle of side $s$ is $A=\dfrac{\sqrt3}{4}s^2$. Differentiating with respect to time, $\dfrac{dA}{dt}=\dfrac{\sqrt3}{2}\,s\,\dfrac{ds}{dt}$. With $s=10$ and $\dfrac{ds}{dt}=2$, this is $\dfrac{\sqrt3}{2}\cdot10\cdot2=10\sqrt3$ cm²/s.

Question 36

A ladder, $5$ metre long, standing on a horizontal floor, leans against a vertical wall. If the top of the ladder slides downwards at the rate of $10$ cm/sec, then the rate at which the angle between the floor and the ladder is decreasing when lower end of ladder is $2$ metres from the wall is:

• (A) $\dfrac{1}{10}$ radian/sec
• (B) $\dfrac{1}{20}$ radian/sec
• (C) $20$ radian/sec
• (D) $10$ radian/sec

Solution. Let the foot be $x$ m from the wall, the top $y$ m up the wall, and $\theta$ the angle with the floor, so $x=5\cos\theta$ and $y=5\sin\theta$. When $x=2$, $\cos\theta=\frac25$ and $y=\sqrt{25-4}=\sqrt{21}$, so $\sin\theta=\frac{\sqrt{21}}{5}$. The top is sliding down at $0.1$ m/s, so $\dfrac{dy}{dt}=-0.1$. From $y=5\sin\theta$ we get $\dfrac{dy}{dt}=5\cos\theta\,\dfrac{d\theta}{dt}$, hence $\dfrac{d\theta}{dt}=\dfrac{-0.1}{5\cdot\frac25}=\dfrac{-0.1}{2}=-\dfrac1{20}$. The angle decreases at $\dfrac1{20}$ radian/sec.

Question 37

The curve $y=x^{\frac15}$ has at $(0,0)$

• (A) a vertical tangent (parallel to $y$-axis)
• (B) a horizontal tangent (parallel to $x$-axis)
• (C) an oblique tangent
• (D) no tangent

Solution. The slope is $\dfrac{dy}{dx}=\dfrac15x^{-4/5}$, which tends to $+\infty$ as $x\to0$. Equivalently, $x=y^5$ gives $\dfrac{dx}{dy}=5y^4=0$ at $y=0$. A zero $\dfrac{dx}{dy}$ means the tangent is vertical, so the curve has a tangent parallel to the $y$-axis at the origin.

Question 38

The equation of normal to the curve $3x^2-y^2=8$ which is parallel to the line $x+3y=8$ is

• (A) $3x-y=8$
• (B) $3x+y+8=0$
• (C) $x+3y\pm8=0$
• (D) $x+3y=0$

Solution. The line $x+3y=8$ has slope $-\frac13$, so the normal must have slope $-\frac13$. From $6x-2y\dfrac{dy}{dx}=0$, the tangent slope is $\dfrac{3x}{y}$, so the normal slope is $-\dfrac{y}{3x}$. Setting $-\dfrac{y}{3x}=-\dfrac13$ gives $y=x$. Substituting into the curve, $3x^2-x^2=8$, so $x=\pm2$ and the points are $(2,2)$ and $(-2,-2)$. The normals there are $y-2=-\frac13(x-2)$, that is $x+3y=8$, and $y+2=-\frac13(x+2)$, that is $x+3y=-8$. Together these are $x+3y\pm8=0$.

Question 39

If the curve $ay+x^2=7$ and $x^3=y$ cut orthogonally at $(1,1)$, then the value of $a$ is:

• (A) $1$
• (B) $0$
• (C) $-6$
• (D) $6$

The printed option D reads “.6”, a misprint for $6$.

Solution. Two curves cut orthogonally when the product of their slopes at the common point is $-1$. For $ay+x^2=7$, $a\dfrac{dy}{dx}+2x=0$, so the slope at $(1,1)$ is $-\dfrac2a$. For $y=x^3$ the slope at $x=1$ is $3$. Then $-\dfrac2a\cdot3=-1$ gives $a=6$.

