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CBSE Sample Paper 2020-21 · Part B: Section IV · 3 marks

Q31: [x][x] is not differentiable at x=1x = 1⁠, or d2ydx2\frac{d^2y}{dx^2} for x=asec⁡θx = a\sec\theta⁠, y=btan⁡θy = b\tan\theta

Sample 2020-21Ch 5 · Continuity and Differentiability
The question

(a) Prove that the greatest integer function defined by f(x)=[x]f(x) = [x]⁠, 0<x<20 < x < 2 is not differentiable at x=1x = 1

OR

(b) If x=asec⁡θx = a\sec\theta⁠, y=btan⁡θy = b\tan\theta find d2ydx2\dfrac{d^2y}{dx^2} at x=π6x = \dfrac{\pi}{6}

Exam answer

(a)

For 0<x<10 < x < 1⁠, [x]=0[x] = 0⁠; for 1≤x<21 \le x < 2⁠, [x]=1[x] = 1⁠. Let h>0h > 0 be small.

Why?

[x][x] is the greatest integer less than or equal to xx⁠. So [1]=1[1] = 1⁠. For a small h>0h > 0⁠, 1+h1 + h lies between 1 and 2, so [1+h]=1[1 + h] = 1⁠; and 1−h1 - h lies between 0 and 1, so [1−h]=0[1 - h] = 0⁠.

RHD=lim⁡h→0f(1+h)−f(1)h=lim⁡h→0[1+h]−[1]h=lim⁡h→01−1h=0\begin{aligned} \text{RHD} &= \lim_{h \to 0}\frac{f(1 + h) - f(1)}{h} \\ &= \lim_{h \to 0}\frac{[1 + h] - [1]}{h} \\ &= \lim_{h \to 0}\frac{1 - 1}{h} = 0 \end{aligned}
Why?

The right-hand derivative approaches 1 from the right, through 1+h1 + h⁠. The numerator is 00 for every small hh⁠, so the fraction is 00 all the way to the limit.

LHD=lim⁡h→0f(1−h)−f(1)−h=lim⁡h→0[1−h]−[1]−h=lim⁡h→00−1−h=lim⁡h→01h=∞\begin{aligned} \text{LHD} &= \lim_{h \to 0}\frac{f(1 - h) - f(1)}{-h} \\ &= \lim_{h \to 0}\frac{[1 - h] - [1]}{-h} \\ &= \lim_{h \to 0}\frac{0 - 1}{-h} = \lim_{h \to 0}\frac{1}{h} = \infty \end{aligned}
Why?

The left-hand derivative approaches 1 from the left, through 1−h1 - h⁠. The change in xx is then −h-h⁠, which is why the denominator is −h-h⁠.

0−1−h=−1−h=1h\begin{aligned} \frac{0 - 1}{-h} &= \frac{-1}{-h} = \frac{1}{h} \end{aligned}

0−1−h=1h\frac{0 - 1}{-h} = \frac{1}{h}⁠, and 1h\frac1h grows without bound as h→0h \to 0⁠. So this limit is not a finite number.

So the LHD does not exist (it is not finite). Since RHD ≠\ne LHD,

Why?

ff is differentiable at a point only if both one-sided derivatives exist, are finite, and are equal. Here the LHD fails, so ff is not differentiable at 1.

Another way to see it: ff jumps from 0 to 1 at x=1x = 1⁠, so it is not even continuous there.

f(x)f(x) is not differentiable at x=1x = 1⁠. Hence proved.

(b)

Here the point is taken as θ=π6\theta = \frac{\pi}{6}⁠, as in the marking scheme.

Why?

