Tangents, Normals, Rolle's Theorem and LMVT for JEE: Notes and PYQs

At a point where a curve is smooth, one number answers two questions: how steep the curve is, and how fast one quantity changes with another. That number is the derivative. This chapter reads it both ways. As a rate it solves related-rates problems; as a slope it gives tangents, normals, the angle between curves and the lengths attached to a tangent. The mean value theorems then turn slopes into statements about values: a curve that returns to its starting height must level off, and some tangent is always parallel to the chord.

This note covers rate of change, approximation by differentials, tangents and normals (explicit, implicit and parametric curves), tangents in special positions, tangents from an outside point, common tangents, normals and shortest distance, the angle between two curves, the lengths of tangent, normal, sub-tangent and sub-normal, Rolle's theorem, counting roots, helper functions, Lagrange's mean value theorem and Cauchy's mean value theorem. The worked examples include JEE Main 2019 to 2025 questions and JEE Advanced and IIT JEE questions; every answer was re-derived from scratch.

The derivative as a slope and as a rate

At a point where $f$ is differentiable,

$$f'(x_1)=\lim_{h\to0}\frac{f(x_1+h)-f(x_1)}{h}$$

• Geometrically, $f'(x_1)$ is the slope of the tangent to $y=f(x)$ at $(x_1,f(x_1))$. If the tangent makes angle $\theta$ with the positive $x$-axis, then $f'(x_1)=\tan\theta$.
• Physically, $f'(x_1)$ is the instantaneous rate at which $y$ changes with $x$.

The secant through $(x_1,f(x_1))$ and $(x_1+h,f(x_1+h))$ has slope equal to the difference quotient; as $h\to0$ the secant turns into the tangent. Every idea in this chapter is one of these two readings.

Rate of change

If $x$ and $y$ both change with time $t$ and are linked by an equation, differentiate the equation with respect to $t$. By the chain rule,

$$\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}$$

Method.

1. Draw the figure and name every changing quantity.
2. Write the equation that links them, true at every instant.
3. Differentiate with respect to $t$.
4. Only now substitute the values at the given instant.

A quantity is increasing when its rate is positive and decreasing when its rate is negative.

Example 1 (IITianForum). The radius of a sphere increases at $2$ cm/s. Find the rates at which its volume and its surface area increase when the radius is $5$ cm.

▸ SHOW SOLUTION

Step 1. Volume

$$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}=4\pi(25)(2)=200\pi$$

Step 2. Surface area

$$\frac{dS}{dt}=8\pi r\frac{dr}{dt}=8\pi(5)(2)=80\pi$$

Answer: $200\pi$ cm$^3$/s and $80\pi$ cm$^2$/s

Example 2 (IITianForum). A ladder $5$ m long leans against a vertical wall. Its foot slides away from the wall at $0.5$ m/s. How fast is the top sliding down when the foot is $3$ m from the wall?

A right triangle formed by a wall, the floor and a ladder of length 5: the foot of the ladder is at distance x from the wall along the floor, the top is at height y up the wall, with arrows showing the foot moving away and the top moving down

▸ SHOW SOLUTION

Step 1. The link

With $x$ the distance of the foot and $y$ the height of the top,

$$x^2+y^2=25$$

Step 2. Differentiate

$$2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$$

Step 3. Substitute

At $x=3$, $y=4$:

$$\frac{dy}{dt}=-\frac{x}{y}\cdot\frac{dx}{dt}=-\frac34(0.5)=-\frac38$$

Answer: the top slides down at $\frac38$ m/s

Example 3 (JEE Main 2019, 9 April Shift 2). A water tank has the shape of an inverted right circular cone whose semi-vertical angle is $\tan^{-1}\left(\frac12\right)$. Water is poured into it at a constant rate of $5$ cubic metre per minute. Then the rate (in m/min) at which the level of water is rising at the instant when the depth of water in the tank is $10$ m is

▸ SHOW SOLUTION

Step 1. Volume in terms of h

$r=h\tan\theta=\frac h2$, so

$$V=\frac13\pi\left(\frac h2\right)^2h=\frac{\pi h^3}{12}$$

Step 2. Differentiate

$$\frac{dV}{dt}=\frac{\pi h^2}{4}\cdot\frac{dh}{dt}$$

Step 3. Substitute

$$5=\frac{\pi(100)}{4}\cdot\frac{dh}{dt}\ \Rightarrow\ \frac{dh}{dt}=\frac1{5\pi}$$

Answer: option (4)

Example 4 (JEE Main 2022, 24 June Shift 1). The surface area of a balloon of spherical shape being inflated increases at a constant rate. If initially the radius of the balloon is $3$ units and after $5$ seconds it becomes $7$ units, then its radius after $9$ seconds is

▸ SHOW SOLUTION

Step 1. Constant rate means linear

$\frac{dS}{dt}=k$ gives $S=S_0+kt$. With $S=4\pi r^2$, this says $r^2$ grows linearly in $t$:

$$r^2=9+mt$$

Step 2. Use t = 5

$$49=9+5m\ \Rightarrow\ m=8$$

Step 3. At t = 9

$$r^2=9+72=81,\qquad r=9$$

Answer: option (A)

Example 5 (JEE Main 2025, 23 January Shift 2). A spherical chocolate ball has a layer of ice-cream of uniform thickness around it. When the thickness of the ice-cream layer is $1$ cm, the ice-cream melts at the rate of $81$ cm$^3$/min and the thickness of the ice-cream layer decreases at the rate of $\frac1{4\pi}$ cm/min. The surface area (in cm$^2$) of the chocolate ball (without the ice-cream layer) is

▸ SHOW SOLUTION

Step 1. Only the outer radius changes

The chocolate ball is fixed, so the ice-cream volume changes exactly as the volume $\frac43\pi R^3$ of the outer sphere, and the thickness changes exactly as $R$:

$$\frac{dV}{dt}=4\pi R^2\frac{dR}{dt}$$

Step 2. Substitute

$$81=4\pi R^2\cdot\frac1{4\pi}=R^2\ \Rightarrow\ R=9$$

Step 3. The chocolate ball

Its radius is $R-1=8$, so its surface area is

$$4\pi(8)^2=256\pi$$

Answer: option (4)

Example 6 (JEE Main 2022, 25 June Shift 2). Water is being filled at the rate of $1$ cm$^3$/s in a right circular conical vessel (vertex downwards) of height $35$ cm and diameter $14$ cm. When the height of the water level is $10$ cm, the rate (in cm$^2$/s) at which the wet conical surface area of the vessel increases is

▸ SHOW SOLUTION

Step 1. Shape of the water

Radius to height is $7:35$, so $r=\frac h5$ and the slant height is

$$l=\sqrt{r^2+h^2}=\frac{\sqrt{26}}5h$$

Step 2. Volume and wet surface

$$V=\frac13\pi\frac{h^2}{25}h=\frac{\pi h^3}{75}$$$$S=\pi rl=\frac{\sqrt{26}\,\pi h^2}{25}$$

Step 3. Rates

$$\frac{dV}{dt}=\frac{\pi h^2}{25}\cdot\frac{dh}{dt}=1$$$$\frac{dS}{dt}=\frac{2\sqrt{26}\,\pi h}{25}\cdot\frac{dh}{dt}=\frac{2\sqrt{26}}{h}$$

At $h=10$ this is $\frac{\sqrt{26}}5$.

Answer: option (C)

Example 7 (IITianForum). $BC$ is a circular arc of variable radius $r$ with centre $A$, and $\angle BAC=45^\circ$. The circle through $A$, $B$, $C$ has radius $R$. Find $\frac{dR}{dr}$.

▸ SHOW SOLUTION

Step 1. The triangle

$AB=AC=r$ and $\angle A=45^\circ$, so $\angle B=\angle C=67.5^\circ$.