Question 40

If $y=x^4-10$ and if $x$ changes from $2$ to $1.99$, what is the change in $y$?

• (A) $0.32$
• (B) $0.032$
• (C) $5.68$
• (D) $5.968$

Solution. This is an approximation by differentials: $dy=\dfrac{dy}{dx}\,dx=4x^3\,dx$. At $x=2$, $dx=1.99-2=-0.01$, so $dy=4\cdot8\cdot(-0.01)=-0.32$. The value of $y$ falls by $0.32$, and the magnitude of the change is the option $0.32$. (The exact change is $1.99^4-16\approx-0.3176$, which is close.)

Question 41

The equation of tangent to the curve $y(1+x^2)=2-x$, where it crosses $x$-axis is:

• (A) $x+5y=2$
• (B) $x-5y=2$
• (C) $5x-y=2$
• (D) $5x+y=2$

Solution. Write $y=\dfrac{2-x}{1+x^2}$. It crosses the $x$-axis where $y=0$, that is $x=2$, at the point $(2,0)$. Differentiating, $\dfrac{dy}{dx}=\dfrac{-(1+x^2)-(2-x)(2x)}{(1+x^2)^2}$; at $x=2$ the second term in the numerator vanishes, leaving $\dfrac{-5}{25}=-\dfrac15$. The tangent is $y=-\frac15(x-2)$, that is $x+5y=2$.

Question 42

The points at which the tangents to the curve $y=x^3-12x+18$ are parallel to $x$-axis are:

• (A) $(2,-2),\ (-2,-34)$
• (B) $(2,34),\ (-2,0)$
• (C) $(0,34),\ (-2,0)$
• (D) $(2,2),\ (-2,34)$

Solution. A tangent is parallel to the $x$-axis where the slope is zero: $\dfrac{dy}{dx}=3x^2-12=0$, so $x=\pm2$. Then $y(2)=8-24+18=2$ and $y(-2)=-8+24+18=34$. The points are $(2,2)$ and $(-2,34)$.

Question 43

The tangent to the curve $y=e^{2x}$ at the point $(0,1)$ meets $x$-axis at:

• (A) $(0,1)$
• (B) $\left(-\dfrac12,0\right)$
• (C) $(2,0)$
• (D) $(0,2)$

Solution. The slope is $\dfrac{dy}{dx}=2e^{2x}$, which is $2$ at $x=0$. The tangent is $y-1=2x$. It meets the $x$-axis where $y=0$, giving $x=-\frac12$.

Question 44

The slope of tangent to the curve $x=t^2+3t-8$, $y=2t^2-2t-5$ at the point $(2,-1)$ is:

• (A) $\dfrac{22}{7}$
• (B) $\dfrac67$
• (C) $\dfrac{-6}{7}$
• (D) $-6$

Solution. First find the parameter at $(2,-1)$. Setting $t^2+3t-8=2$ gives $t^2+3t-10=0$, so $t=2$ or $t=-5$. Only $t=2$ gives $y=2\cdot4-4-5=-1$. Next, $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{4t-2}{2t+3}$, which at $t=2$ is $\dfrac{6}{7}$.

Question 45

The two curves $x^3-3xy^2+2=0$ and $3x^2y-y^3-2=0$ intersect at an angle of

• (A) $\dfrac{\pi}{4}$
• (B) $\dfrac{\pi}{3}$
• (C) $\dfrac{\pi}{2}$
• (D) $\dfrac{\pi}{6}$

Solution. Recognise the real and imaginary parts of $z^3$ with $z=x+iy$: $z^3=(x^3-3xy^2)+i(3x^2y-y^3)$. The curves say $\operatorname{Re}z^3=-2$ and $\operatorname{Im}z^3=2$. Differentiating, the gradient of the first is $\nabla u=(3x^2-3y^2,\,-6xy)$ and of the second is $\nabla v=(6xy,\,3x^2-3y^2)$. Their dot product is $(3x^2-3y^2)(6xy)-6xy(3x^2-3y^2)=0$ at every point, so the gradients, and hence the curves, are perpendicular wherever they meet. The angle is $\dfrac{\pi}{2}$.