The paper prints "at x=π6x = \frac{\pi}{6}", but x=asec⁡θx = a\sec\theta depends on aa⁠, so a value of xx does not fix θ\theta without knowing aa⁠. The value π6\frac{\pi}{6} is meant for the parameter θ\theta⁠.

y=btan⁡θ⇒dydθ=bsec⁡2θ…(i)\begin{aligned} y &= b\tan\theta \\ \Rightarrow \frac{dy}{d\theta} &= b\sec^2\theta \quad\ldots(\text{i}) \end{aligned} x=asec⁡θ⇒dxdθ=asec⁡θtan⁡θ…(ii)\begin{aligned} x &= a\sec\theta \\ \Rightarrow \frac{dx}{d\theta} &= a\sec\theta\tan\theta \quad\ldots(\text{ii}) \end{aligned}
Why?

Both xx and yy are given in terms of the parameter θ\theta⁠, so differentiate each with respect to θ\theta⁠. Standard derivatives:

ddθtan⁡θ=sec⁡2θddθsec⁡θ=sec⁡θtan⁡θ\begin{aligned} \frac{d}{d\theta}\tan\theta &= \sec^2\theta \\ \frac{d}{d\theta}\sec\theta &= \sec\theta\tan\theta \end{aligned}
dydx=dydθdxdθ=bsec⁡2θasec⁡θtan⁡θ=bacosec⁡θ\begin{aligned} \frac{dy}{dx} &= \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{b\sec^2\theta}{a\sec\theta\tan\theta} \\ &= \frac{b}{a}\operatorname{cosec}\theta \end{aligned}
Why?

Cancel one sec⁡θ\sec\theta⁠, then write in sines and cosines:

bsec⁡2θasec⁡θtan⁡θ=b⋅sec⁡θ⋅sec⁡θa⋅sec⁡θ⋅tan⁡θ=ba⋅sec⁡θtan⁡θ\begin{aligned} \frac{b\sec^2\theta}{a\sec\theta\tan\theta} &= \frac{b \cdot \sec\theta \cdot \sec\theta}{a \cdot \sec\theta \cdot \tan\theta} \\ &= \frac{b}{a} \cdot \frac{\sec\theta}{\tan\theta} \end{aligned} sec⁡θtan⁡θ=1cos⁡θ⋅cos⁡θsin⁡θ\frac{\sec\theta}{\tan\theta} = \frac{1}{\cos\theta}\cdot\frac{\cos\theta}{\sin\theta}

which is 1sin⁡θ=cosec⁡θ\frac{1}{\sin\theta} = \operatorname{cosec}\theta⁠.

Differentiating both sides w.r.t. xx⁠, and using (ii),

d2ydx2=−bacosec⁡θcot⁡θ×dθdx=−bacosec⁡θcot⁡θ×1asec⁡θtan⁡θ=−ba2cot⁡3θ\begin{aligned} \frac{d^2y}{dx^2} &= -\frac{b}{a}\operatorname{cosec}\theta\cot\theta \\ &\qquad \times \frac{d\theta}{dx} \\ &= -\frac{b}{a}\operatorname{cosec}\theta\cot\theta \\ &\qquad \times \frac{1}{a\sec\theta\tan\theta} \\ &= -\frac{b}{a^2}\cot^3\theta \end{aligned}
Why?

dydx\frac{dy}{dx} is a function of θ\theta⁠, so differentiating it w.r.t. xx needs the chain rule: differentiate w.r.t. θ\theta⁠, then multiply by dθdx\frac{d\theta}{dx}⁠. Here ddθcosec⁡θ=−cosec⁡θcot⁡θ\frac{d}{d\theta}\operatorname{cosec}\theta = -\operatorname{cosec}\theta\cot\theta⁠, and dθdx\frac{d\theta}{dx} is 11 divided by (ii).

Watch out: forgetting the factor dθdx\frac{d\theta}{dx} is the usual mistake.