Step 2. Sine rule

$$2R=\frac{AB}{\sin C}=\frac{r}{\cos22.5^\circ}$$

Step 3. Differentiate

$$\frac{dR}{dr}=\frac1{2\cos22.5^\circ}$$

Since $2\cos22.5^\circ=\sqrt{2+\sqrt2}$, this is $\frac1{\sqrt{2+\sqrt2}}$.

Answer: $\frac{dR}{dr}=\frac1{\sqrt{2+\sqrt2}}$

Approximation and errors

Near a point where $f$ is differentiable, the tangent line stays close to the curve. For a small change $\Delta x$,

$$f(x+\Delta x)\approx f(x)+f'(x)\,\Delta x$$

• Absolute error in $y$: $\Delta y\approx f'(x)\,\Delta x$.
• Relative error: $\frac{\Delta y}y$. Percentage error: $100\frac{\Delta y}y$.
• For $y=x^n$: $\frac{\Delta y}y\approx n\frac{\Delta x}x$. Taking logarithms first turns any product of powers into a sum of relative errors.

Choose $x$ as the nearest point where $f$ is easy to evaluate exactly. The approximation is the tangent line itself, so it is good only for small $\Delta x$.

Example 8 (IITianForum). (a) Find an approximate value of $\sqrt{25.3}$. (b) The radius of a sphere is measured with an error of $1\%$. Find the approximate percentage errors in its surface area and volume.

▸ SHOW SOLUTION

Step 1. Part (a)

Take $f(x)=\sqrt x$, $x=25$, $\Delta x=0.3$, $f'(x)=\frac1{2\sqrt x}=\frac1{10}$:

$$\sqrt{25.3}\approx5+\frac{0.3}{10}=5.03$$

Step 2. Part (b)

$S=4\pi r^2$ and $V=\frac43\pi r^3$, so

$$\frac{\Delta S}S\approx2\frac{\Delta r}r,\qquad \frac{\Delta V}V\approx3\frac{\Delta r}r$$

Answer: (a) $5.03$; (b) $2\%$ and $3\%$

Tangent and normal

Let $P(x_1,y_1)$ lie on the curve and let $m$ be the slope of the tangent at $P$.

$$\text{tangent: } y-y_1=m(x-x_1)$$ $$\text{normal: } y-y_1=-\frac1m(x-x_1)$$

The normal is the line through $P$ perpendicular to the tangent, so the two slopes multiply to $-1$. If $m=0$ the tangent is $y=y_1$ and the normal is $x=x_1$.

A curve with a point P on it, the tangent line touching the curve at P and the normal line through P perpendicular to it, with the right angle between them marked

Finding m.

For a parametric curve the tangent at parameter $t$ can be written straight away:

$$y-y(t)=\frac{y'(t)}{x'(t)}\big(x-x(t)\big)$$

Example 9 (IITianForum). Find the tangent and the normal to $y=x^3-3x+1$ at $x=2$.

▸ SHOW SOLUTION

Step 1. Point and slope

$y(2)=8-6+1=3$, and $y'=3x^2-3$ gives $m=9$.

Step 2. Tangent

$$y-3=9(x-2)\ \Rightarrow\ y=9x-15$$

Step 3. Normal

$$y-3=-\frac19(x-2)\ \Rightarrow\ x+9y=29$$

Answer: tangent $y=9x-15$; normal $x+9y=29$

Example 10 (JEE Main 2019, 10 January Shift 2). The tangent to the curve $y=xe^{x^2}$ passing through the point $(1,e)$ also passes through the point

▸ SHOW SOLUTION

Step 1. Slope at x = 1

$(1,e)$ lies on the curve.

$$y'=e^{x^2}(1+2x^2),\qquad y'(1)=3e$$

Step 2. Tangent

$$y-e=3e(x-1)\ \Rightarrow\ y=3ex-2e$$

Step 3. Test the options

At $x=\frac43$: $y=4e-2e=2e$.

Answer: option (2)

Example 11 (JEE Advanced 2014, Paper 1). The slope of the tangent to the curve $(y-x^5)^2=x(1+x^2)^2$ at the point $(1,3)$ is ______.

▸ SHOW SOLUTION

Step 1. Differentiate implicitly

$$2(y-x^5)(y'-5x^4)$$$$=(1+x^2)^2+4x^2(1+x^2)$$

Step 2. Put (1, 3)

$$2(2)(y'-5)=4+8=12$$$$y'-5=3\ \Rightarrow\ y'=8$$

Answer: $8$

Example 12 (JEE Main 2022, 25 June Shift 2). If the angle made by the tangent at the point $(x_0,y_0)$ on the curve $x=12(t+\sin t\cos t)$, $y=12(1+\sin t)^2$, $0<t<\frac\pi2$, with the positive $x$-axis is $\frac\pi3$, then $y_0$ is equal to

▸ SHOW SOLUTION

Step 1. The two rates

$$\frac{dx}{dt}=12(1+\cos2t)=24\cos^2t$$$$\frac{dy}{dt}=24(1+\sin t)\cos t$$

Step 2. Slope

$$\frac{dy}{dx}=\frac{1+\sin t}{\cos t}$$

Step 3. Slope = tan 60°

$$\frac{1+\sin t}{\cos t}=\sqrt3$$

$t=\frac\pi6$ works: $\frac{3/2}{\sqrt3/2}=\sqrt3$. The left side is increasing on $\left(0,\frac\pi2\right)$, so this is the only value.

$$y_0=12\left(\frac32\right)^2=27$$

Answer: option (C)

Example 13 (JEE Main 2023, 29 January Shift 2). If the equation of the normal to the curve

$$y=\frac{x-a}{(x+b)(x-2)}$$

at the point $(1,-3)$ is $x-4y=13$, then the value of $a+b$ is equal to ______.

▸ SHOW SOLUTION

Step 1. Point on the curve

$$-3=\frac{1-a}{(1+b)(-1)}\ \Rightarrow\ 1-a=3(1+b)$$

Step 2. Slope of the tangent

The normal has slope $\frac14$, so the tangent has slope $-4$. Write $y=\frac{x-a}{D}$ with $D=(x+b)(x-2)$; at $x=1$, $D=-(1+b)$ and $D'=2x-2+b=b$:

$$y'=\frac{D-(x-a)D'}{D^2}$$$$-4=\frac{-(1+b)-(1-a)b}{(1+b)^2}$$

Step 3. Solve

Put $1-a=3(1+b)$:

$$-4(1+b)^2=-(1+b)-3b(1+b)$$

Cancel $1+b\ne0$: $4(1+b)=1+3b$, so $b=-3$ and $a=1-3(1+b)=7$.

Answer: $a+b=4$

Example 14 (JEE Main 2022, 25 July Shift 2). Let the area enclosed by the $x$-axis and the tangent and normal drawn to the curve

$$4x^3-3xy^2+6x^2-5xy$$ $$-8y^2+9x+14=0$$

at the point $(-2,3)$ be $A$. Then $8A$ is equal to ______.

▸ SHOW SOLUTION

Step 1. Slope by implicit differentiation

$$F_x=12x^2-3y^2+12x-5y+9$$$$F_y=-6xy-5x-16y$$

At $(-2,3)$: $F_x=48-27-24-15+9=-9$ and $F_y=36+10-48=-2$.

$$m=-\frac{F_x}{F_y}=-\frac92$$

Step 2. Where the lines meet the x-axis

Tangent: $0-3=-\frac92(x+2)$ gives $x=-\frac43$.

Normal: $0-3=\frac29(x+2)$ gives $x=-\frac{31}2$.

Step 3. Area

The base is $\frac{31}2-\frac43=\frac{85}6$ and the height is $3$:

$$A=\frac12\cdot\frac{85}6\cdot3=\frac{85}4$$

Answer: $8A=170$

Tangents in special positions

Vertical tangent. If $f$ is continuous at $a$ and $|f'(x)|\to\infty$ as $x\to a$, the tangent at $a$ is the vertical line $x=a$. Fractional powers such as $x^{1/3}$ are the usual source. If instead the two one-sided derivatives are finite and different, there is a corner and no tangent at all. The corner rule for $|f(x)|$ is treated in the Continuity and Differentiability note.