Question 46

The interval on which the function $f(x)=2x^3+9x^2+12x-1$ is decreasing is:

• (A) $[-1,\infty)$
• (B) $[-2,-1]$
• (C) $(-\infty,-2]$
• (D) $[-1,1]$

Solution. $f'(x)=6x^2+18x+12=6(x+1)(x+2)$. This is negative exactly between the roots, for $-2<x<-1$, so $f$ is decreasing on $[-2,-1]$.

Question 47

Let $f:\mathbf{R}\to\mathbf{R}$ be defined by $f(x)=2x+\cos x$, then $f$:

• (A) has a minimum at $x=\pi$
• (B) has a maximum at $x=0$
• (C) is a decreasing function
• (D) is an increasing function

Solution. $f'(x)=2-\sin x$. Since $\sin x\le1$, $f'(x)\ge1>0$ for every real $x$. So $f$ is increasing everywhere and has no maximum or minimum.

Question 48

$y=x(x-3)^2$ decreases for the values of $x$ given by:

• (A) $1<x<3$
• (B) $x<0$
• (C) $x>0$
• (D) $0<x<\dfrac32$

Solution. $\dfrac{dy}{dx}=(x-3)^2+2x(x-3)=(x-3)(3x-3)=3(x-1)(x-3)$. This is negative between the roots, for $1<x<3$, which is where $y$ decreases.

Question 49

The function $f(x)=4\sin^3x-6\sin^2x+12\sin x+100$ is strictly

• (A) increasing in $\left(\pi,\dfrac{3\pi}{2}\right)$
• (B) decreasing in $\left(\dfrac{\pi}{2},\pi\right)$
• (C) decreasing in $\left[\dfrac{-\pi}{2},\dfrac{\pi}{2}\right]$
• (D) decreasing in $\left[0,\dfrac{\pi}{2}\right]$

Solution. $f'(x)=12\sin^2x\cos x-12\sin x\cos x+12\cos x=12\cos x\,(\sin^2x-\sin x+1)$. The quadratic $\sin^2x-\sin x+1$ has discriminant $1-4<0$, so it is always positive. The sign of $f’$ is therefore the sign of $\cos x$. On $\left(\frac{\pi}{2},\pi\right)$, $\cos x<0$, so $f$ is strictly decreasing there. On $\left(\pi,\frac{3\pi}{2}\right)$, $\cos x<0$ too, so $f$ decreases rather than increases (A is false). On $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ and $\left[0,\frac{\pi}{2}\right]$, $\cos x\ge0$, so $f$ increases (C and D are false).

Question 50

Which of the following functions is decreasing on $\left(0,\dfrac{\pi}{2}\right)$?

• (A) $\sin2x$
• (B) $\tan x$
• (C) $\cos x$
• (D) $\cos3x$

Solution. The derivative of $\cos x$ is $-\sin x<0$ on $\left(0,\frac{\pi}{2}\right)$, so $\cos x$ decreases there. $\tan x$ has derivative $\sec^2x>0$. $\sin2x$ rises on $\left(0,\frac\pi4\right)$ then falls, and $\cos3x$ falls on $\left(0,\frac\pi3\right)$ then rises, so neither is decreasing on the whole interval.

Question 51

The function $f(x)=\tan x-x$

• (A) always increases
• (B) always decreases
• (C) never increases
• (D) sometimes increases and sometimes decreases.

Solution. $f'(x)=\sec^2x-1=\tan^2x\ge0$ wherever $f$ is defined, with equality only at isolated points $x=n\pi$. So $f$ always increases (on each interval where it is defined).

Question 52

If $x$ is real, the minimum value of $x^2-8x+17$ is

• (A) $-1$
• (B) $0$
• (C) $1$
• (D) $2$

Solution. Complete the square: $x^2-8x+17=(x-4)^2+1$. A square is never negative, so the least value is $1$, reached at $x=4$.

