JEE Corner — Application of Derivatives for JEE Main and Advanced

What JEE Actually Asks From This Chapter

Here is the current, official picture — worth knowing precisely, because it changed recently:

• JEE Main (per the official NTA syllabus): the Applications of Derivatives portion reads "Rate of change of quantities, monotonic — increasing and decreasing functions, maxima and minima of functions of one variable." That's it. Tangents & normals and Rolle's/Lagrange's Mean Value Theorems were dropped from the JEE Main syllabus in the 2024 rationalisation — they no longer appear in Main papers.
• JEE Advanced: the Differential Calculus syllabus still explicitly includes tangents and normals, increasing and decreasing functions, maxima and minima, Rolle's theorem and Lagrange's mean value theorem with their geometric interpretation.

So your strategy depends on your target:

1. JEE Main aspirants: go deep on monotonicity (especially with parameters) and maxima-minima (especially of non-polynomial functions). Expect 1-2 questions almost every session, frequently MCQ-style versions of optimisation and monotonic-interval problems.
2. JEE Advanced aspirants: everything in this section is testable, including tangents/normals, orthogonal curves, and MVT-based reasoning.

This section covers the full superset, with each topic labelled. Even pure Main aspirants should skim the tangent/normal ideas — they quietly power "minimum distance" and "common point of two curves" problems that ARE in the Main syllabus.

[JEE Tip] In recent Main papers, this chapter's questions increasingly hybridise with other units — monotonicity of $f(x) = \int_0^x g(t)\,dt$, extrema of functions defined with logs and exponentials, and root-counting via shape-of-graph arguments. The last notes block here covers that root-counting technique.

Tangents and Normals (JEE Advanced syllabus; tools useful everywhere)

The derivative at a point IS the slope of the tangent there. For a curve $y = f(x)$ and a point $P(x_0, y_0)$ on it:

• Tangent at $P$: slope $m = f'(x_0)$, equation $y - y_0 = f'(x_0)(x - x_0)$
• Normal at $P$: perpendicular to the tangent, slope $-\frac{1}{f'(x_0)}$ (when $f'(x_0) \neq 0$), equation $y - y_0 = -\frac{1}{f'(x_0)}(x - x_0)$

Special slopes to recognise instantly: tangent parallel to the $x$-axis $\Leftrightarrow f'(x_0) = 0$; tangent parallel to the $y$-axis $\Leftrightarrow f'(x_0)$ is undefined/infinite (for implicit curves, $\frac{dx}{dy} = 0$); tangent parallel to a line of slope $m$ $\Leftrightarrow f'(x_0) = m$; tangent perpendicular to a line of slope $m$ $\Leftrightarrow f'(x_0) = -\frac{1}{m}$.

Angle of intersection of two curves

At a common point, if the tangent slopes are $m_1$ and $m_2$:

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

• Orthogonal curves: $m_1 m_2 = -1$ at every intersection point.
• Touching curves: $m_1 = m_2$ (common tangent at the point).

Key Point: For implicit curves like $x^2 = 4y$ or $xy = c$, differentiate implicitly and evaluate at the actual intersection point — find that point first by solving the two equations together.

[JEE Advanced] Classic result worth caching: the curves $x = y^2$ and $xy = k$ cut at right angles if and only if $8k^2 = 1$ — proved in Example 5 below. Variants of this appear repeatedly.

Rolle's and Lagrange's Mean Value Theorems (JEE Advanced syllabus)

Rolle's Theorem. If $f$ is continuous on $[a, b]$, differentiable on $(a, b)$, and $f(a) = f(b)$, then there exists at least one $c \in (a, b)$ with $f'(c) = 0$.

Geometrically: equal endpoint heights force at least one horizontal tangent in between.

Lagrange's Mean Value Theorem (LMVT). If $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$, then there exists $c \in (a, b)$ with

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

Geometrically: some tangent is parallel to the chord joining the endpoints. (Rolle's is the special case where the chord is horizontal.)