For a curve $F(x,y)=0$, horizontal tangents come from $F_x=0$ and vertical tangents from $F_y=0$, each solved together with the curve. For a parametric curve, horizontal tangents need $y'(t)=0$ with $x'(t)\ne0$, and vertical tangents need $x'(t)=0$ with $y'(t)\ne0$.

Tangent at the origin. If a polynomial curve passes through the origin, the tangents there are found by equating to zero the lowest-degree terms of its equation.

Two graphs side by side: on the left the curve y equals the cube root of x, which passes through the origin with a vertical tangent along the y-axis; on the right the looped curve y squared equals x cubed plus x squared, which crosses itself at the origin with the two tangent lines y equals x and y equals minus x drawn dashed

Example 15 (IITianForum). (a) Find the tangent to $y=x^{1/3}$ at the origin. (b) Find the tangents at the origin to $y^2=x^3+x^2$.

▸ SHOW SOLUTION

Step 1. Part (a)

$$y'=\frac1{3x^{2/3}}\to\infty\ \text{ as } x\to0$$

The curve is continuous at $0$, so the tangent is the vertical line $x=0$.

Step 2. Part (b)

The lowest-degree terms are $y^2-x^2$:

$$y^2-x^2=0\ \Rightarrow\ y=\pm x$$

Answer: (a) $x=0$; (b) $y=x$ and $y=-x$

Example 16 (IITianForum). For the curve $x=t^3-3t^2-9t$, $y=t^2-4t$, $t\in\mathbb R$, let $H$ be the number of points where the tangent is horizontal and $V$ the number where it is vertical. Find $H$ and $V$.

▸ SHOW SOLUTION

Step 1. The two rates

$$x'(t)=3t^2-6t-9=3(t-3)(t+1)$$$$y'(t)=2t-4$$

Step 2. Horizontal

$y'=0$ at $t=2$, where $x'(2)=-9\ne0$. So $H=1$.

Step 3. Vertical

$x'=0$ at $t=3$ and $t=-1$, where $y'=2$ and $-6$, both non-zero. So $V=2$.

Answer: $H=1$, $V=2$

Example 17 (JEE Main 2022, 27 July Shift 1). Let $M$ and $N$ be the number of points on the curve $y^5-9xy+2x=0$ where the tangents to the curve are parallel to the $x$-axis and the $y$-axis respectively. Then the value of $M+N$ equals ______.

▸ SHOW SOLUTION

Step 1. Slope

$$5y^4y'-9y-9xy'+2=0$$$$y'=\frac{9y-2}{5y^4-9x}$$

Step 2. Horizontal tangents

$9y=2$. The curve gives $y^5=x(9y-2)=0$, so $y=0$, which contradicts $y=\frac29$. So $M=0$.

Step 3. Vertical tangents

$9x=5y^4$. Put this in the curve:

$$y^5-5y^5+\frac{10}9y^4=0$$$$y^4\left(\frac{10}9-4y\right)=0$$

$y=0$ gives $(0,0)$ and $y=\frac5{18}$ gives one more point. At both the numerator $9y-2\ne0$. So $N=2$.

Answer: $M+N=2$

Tangent from an outside point

If the point $(h,k)$ is not on the curve $y=f(x)$, take the point of contact as $(t,f(t))$, write the tangent there, and make it pass through $(h,k)$:

$$k-f(t)=f'(t)(h-t)$$

Each real root $t$ of this equation gives one tangent from $(h,k)$.

Meeting the curve again. The tangent at $x=t$ meets a polynomial curve at $x=t$ at least twice. For a cubic, subtract the tangent from the curve: the difference has the factor $(x-t)^2$, and the third factor gives the second meeting point.

The parabola y equals x squared and an outside point A at 1 comma minus 3 below it, with two tangent lines drawn from A touching the parabola at x equals minus 1 and x equals 3

Example 18 (IITianForum). Find the tangents to $y=x^2$ drawn from the point $(1,-3)$.

▸ SHOW SOLUTION

Step 1. Tangent at (t, t²)

$$y=2tx-t^2$$

Step 2. Make it pass through (1, −3)

$$-3=2t-t^2\ \Rightarrow\ (t-3)(t+1)=0$$

Step 3. The tangents

$t=3$: $y=6x-9$. $t=-1$: $y=-2x-1$.

Answer: $y=6x-9$ and $y=-2x-1$

Example 19 (JEE Main 2022, 24 June Shift 1). If the tangent at the point $(x_1,y_1)$ on the curve $y=x^3+3x^2+5$ passes through the origin, then $(x_1,y_1)$ does NOT lie on the curve

▸ SHOW SOLUTION

Step 1. Through the origin

The tangent passes through $(0,0)$ when $y_1=x_1\,y'(x_1)$:

$$x_1^3+3x_1^2+5=3x_1^3+6x_1^2$$$$2x_1^3+3x_1^2-5=0$$

Step 2. Solve

$$(x_1-1)(2x_1^2+5x_1+5)=0$$

The quadratic has $D=25-40<0$, so $x_1=1$, $y_1=9$.

Step 3. Test

(A)$1+1=2$,

(B)$9-1=8$,

(C)$4+5=9$: all hold.

(D)$\frac13-81\ne2$.

Answer: option (D)

Example 20 (JEE Main 2021, 24 February Shift 1). If the tangent to the curve $y=x^3$ at the point $P(t,t^3)$ meets the curve again at $Q$, then the ordinate of the point which divides $PQ$ internally in the ratio $1:2$ is

▸ SHOW SOLUTION

Step 1. Tangent at P

$$y=3t^2x-2t^3$$

Step 2. Meet the curve

$$x^3-3t^2x+2t^3=(x-t)^2(x+2t)$$

So $Q=(-2t,-8t^3)$.

Step 3. Divide PQ in 1 : 2

$$y=\frac{2(t^3)+1(-8t^3)}{3}=-2t^3$$

Answer: option (1)

Example 21 (JEE Main 2025, 4 April Shift 2). A line passing through the point $A(-2,0)$ touches the parabola $P: y^2=x-2$ at the point $B$ in the first quadrant. The area of the region bounded by the line $AB$, the parabola $P$ and the $x$-axis is

▸ SHOW SOLUTION

Step 1. Point of contact

Take $B=(t^2+2,\ t)$, $t>0$. From $2yy'=1$ the tangent slope is $\frac1{2t}$; the slope of $AB$ is $\frac{t}{t^2+4}$:

$$\frac1{2t}=\frac t{t^2+4}\ \Rightarrow\ t^2=4,\ t=2$$

So $B=(6,2)$ and $AB$ is $x=4y-2$.

Step 2. Area with horizontal strips

For $0\le y\le2$ the region lies between the line and the parabola $x=y^2+2$:

$$A=\int_0^2\big[(y^2+2)-(4y-2)\big]dy$$$$=\left[\frac{y^3}3-2y^2+4y\right]_0^2=\frac83$$

Answer: option (3)

Example 22 (IITianForum). The tangent to $y=x^4-8x^2+3x$ at $(\alpha_1,\beta_1)$ touches the curve again at $(\alpha_2,\beta_2)$. Find $|\alpha_1|+|\alpha_2|+|\beta_1|+|\beta_2|$.

▸ SHOW SOLUTION

Step 1. A line touching twice

A line $y=mx+c$ touches the quartic at two points exactly when the difference is a perfect square:

$$x^4-8x^2+3x-(mx+c)$$$$=(x-p)^2(x-q)^2$$

Step 2. Match

$x^4-8x^2+16=(x^2-4)^2$, so $m=3$, $c=-16$ works with points of contact $x=\pm2$.