Question 53

The smallest value of the polynomial $x^3-18x^2+96x$ in $[0,9]$ is

• (A) $126$
• (B) $0$
• (C) $135$
• (D) $160$

Solution. $f'(x)=3x^2-36x+96=3(x-4)(x-8)$, so the critical points in $[0,9]$ are $x=4$ and $x=8$. The smallest value on a closed interval is among the critical values and the end values: $f(0)=0$, $f(4)=64-288+384=160$, $f(8)=512-1152+768=128$, $f(9)=729-1458+864=135$. The smallest is $0$.

Question 54

The function $f(x)=2x^3-3x^2-12x+4$, has

• (A) two points of local maximum
• (B) two points of local minimum
• (C) one maxima and one minima
• (D) no maxima or minima

Solution. $f'(x)=6x^2-6x-12=6(x-2)(x+1)$. The sign of $f’$ is positive for $x<-1$, negative on $(-1,2)$ and positive for $x>2$. So $x=-1$ is a local maximum (sign changes from $+$ to $-$) and $x=2$ is a local minimum (from $-$ to $+$): one of each.

Question 55

The maximum value of $\sin x\cdot\cos x$ is

• (A) $\dfrac14$
• (B) $\dfrac12$
• (C) $\sqrt2$
• (D) $2\sqrt2$

Solution. Rewrite $\sin x\cos x=\frac12\sin2x$. Since $\sin2x\le1$, the maximum is $\frac12$.

Question 56

At $x=\dfrac{5\pi}{6}$, $f(x)=2\sin3x+3\cos3x$ is:

• (A) maximum
• (B) minimum
• (C) zero
• (D) neither maximum nor minimum.

Solution. $f'(x)=6\cos3x-9\sin3x$. At $x=\frac{5\pi}{6}$ we have $3x=\frac{5\pi}{2}$, where $\cos\frac{5\pi}{2}=0$ and $\sin\frac{5\pi}{2}=1$, so $f’\!\left(\frac{5\pi}{6}\right)=-9\ne0$. A point where the derivative is not zero cannot be an extremum, so it is neither a maximum nor a minimum. As a check, $f\!\left(\frac{5\pi}{6}\right)=2\cdot1+3\cdot0=2$, while the largest value of $f$ is $\sqrt{2^2+3^2}=\sqrt{13}$. The printed answer key gives (A) maximum; that is wrong, because the derivative is non-zero and the value $2$ is well below the maximum $\sqrt{13}$.

Question 57

Maximum slope of the curve $y=-x^3+3x^2+9x-27$ is:

• (A) $0$
• (B) $12$
• (C) $16$
• (D) $32$

Solution. The slope is $m=\dfrac{dy}{dx}=-3x^2+6x+9$. To maximise $m$, set $\dfrac{dm}{dx}=-6x+6=0$, so $x=1$, and $\dfrac{d^2m}{dx^2}=-6<0$ confirms a maximum. Then $m=-3+6+9=12$.

Question 58

$f(x)=x^x$ has a stationary point at

• (A) $x=e$
• (B) $x=\dfrac1e$
• (C) $x=1$
• (D) $x=\sqrt e$

Solution. For $x>0$, take logarithms: $\ln f=x\ln x$, so $\dfrac{f’}{f}=\ln x+1$ and $f'(x)=x^x(\ln x+1)$. Since $x^x>0$, the derivative vanishes only when $\ln x=-1$, that is $x=\dfrac1e$.

Question 59

The maximum value of $\left(\dfrac1x\right)^x$ is:

• (A) $e$
• (B) $e^e$
• (C) $e^{\frac1e}$
• (D) $\left(\dfrac1e\right)^{\frac1e}$

Solution. Let $g(x)=\left(\frac1x\right)^x$, so $\ln g=-x\ln x$ and $\dfrac{g’}{g}=-(\ln x+1)$. This is positive for $x<\frac1e$ and negative for $x>\frac1e$, so $g$ has its maximum at $x=\frac1e$. The value is $\left(\dfrac{1}{1/e}\right)^{1/e}=e^{1/e}$.

Question 60

The curves $y=4x^2+2x-8$ and $y=x^3-x+13$ touch each other at the point ______.

