CBSE Sample Paper 2024-25 · Section D · 5 marks

Q34: $f''(x)$ for $f(x) = |x|^3$ ⁠, or $\frac{[1 + (y')^2]^{3/2}}{y''}$ on a circle is constant

Sample 2024-25Ch 5 · Continuity and Differentiability

The question

(a) If $f: R \to R$ is defined by $f(x) = |x|^3$ ⁠, show that $f''(x)$ exists for all real $x$ and find it.

(b) If $(x - a)^2 + (y - b)^2 = c^2$ ⁠, for some $c > 0$ ⁠, prove that

\frac{\left[1 + \left(\dfrac{dy}{dx}\right)^2\right]^{\frac32}}{\dfrac{d^2y}{dx^2}}

is a constant independent of $a$ and $b$ ⁠.

Exam answer

(a)

For $x < 0$ ⁠, $\lvert x\rvert^3 = (-x)^3 = -x^3$ ⁠. So

f(x) = |x|^3 = \begin{cases} x^3, & x \ge 0 \\ -x^3, & x < 0 \end{cases}

Why?

A modulus is handled by removing it piece by piece: $\lvert x\rvert = x$ for $x \ge 0$ and $\lvert x\rvert = -x$ for $x < 0$ ⁠.

Cubing keeps the sign: $(-x)^3 = -x^3$ ⁠.

For $x \ne 0$ ⁠, $f$ is a polynomial on each side, so it is differentiable there. At $x = 0$ ⁠:

Why?

At $x = 0$ the formula changes, so the derivative there must come from the definition. We find the left-hand and right-hand derivatives:

\text{LHD} = \lim_{x \to 0^-}\frac{f(x) - f(0)}{x - 0}

and the same with $x \to 0^+$ for the RHD. Here $f(0) = 0^3 = 0$ ⁠.

\begin{aligned} \text{LHD} &= \lim_{x \to 0^-}\frac{f(x) - f(0)}{x - 0} \\ &= \lim_{x \to 0^-}\frac{-x^3 - 0}{x} \\ &= \lim_{x \to 0^-}(-x^2) = 0 \end{aligned}

\begin{aligned} \text{RHD} &= \lim_{x \to 0^+}\frac{f(x) - f(0)}{x - 0} \\ &= \lim_{x \to 0^+}\frac{x^3 - 0}{x} \\ &= \lim_{x \to 0^+}x^2 = 0 \end{aligned}

Why?

For $x$ to the left of 0 use $f(x) = -x^3$ ⁠; to the right use $f(x) = x^3$ ⁠.

In the limit $x \ne 0$ ⁠, so we may cancel one $x$ ⁠: $\frac{-x^3}{x} = -x^2$ and $\frac{x^3}{x} = x^2$ ⁠. Both tend to 0.

$\therefore$ LHD of $f(x)$ at $x = 0$ $=$ RHD of $f(x)$ at $x = 0$ ⁠.

So, $f(x)$ is differentiable at $x = 0$ ⁠, and the derivative of $f(x)$ is given by

f'(x) = \begin{cases} 3x^2, & x \ge 0 \\ -3x^2, & x < 0 \end{cases}

Why?

On each side differentiate the polynomial: $\frac{d}{dx}x^3 = 3x^2$ and $\frac{d}{dx}(-x^3) = -3x^2$ ⁠.

At 0 the derivative is the common value 0 of LHD and RHD, which agrees with $3(0)^2 = 0$ ⁠. So $x = 0$ can go in the first line.

Again $f'$ is a polynomial on each side of 0. At $x = 0$ ⁠:

\begin{aligned} &\text{LHD of } f'(x) \\ &= \lim_{x \to 0^-}\frac{f'(x) - f'(0)}{x - 0} \\ &= \lim_{x \to 0^-}\frac{-3x^2 - 0}{x} \\ &= \lim_{x \to 0^-}(-3x) = 0 \end{aligned}

\begin{aligned} &\text{RHD of } f'(x) \\ &= \lim_{x \to 0^+}\frac{f'(x) - f'(0)}{x - 0} \\ &= \lim_{x \to 0^+}\frac{3x^2 - 0}{x - 0} \\ &= \lim_{x \to 0^+}3x = 0 \end{aligned}

Why?

The same test, now applied to $f'$ ⁠, with $f'(0) = 0$ ⁠. Cancel one $x$ ⁠: $\frac{-3x^2}{x} = -3x$ and $\frac{3x^2}{x} = 3x$ ⁠. Both tend to 0.

$\therefore$ LHD of $f'(x)$ at $x = 0$ $=$ RHD of $f'(x)$ at $x = 0$ ⁠.

So, $f'(x)$ is differentiable at $x = 0$ ⁠, and $f''(x)$ exists for all real $x$ ⁠.

For $x > 0$ ⁠, $f''(x) = 6x$ ⁠; for $x < 0$ ⁠, $f''(x) = -6x$ ⁠; and $f''(0) = 0$ ⁠.

