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  • If <img alt="\mathrm{f(x)=\left\{\begin{array}{ll}\frac{2 \cos x-\sin 2 x}{(\pi-2 x)^{2}}, & x \leq \frac{\pi}{2} \\ \frac{e^{-\cos x}-1}{8 x-4 \pi}, & x>\frac{\pi}{2}\end{array}\right.}

If \mathrm{f(x)=\left\{\begin{array}{ll}\frac{2 \cos x-\sin 2 x}{(\pi-2 x)^{2}}, & x \leq \frac{\pi}{2} \\ \frac{e^{-\cos x}-1}{8 x-4 \pi}, & x>\frac{\pi}{2}\end{array}\right.} then which of the following holds?

Option: 1

f is continuous at x=\pi / 2


Option: 2

f  has an irremovable discontinuity at x=\pi / 2


Option: 3

f has a removable discontinuity at x=\pi / 2


Option: 4

None of these


Answers (1)

best_answer

\mathrm{f(x)= \begin{cases}\frac{2 \cos x-\sin 2 x}{(\pi-2 x)^{2}}, & x \leq \frac{\pi}{2} \\ \frac{e^{-\cos x}-1}{8 x-4 \pi}, & x>\frac{\pi}{2}\end{cases}}

L.H.L. at \mathrm{x=\pi / 2}
\mathrm{\lim _{h \rightarrow 0} \frac{2 \sin h-\sin 2 h}{4 h^{2}}=\lim _{h \rightarrow 0} \frac{2 \sin h(1-\cos h)}{4 h^{2}}=0}

\mathrm{R.H.L. =\lim _{h \rightarrow 0} \frac{e^{\sin h}-1}{8((\pi / 2)+h)-4 \pi}}
\mathrm{=\lim _{h \rightarrow 0} \frac{e^{\sin h}-1}{8 h} \cdot \frac{\sin h}{\sin h}=\frac{1}{8}}

\mathrm{\Rightarrow \quad h(x)}  has irremovable discontinuity at \mathrm{x=\pi / 2}.

 

Posted by

Kuldeep Maurya

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