Step 3. The points

$$(2,-10),\qquad (-2,-22)$$

Answer: $2+2+10+22=36$

Tangent with a given slope and common tangents

Given slope. For the tangent with slope $m$, solve $f'(x)=m$ for the point of contact, then write the line. A normal parallel to a given line means the tangent has the negative reciprocal slope.

Common tangent. A line is a common tangent of $y=f(x)$ and $y=g(x)$ when the tangent at $(p,f(p))$ and the tangent at $(r,g(r))$ are the same line. Equate their slopes and their $y$-intercepts:

$$f'(p)=g'(r)$$ $$f(p)-pf'(p)=g(r)-rg'(r)$$

Curves touching at a point. Two curves touch at $P$ when $P$ lies on both and the slopes there are equal.

The parabola y equals x squared and the curve y equals minus 64 over x in the fourth quadrant, with the straight line y equals 16x minus 64 touching the parabola at 8 comma 64 and the other curve at 2 comma minus 32

Example 23 (JEE Advanced 2026, Paper 1). Let $P$ be the point on the parabola $y=x^2$ such that the slope of the tangent to the parabola at $P$ is $4$. Let $Q$ be the point in the first quadrant on the circle $x^2+y^2=2$ such that the slope of the tangent to the circle at $Q$ is $-1$. Let $R$ be the point in the first quadrant on the ellipse $x^2+4y^2=8$ such that the slope of the tangent to the ellipse at $R$ is $-\frac12$. Then the radius of the circle passing through $P$, $Q$ and $R$ is

▸ SHOW SOLUTION

Step 1. P

$2x=4$, so $P=(2,4)$.

Step 2. Q

$2x+2yy'=0$ gives $y'=-\frac xy=-1$, so $x=y$ and $2x^2=2$: $Q=(1,1)$.

Step 3. R

$2x+8yy'=0$ gives $y'=-\frac x{4y}=-\frac12$, so $x=2y$ and $8y^2=8$: $R=(2,1)$.

Step 4. The circle

$PR=3$, $QR=1$ and $PQ=\sqrt{10}$, with $PR^2+QR^2=PQ^2$. The angle at $R$ is right, so $PQ$ is a diameter:

$$\text{radius}=\frac{\sqrt{10}}2=\sqrt{\frac52}$$

Answer: option (3)

Example 24 (JEE Main 2023, 30 January Shift 1). The number of points on the curve $y=54x^5-135x^4-70x^3+180x^2+210x$ at which the normal lines are parallel to $x+90y+2=0$ is

▸ SHOW SOLUTION

Step 1. Tangent slope needed

The normal has slope $-\frac1{90}$, so the tangent has slope $90$:

$$270x^4-540x^3-210x^2$$$$+360x+210=90$$$$9x^4-18x^3-7x^2+12x+4=0$$

Step 2. Factor

$x=1$ and $x=2$ are roots:

$$(x-1)(x-2)(9x^2+9x+2)=0$$$$9x^2+9x+2=(3x+1)(3x+2)$$

Step 3. Count

Four distinct real values $1,2,-\frac13,-\frac23$, so four points.

Answer: option (3)

Example 25 (JEE Main 2021, 16 March Shift 1). If the normal to the curve $y(x)=\int_0^x(2t^2-15t+10)\,dt$ at a point $(a,b)$ is parallel to the line $x+3y=-5$, $a>1$, then the value of $|a+6b|$ is equal to ______.

▸ SHOW SOLUTION

Step 1. Slope of the tangent

The normal has slope $-\frac13$, so $y'(a)=3$:

$$2a^2-15a+10=3$$$$(2a-1)(a-7)=0$$

With $a>1$, $a=7$.

Step 2. b

$$b=\int_0^7(2t^2-15t+10)\,dt$$$$b=\frac{686}3-\frac{735}2+70$$$$b=-\frac{413}6$$

Answer: $|7-413|=406$

Example 26 (JEE Main 2022, 29 July Shift 2). If the tangent to the curve $y=x^3-x^2+x$ at the point $(a,b)$ is also tangent to the curve $y=5x^2+2x-25$ at the point $(2,-1)$, then $|2a+9b|$ is equal to ______.

▸ SHOW SOLUTION

Step 1. Tangent at (2, −1)

Slope $10x+2=22$:

$$y=22x-45$$

Step 2. Same slope on the cubic

$$3a^2-2a+1=22$$$$(a-3)(3a+7)=0$$

Step 3. Same line

$a=3$: $b=27-9+3=21$ and $22(3)-45=21$, so the point lies on the line.

$a=-\frac73$: $b=-\frac{553}{27}$, but the line gives $-\frac{289}3$. Rejected.

Answer: $|6+189|=195$

Example 27 (IITianForum). There is a point $(p,q)$ on $f(x)=x^2$ and a point $(r,s)$ on $g(x)=-\frac{64}x$, with $p>0$, $r>0$. The line through these two points is tangent to both curves at these points. Find $p+r$.

▸ SHOW SOLUTION

Step 1. The two tangents

At $(p,p^2)$: $y=2px-p^2$.

At $\left(r,-\frac{64}r\right)$, slope $\frac{64}{r^2}$: $y=\frac{64}{r^2}x-\frac{128}r$.

Step 2. Equate slope and intercept

$$2p=\frac{64}{r^2},\qquad p^2=\frac{128}r$$

Step 3. Solve

From the first, $p=\frac{32}{r^2}$; then $\frac{1024}{r^4}=\frac{128}r$, so $r^3=8$, $r=2$, $p=8$.

Answer: $p+r=10$

Example 28 (IITianForum). The parabolas $y=x^2+ax+b$ and $y=x(c-x)$ touch each other at the point $(2,0)$. Find $a+b+c$.

▸ SHOW SOLUTION

Step 1. Second curve through (2, 0)

$$0=2(c-2)\ \Rightarrow\ c=2$$

Step 2. Equal slopes at x = 2

$$2x+a=c-2x$$$$4+a=-2,\qquad a=-6$$

Step 3. First curve through (2, 0)

$$4-12+b=0\ \Rightarrow\ b=8$$

Answer: $a+b+c=4$

Normals and shortest distance

Normal through a given point. The normal at $(t,f(t))$ passes through $(h,k)$ when

$$(h-t)+\big(k-f(t)\big)f'(t)=0$$

Shortest distance. If $P$ on a curve is nearest to a point $A$ off the curve, then $AP$ is along the normal at $P$. If $P$ is nearest to a line, the tangent at $P$ is parallel to that line. For two curves, the shortest segment is along a common normal.

Example 29 (JEE Main 2019, 10 January Shift 2). A helicopter is flying along the curve given by $y-x^{3/2}=7$, $(x\ge0)$. A soldier positioned at the point $\left(\frac12,7\right)$ wants to shoot down the helicopter when it is nearest to him. Then this nearest distance is

▸ SHOW SOLUTION

Step 1. Normal through the soldier

At $P\left(x,x^{3/2}+7\right)$, $y'=\frac32x^{1/2}$. The normal at $P$ passes through $\left(\frac12,7\right)$:

$$\left(\frac12-x\right)-x^{3/2}\cdot\frac32x^{1/2}=0$$$$3x^2+2x-1=0\ \Rightarrow\ x=\frac13$$

Step 2. Distance

$$d^2=\left(\frac13-\frac12\right)^2+\left(\frac13\right)^3$$$$d^2=\frac1{36}+\frac1{27}=\frac7{108}$$$$d=\frac16\sqrt{\frac73}$$

Answer: option (3)

Example 30 (JEE Main 2025, 8 April Shift 2). Let $r$ be the radius of the circle which touches the $x$-axis at the point $(a,0)$, $a<0$, and the parabola $y^2=9x$ at the point $(4,6)$. Then $r$ is equal to ______.