Why?

Away from 0, differentiate each piece of $f'$ ⁠: $\frac{d}{dx}(3x^2) = 6x$ and $\frac{d}{dx}(-3x^2) = -6x$ ⁠. At 0, $f''(0)$ is the common value 0 found above, and $6(0) = 0$ ⁠.

In short, $f''(x) = 6\lvert x\rvert$ for every real $x$ ⁠.

Hence, $f''(x) = \begin{cases} 6x, & x \ge 0 \\ -6x, & x < 0 \end{cases}$

(b)

Given $(x - a)^2 + (y - b)^2 = c^2$ ⁠, $c > 0$ ⁠.

Let $x - a = c\cos\theta$ and $y - b = c\sin\theta$ ( $\sin\theta \ne 0$ ⁠).

Why?

The curve is a circle with centre $(a, b)$ and radius $c$ ⁠. This substitution fits it, because

\begin{aligned} &(c\cos\theta)^2 + (c\sin\theta)^2 \\ &= c^2(\cos^2\theta + \sin^2\theta) = c^2 \end{aligned}

We need $\sin\theta \ne 0$ so that $\frac{dx}{d\theta} \ne 0$ below. These excluded points are where the tangent is vertical.

\therefore\ \frac{dx}{d\theta} = -c\sin\theta \ \text{ and } \ \frac{dy}{d\theta} = c\cos\theta

Why?

$x = a + c\cos\theta$ and $y = b + c\sin\theta$ ⁠, where $a$ ⁠, $b$ ⁠, $c$ are constants. Use $\frac{d}{d\theta}\cos\theta = -\sin\theta$ and $\frac{d}{d\theta}\sin\theta = \cos\theta$ ⁠.

\therefore\ \frac{dy}{dx} = \frac{c\cos\theta}{-c\sin\theta} = -\cot\theta

Why?

For a curve given in terms of a parameter,

\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}

The $c$ cancels, and $\frac{\cos\theta}{\sin\theta} = \cot\theta$ ⁠.

Differentiating both sides w.r.t. $\theta$ ⁠,

\frac{d}{d\theta}\left(\frac{dy}{dx}\right) = \frac{d}{d\theta}(-\cot\theta)

\begin{aligned} &\text{Or, } \frac{d}{dx}\left(\frac{dy}{dx}\right)\frac{dx}{d\theta} = \operatorname{cosec}^2\theta \\ &\text{Or, } \frac{d^2y}{dx^2}(-c\sin\theta) = \operatorname{cosec}^2\theta \end{aligned}

Why?

Watch out: $\frac{d^2y}{dx^2}$ is the derivative of $\frac{dy}{dx}$ with respect to $x$ ⁠, not $\theta$ ⁠. The chain rule links them:

\frac{d}{d\theta}(\ldots) = \frac{d}{dx}(\ldots)\cdot\frac{dx}{d\theta}

On the right, $\frac{d}{d\theta}\cot\theta = -\operatorname{cosec}^2\theta$ ⁠, so $-\cot\theta$ gives $+\operatorname{cosec}^2\theta$ ⁠.

\therefore\ \frac{d^2y}{dx^2} = -\frac{\operatorname{cosec}^3\theta}{c}

Why?

Divide both sides by $-c\sin\theta$ ⁠. Since $\frac{1}{\sin\theta} = \operatorname{cosec}\theta$ ⁠,

\frac{\operatorname{cosec}^2\theta}{-c\sin\theta} = -\frac{\operatorname{cosec}^3\theta}{c}

In steps:

\begin{aligned} \frac{d^2y}{dx^2} &= \frac{\operatorname{cosec}^2\theta}{-c\sin\theta} \\ &= -\frac1c \cdot \operatorname{cosec}^2\theta \cdot \frac{1}{\sin\theta} \\ &= -\frac1c \cdot \operatorname{cosec}^2\theta \cdot \operatorname{cosec}\theta \\ &= -\frac{\operatorname{cosec}^3\theta}{c} \end{aligned}

Also $1 + \left(\frac{dy}{dx}\right)^2 = 1 + \cot^2\theta = \operatorname{cosec}^2\theta$ ⁠, so

\left[1 + \left(\frac{dy}{dx}\right)^2\right]^{\frac32} = \lvert\operatorname{cosec}\theta\rvert^3

Why?

The identity $1 + \cot^2\theta = \operatorname{cosec}^2\theta$ comes from dividing $\sin^2\theta + \cos^2\theta = 1$ by $\sin^2\theta$ ⁠.

\begin{aligned} \frac{\sin^2\theta}{\sin^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} &= \frac{1}{\sin^2\theta} \\ 1 + \cot^2\theta &= \operatorname{cosec}^2\theta \end{aligned}

And $(-\cot\theta)^2 = \cot^2\theta$ ⁠, so $1 + \left(\frac{dy}{dx}\right)^2 = 1 + \cot^2\theta$ ⁠.

