If limx→π/4 42−(cos⁡x+sin⁡x)51−sin⁡2x=

limx→π/4 42−(cos⁡x+sin⁡x)51−sin⁡2x=limx→π/4 (cos⁡x+sin⁡x)25/2−(2)5/2(1+sin⁡2x)−2=limx→π/4 (1+sin⁡2x)5/2−25/2(1+sin⁡2x)−2=limy→2 y5/2−25/2y−2, where y=1+sin⁡2x=52×25/2−1=52

If limx→π/4 42−(cos⁡x+sin⁡x)51−sin⁡2x=

limx→π/4 42−(cos⁡x+sin⁡x)51−sin⁡2x=limx→π/4 (cos⁡x+sin⁡x)25/2−(2)5/2(1+sin⁡2x)−2=limx→π/4 (1+sin⁡2x)5/2−25/2(1+sin⁡2x)−2=limy→2 y5/2−25/2y−2, where y=1+sin⁡2x=52×25/2−1=52

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

If $lim_{x \to π / 4} \frac{4 \sqrt{2} - (\cos x + \sin x)^{5}}{1 - \sin 2 x} =$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

none of these

$5 \sqrt{2}$

$3 \sqrt{2}$

$\sqrt{2}$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} lim_{x \to π / 4} \frac{4 \sqrt{2} - (\cos x + \sin x)^{5}}{1 - \sin 2 x} \\ = lim_{x \to π / 4} \frac{{\{(\cos x + \sin x)^{2}\}}^{5 / 2} - (2)^{5 / 2}}{(1 + \sin 2 x) - 2} \end{array}$