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NDA Math Limit, Continuity and Differentiability

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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 : 68 of 105

Marks: +1, -0

Consider the following statements :

1. The function

f(x)

is continuous at

x=0

2. The function

f(x)

is continuous 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:

Given

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}

For continuity

f(a)=L.H.L=R.H.L

At

x=0

f(0)=\pi\cos0=\pi

L.H.L

=\lim\limits_{x\to0^{-}}f(x-h)

=\lim\limits_{x\to0^{-}}f(-h)

=\lim\limits_{h\to0}(-h+\pi)=\pi

R.H.L

=\lim\limits_{x\to0^{+}}f(x+h)

=\lim\limits_{h\to0}f(0+h)

=\lim\limits_{h\to0}\pi\cos h

=\pi\cos0=\pi

f(0) = L.HL. = R.H .L

Hence function is continuous at

x=0

Statement (1) is correct

At

x=\frac{\pi}{2}

L.H.L

=\lim\limits_{x\to\frac{\pi^{-}}{2}}f(x-h)

=\lim\limits_{h\to0}f\left(\frac{\pi}{2}-h\right)

=\lim\limits_{h\to0}\pi\cos\left(\frac{\pi}{2}-h\right)

=\pi\cos\frac{\pi}{2}=0

R.H.L

=\lim\limits_{x\to\frac{\pi^{+}}{2}}f(x+h)

=\lim\limits_{h\to0}f\left(\frac{\pi}{2}+h\right)

=\lim\limits_{h\to0}\left(\frac{\pi}{2}+h-\frac{\pi}{2}\right)^{2}=0

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

L.HL. =RH.L

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

Hence function is continuous at

x=\frac{\pi}{2}

Statement (2) is correct

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