The power $\frac32$ means: square root first, then cube. A square root is never negative, so

\sqrt{\operatorname{cosec}^2\theta} = \lvert\operatorname{cosec}\theta\rvert

\begin{aligned} \therefore\ &\frac{\left[1 + \left(\dfrac{dy}{dx}\right)^2\right]^{\frac32}}{\dfrac{d^2y}{dx^2}} \\ &= \frac{\lvert\operatorname{cosec}\theta\rvert^3}{-\dfrac{\operatorname{cosec}^3\theta}{c}} \\ &= -c\,\frac{\lvert\operatorname{cosec}\theta\rvert^3}{\operatorname{cosec}^3\theta} \end{aligned}

Why?

Dividing by $-\frac{\operatorname{cosec}^3\theta}{c}$ is the same as multiplying by $-\frac{c}{\operatorname{cosec}^3\theta}$ ⁠.

\begin{aligned} &\lvert\operatorname{cosec}\theta\rvert^3 \times \left(-\frac{c}{\operatorname{cosec}^3\theta}\right) \\ &= -c\,\frac{\lvert\operatorname{cosec}\theta\rvert^3}{\operatorname{cosec}^3\theta} \end{aligned}

This is $-c$ when $\sin\theta > 0$ (that is $y > b$ ⁠) and $c$ when $\sin\theta < 0$ ( $y < b$ ⁠).

Why?

$\frac{\lvert u\rvert^3}{u^3}$ is 1 when $u > 0$ and $-1$ when $u < 0$ ⁠. Here $u = \operatorname{cosec}\theta$ has the same sign as $\sin\theta$ ⁠, and $\sin\theta = \frac{y - b}{c}$ has the sign of $y - b$ ⁠.

\begin{aligned} u > 0&: \frac{\lvert u\rvert^3}{u^3} = \frac{u^3}{u^3} = 1 \\ &\Rightarrow -c(1) = -c \\[4pt] u < 0&: \frac{\lvert u\rvert^3}{u^3} = \frac{(-u)^3}{u^3} = \frac{-u^3}{u^3} \\ &= -1 \Rightarrow -c(-1) = c \end{aligned}

Either way the value depends only on the radius $c$ ⁠, not on the centre $(a, b)$ ⁠.

$\therefore$ The expression equals $-c$ (or $c$ on the lower half of the circle), which is a constant and is independent of $a$ and $b$ ⁠. Hence proved.

The official marking scheme writes $(\operatorname{cosec}^2\theta)^{\frac32} = \operatorname{cosec}^3\theta$ and gets $-c$ ⁠; that holds where $\sin\theta > 0$ ⁠, that is $y > b$ ⁠.

Where this comes from

Chapter 5, Continuity and Differentiability: Second derivative of |x|³; parametric form.

Question type: (a) Is $f$ differentiable at a point? Compare LHD and RHD; (b) Parametric form + Second derivative (combined)

How to spot it: click to see

(a) Type 3. Is $f$ differentiable at a point? Compare LHD and RHD

Spot it: a modulus such as $|x - 1|$ ⁠, or a function in pieces, and the word "differentiable".

Here: $|x|^3$ is $x^3$ for $x \ge 0$ and $-x^3$ for negative $x$ ⁠. At 0, LHD and RHD are both 0, so $f'$ exists; do the same check on $f' = \pm3x^2$ to show $f''$ exists too.

(b) Type 8. Parametric form: $x$ and $y$ given through $t$ or $\theta$

Spot it: two equations, one for $x$ and one for $y$ ⁠, both in the same third letter.

Here: First, the circle has no parameter, so make one: $x - a = c\cos\theta$ ⁠, $y - b = c\sin\theta$ ⁠. Then $\frac{dy}{dx} = \frac{c\cos\theta}{-c\sin\theta} = -\cot\theta$ ⁠.

(b) Type 9. Second derivative: differentiate twice, then prove

Spot it: $\frac{d^2y}{dx^2}$ or $y''$ in the question, often "prove that $\frac{d^2y}{dx^2} + y = 0$ ".

Here: Then differentiate $-\cot\theta$ again, times $\frac{d\theta}{dx}$ ⁠, to get $-\frac{\operatorname{cosec}^3\theta}{c}$ ⁠, and divide into $(1 + \cot^2\theta)^{\frac32}$ ⁠: $a$ and $b$ are gone.

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Continuity and Differentiability in past papers: question 26 of 35 · see them all

Q34: f′′(x)f''(x)f′′(x) for f(x)=∣x∣3f(x) = |x|^3f(x)=∣x∣3⁠, or [1+(y′)2]3/2y′′\frac{[1 + (y')^2]^{3/2}}{y''}y′′[1+(y′)2]3/2​ on a circle is constant

Q34: $f''(x)$ for $f(x) = |x|^3$ ⁠, or $\frac{[1 + (y')^2]^{3/2}}{y''}$ on a circle is constant