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UPSC NDA I 2016 Mathematics Solved Paper

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For the next two (2) items that follow :

A function f(x) is defined as follows:

f(x) = \begin{cases} x + \pi & \text{for } x \in [-\pi, 0) \\ \pi\; \cos x & \text{for } x \in [0, \pi/2] \\ (x - \pi/2)^2 & \text{for } x \in (\pi/2, \pi] \end{cases}

[NDA I 2016]

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Question : 55 of 120

Marks: +1, -0

Consider the following statements

1.The function f(x) is differentiable at

x=0

2. The function f(x) is differentiable at

x=\frac{\pi}{2}

Which of the above statements is/are correct?

1 only
2 only
Both 1 and 2
Neither 1 nor 2

Solution:

For differentiability,

L.HD. = RH.D.

Thus at

x=0

L.H.D

=\lim\limits_{h\to 0}\frac{f(0-h)-f(0)}{-h}

=\lim\limits_{h\to 0}\frac{f(-h)-f(0)}{-h}

=\lim\limits_{h\to 0}\frac{(-h+\pi)-(\pi\cos 0)}{-h}=1

L.HD. = 1

R.H.D

=\lim\limits_{h\to 0}\frac{f(0+h)-f(0)}{h}

=\lim\limits_{h\to 0}\frac{f(h)-f(0)}{h}

=\lim\limits_{h\to 0}\frac{\pi\cos h-\pi\cos 0}{h}

=\lim\limits_{h\to 0}\frac{\pi[\cos h-1]}{h}

=\lim\limits_{h\to 0}\frac{\pi\left[1-\frac{h^{2}}{2!}+\frac{h^{4}}{4!}\cdots-1\right]}{h}

=\lim\limits_{h\to 0}\frac{\pi\left[1-\frac{1}{2}h^{2}+\frac{1}{24}h^{4}\cdot\ldots-1\right]}{h}

=\lim\limits_{h\to 0}\frac{\pi\left[-\frac{1}{2}h^{2}+\frac{1}{24}h^{4}\cdot\ldots\right]}{h}=0

L.H.D≠R.H.D.

So at

x=0

function is not differentiable.

Statement (1) is not correct.

At

x=\frac{\pi}{2}

f\left(\frac{\pi}{2}\right)=\pi\cos\frac{\pi}{2}=0

R.H.D.

=\lim\limits_{h\to 0}\frac{f\left(\frac{\pi}{2}+h\right)-f\left(\frac{\pi}{2}\right)}{-h}

=\lim\limits_{h\to 0}\frac{\left(\frac{\pi}{2}+h-\frac{\pi}{2}\right)^{2}-0}{h}

=\lim\limits_{h\to 0}\frac{h^{2}}{h}=0

L.H.D.

=\lim\limits_{h\to 0}\frac{f\left(\frac{\pi}{2}-h\right)-f\left(\frac{\pi}{2}\right)}{-h}

=\lim\limits_{h\to 0}\frac{\pi\cos\left(\frac{\pi}{2}-h\right)-0}{-h}

=\lim\limits_{h\to 0}\pi\left(\frac{\sinh}{-h}\right)

=-\pi\lim\limits_{h\to 0}\left(\frac{\sinh}{h}\right)=-\pi(1)=-\pi

L.H.D≠R.H.D

Hence function is not differentiable at

x=\frac{\pi}{2}

Statement (2) is not correct.

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