Q99 / 160 · AP EAMCET 2024 (20 May Shift 2) · Mathematics

$\lim _{x \rightarrow \frac{\pi}{4}} \frac{4 \sqrt{2}-(\cos x+\sin x)^5}{1-\sin 2 x}=$

  • Correct answer
    A
    $5 \sqrt{2}$
  • B
    $3 \sqrt{2}$
  • C
    $2 \sqrt{2}$
  • D
    $\sqrt{2}$

Solution

$\lim _{x \rightarrow \frac{\pi}{4}} \frac{4 \sqrt{2}-(\cos x+\sin x)^5}{1-\sin 2 x}\left(\frac{0}{0}\right.$ form $)$ $\begin{aligned} & \Rightarrow \lim _{x \rightarrow \frac{\pi}{4}} \frac{-5(\cos x+\sin x)^4(\cos x-\sin x)}{-2 \cos 2 x}\left(\frac{0}{0} \text { form }\right) \\ & \Rightarrow \lim _{x \rightarrow \frac{\pi}{4}} \frac{-20(\cos x+\sin x)^3(\cos x-\sin x)+}{4 \sin 2 x} \\ & =\frac{20 \sqrt{2}}{4}=5 \sqrt{2}\end{aligned}$

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

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