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Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability1 Mark
MCQ
If $\text{f(x)}=\begin{cases}\frac{\sin(\cos\text{x})-\cos\text{x}}{(\pi-2\text{x})^2},&\text{x}\neq\frac{\pi}{2}\\\text{k},&\text{x}=\frac{\pi}{2}\end{cases}$ is continuous at $\text{x}=\frac{\pi}{2},$ then $k$ is equal to :
  • ✓
    $0$
  • B
    $\frac{1}{2}$
  • C
    $1$
  • D
    $-1$
✓

Answer

Correct option: A.
$0$
$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}}\frac{\sin(\cos\text{x})-\cos\text{x}}{(\pi-2\text{x})^2}$
$\text{x}\rightarrow\frac{\pi}{2}+\text{h}$
$\text{x}\rightarrow\frac{\pi}{2}$
$\Rightarrow\text{h}\rightarrow0$
$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{h}\rightarrow0}\frac{\sin\Big(\cos\big(\frac{\pi}{2}+\text{h}\big)\Big)-\cos\big(\frac{\pi}{2}+\text{h}\big)}{\Big(\pi-2\big(\frac{\pi}{2}+\text{h}\big)\Big)^2}$
$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{h}\rightarrow0}\frac{\sin(-\sin\text{h})+\sin\text{h}}{4\text{h}^2}$
$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{h}\rightarrow0}\frac{\sin\text{h}-\sin(\sin\text{h})}{4\text{h}^2}$
$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{h}\rightarrow0}\frac{\sin\text{h}-\sin(\sin\text{h})}{4\text{h}^2}$
$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{h}\rightarrow0}\frac{2\cos\big(\frac{\text{h}+\sin\text{h}}{2}\big)\sin\big(\frac{\text{h}-\sin\text{h}}{2}\big)}{4\text{h}^2}$
$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{h}\rightarrow0}\frac{2\cos\big(\frac{\text{h}+\sin\text{h}}{2}\big)}{4\text{h}^2}\times\frac{\sin\big(\frac{\text{h}-\sin\text{h}}{2}\big)}{\frac{\text{h}-\sin\text{h}}{2}}\times\frac{\text{h}-\sin\text{h}}{2}$
$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{h}\rightarrow0}\frac{2\cos\big(\frac{\text{h}+\sin\text{h}}{2}\big)}{4\text{h}^2}\times\frac{\text{h}-\sin\text{h}}{2}$
$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{h}\rightarrow0}\frac{2\cos\big(\frac{\text{h}+\sin\text{h}}{2}\big)}{4\text{h}^2}\times(\text{h}-\sin\text{h})$
$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{h}\rightarrow0}\frac{1}{4}\frac{\cos\big(\frac{\text{h}+\sin\text{h}}{2}\big)}{\text{h}^2}\times(\text{h}-\sin\text{h})=0$
$\text{k}=0$

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