▸ SHOW SOLUTION

Step 1. The centre lies on the normal at (4, 6)

$2yy'=9$ gives tangent slope $\frac34$, so the normal slope is $-\frac43$. The centre is $(a,r)$:

$$r-6=-\frac43(a-4)$$

Step 2. Distance from the centre to (4, 6) is r

Put $u=a-4<0$. Then $(r-6)^2=\frac{16}9u^2$ and

$$u^2+\frac{16}9u^2=r^2\ \Rightarrow\ r=-\frac53u$$

Step 3. Solve

$$-\frac53u=6-\frac43u\ \Rightarrow\ u=-18$$

So $a=-14$ and $r=30$.

Answer: $30$

Example 31 (IITianForum). At the point $P(a,3a^n)$, $a>0$, on $y=3x^n$, $n\in\mathbb N$, the normal is drawn; it meets the $y$-axis at $(0,b)$. If $\lim_{a\to0^+}b=\frac16$, find $n$.

▸ SHOW SOLUTION

Step 1. The normal

Slope of the tangent is $3na^{n-1}$:

$$y-3a^n=-\frac1{3na^{n-1}}(x-a)$$

Step 2. y-intercept

Put $x=0$:

$$b=3a^n+\frac{a^{2-n}}{3n}$$

Step 3. The limit

$n=1$: $b=3a+\frac a3\to0$. $n=2$: $b=3a^2+\frac16\to\frac16$. $n\ge3$: $\frac{a^{2-n}}{3n}\to\infty$.

Answer: $n=2$

Example 32 (JEE Main 2025, 3 April Shift 1). The radius of the smallest circle which touches the parabolas $y=x^2+2$ and $x=y^2+2$ is

The parabola y equals x squared plus 2 opening upwards and its mirror image x equals y squared plus 2 opening to the right, symmetric about the dashed line y equals x, with a small circle between them touching both at the points B and A that are mirror images

▸ SHOW SOLUTION

Step 1. Symmetry

The parabolas are mirror images in $y=x$. The nearest points $B$ and $A$ are mirror images, and the smallest touching circle has $AB$ as diameter.

Step 2. Nearest point to the line y = x

The tangent at $B$ is parallel to $y=x$:

$$2x=1,\qquad B=\left(\frac12,\frac94\right)$$$$A=\left(\frac94,\frac12\right)$$

Step 3. Radius

$$AB=\sqrt2\left(\frac94-\frac12\right)=\frac{7\sqrt2}4$$

Answer: radius $\frac{7\sqrt2}8$, option (4)

Angle between two curves

The angle between two curves at a common point is the angle between their tangents there. With tangent slopes $m_1$ and $m_2$,

$$\tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right|$$

• Touch: $m_1=m_2$, so $\theta=0$.
• Cut orthogonally: $m_1m_2=-1$, or one tangent horizontal and the other vertical.

Two curves may meet at several points and the angle can differ from point to point. Find the points of intersection first.

The curves y equals x squared and y equals x cubed meeting at the point 1 comma 1, with their tangent lines of slopes 2 and 3 drawn through that point and the small angle theta between the tangents marked

Example 33 (IITianForum). Find the angle between $y=x^2$ and $y=x^3$ at each point where they meet.

▸ SHOW SOLUTION

Step 1. Points

$x^2=x^3$ gives $x=0$ or $x=1$: the points $(0,0)$ and $(1,1)$.

Step 2. At (0, 0)

Both slopes ($2x$ and $3x^2$) are $0$. The curves touch: $\theta=0$.

Step 3. At (1, 1)

$m_1=2$, $m_2=3$:

$$\tan\theta=\frac{3-2}{1+6}=\frac17$$

Answer: $0$ at the origin; $\tan^{-1}\frac17$ at $(1,1)$

Example 34 (JEE Main 2018). If the curves $y^2=6x$ and $9x^2+by^2=16$ intersect each other at right angles, then the value of $b$ is

The parabola y squared equals 6x and the ellipse 9x squared plus 9 over 2 y squared equals 16 crossing at two points, with the tangent lines of both curves drawn at the upper crossing point meeting at a right angle

▸ SHOW SOLUTION

Step 1. Slopes at a common point (x, y)

$$y^2=6x:\quad m_1=\frac3y$$$$9x^2+by^2=16:\quad m_2=-\frac{9x}{by}$$

Step 2. Right angle

$$m_1m_2=-\frac{27x}{by^2}=-1$$$$27x=by^2=6bx$$

At a common point $x\ne0$ (at $x=0$ the second curve needs $by^2=16$, but the first gives $y=0$). So $b=\frac92$.

Answer: option (2)

Example 35 (JEE Main 2021, 25 February Shift 2). If the curves $x=y^4$ and $xy=k$ cut at right angles, then $(4k)^6$ is equal to ______.

▸ SHOW SOLUTION

Step 1. Slopes

$$x=y^4:\quad \frac{dy}{dx}=\frac1{4y^3}$$$$xy=k:\quad \frac{dy}{dx}=-\frac yx$$

Step 2. Right angle

$$\frac1{4y^3}\cdot\frac yx=1\ \Rightarrow\ 4xy^2=1$$

With $x=y^4$: $4y^6=1$.

Step 3. k

$k=xy=y^5$, so

$$(4k)^6=4^6y^{30}=4^6\left(\frac14\right)^5=4$$

Answer: $4$

Lengths of tangent, normal, sub-tangent and sub-normal

Let the tangent at $P(x_1,y_1)$ meet the $x$-axis at $T$, the normal meet it at $N$, and let $M$ be the foot of the perpendicular from $P$ to the $x$-axis. With $m$ the slope at $P$:

A curve with a point P, the tangent from P meeting the x-axis at T, the normal from P meeting the x-axis at N, and the foot M of the perpendicular from P, with TM marked as the sub-tangent, MN as the sub-normal, PT as the tangent length and PN as the normal length

Each comes from the right triangles $PMT$ and $PMN$, with $PM=|y_1|$ and $\tan\angle PTM=|m|$.

Two standard facts: for $y^2=4ax$ the sub-normal is the constant $2a$; for $xy=c^2$ the sub-tangent at any point equals $|x|$.

Example 36 (IITianForum). For $y=x^2$ at $(2,4)$, find the sub-tangent, sub-normal, length of tangent and length of normal.

▸ SHOW SOLUTION

Step 1. Slope

$m=2x=4$ and $y_1=4$.

Step 2. The four lengths

$$TM=\frac44=1,\qquad MN=4\cdot4=16$$$$PT=1\cdot\sqrt{17},\qquad PN=4\sqrt{17}$$

Answer: $1$, $16$, $\sqrt{17}$, $4\sqrt{17}$

Example 37 (IITianForum). For which of these curves is the length of the sub-normal the same at every point?

▸ SHOW SOLUTION

Step 1. Sub-normal = |y y′|

(A) $yy'=x^2\cdot2x=2x^3$: varies.

(B) $2yy'=5$, so $yy'=\frac52$: constant.

(C) $y'=-\frac4{x^2}$, $yy'=-\frac{16}{x^3}$: varies.

(D) $yy'=e^{2x}$: varies.

Answer: option (B)

For $y=e^x$ it is the sub-tangent $\left\lvert\frac y{y'}\right\rvert=1$ that is constant.

Rolle's theorem

If $f$ is

1. continuous on $[a,b]$,
2. differentiable on $(a,b)$, and
3. $f(a)=f(b)$,

then there is at least one $c\in(a,b)$ with

$$f'(c)=0$$

Geometric meaning. A smooth curve that returns to its starting height has a horizontal tangent somewhere in between.

Why. A continuous $f$ on $[a,b]$ has a greatest and a least value. If both occur at the ends, $f$ is constant and $f'=0$ everywhere. Otherwise one of them occurs at an inside point $c$, where the curve turns, so $f'(c)=0$.