Solution. Two curves touch where they share a point and have the same slope there. Equal slopes need $8x+2=3x^2-1$, that is $3x^2-8x-3=0$, so $(3x+1)(x-3)=0$ and $x=3$ or $x=-\frac13$. At $x=3$ the first curve gives $y=36+6-8=34$ and the second gives $27-3+13=37$; at $x=-\frac13$ the two $y$-values also differ. So as printed, the curves never touch. The question has a misprint: with $y=x^3-x+10$ the second value at $x=3$ is $27-3+10=34$, so the curves meet and share a slope at $(3,34)$. This is almost certainly the intended question, and it matches the printed answer key.

Question 61

The equation of normal to the curve $y=\tan x$ at $(0,0)$ is ______.

Solution. The slope of the tangent is $\sec^2x=1$ at $x=0$. The normal is perpendicular, so its slope is $-1$, and through the origin it is $y=-x$.

Question 62

The values of $a$ for which the function $f(x)=\sin x-ax+b$ increases on $\mathbf{R}$ are ______.

Solution. $f'(x)=\cos x-a$. For $f$ to increase on $\mathbf{R}$ we need $\cos x-a\ge0$ for all $x$, that is $a\le\cos x$ for every $x$, and the smallest value of $\cos x$ is $-1$. So $a\le-1$. The end value $a=-1$ is allowed: then $f'(x)=1+\cos x$ vanishes only at the isolated points $x=(2n+1)\pi$, so $f$ is still strictly increasing. The printed answer key gives the open interval $(-\infty,-1)$; that is incomplete, because it leaves out $a=-1$.

Question 63

The function $f(x)=\dfrac{2x^2-1}{x^4}$, $x>0$, decreases in the interval ______.

Solution. Write $f(x)=2x^{-2}-x^{-4}$. Then $f'(x)=-4x^{-3}+4x^{-5}=\dfrac{-4x^2+4}{x^5}=\dfrac{-4(x-1)(x+1)}{x^5}$. For $x>0$, $x+1>0$ and $x^5>0$, so $f'<0$ exactly when $x-1>0$. So $f$ decreases on $(1,\infty)$.

Question 64

The least value of the function $f(x)=ax+\dfrac bx$ $(a>0,\ b>0,\ x>0)$ is ______.

Solution. $f'(x)=a-\dfrac{b}{x^2}=0$ gives $x=\sqrt{\dfrac ba}$ for $x>0$, and $f”(x)=\dfrac{2b}{x^3}>0$ shows this is a minimum. The value is $a\sqrt{\dfrac ba}+b\sqrt{\dfrac ab}=\sqrt{ab}+\sqrt{ab}=2\sqrt{ab}$. (The same follows from the AM-GM inequality: $ax+\frac bx\ge2\sqrt{ab}$.)

Common mistakes

• Forgetting to check the sign of the derivative. In Question 56 the derivative is not zero at $x=\frac{5\pi}{6}$, so the point cannot be an extremum, whatever the value there looks like. A stationary point needs $f’=0$ first.
• Dropping the end value of an interval. In Question 62, $a=-1$ makes $f’=1+\cos x$ vanish only at isolated points, which does not stop $f$ increasing. Isolated zeros of $f’$ never break strict monotonicity.
• Using the tangent slope for a normal. The normal slope is the negative reciprocal. Questions 38 and 61 ask for normals, and Question 38 needs both signs, $x+3y=\pm8$.
• Finding the point from one coordinate only. In Questions 44 and 60 the parameter or $x$-value must satisfy both curves’ equations. For Question 44, $t=-5$ also gives $x=2$ but not $y=-1$.
• Reading monotonicity from a single piece of the interval. Questions 49 and 50 need the sign of the derivative over the whole stated interval, not just near one end.

Practise next

• The first two bands of this unit: Application of Derivatives, short answer (Q1-24) and long answer (Q25-34).
• This band again: Application of Derivatives, objective and fill in the blanks (Q35-64).
• Textbook practice: Exercise 6.1, rate of change, Exercise 6.2, increasing and decreasing functions, Exercise 6.3, maxima and minima and the Miscellaneous Exercise.