Chapter 6: Application of Derivatives

Board | JEE Mains | NDA | CDS — All Concepts & All Problem Types with Detailed Solutions

📖 To the Student: AOD is the most application-rich chapter of Class 12. Board exams guarantee 5–6 mark optimization and tangent-normal questions. JEE Mains asks 2–3 questions on monotonicity and local maxima. NDA/CDS tests rate-of-change and kinematics. This file covers every concept and every type of problem you will encounter. Read theory first, then solve each practice problem fully before looking at the solution.

1. Rate of Change of Quantities

Core Idea (NCERT §6.1): The derivative $\dfrac{dy}{dx}$ is the instantaneous rate of change of $y$ with respect to $x$. In physical problems, quantities change with time $t$, so we use the Chain Rule.

The Chain Rule for Rates

Master Formula If $y$ depends on $x$, and $x$ depends on time $t$: $$\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$$ Sign Convention:
• $\dfrac{dy}{dt} > 0$ → $y$ is increasing with time.
• $\dfrac{dy}{dt} < 0$ → $y$ is decreasing with time.

Algorithm for Related-Rates Problems:

Read carefully: identify the given rate and the required rate.
Write the geometric formula relating the variables (Area, Volume, etc.).
Differentiate both sides with respect to time $t$.
Substitute all known values. Solve for the unknown rate.
Attach correct units to the answer.

Type A — Geometric Bodies (Sphere, Cube, Cylinder, Cone)

Problem 1 — Sphere Board Q: The radius of a spherical balloon increases at 2 cm/s. Find the rate of increase of (a) Volume and (b) Surface Area when radius is 5 cm.

Given: $\dfrac{dr}{dt} = 2$ cm/s, $r = 5$ cm.
(a) $V = \dfrac{4}{3}\pi r^3 \Rightarrow \dfrac{dV}{dt} = 4\pi r^2 \dfrac{dr}{dt} = 4\pi(25)(2) = \boxed{200\pi \text{ cm}^3/\text{s}}$
(b) $S = 4\pi r^2 \Rightarrow \dfrac{dS}{dt} = 8\pi r \dfrac{dr}{dt} = 8\pi(5)(2) = \boxed{80\pi \text{ cm}^2/\text{s}}$

Problem 2 — Ripple / Circle Board NDA Q: A stone dropped into a pond creates circular ripples. The radius increases at 5 cm/s. How fast is the area increasing when radius = 12 cm?

$A = \pi r^2 \Rightarrow \dfrac{dA}{dt} = 2\pi r \dfrac{dr}{dt} = 2\pi(12)(5) = \boxed{120\pi \text{ cm}^2/\text{s}}$

Problem 3 — Cube Board Q: The side of a cube is increasing at 3 cm/s. Find the rate of increase of (a) volume and (b) total surface area when the side is 10 cm.

Let side $= a$ cm. Given $\dfrac{da}{dt} = 3$ cm/s, $a = 10$.
(a) $V = a^3 \Rightarrow \dfrac{dV}{dt} = 3a^2\dfrac{da}{dt} = 3(100)(3) = \boxed{900 \text{ cm}^3/\text{s}}$
(b) $S = 6a^2 \Rightarrow \dfrac{dS}{dt} = 12a\dfrac{da}{dt} = 12(10)(3) = \boxed{360 \text{ cm}^2/\text{s}}$

Problem 4 — Water Tank (Inverted Cone) Board JEE Q: Water is draining from an inverted conical tank (vertex down) at 2 m³/min. The tank has height 10 m and radius 5 m at the top. How fast is the water level falling when the water depth is 6 m?

By similar triangles: $\dfrac{r}{h} = \dfrac{5}{10} = \dfrac{1}{2}$, so $r = \dfrac{h}{2}$.
Volume: $V = \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3}\pi \left(\dfrac{h}{2}\right)^2 h = \dfrac{\pi h^3}{12}$
$\dfrac{dV}{dt} = \dfrac{\pi h^2}{4} \dfrac{dh}{dt}$
Given $\dfrac{dV}{dt} = -2$ m³/min (draining). At $h = 6$:
$-2 = \dfrac{\pi(36)}{4}\dfrac{dh}{dt} \Rightarrow \dfrac{dh}{dt} = \dfrac{-8}{36\pi} = \boxed{-\dfrac{2}{9\pi} \text{ m/min}}$
The water level is falling at $\dfrac{2}{9\pi}$ m/min.

Type B — Shadow, Ladder & Distance Problems

Problem 5 — Ladder Problem Board Q: A 5 m ladder leans against a wall. Its foot slides away at 1 m/s. How fast is the top sliding down when the foot is 3 m from the wall?

$x^2 + y^2 = 25$. Differentiate: $2x\dfrac{dx}{dt} + 2y\dfrac{dy}{dt} = 0$
When $x = 3$: $y = 4$. Given $\dfrac{dx}{dt} = 1$.
$2(3)(1) + 2(4)\dfrac{dy}{dt} = 0 \Rightarrow \dfrac{dy}{dt} = \boxed{-\dfrac{3}{4} \text{ m/s}}$ (sliding down)

Problem 6 — Shadow Problem Board NDA Q: A man 2 m tall walks at 5 m/min away from a lamp post 8 m high. Find the rate at which (a) his shadow lengthens, and (b) the tip of his shadow moves.

Let the man be $x$ m from the post and his shadow have length $s$ m.
By similar triangles: $\dfrac{8}{x+s} = \dfrac{2}{s} \Rightarrow 8s = 2(x+s) \Rightarrow 6s = 2x \Rightarrow s = \dfrac{x}{3}$
(a) $\dfrac{ds}{dt} = \dfrac{1}{3}\dfrac{dx}{dt} = \dfrac{1}{3}(5) = \boxed{\dfrac{5}{3} \text{ m/min}}$ (shadow lengthens)
(b) Tip of shadow is at distance $x + s = x + x/3 = \dfrac{4x}{3}$ from post.
$\dfrac{d}{dt}(x+s) = \dfrac{dx}{dt} + \dfrac{ds}{dt} = 5 + \dfrac{5}{3} = \boxed{\dfrac{20}{3} \text{ m/min}}$

Problem 7 — Two Cars Moving NDA Q: Two cars start from the same point. One travels east at 60 km/h and the other north at 80 km/h. How fast is the distance between them increasing after 1 hour?

After $t$ hours: $x = 60t$ (east), $y = 80t$ (north).
Distance: $D = \sqrt{x^2 + y^2} = \sqrt{3600t^2 + 6400t^2} = 100t$.
$\dfrac{dD}{dt} = 100$ km/h. After 1 hour: $D = 100$ km, $\dfrac{dD}{dt} = \boxed{100 \text{ km/h}}$
Alternatively by formula: $\dfrac{dD}{dt} = \dfrac{x\frac{dx}{dt} + y\frac{dy}{dt}}{D} = \dfrac{60(60) + 80(80)}{100} = \dfrac{3600+6400}{100} = 100$ km/h ✔

Type C — Marginal Cost, Revenue & Profit

Economic Applications (R.S. Aggarwal Chapter) If $C(x)$, $R(x)$, $P(x)$ = total cost, revenue, profit for $x$ units:
• Marginal Cost: $MC = C'(x)$ — cost of one extra unit
• Marginal Revenue: $MR = R'(x)$ — revenue from one extra unit sold
• Profit: $P(x) = R(x) - C(x)$; maximized when $R'(x) = C'(x)$ i.e., $MR = MC$
• Approximate cost/revenue of $(n+1)$th unit $\approx$ value of $MC$/$MR$ at $x = n$

Problem 8 — Marginal Cost Board Q: $C(x) = 0.005x^3 - 0.02x^2 + 30x + 5000$. Find the marginal cost when $x = 3$ and interpret.

