A function f(x) is defined as follows: f(x)=\ array ccc x+π & for & x ∈[-π, 0) π x & for & x ∈[0, (π)/(2)] (x-(π)/(2))^2 & for & x ∈((π)/(2), π] array . Consider the following statements: 1. The function f(x) is

Correct answer: C. Both 1 and 2 Given f(x)=\ array cc (x+π) & for x ∈[-π, 0) π x & for x ∈[0, (π)/(2)] (x-(π)/(2))^2 & for x ∈((π)/(2), π] array . For continuity, aligned & f(a)= L.H.L. = R.H.L. & aligned & At x=0 & f(0)=π 0=π & L.H.L. = _ x arrow 0^ - f(x-h) &= _ h arrow 0 f(-h) &= _ h arrow 0 (-h+π)=π aligned aligned aligned R.H.L. & = _ x arrow 0^ + f(x+h) & = _ h arrow 0 f(0+h) & = _ h arrow 0 π h & =π 0=π aligned f(0)= L.H.L. = R.H.L. Hence function is continuous at x=0 . Statement (1) is correct. At x=(π)/(2) aligned L.H.L. & = _ x arrow π^ - 2 f(x-h) & = _ h arrow 0 f((π)/(2)-h) & = _ h arrow 0 π ((π)/(2)-h) & =π (π)/(2)=0 aligned aligned RH.L. & = _ x arrow π^ + 2 f(x+h) & = _ h arrow 0 f((π)/(2)+h) & = _ h arrow 0 ((π)/(2)+h-(π)/(2))^2=0 aligned aligned & f((π)/(2))=π (π)/(2)=0 & L.H.L. = R.H.L. =f((π)/(2)) aligned Hence function is continuous at x=(π)/(2) Statement (2) is correct.

NDA 2016

Exam: NDA
Year: 2016
Subject: Mathematics
Chapter: Continuity and Differentiability

NDA 2016 · Mathematics · Continuity and Differentiability

A function f(x) is defined as follows: f(x)=\ array ccc x+π & for &...

A function

\mathrm{f}(\mathrm{x})

is defined as follows:

f(x)=\left\{

\begin{array}{ccc} x+\pi & \text{ for } & x \in[-\pi, 0) \\ \pi \cos x & \text{ for } & x \in\left[0, \frac{\pi}{2}\right] \\ \left(x-\frac{\pi}{2}\right)^2 & \text{ for } & x \in\left(\frac{\pi}{2}, \pi\right] \end{array}

\right.

Consider the following statements: 1. The function

\mathrm{f}(\mathrm{x})

is continuous at

\mathrm{x}=0

. 2. The function

f(x)

is continuous at

x=\frac{\pi}{2}

. Which of the above statements is/are correct?

Options

A1 only
B2 only
CBoth 1 and 2
DNeither 1 nor 2

Correct answer

C. Both 1 and 2

Step-by-step solution

Given

f(x)=\left\{

\begin{array}{cc} (x+\pi) & \text{ for } x \in[-\pi, 0) \\ \pi \cos x & \text{ for } x \in\left[0, \frac{\pi}{2}\right] \\ \left(x-\frac{\pi}{2}\right)^2 & \text{ for } x \in\left(\frac{\pi}{2}, \pi\right] \end{array}

\right.

For continuity,

\begin{aligned} & f(a)=\text{ L.H.L. }=\text{ R.H.L. } \\ & \begin{aligned} & \text{ At } x=0 \\ & f(0)=\pi \cos 0=\pi \\ & \text{ L.H.L. }=\lim _{x \rightarrow 0^{-}} f(x-h) \\ &=\lim _{h \rightarrow 0} f(-h) \\ &=\lim _{h \rightarrow 0}(-h+\pi)=\pi \end{aligned}

\end{aligned}

\begin{aligned} \text{ R.H.L. } & =\lim _{x \rightarrow 0^{+}} f(x+h) \\ & =\lim _{h \rightarrow 0} f(0+h) \\ & =\lim _{h \rightarrow 0} \pi \cos h \\ & =\pi \cos 0=\pi \end{aligned}

f(0)=\text{ L.H.L. }=\text{ R.H.L. }

Hence function is continuous at

x=0

. Statement (1) is correct. At

x=\frac{\pi}{2}

\begin{aligned} \text{ L.H.L. } & =\lim _{x \rightarrow \frac{\pi^{-}}{2}} f(x-h) \\ & =\lim _{h \rightarrow 0} f\left(\frac{\pi}{2}-h\right) \\ & =\lim _{h \rightarrow 0} \pi \cos \left(\frac{\pi}{2}-h\right) \\ & =\pi \cos \frac{\pi}{2}=0 \end{aligned}

\begin{aligned} \text{ RH.L. } & =\lim _{x \rightarrow \frac{\pi^{+}}{2}} f(x+h) \\ & =\lim _{h \rightarrow 0} f\left(\frac{\pi}{2}+h\right) \\ & =\lim _{h \rightarrow 0}\left(\frac{\pi}{2}+h-\frac{\pi}{2}\right)^2=0 \end{aligned}

\begin{aligned} & f\left(\frac{\pi}{2}\right)=\pi \cos \frac{\pi}{2}=0 \\ & \text{ L.H.L. }=\text{ R.H.L. }=f\left(\frac{\pi}{2}\right) \end{aligned}

Hence function is continuous at

x=\frac{\pi}{2}$ Statement (2) is correct.

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