The three JEE application patterns

1. Find the $c$: verify hypotheses, set $f'(c)$ equal to the average slope, solve. (Answers must lie strictly inside $(a,b)$ — check!)
2. Root existence: if a differentiable $f$ has two zeros, Rolle's gives a zero of $f'$ between them. Contrapositive: if $f'$ never vanishes on an interval, $f$ has at most one root there — the standard "exactly one real root" argument when combined with a sign change.
3. Inequalities: LMVT bounds a difference by a derivative bound: $|f(b) - f(a)| \leq M|b - a|$ where $M = \max|f'|$. Example: $|\sin a - \sin b| \leq |a - b|$ since $|\cos c| \leq 1$.

Key Point: MVT questions are about existence, not computation. The theorem never tells you what $c$ is in general — only that it exists. When a question asks for $c$ explicitly, the function is deliberately chosen so the equation solves cleanly.

[JEE Advanced] Watch the hypotheses. $f(x) = |x|$ on $[-1, 1]$ has $f(-1) = f(1)$ but NO horizontal tangent — Rolle's does not apply because $f$ is not differentiable at 0. Options that ignore a hypothesis are planted traps.

Monotonicity, JEE Style (core JEE Main syllabus)

Board questions ask "find the intervals". JEE questions hide monotonicity inside something else:

1. Parameter ranges

For what $a$ is $f(x) = x^3 + ax^2 + 3x + 100$ increasing on all of $\mathbb{R}$? Need $f'(x) = 3x^2 + 2ax + 3 \geq 0$ for all $x$: a positive-leading-coefficient quadratic is non-negative everywhere iff its discriminant $\leq 0$:

$$4a^2 - 36 \leq 0 \quad \Rightarrow \quad -3 \leq a \leq 3$$

Memorise the template: $px^2 + qx + r \geq 0\ \forall x \iff p > 0$ and $q^2 - 4pr \leq 0$ (allowing equality at isolated points).

2. Proving inequalities

To prove $g(x) > h(x)$ for $x > 0$: form $F = g - h$, show $F(0) = 0$ and $F'(x) > 0$ for $x > 0$. Classics every JEE aspirant should have done once:

• $\frac{x}{1+x} < \log(1+x) < x$ for $x > 0$
• $\sin x < x < \tan x$ for $0 < x < \frac{\pi}{2}$
• $e^x > 1 + x$ for $x \neq 0$

3. Root counting

A strictly monotonic continuous function has at most one real root; a sign change gives exactly one. For cubics with two turning points, compare the local max and local min values with 0: three distinct real roots iff $f(\text{local max}) \cdot f(\text{local min}) < 0$; a repeated root iff one of them is exactly 0.

Key Point: "Number of real solutions" questions in JEE are graph-shape questions. Find the turning values, sketch mentally, count crossings — no root formulas needed.

[JEE Main] When $f'$ has a squared factor — e.g. $f'(x) = (x-2)^2(x+1)$ — the function does NOT change monotonicity at the squared root. Monotonic-interval options that split at $x = 2$ are the trap.

Maxima–Minima, JEE Style (core JEE Main syllabus)

Beyond the Board-style optimisation of Section 4, JEE adds these recurring shapes:

1. Standard bounds — quote, don't derive

• $a\sin x + b\cos x \in \left[-\sqrt{a^2+b^2},\ \sqrt{a^2+b^2}\right]$
• $x + \frac{a^2}{x} \geq 2a$ for $x > 0$ (AM-GM; equality at $x = a$)
• $\frac{\log x}{x}$ peaks at $x = e$ (value $\frac{1}{e}$); $x^{1/x}$ peaks at $x = e$ (value $e^{1/e}$); $x^x$ bottoms out at $x = \frac{1}{e}$ (value $e^{-1/e}$)

2. Minimum distance from a point to a curve

Minimise the squared distance; at the optimum, the join $PQ$ is normal to the curve — a fact that can shortcut MCQs entirely.

3. Extrema of rational functions

For $y = \frac{x^2 - x + 1}{x^2 + x + 1}$-type functions, two clean methods: differentiate, or treat it as the range problem — cross-multiply to a quadratic in $x$ and force its discriminant $\geq 0$ for real $x$; the resulting inequality in $y$ gives the max and min directly.

4. Critical-point bookkeeping

From a factorised derivative like $f'(x) = x^{m}(x-a)^{n}\cdots$, count sign changes only at odd-multiplicity roots. Number of local extrema = number of sign changes of $f'$, not the number of roots of $f'$.