$MC = C'(x) = 0.015x^2 - 0.04x + 30$.
At $x = 3$: $MC = 0.015(9) - 0.04(3) + 30 = 0.135 - 0.12 + 30 = \boxed{30.015}$ ₹/unit.
Interpretation: Producing the 4th unit will cost approximately ₹30.015.

Problem 9 — Max Profit Board Q: Revenue $R(x) = 13x^2 - 2x^3$ and Cost $C(x) = x^3 - 3x^2 + 7x$. Find the output $x$ that maximizes profit.

$P(x) = R(x) - C(x) = 13x^2 - 2x^3 - x^3 + 3x^2 - 7x = -3x^3 + 16x^2 - 7x$
$P'(x) = -9x^2 + 32x - 7 = -(9x-1)(x-7) = 0 \Rightarrow x = 7$ or $x = \dfrac{1}{9}$
$P''(x) = -18x + 32$. At $x = 7$: $P''(7) = -126 + 32 = -94 < 0$ ✔ Maximum.
Profit is maximized at $x = \boxed{7}$ units.

2. Increasing and Decreasing Functions (Monotonicity)

Monotonicity — increasing and decreasing functions

Fig 2.1 — Sign of f'(x) determines intervals of increase/decrease. Critical points divide the domain into sign intervals.

Type	Derivative Condition	Set Definition
Strictly Increasing	$f'(x) > 0$ on $(a,b)$	$x_1 < x_2 \Rightarrow f(x_1) < f(x_2)$
Non-decreasing	$f'(x) \ge 0$, zero only at isolated pts	$x_1 < x_2 \Rightarrow f(x_1) \le f(x_2)$
Strictly Decreasing	$f'(x) < 0$ on $(a,b)$	$x_1 < x_2 \Rightarrow f(x_1) > f(x_2)$
Constant	$f'(x) = 0$ throughout	$f(x_1) = f(x_2)$ always

The Wavy Curve Method (Fastest Algorithm)

Wavy Curve Method (R.D. Sharma) For $f'(x) = a(x-r_1)^{p_1}(x-r_2)^{p_2}\cdots$ with $r_1 < r_2 < \cdots$:
1. Plot roots $r_1, r_2, \ldots$ on number line.
2. Rightmost region: sign = sign of leading coefficient $a$.
3. Crossing each root: sign flips if odd power, stays same if even power.
4. Regions with $+$ sign $\Rightarrow$ increasing; $-$ sign $\Rightarrow$ decreasing.

Problem 10 — Polynomial Board Q: Find intervals where $f(x) = x^4 - 8x^3 + 22x^2 - 24x + 21$ is increasing/decreasing.

$f'(x) = 4x^3 - 24x^2 + 44x - 24 = 4(x-1)(x-2)(x-3)$
Roots: $x = 1, 2, 3$. Leading coeff $> 0$: rightmost sign is $+$.
Sign pattern (right to left): $+, -, +, -$
Increasing: $(1,2) \cup (3,\infty)$ Decreasing: $(-\infty,1) \cup (2,3)$

Problem 11 — Repeated Root (Even Power) JEE Q: Find intervals of monotonicity of $f(x) = x^3(x-1)^2$.

$f'(x) = 3x^2(x-1)^2 + x^3 \cdot 2(x-1) = x^2(x-1)[3(x-1) + 2x] = x^2(x-1)(5x-3)$
Roots: $x = 0$ (even power — sign doesn't change), $x = 3/5$, $x = 1$ (both odd power).
Number line: mark $0, 3/5, 1$. Rightmost sign = $+$ (leading coeff of $f'$ is $5 > 0$).
Sign: $\quad (+) \xrightarrow{1} (-) \xrightarrow{3/5} (+) \xrightarrow{0} (+)$ (0 doesn't change!)
Increasing: $(-\infty, 0) \cup (0, 3/5) \cup (1, \infty)$, simplified as $(-\infty, 3/5) \cup (1, \infty)$
Decreasing: $(3/5, 1)$

Monotonicity of Trig, Log & Exponential Functions

Problem 12 — Trig Function Board JEE Q: Find intervals where $f(x) = \sin x - \cos x$ is increasing for $x \in [0, 2\pi]$.

$f'(x) = \cos x + \sin x = \sqrt{2}\sin\!\left(x + \dfrac{\pi}{4}\right) > 0$
$\Rightarrow 0 < x + \dfrac{\pi}{4} < \pi \Rightarrow x \in \left(-\dfrac{\pi}{4}, \dfrac{3\pi}{4}\right)$
In $[0, 2\pi]$: Increasing on $\left[0, \dfrac{3\pi}{4}\right) \cup \left(\dfrac{7\pi}{4}, 2\pi\right]$
Decreasing on $\left(\dfrac{3\pi}{4}, \dfrac{7\pi}{4}\right)$

Problem 13 — Log Function JEE Q: Show that $f(x) = x - \ln x$ is decreasing on $(0,1)$ and increasing on $(1, \infty)$.

$f'(x) = 1 - \dfrac{1}{x} = \dfrac{x-1}{x}$
For $x \in (0,1)$: $x - 1 < 0$, $x > 0$, so $f'(x) < 0$ → Decreasing.
For $x \in (1, \infty)$: $x - 1 > 0$, $x > 0$, so $f'(x) > 0$ → Increasing.
Critical point: $x = 1$ is a global minimum. $f(1) = 1 - 0 = 1$. $\quad\blacksquare$

Problem 14 — Proving Strictly Increasing Board Q: Prove that $f(x) = x^3$ is strictly increasing on $\mathbb{R}$.

$f'(x) = 3x^2 \ge 0$ for all $x \in \mathbb{R}$.
$f'(x) = 0$ only at $x = 0$ — which is an isolated point, not an interval.
Since $f'(x) \ge 0$ and vanishes only at one isolated point, $f$ is strictly increasing on $\mathbb{R}$. $\quad\blacksquare$

Using Monotonicity to Prove Inequalities

Inequality Proof Method To prove $f(x) > g(x)$ for $x > a$: let $h(x) = f(x) - g(x)$.
Show $h(a) \ge 0$ and $h'(x) > 0$ for $x > a$.
Since $h$ is increasing and starts non-negative, $h(x) > h(a) \ge 0$ for $x > a$. $\quad\blacksquare$

Problem 15 — Prove Inequality JEE Board Q: Prove that $\sin x < x < \tan x$ for $x \in \left(0, \dfrac{\pi}{2}\right)$.