A smooth curve rising from height f of a at x equals a to a peak and coming back down to the same height at x equals b, with the horizontal tangent at the peak above the point c and the equal end heights joined by a dashed line

Algebraic meaning. With $f(a)=f(b)=0$: between two roots of $f$ lies a root of $f'$.

All three conditions are needed, and the theorem says nothing about how many such $c$ there are or where they lie.

Three small graphs where Rolle's theorem fails: on the left a function with a jump inside the interval and equal end values but no horizontal tangent; in the middle y equals the modulus of x on minus 1 to 1 with a corner at the origin; on the right an increasing line whose end values differ

Example 38 (IITianForum). Verify Rolle's theorem for $f(x)=x^3-6x^2+11x-6$ on $[1,3]$ and find $c$.

▸ SHOW SOLUTION

Step 1. Conditions

$f$ is a polynomial, so continuous and differentiable. $f(1)=0$ and $f(3)=27-54+33-6=0$.

Step 2. Solve f′(c) = 0

$$3c^2-12c+11=0\ \Rightarrow\ c=2\pm\frac1{\sqrt3}$$

Both lie in $(1,3)$.

Answer: $c=2-\frac1{\sqrt3}$ and $c=2+\frac1{\sqrt3}$

Example 39 (IITianForum). For $x\in[-1,3]$ let

$$f(x)=\frac{x^3-4x^2+x+6}{x-2}\quad(x\ne2)$$

and $f(2)=-3$.

Then (A) $f$ is discontinuous at $x=2$, so Rolle's theorem cannot be applied; (B) $f(-1)\ne f(3)$, so Rolle's theorem cannot be applied; (C) Rolle's theorem applies and $c=1$; (D) Rolle's theorem applies and $c=\frac12$.

▸ SHOW SOLUTION

Step 1. Simplify

$$x^3-4x^2+x+6=(x-2)(x^2-2x-3)$$

So $f(x)=x^2-2x-3$ for $x\ne2$, and this also gives $f(2)=-3$. Hence $f(x)=x^2-2x-3$ on all of $[-1,3]$.

Step 2. Conditions

A polynomial is continuous and differentiable; $f(-1)=0=f(3)$.

Step 3. c

$f'(c)=2c-2=0$, so $c=1$.

Answer: option (C)

Example 40 (JEE Main 2020, 8 January Shift 1). If $c$ is a point at which Rolle's theorem holds for the function $f(x)=\log_e\left(\frac{x^2+\alpha}{7x}\right)$ in the interval $[3,4]$, where $\alpha\in\mathbb R$, then $f''(c)$ is equal to

▸ SHOW SOLUTION

Step 1. f(3) = f(4)

$$\frac{9+\alpha}{21}=\frac{16+\alpha}{28}\ \Rightarrow\ \alpha=12$$

Step 2. f′(c) = 0

$f(x)=\ln(x^2+12)-\ln7-\ln x$:

$$f'(x)=\frac{2x}{x^2+12}-\frac1x=0\ \Rightarrow\ x^2=12$$

So $c=2\sqrt3\in(3,4)$.

Step 3. f″(c)

$$f''(x)=\frac{24-2x^2}{(x^2+12)^2}+\frac1{x^2}$$

At $x^2=12$ the first part is $0$, so $f''(c)=\frac1{12}$.

Answer: option (A)

Example 41 (JEE Main 2014). If $f$ and $g$ are differentiable functions in $[0,1]$ satisfying $f(0)=2=g(1)$, $g(0)=0$ and $f(1)=6$, then for some $c\in(0,1)$

▸ SHOW SOLUTION

Step 1. Build equal end values

Try $h=f-2g$:

$$h(0)=2-0=2,\qquad h(1)=6-4=2$$

Step 2. Rolle

$h(0)=h(1)$, so $h'(c)=0$ for some $c\in(0,1)$:

$$f'(c)=2g'(c)$$

Answer: option (2)

Example 42 (JEE Main 2020, 8 January Shift 2). Let $S$ be the set of all functions $f:[0,1]\to\mathbb R$ which are continuous on $[0,1]$ and differentiable on $(0,1)$. Then for every $f$ in $S$ there exists a $c\in(0,1)$, depending on $f$, such that

▸ SHOW SOLUTION

Step 1. (A) and (B)

Take $f$ constant. Both sides are $0$ and the strict inequality $0<0$ fails.

Step 2. (C)

Take $f(x)=x^2$:

$$\frac{1-c^2}{1-c}=1+c=2c\ \Rightarrow\ c=1$$

which is not in $(0,1)$.

Answer: option (D)

Lagrange's theorem on $[c,1]$ gives a point $d$ with $f(1)-f(c)=(1-c)f'(d)$; the point $d$ need not be $c$.

Example 43 (JEE Advanced 2015, Paper 2). Let $f,g:[-1,2]\to\mathbb R$ be continuous functions which are twice differentiable on $(-1,2)$. The values of $f$ and $g$ at $-1$, $0$ and $2$ are

In each of the intervals $(-1,0)$ and $(0,2)$ the function $(f-3g)''$ never vanishes. Then the correct statement(s) is (are)

▸ SHOW SOLUTION

Step 1. h = f − 3g

$$h(-1)=3,\qquad h(0)=6-3=3$$$$h(2)=0+3=3$$

Step 2. At least one in each interval

Rolle on $[-1,0]$ and on $[0,2]$ gives a zero of $h'$ in each interval.

Step 3. At most one in each

If $h'$ had two zeros in one interval, Rolle applied to $h'$ would give a zero of $h''$ there, which is not allowed.

Answer: (B) and (C)

Example 44 (JEE Main 2022, 28 July Shift 1). Let $f:[0,1]\to\mathbb R$ be a twice differentiable function in $(0,1)$ such that $f(0)=3$ and $f(1)=5$. If the line $y=2x+3$ intersects the graph of $f$ at only two distinct points in $(0,1)$, then the least number of points $x\in(0,1)$ at which $f''(x)=0$ is ______.

▸ SHOW SOLUTION

Step 1. g = f − (2x + 3)

$g(0)=3-3=0$ and $g(1)=5-5=0$. The two meeting points $x_1<x_2$ in $(0,1)$ are zeros of $g$ too:

$$0<x_1<x_2<1,\quad g=0 \text{ at all four}$$

Step 2. Rolle twice

Four zeros of $g$ give at least three zeros of $g'$, and these give at least two zeros of $g''$.

Step 3. Back to f

$g''=f''$, so $f''=0$ at least twice. Two is reached by $f(x)=2x+3+x(3x-1)(3x-2)(x-1)$, whose $f''$ is a quadratic with two roots in $(0,1)$.

Answer: $2$

Counting roots with Rolle

• If $f$ has $n$ distinct real roots, $f'$ has at least $n-1$, one between each consecutive pair.
• So if $f'$ has $k$ real roots, $f$ has at most $k+1$.
• If $f'$ has no real root, $f$ has at most one; a sign change of $f$ then gives exactly one.
• If $f''$ never vanishes, $f'$ has at most one root and $f$ at most two.

To count exactly: bound the number from above with Rolle, then find that many sign changes of $f$.

A wavy curve crossing the x-axis at three points, with the derivative's two zeros marked as horizontal tangents lying one between each consecutive pair of roots

Example 45 (IIT JEE 2011, Paper 2). The number of distinct real roots of $x^4-4x^3+12x^2+x-1=0$ is ______.

▸ SHOW SOLUTION

Step 1. f″ never vanishes

$$f''(x)=12x^2-24x+24$$$$f''(x)=12\big((x-1)^2+1\big)>0$$

So $f'$ has at most one root and $f$ at most two.

Step 2. Two sign changes

$f(0)=-1<0$, while $f(x)\to+\infty$ as $x\to\pm\infty$. So $f$ has a root on each side of $0$.