Key Point: JEE MCQs rarely need the full second-derivative ritual. Sign-of-$f'$ reasoning plus the standard bounds above answer most questions in under two minutes.

[JEE Main] A favourite twist: extrema of $f(|x|)$ or $|f(x)|$. Remember $|f|$ has possible extra minima where $f = 0$, and even functions automatically have a critical point at $x = 0$ — check corners separately, the derivative machinery alone misses them.

Solved Examples (JEE Pattern)

Example 1: Where the slope equals the height [JEE Main pattern]

Find the points on the curve $y = x^3$ at which the slope of the tangent equals the $y$-coordinate of the point.

Solution:

1. Slope $= \frac{dy}{dx} = 3x^2$; condition: $3x^2 = y = x^3$.
2. $x^3 - 3x^2 = 0 \Rightarrow x^2(x - 3) = 0 \Rightarrow x = 0$ or $x = 3$.
3. Points: $(0, 0)$ and $(3, 27)$.

Final Answer: $(0, 0)$ and $(3, 27)$.

Example 2: Tangent parallel to a given line [JEE Main pattern]

Find the equation of the tangent to $y^2 = 4x$ that is parallel to the line $y = 2x + 3$.

Solution:

1. Implicit differentiation: $2y\frac{dy}{dx} = 4 \Rightarrow \frac{dy}{dx} = \frac{2}{y}$.
2. Match slopes: $\frac{2}{y} = 2 \Rightarrow y = 1$, then $x = \frac{y^2}{4} = \frac{1}{4}$.
3. Tangent at $\left(\frac{1}{4}, 1\right)$: $y - 1 = 2\left(x - \frac{1}{4}\right)$, i.e. $y = 2x + \frac{1}{2}$.

Final Answer: $y = 2x + \frac{1}{2}$ (i.e. $4x - 2y + 1 = 0$).

Takeaway: For $y^2 = 4ax$, the tangent of slope $m$ is always $y = mx + \frac{a}{m}$ — here $a = 1, m = 2$ gives $y = 2x + \frac{1}{2}$ instantly. Knowing the conic shortcut converts a 3-minute problem into a 15-second one.

Example 3: Equation of a normal [JEE Advanced pattern]

Find the equation of the normal to $y = x^2 - 4x + 3$ at the point where the curve crosses the $x$-axis with the smaller abscissa.

Solution:

1. Crossing points: $x^2 - 4x + 3 = (x-1)(x-3) = 0 \Rightarrow x = 1, 3$; the smaller is $(1, 0)$.
2. Tangent slope: $y' = 2x - 4$, so at $x = 1$: $m = -2$.
3. Normal slope: $-\frac{1}{m} = \frac{1}{2}$; equation: $y - 0 = \frac{1}{2}(x - 1)$.

Final Answer: $x - 2y - 1 = 0$.

Example 4: Angle between two parabolas [JEE Advanced pattern]

Find the angle of intersection of the curves $y = x^2$ and $x = y^2$ at their intersection point other than the origin.

Solution:

1. Intersection: $y = x^2$ and $x = y^2$ give $x = x^4 \Rightarrow x = 0, 1$; the non-origin point is $(1, 1)$.
2. Slopes at $(1,1)$: for $y = x^2$: $m_1 = 2x = 2$. For $x = y^2$: $1 = 2y y' \Rightarrow m_2 = \frac{1}{2y} = \frac{1}{2}$.
3. Angle: $\tan\theta = \left|\frac{2 - \frac{1}{2}}{1 + 2 \cdot \frac{1}{2}}\right| = \frac{3/2}{2} = \frac{3}{4}$.

Final Answer: $\theta = \tan^{-1}\left(\frac{3}{4}\right)$.

Takeaway: At the origin these two curves meet at right angles (tangents along the axes) — a two-intersection curve pair can meet at different angles at different points.

Example 5: Orthogonality condition (a classic) [JEE Advanced pattern]

Show that the curves $x = y^2$ and $xy = k$ cut at right angles if $8k^2 = 1$.