Part 1: $x > \sin x$
Let $h_1(x) = x - \sin x$. $h_1(0) = 0$.
$h_1'(x) = 1 - \cos x \ge 0$ for all $x$, and $> 0$ for $x \ne 0$.
So $h_1$ is increasing, $h_1(x) > h_1(0) = 0$ for $x > 0$. $\Rightarrow x > \sin x$. ✔

Part 2: $\tan x > x$
Let $h_2(x) = \tan x - x$. $h_2(0) = 0$.
$h_2'(x) = \sec^2 x - 1 = \tan^2 x > 0$ for $x \in (0, \pi/2)$.
So $h_2$ is increasing, $h_2(x) > h_2(0) = 0$ for $x > 0$. $\Rightarrow \tan x > x$. ✔
Combining: $\sin x < x < \tan x$ for $x \in (0, \pi/2)$. $\quad\blacksquare$

3. Tangents and Normals

Tangent and Normal to a curve at a point

Fig 3.1 — Tangent (slope = m_T) and Normal (slope = m_N = -1/m_T) are perpendicular to each other at point P.

Key Formulas & Special Cases

Core Formulas At point $P(x_1, y_1)$ on curve $y = f(x)$:

Slope of Tangent: $m_T = \left(\dfrac{dy}{dx}\right)_{(x_1,y_1)}$
Slope of Normal: $m_N = -\dfrac{1}{m_T}$
Equation of Tangent: $y - y_1 = m_T(x - x_1)$
Equation of Normal: $y - y_1 = m_N(x - x_1)$

Special Cases:
• Horizontal tangent (parallel to x-axis): $m_T = 0 \Rightarrow f'(x_1) = 0$
• Vertical tangent (parallel to y-axis): $m_T \to \infty \Rightarrow \left(\dfrac{dx}{dy}\right)_{P} = 0$
• Tangent passes through origin: $y_1 = m_T x_1$
• Normal parallel to x-axis: $m_N = 0 \Rightarrow m_T = \infty$ (vertical tangent)

Problem 16 — Basic Tangent & Normal Board Q: Find the tangent and normal to $y = x^3 - 3x + 2$ at $(2, 4)$.

$f'(x) = 3x^2 - 3$. At $x = 2$: $m_T = 3(4)-3 = 9$, $m_N = -1/9$.
Tangent: $y - 4 = 9(x-2) \Rightarrow \boxed{9x - y - 14 = 0}$
Normal: $y - 4 = -\dfrac{1}{9}(x-2) \Rightarrow \boxed{x + 9y - 38 = 0}$

Problem 17 — Tangent Parallel to Given Line Board NDA Q: Find points on $y = x^3 - 3x$ where the tangent is parallel to $y = 2x - 4$.

Need $f'(x) = 2$: $3x^2 - 3 = 2 \Rightarrow x^2 = 5/3 \Rightarrow x = \pm\sqrt{5/3}$
At $x = \sqrt{5/3}$: $y = (5/3)^{3/2} - 3\sqrt{5/3} = -\dfrac{4}{3}\sqrt{\dfrac{5}{3}}$. Similarly for the negative root.
Points: $\boxed{\left(\pm\sqrt{\dfrac{5}{3}},\, \mp\dfrac{4}{3}\sqrt{\dfrac{5}{3}}\right)}$

Problem 18 — Tangent to Ellipse / Implicit Curve Board Q: Find the tangent to $\sqrt{x} + \sqrt{y} = a$ at the point $\left(\dfrac{a^2}{4}, \dfrac{a^2}{4}\right)$.

Differentiate implicitly: $\dfrac{1}{2\sqrt{x}} + \dfrac{1}{2\sqrt{y}}\dfrac{dy}{dx} = 0 \Rightarrow \dfrac{dy}{dx} = -\sqrt{\dfrac{y}{x}}$
At $\left(\dfrac{a^2}{4}, \dfrac{a^2}{4}\right)$: $m_T = -1$.
Tangent: $y - \dfrac{a^2}{4} = -1\left(x - \dfrac{a^2}{4}\right) \Rightarrow \boxed{x + y = \dfrac{a^2}{2}}$

Problem 19 — Horizontal Tangent Board Q: Find the points on the curve $y = x^3 - 3x^2 - 9x + 7$ where the tangent is horizontal.

$f'(x) = 3x^2 - 6x - 9 = 3(x^2 - 2x - 3) = 3(x-3)(x+1) = 0$
$x = 3$ or $x = -1$.
At $x = 3$: $y = 27 - 27 - 27 + 7 = -20$. At $x = -1$: $y = -1 - 3 + 9 + 7 = 12$.
Horizontal tangents at $\boxed{(3, -20)}$ and $\boxed{(-1, 12)}$.

Tangent and Normal to Parametric Curves

Parametric Form If $x = f(t)$, $y = g(t)$, then:
$$m_T = \frac{dy/dt}{dx/dt} = \frac{g'(t)}{f'(t)}$$ The tangent at $t = t_0$: $y - g(t_0) = m_T\big[x - f(t_0)\big]$

Problem 20 — Parametric Tangent Board NDA Q: For $x = a\cos^3\theta$, $y = a\sin^3\theta$ (astroid), find the equation of the tangent at $\theta = \pi/4$.

$\dfrac{dx}{d\theta} = -3a\cos^2\theta\sin\theta$, $\quad\dfrac{dy}{d\theta} = 3a\sin^2\theta\cos\theta$
$m_T = \dfrac{3a\sin^2\theta\cos\theta}{-3a\cos^2\theta\sin\theta} = -\tan\theta$
At $\theta = \pi/4$: $m_T = -1$, $x_0 = a/2\sqrt{2}$, $y_0 = a/2\sqrt{2}$.
Tangent: $y - \dfrac{a}{2\sqrt{2}} = -1\left(x - \dfrac{a}{2\sqrt{2}}\right) \Rightarrow \boxed{x + y = \dfrac{a}{\sqrt{2}}}$
Note: The tangent cuts equal intercepts $a/\sqrt{2}$ on both axes — a beautiful property of the astroid!

Problem 21 — Cycloid Tangent JEE Q: For $x = a(\theta - \sin\theta)$, $y = a(1 - \cos\theta)$, find the tangent at $\theta = \pi/2$.

$\dfrac{dx}{d\theta} = a(1-\cos\theta)$, $\dfrac{dy}{d\theta} = a\sin\theta$
$m_T = \dfrac{\sin\theta}{1-\cos\theta} = \dfrac{2\sin(\theta/2)\cos(\theta/2)}{2\sin^2(\theta/2)} = \cot(\theta/2)$
At $\theta = \pi/2$: $m_T = \cot(\pi/4) = 1$.
Point: $x_0 = a(\pi/2 - 1)$, $y_0 = a(1 - 0) = a$.
Tangent: $y - a = 1\cdot(x - a(\pi/2-1)) \Rightarrow \boxed{y = x + a(2 - \pi/2)}$

Length of Tangent, Normal, Subtangent & Subnormal

Fig 3.2 — Geometric interpretation of Lengths of Tangent, Normal, Subtangent and Subnormal at point P(x₁,y₁)

Four Length Formulas (NDA/JEE Frequently Tested) Let $P(x_1, y_1)$ be a point on $y = f(x)$ and let $m = \left(\dfrac{dy}{dx}\right)_{P}$. Then:

Length of Tangent (PT) $= |y_1|\sqrt{1 + \dfrac{1}{m^2}} = \dfrac{|y_1|\sqrt{1+m^2}}{|m|}$

Length of Normal (PN) $= |y_1|\sqrt{1 + m^2}$

Length of Subtangent (TM) $= \left|\dfrac{y_1}{m}\right|$

Length of Subnormal (MN) $= |y_1 m|$

Memory Aid: Subtangent ÷ Subnormal $= \dfrac{1}{m^2} = \left(\dfrac{dx}{dy}\right)^2$

Problem 22 — Length of Normal & Subnormal NDA Q: For the parabola $y^2 = 4ax$, show that the subnormal is constant and equals $2a$.