Answer: $2$

Example 46 (JEE Advanced 2013, Paper 1). The number of points in $(-\infty,\infty)$ for which $x^2-x\sin x-\cos x=0$ is

▸ SHOW SOLUTION

Step 1. Derivative

$$f'(x)=2x-\sin x-x\cos x+\sin x$$$$f'(x)=x(2-\cos x)$$

$2-\cos x>0$, so $f'$ has the sign of $x$: $f$ falls on $(-\infty,0)$ and rises on $(0,\infty)$.

Step 2. Values

$f(0)=-1<0$ and $f(x)\ge x^2-|x|-1\to\infty$ as $|x|\to\infty$.

So $f$ crosses zero exactly once on each side.

Answer: option (C)

Example 47 (JEE Advanced 2014, Paper 1). Let $a\in\mathbb R$ and let $f:\mathbb R\to\mathbb R$ be given by $f(x)=x^5-5x+a$. Then

▸ SHOW SOLUTION

Step 1. At most three roots

$f'(x)=5(x^4-1)$ vanishes only at $x=\pm1$, so $f$ has at most three real roots.

Step 2. The values at ±1

$$f(-1)=a+4,\qquad f(1)=a-4$$

$f$ rises on $(-\infty,-1)$, falls on $(-1,1)$, rises on $(1,\infty)$.

Step 3. Cases

$-4<a<4$: $f(-1)>0>f(1)$, three sign changes, three roots. $a>4$: $f(1)>0$, so only the root left of $-1$. $a<-4$: $f(-1)<0$, so only the root right of $1$.

Answer: (B) and (D)

Helper functions

Rolle applied to a well-chosen product shows that an equation involving $f$ and $f'$ has a root.

The reason: $h'(x)=e^{kx}\big(f'(x)+kf(x)\big)$ and $e^{kx}>0$, so a zero of $h'$ is a zero of $f'+kf$. The same helper also turns an inequality such as $f'>kf$ into "$h$ is increasing".

Example 48 (IITianForum). $f$ is differentiable on $[a,b]$ and $f(a)=f(b)=0$. Show that $f'(c)+3f(c)=0$ for some $c\in(a,b)$.

▸ SHOW SOLUTION

Step 1. Helper

$h(x)=e^{3x}f(x)$ is differentiable and $h(a)=h(b)=0$.

Step 2. Rolle

$h'(c)=0$ for some $c\in(a,b)$:

$$e^{3c}\big(f'(c)+3f(c)\big)=0$$

Since $e^{3c}>0$, $f'(c)+3f(c)=0$.

Answer: proved

Example 49 (JEE Advanced 2017, Paper 2). If $f:\mathbb R\to\mathbb R$ is a differentiable function such that $f'(x)>2f(x)$ for all $x\in\mathbb R$, and $f(0)=1$, then

▸ SHOW SOLUTION

Step 1. Helper

$h(x)=e^{-2x}f(x)$:

$$h'(x)=e^{-2x}\big(f'(x)-2f(x)\big)>0$$

Step 2. Compare with h(0)

$h(0)=1$ and $h$ is increasing, so for $x>0$, $h(x)>1$, i.e. $f(x)>e^{2x}$. This is (A).

Step 3. Monotonic

For $x>0$: $f'(x)>2f(x)>2e^{2x}>0$. So $f$ is increasing, and also $f'(x)>e^{2x}$, so (B) is false.

Answer: (A) and (C)

Example 50 (JEE Advanced 2013, Paper 2). Let $f:[0,1]\to\mathbb R$ be twice differentiable with $f(0)=f(1)=0$ and $f''(x)-2f'(x)+f(x)\ge e^x$ for $x\in[0,1]$.

(i) Which is true for $0<x<1$? (A) $0<f(x)<\infty$ (B) $-\frac12<f(x)<\frac12$ (C) $-\frac14<f(x)<1$ (D) $-\infty<f(x)<0$

(ii) If $e^{-x}f(x)$ has its minimum in $[0,1]$ at $x=\frac14$, which is true? (A) $f'(x)<f(x)$, $\frac14<x<\frac34$ (B) $f'(x)>f(x)$, $0<x<\frac14$ (C) $f'(x)<f(x)$, $0<x<\frac14$ (D) $f'(x)<f(x)$, $\frac34<x<1$

▸ SHOW SOLUTION

Step 1. Helper

$g(x)=e^{-x}f(x)$. Then $g(0)=g(1)=0$ and

$$g'(x)=e^{-x}\big(f'(x)-f(x)\big)$$$$g''(x)=e^{-x}\big(f''-2f'+f\big)\ge1$$

So $g'$ is strictly increasing on $[0,1]$.

Step 2. Part (i)

Suppose $g(x_0)\ge0$ for some $x_0\in(0,1)$. Lagrange's theorem on $[0,x_0]$ and $[x_0,1]$ gives $p<x_0<q$ with

$$g'(p)=\frac{g(x_0)}{x_0}\ge0$$$$g'(q)=-\frac{g(x_0)}{1-x_0}\le0$$

Then $g'(q)\le g'(p)$ with $p<q$, against $g'$ strictly increasing. So $g<0$, hence $f<0$, on $(0,1)$.

Step 3. Part (ii)

The minimum is at $\frac14$, so $g'\left(\frac14\right)=0$; with $g'$ increasing, $g'<0$ on $\left(0,\frac14\right)$. There $f'(x)<f(x)$.

Answer: (i) option (D); (ii) option (C)

Lagrange's mean value theorem

If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, there is a $c\in(a,b)$ with

$$f'(c)=\frac{f(b)-f(a)}{b-a}$$

Geometric meaning. Some tangent is parallel to the chord joining the end points.

Why. The function $h(x)=f(x)-\frac{f(b)-f(a)}{b-a}(x-a)$ satisfies $h(a)=h(b)=f(a)$; Rolle gives $h'(c)=0$, which is the statement. Rolle's theorem is the special case $f(a)=f(b)$.

The curve y equals the square root of x between x equals 1 and x equals 4, the chord joining the end points, and the tangent at x equals 9 over 4 drawn parallel to the chord

Working form.

$$f(b)-f(a)=(b-a)f'(c)$$

Example 51 (IITianForum). Find $c$ of Lagrange's theorem for $f(x)=\sqrt x$ on $[1,4]$.

▸ SHOW SOLUTION

Step 1. Chord slope

$$\frac{2-1}{4-1}=\frac13$$

Step 2. Solve

$$\frac1{2\sqrt c}=\frac13\ \Rightarrow\ c=\frac94\in(1,4)$$

Answer: $c=\frac94$

Example 52 (JEE Main 2020, 6 September Shift 2). If the tangent to the curve $y=f(x)=x\log_ex$, $x>0$, at a point $(c,f(c))$ is parallel to the line segment joining the points $(1,0)$ and $(e,e)$, then $c$ is equal to

▸ SHOW SOLUTION

Step 1. Chord slope

$f(1)=0$ and $f(e)=e$:

$$\frac{e-0}{e-1}=\frac e{e-1}$$

Step 2. Solve f′(c)

$$\ln c+1=\frac e{e-1}\ \Rightarrow\ \ln c=\frac1{e-1}$$

Answer: option (D)

Example 53 (JEE Main 2019, 9 April Shift 1). Let $S$ be the set of all values of $x$ for which the tangent to the curve $y=f(x)=x^3-x^2-2x$ at $(x,y)$ is parallel to the line segment joining the points $(1,f(1))$ and $(-1,f(-1))$. Then $S$ is equal to

▸ SHOW SOLUTION

Step 1. Chord slope

$f(1)=-2$ and $f(-1)=0$, so the slope is $\frac{-2-0}{2}=-1$.

Step 2. Solve

$$3x^2-2x-2=-1$$$$(3x+1)(x-1)=0$$

The question asks for every $x$ on the curve, not only those inside $(-1,1)$.

