_ x 4 4 2-( x+ x)^5 1- 2 x = - Vidyadip

MCQ

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

✓
$5 \sqrt{2}$
B
$3 \sqrt{2}$
C
$\sqrt{2}$
D
$\frac{5 \sqrt{2}}{2}$

✓

Answer

Correct option: A.

$5 \sqrt{2}$

(A)
$\lim _{x \rightarrow \pi / 4} \frac{4 \sqrt{2}-(\cos x+\sin x)^5}{1-\sin 2 x}$
$=\lim _{x \rightarrow \pi / 4} \frac{\left[(\cos x+\sin x)^2\right]^{\frac{5}{2}}-(2)^{\frac{5}{2}}}{(1+\sin 2 x) 2}$
$=\lim _{x \rightarrow \pi / 4} \frac{(1+\sin 2 x)^{\frac{5}{2}}-2^{\frac{5}{2}}}{(1+\sin 2 x)-2}$
$=\lim _{y \rightarrow 2} \frac{y^{\frac{5}{2}}-2^{\frac{5}{2}}}{y-2}$, where $y=1+\sin 2 x$
$=\frac{5}{2} \times 2^{\frac{5}{2}-1}=5 \sqrt{2}$

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