Solution:

1. Intersection point: substituting $x = y^2$ into $xy = k$: $y^3 = k \Rightarrow y = k^{1/3}$, $x = k^{2/3}$.
2. Slopes: from $x = y^2$: $m_1 = \frac{1}{2y}$. From $xy = k$: $y + xy' = 0 \Rightarrow m_2 = -\frac{y}{x}$.
3. Orthogonality: $m_1 m_2 = \frac{1}{2y} \cdot \left(-\frac{y}{x}\right) = -\frac{1}{2x} = -1 \Rightarrow x = \frac{1}{2}$.
4. Convert to $k$: $x = k^{2/3} = \frac{1}{2} \Rightarrow k^2 = \frac{1}{8} \Rightarrow 8k^2 = 1$. ∎

Takeaway: Orthogonal-curves problems always reduce to $m_1m_2 = -1$ at the intersection point — never at a general point.

Example 6: Rolle's theorem, then a root count [JEE Advanced pattern]

(a) Verify Rolle's theorem for $f(x) = x^2 - 4x + 3$ on $[1, 3]$. (b) Show that $g(x) = x^3 + x - 1$ has exactly one real root.

Solution:

1. (a) Hypotheses: $f$ is a polynomial (continuous, differentiable everywhere) and $f(1) = 0 = f(3)$. ✓
2. Find $c$: $f'(x) = 2x - 4 = 0 \Rightarrow c = 2 \in (1, 3)$. Rolle's verified.
3. (b) At most one root: $g'(x) = 3x^2 + 1 > 0$ always, so $g$ is strictly increasing — it can cross zero at most once. (Equivalently: two roots would force $g'(c) = 0$ somewhere by Rolle's — impossible.)
4. At least one root: $g(0) = -1 < 0$ and $g(1) = 1 > 0$, so a root lies in $(0, 1)$ by continuity.

Final Answer: (a) $c = 2$; (b) exactly one real root (it lies in $(0,1)$). ∎

Example 7: LMVT — find the point, prove the bound [JEE Advanced pattern]

(a) Find the value of $c$ guaranteed by Lagrange's MVT for $f(x) = \log x$ on $[1, e]$. (b) Using LMVT, prove $|\sin a - \sin b| \leq |a - b|$ for all reals $a, b$.

Solution:

1. (a) Average slope $= \frac{\log e - \log 1}{e - 1} = \frac{1}{e - 1}$. Setting $f'(c) = \frac{1}{c} = \frac{1}{e-1}$ gives $c = e - 1$.
2. Check: $e - 1 \approx 1.72 \in (1, e)$. ✓
3. (b) For $a \neq b$, LMVT applied to $\sin x$ on $[a, b]$ gives $c$ with $\frac{\sin a - \sin b}{a - b} = \cos c$.
4. Taking absolute values: $\left|\frac{\sin a - \sin b}{a - b}\right| = |\cos c| \leq 1$, so $|\sin a - \sin b| \leq |a - b|$. ∎

Takeaway: Pattern (b) is the single most reused MVT argument in JEE Advanced — any function with a bounded derivative is Lipschitz: $|f(a) - f(b)| \leq M|a-b|$.

Example 8: The log sandwich inequality [JEE Main pattern]

Prove that $\frac{x}{1+x} < \log(1+x) < x$ for all $x > 0$.

Solution:

1. Right half: let $F(x) = x - \log(1+x)$. Then $F(0) = 0$ and $F'(x) = 1 - \frac{1}{1+x} = \frac{x}{1+x} > 0$ for $x > 0$, so $F(x) > 0$: $\log(1+x) < x$.
2. Left half: let $G(x) = \log(1+x) - \frac{x}{1+x}$. Then $G(0) = 0$ and $G'(x) = \frac{1}{1+x} - \frac{1}{(1+x)^2} = \frac{x}{(1+x)^2} > 0$ for $x > 0$, so $G(x) > 0$.

Final Answer: Both inequalities hold for every $x > 0$. ∎

Takeaway: The template — difference function, value 0 at the endpoint, positive derivative — proves nearly every JEE inequality of this type. Practise until it is automatic.

Example 9: Parameter range for monotonicity [JEE Main pattern]

Find all values of $a$ for which $f(x) = x^3 + ax^2 + 3x + 100$ is increasing on $\mathbb{R}$.