Differentiate $y^2 = 4ax$: $2y\dfrac{dy}{dx} = 4a \Rightarrow \dfrac{dy}{dx} = \dfrac{2a}{y} = m$.
Subnormal $= |y \cdot m| = \left|y \cdot \dfrac{2a}{y}\right| = |2a| = 2a$. (constant!) $\quad\blacksquare$

Problem 23 — Length of Tangent JEE Q: Find the length of the tangent from the point $(3, 4)$ to the curve $y^2 = 8x$.

First verify $(3,4)$ is on curve: $16 = 24$? No. So find tangent slope differently.
$y^2 = 8x \Rightarrow m = \dfrac{dy}{dx} = \dfrac{4}{y}$. At a general point $(at^2, 2at)$ with $a=2$: point $(2t^2, 4t)$.
Tangent at $(2t^2, 4t)$: $y \cdot 4t = 2 \cdot 2(x + 2t^2)/2$... use standard form $ty = x + 2t^2$.
Passing through $(3, 4)$: $4t = 3 + 2t^2 \Rightarrow 2t^2 - 4t + 3 = 0$.
Discriminant $= 16 - 24 < 0$: no real tangent from $(3,4)$ — point is inside the parabola.
If problem asks for length of tangent to $y^2 = 8x$ from $(6, 0)$:
Tangent $ty = x + 2t^2$ through $(6,0)$: $0 = 6 + 2t^2 \Rightarrow t^2 = -3$. Still no real tangent (point inside parabola for parabolas opening right if $x_0 > 0$, $y_0^2 < 8x_0$).

Angle of Intersection and Orthogonal Curves

Angle Between Two Curves At the point of intersection, if slopes of tangents are $m_1$ and $m_2$: $$\tan\theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|$$ Orthogonal Curves ($\theta = 90°$): $m_1 m_2 = -1$

Problem 24 — Orthogonal Trajectories Board Q: Show that the curves $x^2 + y^2 = r^2$ and $y = mx$ (any line through origin) are orthogonal.

Curve 1 ($x^2 + y^2 = r^2$): Differentiating: $m_1 = -\dfrac{x}{y}$
Curve 2 ($y = mx$): $m_2 = m$
At intersection: $y = mx \Rightarrow m_1 = -\dfrac{x}{mx} = -\dfrac{1}{m}$.
$m_1 \cdot m_2 = -\dfrac{1}{m} \cdot m = -1$ ✔ → Orthogonal at all intersection points. $\quad\blacksquare$

Problem 25 — Angle of Intersection Board Q: Find the angle of intersection of $y = x^2$ and $y = x^3$.

Intersection: $x^2 = x^3 \Rightarrow x^2(x-1) = 0 \Rightarrow x = 0$ or $x = 1$.
At $(1,1)$: $m_1 = 2x|_{x=1} = 2$, $m_2 = 3x^2|_{x=1} = 3$.
$\tan\theta = \left|\dfrac{2-3}{1+6}\right| = \dfrac{1}{7} \Rightarrow \theta = \tan^{-1}\!\left(\dfrac{1}{7}\right)$
At $(0,0)$: $m_1 = 0$, $m_2 = 0$. Both tangents are the x-axis. Angle = $0°$.

4. Approximations Using Differentials

Approximation Formula For a small change $\Delta x$ in $x$: $$\boxed{f(x + \Delta x) \approx f(x) + f'(x)\cdot \Delta x}$$ The differential: $dy = f'(x)\, dx$. The actual change $\Delta y \approx dy$ when $\Delta x$ is small.

Error Analysis — Three Types

Error Type	Formula	Example
Absolute Error in $y$	$\Delta y \approx dy = f'(x)\, dx$	Error in volume when radius changes by $0.01$
Relative Error in $y$	$\dfrac{\Delta y}{y} \approx \dfrac{dy}{y}$	$\dfrac{dV}{V}$
Percentage Error in $y$	$\dfrac{\Delta y}{y} \times 100$	$\dfrac{dV}{V} \times 100\%$

Problem 26 — Approximate Value Board Q: Using differentials, find approximate values of (a) $\sqrt{49.5}$ (b) $\sqrt[3]{26}$ (c) $\cos 61°$.

(a) $f(x) = \sqrt{x}$, $x = 49$, $\Delta x = 0.5$. $f'(49) = \dfrac{1}{14}$.
$\sqrt{49.5} \approx 7 + \dfrac{1}{14}(0.5) = \boxed{7.0357}$

(b) $f(x) = x^{1/3}$, $x = 27$, $\Delta x = -1$. $f'(27) = \dfrac{1}{3(27)^{2/3}} = \dfrac{1}{27}$.
$\sqrt[3]{26} \approx 3 + \dfrac{1}{27}(-1) = \boxed{\dfrac{80}{27} \approx 2.963}$

(c) $f(x) = \cos x$, $x = 60° = \dfrac{\pi}{3}$, $\Delta x = 1° = \dfrac{\pi}{180}$. $f'\!\left(\dfrac{\pi}{3}\right) = -\sin 60° = -\dfrac{\sqrt{3}}{2}$.
$\cos 61° \approx \cos 60° + \left(-\dfrac{\sqrt{3}}{2}\right)\dfrac{\pi}{180} = 0.5 - \dfrac{\sqrt{3}\pi}{360} \approx \boxed{0.4849}$

Problem 27 — Percentage Error Board Q: If the radius of a sphere is measured as 9 cm with an error of 0.03 cm, find the percentage error in the calculated surface area.

$S = 4\pi r^2 \Rightarrow \dfrac{dS}{S} = \dfrac{8\pi r\, dr}{4\pi r^2} = \dfrac{2\, dr}{r}$
$= \dfrac{2 \times 0.03}{9} = \dfrac{0.06}{9} = \dfrac{1}{150}$
Percentage error $= \dfrac{1}{150} \times 100 = \boxed{\dfrac{2}{3}\%}$

Problem 28 — Volume Error NDA Q: The radius of a cylinder is 5 cm and height is 10 cm. If the radius increases by 0.1 cm while height decreases by 0.2 cm, find the approximate change in volume.

$V = \pi r^2 h$. Total differential: $dV = 2\pi rh\, dr + \pi r^2\, dh$
$= 2\pi(5)(10)(0.1) + \pi(25)(-0.2) = 10\pi - 5\pi = \boxed{5\pi \text{ cm}^3}$

5. Maxima and Minima

Local maxima, minima and point of inflection

Fig 5.1 — Local max (f''<0), local min (f''>0), and inflection (f''=0 with sign change). Critical points are where f'=0.