Answer: $S=\left\{-\frac13,1\right\}$, option (4)

Example 54 (IITianForum). $f$ is differentiable on $\mathbb R$ with $f(1)=3$ and $f'(x)\ge5$ for $x\in[1,4]$. Which is true? (A) $f(4)<18$ (B) $f(4)\ge18$ (C) $f(4)\ge20$ (D) none of these

▸ SHOW SOLUTION

Step 1. Lagrange on [1, 4]

$$f(4)-f(1)=3f'(c)\ge15$$

Step 2. Conclude

$f(4)\ge18$. The value $18$ is reached by $f(x)=5x-2$, so (C) need not hold.

Answer: option (B)

Example 55 (IITianForum). $f$ is differentiable on $\mathbb R$ with $f(0)=2$, $f(4)=9$ and $f'(x)\ge\frac32$ for all $x$. A possible value of $f(2)$ is

▸ SHOW SOLUTION

Step 1. On [0, 2]

$$f(2)-2=2f'(c_1)\ge3\ \Rightarrow\ f(2)\ge5$$

Step 2. On [2, 4]

$$9-f(2)=2f'(c_2)\ge3\ \Rightarrow\ f(2)\le6$$

Step 3. Options

Only $5.5$ lies in $[5,6]$. It is reached by $f(x)=2+\frac74x$, which has $f'=\frac74\ge\frac32$.

Answer: option (B)

Example 56 (IITianForum). Prove that for $0<a<b$,

$$\frac{b-a}b<\ln\frac ba<\frac{b-a}a$$

and deduce bounds for $\ln1.2$.

▸ SHOW SOLUTION

Step 1. Lagrange for ln x on [a, b]

$$\ln b-\ln a=\frac{b-a}c,\qquad a<c<b$$

Step 2. Bound 1/c

$\frac1b<\frac1c<\frac1a$, which gives the result.

Step 3. a = 5, b = 6

$$\frac16<\ln1.2<\frac15$$

Answer: $0.1667<\ln1.2<0.2$ (actual value $0.1823$)

Example 57 (IIT JEE 2009, Paper 2). For the function $f(x)=x\cos\frac1x$, $x\ge1$,

▸ SHOW SOLUTION

Step 1. f′ and f″

$$f'(x)=\cos\frac1x+\frac1x\sin\frac1x$$$$f''(x)=-\frac1{x^3}\cos\frac1x$$

For $x\ge1$, $0<\frac1x\le1$, so $\cos\frac1x>0$ and $f''<0$: (D) is true.

Step 2. Limit

As $x\to\infty$, $\frac1x\to0$, so $f'(x)\to1+0=1$: (B) is true.

Step 3. Lagrange on [x, x + 2]

$f'$ decreases to the limit $1$, so $f'>1$ on $[1,\infty)$:

$$f(x+2)-f(x)=2f'(c)>2$$

So (C) is true and (A) is false.

Answer: (B), (C) and (D)

Example 58 (JEE Advanced 2017, Paper 2). If $f:\mathbb R\to\mathbb R$ is a twice differentiable function such that $f''(x)>0$ for all $x\in\mathbb R$, and $f\left(\frac12\right)=\frac12$, $f(1)=1$, then

▸ SHOW SOLUTION

Step 1. Lagrange on [½, 1]

$$f'(c)=\frac{1-\frac12}{1-\frac12}=1,\qquad \frac12<c<1$$

Step 2. f′ is increasing

$f''>0$, so $f'(1)>f'(c)=1$.

Answer: option (B)

Example 59 (JEE Advanced 2018, Paper 1). For every twice differentiable function $f:\mathbb R\to[-2,2]$ with $(f(0))^2+(f'(0))^2=85$, which of the following statement(s) is (are) TRUE?

▸ SHOW SOLUTION

Step 1. (A)

$(f'(0))^2\ge85-4=81$, so $f'(0)\ne0$. As $f'$ is continuous, it keeps one sign on some $(r,s)$ around $0$, where $f$ is strictly monotonic, so one-one.

Step 2. (B)

Lagrange on $[-4,0]$:

$$|f'(x_0)|=\frac{|f(0)-f(-4)|}4\le\frac44=1$$

Step 3. (C)

$f(x)=2\sin\frac{\sqrt{85}\,x}2$ satisfies every condition: its values lie in $[-2,2]$, $f(0)=0$ and $f'(0)^2=85$. It has no limit at infinity, so (C) is false.

Step 4. (D)

Let $g=f^2+(f')^2$, so $g(0)=85$. By Step 2 there is $x_1\in(-4,0)$ with $g(x_1)\le4+1=5$; in the same way there is $x_2\in(0,4)$ with $g(x_2)\le5$. So $g$ has its greatest value on $[x_1,x_2]$ at an inside point $\alpha$, where

$$g'(\alpha)=2f'(\alpha)\big(f(\alpha)+f''(\alpha)\big)=0$$

There $g(\alpha)\ge85$, so $(f'(\alpha))^2\ge81$ and $f'(\alpha)\ne0$. Hence $f(\alpha)+f''(\alpha)=0$.

Answer: (A), (B) and (D)

Cauchy's mean value theorem

If $f$ and $g$ are continuous on $[a,b]$, differentiable on $(a,b)$, and $g'(x)\ne0$ on $(a,b)$, then there is a $c\in(a,b)$ with

$$\frac{f(b)-f(a)}{g(b)-g(a)}=\frac{f'(c)}{g'(c)}$$

Why. Apply Rolle to $h(x)=f(x)-\frac{f(b)-f(a)}{g(b)-g(a)}g(x)$, which has $h(a)=h(b)$. Taking $g(x)=x$ gives Lagrange's theorem.

Example 60 (IITianForum). Find $c$ of Cauchy's theorem for $f(x)=x^2$, $g(x)=x^3$ on $[1,2]$.

▸ SHOW SOLUTION

Step 1. Left side

$$\frac{4-1}{8-1}=\frac37$$

Step 2. Right side

$$\frac{2c}{3c^2}=\frac2{3c}=\frac37\ \Rightarrow\ c=\frac{14}9$$

$\frac{14}9\in(1,2)$.

Answer: $c=\frac{14}9$

Which method to use

A decision chart: from the question asked, arrows lead to eight situations, each paired with its method: a rate, differentiate the link in t and substitute after; tangent at a point, slope from the derivative; tangent from an outside point, contact point t and pass through the point; given slope or parallel line, solve f prime equals m; nearest point or common normal, normal through the point; angle or orthogonal curves, the tan theta formula; equal values or roots of f prime, Rolle or a helper function; bound on a difference, Lagrange

What JEE asks

Across the JEE Main papers from 2019 to 2025 and the JEE Advanced and IIT JEE papers, the chapter appears in these forms.

1. Tangent or normal with a condition: through a point, parallel to a line, at a given angle, or meeting the curve again (Examples 9 to 28). This is the most common form in JEE Main.
2. Related rates with spheres and cones (Examples 1 to 7).
3. Nearest point and angle between curves (Examples 29 to 35).
4. Rolle and helper functions, usually with equal values built from given data (Examples 38 to 50). This is the common form in JEE Advanced.
5. Lagrange's theorem to find $c$, to bound a difference, or inside a multi-statement question (Examples 51 to 59).

JEE Advanced 2026 asked for points of contact with given slopes on three curves (Example 23). Earlier JEE Advanced and IIT JEE papers used implicit slopes (Example 11), Rolle on $f-3g$ (Example 43), root counting (Examples 45 to 47), helper functions (Examples 49 and 50) and Lagrange inside statement questions (Examples 57 to 59).

Quick recall

Practise these methods on real exam questions in the doMath Tangents and Normals, Rolle's and Lagrange's Mean Value Theorem and Rate of Change chapters. The sign of $f'$ and the tests for extreme values are developed in the Monotonic Functions and Maxima and Minima notes, and corners and vertical tangents in the Continuity and Differentiability note.