To simplify, use cosec⁡θsec⁡θ=cos⁡θsin⁡θ=cot⁡θ\frac{\operatorname{cosec}\theta}{\sec\theta} = \frac{\cos\theta}{\sin\theta} = \cot\theta and cot⁡θtan⁡θ=cot⁡2θ\frac{\cot\theta}{\tan\theta} = \cot^2\theta⁠:

cosec⁡θcot⁡θsec⁡θtan⁡θ=cot⁡3θ\frac{\operatorname{cosec}\theta\cot\theta}{\sec\theta\tan\theta} = \cot^3\theta

Step by step:

cosec⁡θcot⁡θsec⁡θtan⁡θ=cosec⁡θsec⁡θ⋅cot⁡θtan⁡θ=cot⁡θ⋅cot⁡2θ=cot⁡3θ\begin{aligned} &\frac{\operatorname{cosec}\theta\cot\theta}{\sec\theta\tan\theta} \\ &= \frac{\operatorname{cosec}\theta}{\sec\theta} \cdot \frac{\cot\theta}{\tan\theta} \\ &= \cot\theta \cdot \cot^2\theta = \cot^3\theta \end{aligned}

Here cot⁡θtan⁡θ=cot⁡θ⋅1tan⁡θ=cot⁡θ⋅cot⁡θ\frac{\cot\theta}{\tan\theta} = \cot\theta \cdot \frac{1}{\tan\theta} = \cot\theta \cdot \cot\theta⁠. The numbers in front multiply as −ba×1a=−ba2-\frac{b}{a} \times \frac{1}{a} = -\frac{b}{a^2}⁠.

d2ydx2]θ=π6=−ba2[cot⁡π6]3=−ba2(3)3\begin{aligned} \left.\frac{d^2y}{dx^2}\right]_{\theta = \frac{\pi}{6}} &= -\frac{b}{a^2}\left[\cot\frac{\pi}{6}\right]^3 \\ &= -\frac{b}{a^2}(\sqrt3)^3 \end{aligned}
Why?

tan⁡π6=13\tan\frac{\pi}{6} = \frac{1}{\sqrt3}⁠, so cot⁡π6=3\cot\frac{\pi}{6} = \sqrt3⁠. Then

(3)3=3×3×3=3×3=33−ba2(33)=−33 ba2\begin{aligned} (\sqrt3)^3 &= \sqrt3 \times \sqrt3 \times \sqrt3 \\ &= 3 \times \sqrt3 = 3\sqrt3 \\[4pt] -\frac{b}{a^2}(3\sqrt3) &= -\frac{3\sqrt3\,b}{a^2} \end{aligned}

∴ d2ydx2=−33 ba2\therefore\ \dfrac{d^2y}{dx^2} = -\dfrac{3\sqrt3\,b}{a^2} at θ=π6\theta = \dfrac{\pi}{6}

Where this comes from

Chapter 5, Continuity and Differentiability: Differentiability; parametric second derivative.

Question type: (a) Is ff 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 ff differentiable at a point? Compare LHD and RHD

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

Here: [x][x] jumps at x=1x = 1⁠: it is 0 just left of 1 and 1 from 1 on. RHD is 0, but LHD is lim⁡0−1−h\lim\frac{0 - 1}{-h}⁠, which is not finite, so [x][x] is not differentiable at 1.

(b) Type 8. Parametric form: xx and yy given through tt or θ\theta

Spot it: two equations, one for xx and one for yy⁠, both in the same third letter.

Here: First, x=asec⁡θx = a\sec\theta and y=btan⁡θy = b\tan\theta share θ\theta⁠: divide to get dydx=bacosec⁡θ\frac{dy}{dx} = \frac{b}{a}\operatorname{cosec}\theta⁠.

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

Spot it: d2ydx2\frac{d^2y}{dx^2} or y′′y'' in the question, often "prove that d2ydx2+y=0\frac{d^2y}{dx^2} + y = 0".

Here: Then differentiate bacosec⁡θ\frac{b}{a}\operatorname{cosec}\theta with respect to xx⁠, times dθdx=1asec⁡θtan⁡θ\frac{d\theta}{dx} = \frac{1}{a\sec\theta\tan\theta}⁠, and put θ=π6\theta = \frac\pi6⁠.

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