Definitions Local Maximum: $f(c) \ge f(x)$ for all $x$ near $c$.
Local Minimum: $f(c) \le f(x)$ for all $x$ near $c$.
Critical (Stationary) Point: $c$ where $f'(c) = 0$ or $f'(c)$ does not exist.
Key: Every local extremum is a critical point. Not every critical point is an extremum.

First Derivative Test (FDT)

FDT — Sign Change Rule

Sign of $f'(x)$ as $x$ passes $c$	Conclusion
$+$ to $-$ (↗↘)	Local Maximum
$-$ to $+$ (↘↗)	Local Minimum
No sign change ($+$to$+$ or $-$to$-$)	Point of Inflection

Second Derivative Test (SDT)

SDT — Faster for Board Exams Solve $f'(c) = 0$. Then compute $f''(c)$:
• $f''(c) < 0$ → Local Maximum (concave down = "frown") ☹
• $f''(c) > 0$ → Local Minimum (concave up = "smile") ☺
• $f''(c) = 0$ → SDT Fails! Use FDT instead.

nth Derivative Test (When SDT Fails Repeatedly)

nth Derivative Test (Higher Order) If $f'(c) = f''(c) = \cdots = f^{(n-1)}(c) = 0$ but $f^{(n)}(c) \ne 0$:
• If $n$ is odd: $c$ is a point of inflection (neither max nor min).
• If $n$ is even and $f^{(n)}(c) < 0$: Local Maximum.
• If $n$ is even and $f^{(n)}(c) > 0$: Local Minimum.

Problem 29 — SDT Fails, Use nth Test JEE Q: Find the nature of the critical point of $f(x) = x^4$ at $x = 0$.

$f'(0) = 0$, $f''(0) = 0$. SDT fails twice.
$f'''(0) = 0$, $f^{(4)}(0) = 24 > 0$.
$n = 4$ (even), $f^{(4)}(0) > 0$ → Local (and Global) Minimum at $x = 0$.

Problem 30 — Complete Analysis Board JEE Q: For $f(x) = 2x^3 - 9x^2 + 12x - 3$, find all: (a) intervals of monotonicity, (b) local extrema, (c) inflection points.

$f'(x) = 6x^2 - 18x + 12 = 6(x-1)(x-2)$ Critical: $x = 1, 2$.
$f''(x) = 12x - 18$.
(a) Increasing on $(-\infty,1) \cup (2,\infty)$; Decreasing on $(1,2)$.
(b) At $x=1$: $f''(1) = -6 < 0$ → Local Max $= f(1) = 2$.
At $x=2$: $f''(2) = 6 > 0$ → Local Min $= f(2) = 1$.
(c) $f''(x) = 0 \Rightarrow x = 3/2$. Sign changes. Inflection at $\left(\dfrac{3}{2}, \dfrac{3}{2}\right)$.

Absolute Extrema on a Closed Interval [a, b]

Closed Interval Method (EVT) By Extreme Value Theorem, continuous $f$ on $[a,b]$ attains its global max and min.
Algorithm:
1. Find all critical points in $(a, b)$ (solve $f'(x) = 0$).
2. Evaluate $f$ at: all critical points + both endpoints $a$ and $b$.
3. Largest value = Absolute Max; Smallest = Absolute Min.

Problem 31 — Closed Interval Board Q: Find the absolute max and min of $f(x) = 2x^3 - 15x^2 + 36x + 1$ on $[1, 5]$.

$f'(x) = 6x^2 - 30x + 36 = 6(x-2)(x-3) = 0 \Rightarrow x = 2, 3$ (both in $[1,5]$).
$f(1) = 2 - 15 + 36 + 1 = 24$
$f(2) = 16 - 60 + 72 + 1 = 29$
$f(3) = 54 - 135 + 108 + 1 = 28$
$f(5) = 250 - 375 + 180 + 1 = 56$
Absolute Maximum = 56 at $x = 5$; Absolute Minimum = 24 at $x = 1$.

6. Applied Optimization Problems

Fig 6.1 — Classic optimization: given constraint (surface area or volume), maximize/minimize the other quantity.

6-Step Optimization Framework (Board Guarantee) Step 1 — Draw & Label: Diagram with all variables named.
Step 2 — Constraint: The equation given as fixed (e.g., perimeter = 20 m).
Step 3 — Objective: Quantity to maximize or minimize.
Step 4 — Reduce to 1 variable: Eliminate one variable using constraint.
Step 5 — Differentiate & set to zero.
Step 6 — Verify: Check $d^2O/dv^2$: $< 0$ = MAX, $> 0$ = MIN.

Classic Geometry Optimization

Problem 32 — Cylinder (Total SA) Board ★★★ Q: Show that the right circular cylinder of given total surface area $S$ and maximum volume has height equal to diameter.

$S = 2\pi r^2 + 2\pi rh$ (fixed). $\Rightarrow h = \dfrac{S - 2\pi r^2}{2\pi r}$
$V = \pi r^2 h = \dfrac{r(S-2\pi r^2)}{2} = \dfrac{Sr - 2\pi r^3}{2}$
$\dfrac{dV}{dr} = \dfrac{S - 6\pi r^2}{2} = 0 \Rightarrow S = 6\pi r^2$
Substituting: $6\pi r^2 = 2\pi r^2 + 2\pi rh \Rightarrow h = 2r$ = diameter. ✔
$\dfrac{d^2V}{dr^2} = -6\pi r < 0$ ✔ Maximum. $\quad\blacksquare$

Problem 33 — Rectangle in Semicircle Board ★★ Q: A rectangle is inscribed in a semicircle of radius $r$ with its base on the diameter. Find dimensions for maximum area.

Let half-width $= x$, height $= y$. Constraint: $x^2 + y^2 = r^2 \Rightarrow y = \sqrt{r^2-x^2}$.
Area $A = 2xy$. Maximize $A^2 = 4x^2(r^2-x^2)$.
$\dfrac{d(A^2)}{dx} = 4(2xr^2 - 4x^3) = 8x(r^2-2x^2) = 0 \Rightarrow x = \dfrac{r}{\sqrt{2}}$
Width $= r\sqrt{2}$, Height $= \dfrac{r}{\sqrt{2}}$. Max Area $= r^2$.

Problem 34 — Square in Circle Board Q: Show that the rectangle of maximum area inscribed in a circle is a square.

Diagonal = $2R$: $x^2+y^2=4R^2$. Maximize $A=xy$.
$\dfrac{d(A^2)}{dx} = 4x(2R^2-x^2) = 0 \Rightarrow x = R\sqrt{2} = y$. Square! $\quad\blacksquare$

Problem 35 — Cone in Sphere JEE ★★★ Q: Show that the cone of maximum volume inscribed in a sphere of radius $R$ has height $h = \dfrac{4R}{3}$.

$r^2 = R^2 - (h-R)^2 = 2hR - h^2$. $V = \dfrac{\pi h(2hR-h^2)}{3}$
$\dfrac{dV}{dh} = \dfrac{\pi h(4R-3h)}{3} = 0 \Rightarrow h = \dfrac{4R}{3}$ ✔
$\dfrac{d^2V}{dh^2}\bigg|_{4R/3} = \dfrac{\pi(4R-8R)}{3} < 0$ ✔ Maximum. $\quad\blacksquare$

Problem 36 — Window Problem Board ★★★ Q: A window is a rectangle surmounted by a semicircle. Perimeter = 10 m. Find dimensions for maximum area.

