Q204 / 270 · NDA 2016 (Phase 1) · Mathematics

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?

  • A
    1 only
  • B
    2 only
  • Correct answer
    C
    Both 1 and 2
  • D
    Neither 1 nor 2

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.

Hence, the correct answer is option: 'C'.

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