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Consider the following statements:
1- The function $\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=0$.
2- The function $\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=\frac{\pi}{2}$. Which of the above statements is/are correct?
A function $\mathrm{f}(\mathrm{x})$ is defined as follows: $\quad[2016-\mathrm{I}]$ $\mathrm{f}(\mathrm{x})=\left\{\begin{array}{lll}\mathrm{x}+\pi & \text { for } & \mathrm{x} \in[-\pi, 0) \\ \pi \cos \mathrm{x} & \text { for } & \mathrm{x} \in\left[0, \frac{\pi}{2}\right] \\ \left(\mathrm{x}-\frac{\pi}{2}\right)^{2} & \text { for } & \mathrm{x} \in\left(\frac{\pi}{2}, \pi\right]\end{array}\right.$
Options:
1- The function $\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=0$.
2- The function $\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=\frac{\pi}{2}$. Which of the above statements is/are correct?
A function $\mathrm{f}(\mathrm{x})$ is defined as follows: $\quad[2016-\mathrm{I}]$ $\mathrm{f}(\mathrm{x})=\left\{\begin{array}{lll}\mathrm{x}+\pi & \text { for } & \mathrm{x} \in[-\pi, 0) \\ \pi \cos \mathrm{x} & \text { for } & \mathrm{x} \in\left[0, \frac{\pi}{2}\right] \\ \left(\mathrm{x}-\frac{\pi}{2}\right)^{2} & \text { for } & \mathrm{x} \in\left(\frac{\pi}{2}, \pi\right]\end{array}\right.$
Solution:
2643 Upvotes
Verified Answer
The correct answer is:
Both 1 and 2
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, $f(a)=$ L.H.L. $=$ R.H.L.
At $x=0$
$f(0)=\pi \cos 0=\pi$
L.H.L. $=\lim _{x \rightarrow 0^{-}} f(x-h)$
$=\lim _{h \rightarrow 0} f(-h)$
$=\lim _{h \rightarrow 0}(-h+\pi)=\pi$
R.H.L. $=\lim _{x \rightarrow 0^{+}} f(x+h)$
$=\lim _{h \rightarrow 0} f(0+h)$
$=\lim _{h \rightarrow 0} \pi \cos h$
$\quad=\pi \cos 0=\pi$
$f(0)=$ L.H.L. $=$ R.H.L.
Hence function is continuous at $x=0$.
Statement (1) is correct.
At $x=\frac{\pi}{2}$
$\begin{aligned} \text { L.H.L. } &=\lim _{x \rightarrow \frac{\pi^{-}}{2}} f(x-h) \\ &=\lim _{h \rightarrow 0} f\left(\frac{\pi}{2}-h\right) \end{aligned}$
$$
\begin{array}{l}
=\lim _{h \rightarrow 0} \pi \cos \left(\frac{\pi}{2}-h\right) \\
=\pi \cos \frac{\pi}{2}=0 \\
\mathrm{RH} . \mathrm{L} .=\lim _{x \rightarrow \frac{\pi^{+}}{2}} f(x+h) \\
=\lim _{h \rightarrow 0} f\left(\frac{\pi}{2}+h\right) \\
=\lim _{h \rightarrow 0}\left(\frac{\pi}{2}+h-\frac{\pi}{2}\right)^{2}=0 \\
f\left(\frac{\pi}{2}\right)=\pi \cos \frac{\pi}{2}=0 \\
\text { L.H.L. }=\mathrm{R} . \mathrm{H.L.}=f\left(\frac{\pi}{2}\right)
\end{array}
$$
Hence function is continuous at $x=\frac{\pi}{2}$
Statement $(2)$ is correct.
$f(x)=\left\{\begin{array}{cc}(x+\pi) & \text { for } x \in[-\pi, 0) \\ \pi \cos x & \text { for } x \in\left[0, \frac{\pi}{2}\right] \\ \left(x-\frac{\pi}{2}\right)^{2} & \text { for } x \in\left(\frac{\pi}{2}, \pi\right]\end{array}\right.$
For continuity, $f(a)=$ L.H.L. $=$ R.H.L.
At $x=0$
$f(0)=\pi \cos 0=\pi$
L.H.L. $=\lim _{x \rightarrow 0^{-}} f(x-h)$
$=\lim _{h \rightarrow 0} f(-h)$
$=\lim _{h \rightarrow 0}(-h+\pi)=\pi$
R.H.L. $=\lim _{x \rightarrow 0^{+}} f(x+h)$
$=\lim _{h \rightarrow 0} f(0+h)$
$=\lim _{h \rightarrow 0} \pi \cos h$
$\quad=\pi \cos 0=\pi$
$f(0)=$ L.H.L. $=$ R.H.L.
Hence function is continuous at $x=0$.
Statement (1) is correct.
At $x=\frac{\pi}{2}$
$\begin{aligned} \text { L.H.L. } &=\lim _{x \rightarrow \frac{\pi^{-}}{2}} f(x-h) \\ &=\lim _{h \rightarrow 0} f\left(\frac{\pi}{2}-h\right) \end{aligned}$
$$
\begin{array}{l}
=\lim _{h \rightarrow 0} \pi \cos \left(\frac{\pi}{2}-h\right) \\
=\pi \cos \frac{\pi}{2}=0 \\
\mathrm{RH} . \mathrm{L} .=\lim _{x \rightarrow \frac{\pi^{+}}{2}} f(x+h) \\
=\lim _{h \rightarrow 0} f\left(\frac{\pi}{2}+h\right) \\
=\lim _{h \rightarrow 0}\left(\frac{\pi}{2}+h-\frac{\pi}{2}\right)^{2}=0 \\
f\left(\frac{\pi}{2}\right)=\pi \cos \frac{\pi}{2}=0 \\
\text { L.H.L. }=\mathrm{R} . \mathrm{H.L.}=f\left(\frac{\pi}{2}\right)
\end{array}
$$
Hence function is continuous at $x=\frac{\pi}{2}$
Statement $(2)$ is correct.
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