Width $= 2r$, height $= h$. Perimeter: $2r + 2h + \pi r = 10 \Rightarrow h = 5 - r - \dfrac{\pi r}{2}$
Area $= 2rh + \dfrac{\pi r^2}{2} = 10r - 2r^2 - \dfrac{\pi r^2}{2}$
$\dfrac{dA}{dr} = 10 - r(4+\pi) = 0 \Rightarrow r = \dfrac{10}{4+\pi}$
Width $= \dfrac{20}{4+\pi}$ m, Height $= \dfrac{10}{4+\pi}$ m. Max area $= \dfrac{50}{4+\pi}$ m².

Problem 37 — Triangle of Maximum Area JEE Q: Find the triangle of maximum area inscribed in a circle of radius $R$.

By symmetry, the maximum-area triangle is equilateral. Let's verify.
Fix one vertex at top of circle. Side lengths depend on angle. Using parametric approach:
Area of triangle inscribed = $\dfrac{1}{2}|AB||AC|\sin A$. By optimization (advanced), the maximum occurs when $A = B = C = 60°$ — equilateral triangle.
Side $= R\sqrt{3}$. Max Area $= \dfrac{3\sqrt{3}}{4}R^2$.

Open Box and Wire-Cutting Problems

Problem 38 — Open Box from Square Sheet Board JEE ★★★ Q: A square piece of tin of side 18 cm is to be made into a box (without a lid) by cutting a square of side $x$ cm from each corner and folding up the flaps. Find $x$ so that volume is maximum.

After cutting, base = $(18-2x) \times (18-2x)$, height = $x$.
$V = x(18-2x)^2 = x(324 - 72x + 4x^2) = 324x - 72x^2 + 4x^3$ [valid for $0 < x < 9$]
$\dfrac{dV}{dx} = 324 - 144x + 12x^2 = 12(x^2 - 12x + 27) = 12(x-3)(x-9) = 0$
$x = 3$ or $x = 9$ (rejected, as then box has zero base).
$\dfrac{d^2V}{dx^2} = 24x - 144$. At $x=3$: $= 72 - 144 = -72 < 0$ ✔ Maximum.
Cut squares of side $x = \boxed{3}$ cm. Max volume $= 3(12)^2 = 432 \text{ cm}^3$.

Problem 39 — Wire Cut into Circle & Square Board ★★★ Q: A wire of length 36 cm is cut into two pieces. One piece is bent into a circle, the other into a square. Find how to cut the wire so that the sum of the areas is minimum.

Let length for circle $= x$ cm, for square $= (36-x)$ cm.
Circle radius $r = \dfrac{x}{2\pi}$. Square side $= \dfrac{36-x}{4}$.
Total Area: $A = \pi r^2 + \left(\dfrac{36-x}{4}\right)^2 = \dfrac{x^2}{4\pi} + \dfrac{(36-x)^2}{16}$
$\dfrac{dA}{dx} = \dfrac{x}{2\pi} - \dfrac{36-x}{8} = 0$
$\dfrac{4x}{\pi} = 36-x \Rightarrow 4x = 36\pi - \pi x \Rightarrow x(4+\pi) = 36\pi \Rightarrow x = \dfrac{36\pi}{4+\pi}$
$\dfrac{d^2A}{dx^2} = \dfrac{1}{2\pi} + \dfrac{1}{8} > 0$ ✔ Minimum.
Circle piece $= \dfrac{36\pi}{4+\pi}$ cm, Square piece $= \dfrac{144}{4+\pi}$ cm.

Problem 40 — Minimum Cost Box JEE Q: A box with a square base and open top must have volume 4000 cm³. The base costs ₹5/cm² and sides cost ₹3/cm². Find the dimensions that minimize the total cost.

Let base side $= x$, height $= h$. Volume: $x^2 h = 4000 \Rightarrow h = 4000/x^2$.
Cost: $C = 5x^2 + 4(xh)(3) = 5x^2 + \dfrac{12x \cdot 4000}{x^2} = 5x^2 + \dfrac{48000}{x}$
$\dfrac{dC}{dx} = 10x - \dfrac{48000}{x^2} = 0 \Rightarrow x^3 = 4800 \Rightarrow x = \sqrt[3]{4800} \approx 16.87$ cm
$h = \dfrac{4000}{x^2}$. $\dfrac{d^2C}{dx^2} = 10 + \dfrac{96000}{x^3} > 0$ ✔ Minimum cost.

Fencing and Area Problems

Problem 41 — Fencing a Field Board ★★ Q: A farmer has 100 m of fencing to enclose a rectangular garden along a river (one side needs no fence). Find dimensions for maximum area.

Let the side parallel to river = $x$ m, perpendicular sides = $y$ m each.
Constraint: $x + 2y = 100 \Rightarrow x = 100 - 2y$.
Area: $A = xy = (100-2y)y = 100y - 2y^2$
$\dfrac{dA}{dy} = 100 - 4y = 0 \Rightarrow y = 25$ m, $x = 50$ m.
$\dfrac{d^2A}{dy^2} = -4 < 0$ ✔ Maximum. Max Area $= 50 \times 25 = \boxed{1250 \text{ m}^2}$.

Problem 42 — Shortest Distance from Point to Curve JEE ★★ Q: Find the point on the parabola $y = x^2$ nearest to the point $(3, 0)$.

Point on parabola: $(a, a^2)$. Distance squared: $D = (a-3)^2 + a^4$.
$\dfrac{dD}{da} = 2(a-3) + 4a^3 = 4a^3 + 2a - 6 = 2(2a^3 + a - 3) = 0$
$a = 1$ satisfies $2 + 1 - 3 = 0$ ✔. Factor: $(a-1)(2a^2+2a+3) = 0$. Only $a = 1$.
$\dfrac{d^2D}{da^2} = 2 + 12a^2 = 14 > 0$ ✔ Minimum. Nearest point: $\boxed{(1,1)}$.

Problem 43 — Ellipse Inscribed Rectangle Board Q: Find the area of the largest rectangle inscribed in the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$.

Parametric point: $(a\cos\theta, b\sin\theta)$. Rectangle area $= 4ab\cos\theta\sin\theta = 2ab\sin 2\theta$.
Maximum when $\sin 2\theta = 1 \Rightarrow \theta = \pi/4$. Max area $= \boxed{2ab}$.

7. Kinematics: Displacement, Velocity and Acceleration

Fig 7.1 — Derivatives connect Displacement → Velocity → Acceleration. Integration reverses the direction.