Solution:

1. $f'(x) = 3x^2 + 2ax + 3$ — an upward parabola in $x$.
2. Need $f'(x) \geq 0$ for all $x$: discriminant $\leq 0$.
3. $(2a)^2 - 4(3)(3) \leq 0 \Rightarrow 4a^2 \leq 36 \Rightarrow -3 \leq a \leq 3$.

Final Answer: $a \in [-3, 3]$.

Takeaway: At $a = \pm 3$, $f'$ vanishes at one isolated point only — monotonicity survives, so the endpoints are included.

Example 10: Extrema of a rational function, two ways [JEE Main pattern]

Find the local maximum and minimum values of $f(x) = \frac{x}{x^2 + 4}$.

Solution:

1. Derivative method: $f'(x) = \frac{(x^2+4) - x(2x)}{(x^2+4)^2} = \frac{4 - x^2}{(x^2+4)^2}$.
2. Critical points $x = \pm 2$; $f'$ goes $(-) \to (+)$ through $-2$ (local min) and $(+) \to (-)$ through $2$ (local max).
3. Values: $f(2) = \frac{2}{8} = \frac{1}{4}$; $f(-2) = -\frac{1}{4}$.
4. Range cross-check: $y = \frac{x}{x^2+4} \Rightarrow yx^2 - x + 4y = 0$; real $x$ needs $1 - 16y^2 \geq 0 \Rightarrow -\frac{1}{4} \leq y \leq \frac{1}{4}$ — matching extremes.

Final Answer: Local maximum $\frac{1}{4}$ at $x = 2$; local minimum $-\frac{1}{4}$ at $x = -2$.

Takeaway: The discriminant/range method both verifies the answer and shows these are in fact global extremes — a free upgrade in an MCQ.

Example 11: Minimum distance to a parabola [JEE Main pattern]

Find the point(s) on $y = x^2$ nearest to the point $\left(0, \frac{3}{2}\right)$, and the minimum distance.

Solution:

1. Squared distance to $(x, x^2)$: $D^2(x) = x^2 + \left(x^2 - \frac{3}{2}\right)^2$.
2. Differentiate: $\frac{d(D^2)}{dx} = 2x + 2\left(x^2 - \frac{3}{2}\right)(2x) = 2x(2x^2 - 2) = 4x(x-1)(x+1)$.
3. Candidates: $x = 0, \pm 1$. Values: $D^2(0) = \frac{9}{4}$; $D^2(\pm 1) = 1 + \frac{1}{4} = \frac{5}{4}$.
4. Compare: minimum at $x = \pm 1$, i.e. the points $(\pm 1, 1)$, distance $\sqrt{\frac{5}{4}} = \frac{\sqrt 5}{2}$.

Final Answer: Nearest points $(1, 1)$ and $(-1, 1)$; minimum distance $\frac{\sqrt 5}{2}$.

Takeaway: By symmetry there are TWO nearest points here — $x = 0$ is a local maximum of closeness (a critical point that is not the answer). Evaluate all candidates.

Example 12: Root counting via turning values [JEE Main pattern]

Find the number of distinct real roots of $f(x) = 2x^3 - 9x^2 + 12x - 5 = 0$.

Solution:

1. Turning points: $f'(x) = 6x^2 - 18x + 12 = 6(x-1)(x-2)$: local max at $x = 1$, local min at $x = 2$.
2. Turning values: $f(1) = 2 - 9 + 12 - 5 = 0$; $f(2) = 16 - 36 + 24 - 5 = -1$.
3. Read the shape: the local maximum value is exactly 0, so the graph touches the axis at $x = 1$ (double root) and, since $f(2) < 0$ with $f \to +\infty$, crosses once beyond $x = 2$.
4. Confirm by factoring: $f(x) = (x-1)^2(2x - 5)$ — roots $x = 1$ (double) and $x = \frac{5}{2}$.

Final Answer: 2 distinct real roots ($x = 1$ and $x = \frac{5}{2}$).

Takeaway: local max value $= 0$ or local min value $= 0$ signals a repeated root — the graph kisses the axis at a turning point. This shape-reading is exactly what JEE root-counting questions test.