MCQ
Choose the correct answers from the given four options:
If $\text{f(x)}=\begin{cases}\text{mx}+1,&\text{if x}\leq\frac{\pi}{2}\\\sin\text{x}+\text{n},&\text{if x}>\frac{\pi}{2}\end{cases},$ is continuous at $\text{x}=\frac{\pi}{2},$ then:
  • A
    $\text{m}=1,\text{n}=0$
  • B
    $\text{m}=\frac{\text{n}\pi}{2}+1$
  • $\text{n}=\frac{\text{m}\pi}{2}$
  • D
    $\text{m}=\text{n}=\frac{\pi}{2}$

Answer

Correct option: C.
$\text{n}=\frac{\text{m}\pi}{2}$
We have, $\text{f(x)}=\begin{cases}\text{mx}+1,&\text{if x}\leq\frac{\pi}{2}\\\sin\text{x}+\text{n},&\text{if x}>\frac{\pi}{2}\end{cases},$ is continuous at $\text{x}=\frac{\pi}{2}$
$\therefore\ \text{L.H.L}=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^-}(\text{mx}+1)$ $=\lim\limits_{\text{h}\rightarrow0}\bigg[\text{m}\Big(\frac{\pi}{2}-\text{h}\Big)+1\bigg]=\frac{\text{m}\pi}{2}+1$

and $\text{R.H.L}=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^+}(\sin\text{x}+\text{n})$ $\lim\limits_{\text{h}\rightarrow0}\bigg[\sin\Big(\frac{\pi}{2}+\text{h}\Big)+\text{n}\bigg]$

$\lim\limits_{\text{h}\rightarrow0}[\cos\text{h}+\text{n}]=1+\text{n}$

We must have L.H.L = R.H.L

$\Rightarrow\ \text{m}\frac{\pi}{2}+1=\text{n}+1$

$\therefore\ \text{n}=\text{m}\frac{\pi}{2}$

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