Kinematics Definitions (NDA/CDS Essential) If $s = f(t)$ is the displacement (position) of a particle at time $t$:
$$\text{Velocity: } v = \frac{ds}{dt} = f'(t) \qquad \text{Acceleration: } a = \frac{dv}{dt} = \frac{d^2s}{dt^2} = f''(t)$$ Key Interpretations:
• Particle at rest: $v = 0$
• Moving in positive direction: $v > 0$; Negative direction: $v < 0$
• Speeding up: $v$ and $a$ have the same sign
• Slowing down: $v$ and $a$ have opposite signs
• Maximum displacement: when $v = 0$ (and $a \ne 0$)
• $a = v\dfrac{dv}{ds}$ (acceleration in terms of displacement)

Problem 44 — Standard Kinematics NDA Board Q: A particle moves along a line with displacement $s = 2t^3 - 9t^2 + 12t - 5$ (metres, seconds). Find: (a) velocity when $t=2$, (b) when is it at rest?, (c) acceleration when $t=1$, (d) when does it reverse direction?

$v = \dfrac{ds}{dt} = 6t^2 - 18t + 12 = 6(t-1)(t-2)$ $a = \dfrac{dv}{dt} = 12t - 18$
(a) $v(2) = 6(1)(0) = \boxed{0}$ m/s
(b) $v = 0$: $t = 1$ s and $t = 2$ s.
(c) $a(1) = 12 - 18 = \boxed{-6}$ m/s² (decelerating)
(d) Direction reverses at $t = 1$ (FDT: $v$ changes $+ \to -$) and $t = 2$ ($v$ changes $- \to +$).

Problem 45 — SHM Type NDA Q: A particle executes SHM. Displacement $s = A\sin(\omega t + \phi)$. Show that acceleration $= -\omega^2 s$.

$v = \dfrac{ds}{dt} = A\omega\cos(\omega t + \phi)$
$a = \dfrac{dv}{dt} = -A\omega^2\sin(\omega t + \phi) = -\omega^2 \cdot [A\sin(\omega t + \phi)] = \boxed{-\omega^2 s}$ $\quad\blacksquare$
This is the defining equation of Simple Harmonic Motion!

Problem 46 — Time for Max Velocity Board Q: A particle's velocity is $v = 3t^2 - 6t + 4$ m/s. Find when velocity is minimum and find the minimum velocity.

$\dfrac{dv}{dt} = 6t - 6 = 0 \Rightarrow t = 1$ s.
$\dfrac{d^2v}{dt^2} = 6 > 0$ ✔ Minimum velocity.
Min velocity $= 3(1) - 6(1) + 4 = \boxed{1}$ m/s.

8. Concavity and Points of Inflection

Concavity Rules Concave Up ("smile"): $f''(x) > 0$ — the curve bends upward; a tangent lies below the curve.
Concave Down ("frown"): $f''(x) < 0$ — the curve bends downward; a tangent lies above the curve.

Point of Inflection: where concavity changes. Necessary condition: $f''(c) = 0$.
Sufficient condition: $f''$ must actually change sign at $c$.
Warning: $f''(c) = 0$ alone is NOT enough! Example: $f(x) = x^4$ at $x=0$: $f''(0) = 0$ but no inflection.

Problem 47 — Finding Inflection Points JEE Q: Find all inflection points of $f(x) = 3x^5 - 40x^3 + 3$.

$f'(x) = 15x^4 - 120x^2$
$f''(x) = 60x^3 - 240x = 60x(x^2 - 4) = 60x(x-2)(x+2)$
$f''(x) = 0$: $x = 0, \pm 2$. Check sign changes:
$\quad x < -2$: $f'' < 0$; $-2 < x < 0$: $f'' > 0$; $0 < x < 2$: $f'' < 0$; $x > 2$: $f'' > 0$.
Sign changes at $x = -2, 0, 2$ ✔ — all three are inflection points.
Inflection points: $(-2, f(-2))$, $(0, 3)$, $(2, f(2))$.

9. Proving Inequalities Using Derivatives

Standard Technique (NCERT Miscellaneous / JEE) To prove $f(x) > g(x)$ for $x > a$:
Step 1: Let $h(x) = f(x) - g(x)$.
Step 2: Compute $h(a)$. (Should be $\ge 0$.)
Step 3: Show $h'(x) > 0$ for $x > a$ (i.e., $h$ is increasing).
Step 4: Conclude $h(x) > h(a) \ge 0$ for $x > a$. $\quad\blacksquare$

Problem 48 — Classic Inequality Board JEE Q: Prove that $e^x \ge 1 + x$ for all $x \in \mathbb{R}$.

Let $h(x) = e^x - 1 - x$. $h(0) = 0$.
$h'(x) = e^x - 1$.
For $x > 0$: $e^x > 1 \Rightarrow h'(x) > 0$ → $h$ increasing → $h(x) > h(0) = 0$. ✔
For $x < 0$: $e^x < 1 \Rightarrow h'(x) < 0$ → $h$ decreasing → $h(x) > h(0) = 0$. ✔
At $x = 0$: equality. Therefore $e^x \ge 1 + x$ for all $x \in \mathbb{R}$. $\quad\blacksquare$

Problem 49 — Log Inequality JEE Q: Prove that $\ln(1+x) < x$ for all $x > 0$.

Let $h(x) = x - \ln(1+x)$. $h(0) = 0$.
$h'(x) = 1 - \dfrac{1}{1+x} = \dfrac{x}{1+x} > 0$ for $x > 0$.
So $h$ is increasing for $x > 0$, thus $h(x) > h(0) = 0 \Rightarrow x > \ln(1+x)$. $\quad\blacksquare$

Problem 50 — AM-GM via Derivative JEE Q: Using calculus, prove that for $a, b > 0$: $\dfrac{a+b}{2} \ge \sqrt{ab}$.

Let $f(x) = x + \dfrac{a^2}{x}$ for $x = b > 0$ (fix $a$, optimize over $b$). This equals $f = b + \dfrac{a^2}{b}$.
$f'(b) = 1 - \dfrac{a^2}{b^2} = 0 \Rightarrow b = a$. $f''(b) = \dfrac{2a^2}{b^3} > 0$ ✔ Minimum.
Minimum of $b + a^2/b = a + a = 2a$. So $b + a^2/b \ge 2a \Rightarrow a^2 + b^2 \ge 2ab$...
More directly: $(\sqrt{a}-\sqrt{b})^2 \ge 0 \Rightarrow a - 2\sqrt{ab} + b \ge 0 \Rightarrow \dfrac{a+b}{2} \ge \sqrt{ab}$. $\quad\blacksquare$

10. Quick Reference Summary Table

Topic	Key Formula / Test	Result
Related Rate	$\frac{dy}{dt} = \frac{dy}{dx}\cdot\frac{dx}{dt}$	Chain rule for rates
Strictly Increasing	$f'(x) > 0$	$f(x_1) < f(x_2)$ for $x_1 < x_2$
Strictly Decreasing	$f'(x) < 0$	$f(x_1) > f(x_2)$ for $x_1 < x_2$
Tangent slope	$m_T = (dy/dx)_P$	Slope at point P
Normal slope	$m_N = -1/m_T$	Perpendicular to tangent
Length of Tangent	$\|y\|\sqrt{1+1/m^2}$	PT from P to x-axis via tangent
Length of Normal	$\|y\|\sqrt{1+m^2}$	PN from P to x-axis via normal
Subtangent	$\|y/m\|$	Intercept of tangent on x-axis
Subnormal	$\|ym\|$	Intercept of normal on x-axis
Angle of curves	$\tan\theta=\|m_1-m_2\|/\|1+m_1m_2\|$	Between tangents
Orthogonal	$m_1 m_2 = -1$	Cut at 90°
Approximation	$f(x+\Delta x) \approx f(x) + f'\Delta x$	Linearization
Local Max (SDT)	$f'(c)=0$, $f''(c)<0$	Concave down
Local Min (SDT)	$f'(c)=0$, $f''(c)>0$	Concave up
Abs. Max/Min on [a,b]	Evaluate at crit. pts + endpoints	Pick largest/smallest
Inflection Point	$f''(c)=0$ AND sign changes	Concavity changes
Velocity	$v = ds/dt$	Rate of position change
Acceleration	$a = dv/dt = d^2s/dt^2$	Rate of velocity change

11. Previous Year & Mixed Practice Problems

PYQ 2023 Board — Rate of Change Board Q: Water flows into a conical vessel (vertex up) at 5 L/min. The vessel has height 20 cm and base radius 10 cm. How fast is the water level rising when water level = 8 cm?

By similar triangles: $r/h = 10/20 = 1/2 \Rightarrow r = h/2$.
$V = \dfrac{1}{3}\pi r^2 h = \dfrac{\pi h^3}{12}$. $\dfrac{dV}{dt} = \dfrac{\pi h^2}{4}\dfrac{dh}{dt}$
Convert: $5$ L/min $= 5000$ cm³/min.
$5000 = \dfrac{\pi(64)}{4}\dfrac{dh}{dt} \Rightarrow \dfrac{dh}{dt} = \dfrac{5000}{16\pi} = \boxed{\dfrac{625}{2\pi} \approx 99.5 \text{ cm/min}}$

PYQ 2022 Board — Optimization Board Q: Find two positive numbers $x$ and $y$ such that $x + y = 60$ and $xy^3$ is maximum.

$y = 60 - x$. $P = xy^3 = x(60-x)^3$.
$\dfrac{dP}{dx} = (60-x)^3 + x \cdot 3(60-x)^2(-1) = (60-x)^2[(60-x) - 3x] = (60-x)^2(60-4x) = 0$
$x = 60$ (rejected) or $x = 15$. $y = 45$.
$\dfrac{d^2P}{dx^2}\bigg|_{x=15} < 0$ ✔ Maximum. $x = 15$, $y = 45$.

JEE Mains Style — Parametric Tangent JEE Q: If the tangent at the point $P(x_1, y_1)$ to the parabola $y^2 = 4ax$ meets the parabola $y^2 = 4a(x+b)$ in $Q$ and $R$, then the midpoint of $QR$ is $(x_1 + b, y_1)$.

Tangent to $y^2 = 4ax$ at $(at^2, 2at)$: $ty = x + at^2$.
Substitute in $y^2 = 4a(x+b)$: $y^2 = 4a(ty - at^2 + b) = 4aty - 4a^2t^2 + 4ab$
$y^2 - 4aty + 4a^2t^2 - 4ab = 0$
Sum of roots: $y_Q + y_R = 4at = 2y_1 \Rightarrow$ midpoint $y$-coordinate $= 2at = y_1$.
Midpoint $x$: from $ty = x + at^2$: $x = ty - at^2$. At $y = y_1 = 2at$: $x_{mid} = 2at^2 - at^2 + b = at^2 + b = x_1 + b$. $\quad\blacksquare$

NDA Pattern — Complete Motion Analysis NDA Q: $s = t^3 - 6t^2 + 9t + 2$. Find: (a) when at rest, (b) distance covered in first 4 seconds, (c) max. displacement from $t=0$ to $t=4$.

$v = 3t^2 - 12t + 9 = 3(t-1)(t-3)$. $a = 6t - 12$.
(a) At rest: $v = 0 \Rightarrow t = 1$ s and $t = 3$ s.
(b) $s(0) = 2$, $s(1) = 1-6+9+2 = 6$, $s(3) = 27-54+27+2 = 2$, $s(4) = 64-96+36+2 = 6$.
Particle goes: $2 \to 6$ (+4), then $6 \to 2$ (-4), then $2 \to 6$ (+4). Total distance $= 4+4+4 = \boxed{12}$ m.
(c) Max displacement value = 6 m (at $t=1$ and $t=4$).

Miscellaneous — Minimum Sum of Squares Board Q: Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.

Let numbers be $x$ and $16 - x$. $S = x^3 + (16-x)^3$.
$\dfrac{dS}{dx} = 3x^2 - 3(16-x)^2 = 3(x^2 - (16-x)^2) = 3(2x-16)(16) \cdot \dfrac{1}{2}$...
Simplify: $= 3[x^2 - (16-x)^2] = 3(x-(16-x))(x+(16-x)) = 3(2x-16)(16) = 96(x-8) = 0$
$x = 8$. Both numbers are $8$.
$\dfrac{d^2S}{dx^2} = 6x + 6(16-x) \cdot 1 = 96 > 0$ ✔ Minimum. Both numbers = 8.

Hard — Three Concurrent Tangents JEE Q: If the tangents to the curve $y = x^2 + bx + c$ at points where $x = 1$ and $x = 2$ are parallel to the tangent at $x = 3/2$, prove $b = 3/2 \cdot 2 - 3 = $ (this is always true for any parabola — verify with a specific case).

$f'(x) = 2x + b$. Slope at $x = 3/2$: $m_0 = 3 + b$.
Slope at $x=1$: $m_1 = 2 + b$. Slope at $x=2$: $m_2 = 4 + b$.
Average of slopes at 1 and 2: $\dfrac{m_1 + m_2}{2} = \dfrac{(2+b)+(4+b)}{2} = 3 + b = m_0$. ✔
This shows the tangent at the midpoint of any two x-values on a parabola has slope equal to the average of the slopes at those two points — a beautiful property of parabolas!

12. Common Mistakes & Exam Strategy

⚠️ Most Common Mistakes to Avoid 1. In Rate of Change — forgetting to differentiate w.r.t. time $t$, not $x$.
2. In Tangent/Normal — not verifying the point actually lies on the curve before substituting.
3. In Optimization — not checking whether the critical point gives a MAX or MIN using 2nd derivative.
4. In Closed Interval — forgetting to evaluate $f$ at the endpoints $a$ and $b$.
5. In SDT — when $f''(c) = 0$, the test FAILS — must use FDT.
6. In Increasing/Decreasing — writing closed brackets $[$ for monotonicity intervals is acceptable only for continuous functions; for strict monotonicity in standard NCERT context, open brackets $($ are preferred.

Exam Strategy — Last Day Revision 📌 Board 5-6 mark Optimization: Always draw diagram, write constraint + objective clearly, show verification with 2nd derivative. Copy the conclusion statement clearly.
📌 JEE Monotonicity: Factorize $f'(x)$ fully. Use Wavy Curve for speed.
📌 NDA Rate of Change: Write the geometric formula first, then differentiate. Attach units.
📌 Tangent problems: Compute slope → write equation → simplify to standard form.
📌 AM-GM Shortcut: Minimum of $x + \dfrac{k}{x}$ is $2\sqrt{k}$ at $x = \sqrt{k}$. Saves time in optimization.
📌 Approximation: Always write $f(x) + f'(x)\Delta x$. Never forget the sign of $\Delta x$.