limx→(π)/4(8√2-cosx+sinx^7)/(√2-√2sin2x) is equal to

Correct answer: (A) 14. limx→(π)/4(8√2-cosx+sinx^7)/(√2-√2sin2x) 0/0form =limx→(π)/4(-7cosx+sinx^6-sinx+cosx)/(-2√2cos2x) (using L'Hospital Rule) =limx→(π)/4(56cosx-sinx)/(2√2cos2x) 0/0 form =limx→(π)/4(-56sinx+cosx)/(-4√2sin2x) (using L'Hospital Rule) =(56√2)/(4√2)=14

Q36 / 90 · JEE Main 2022 (25 Jul Shift 2) · Mathematics

$\lim_{x \to \frac{π}{4}} \frac{8 \sqrt{2} - {(\cos x + \sin x)}^{7}}{\sqrt{2} - \sqrt{2} \sin 2 x}$ is equal to

Correct answer
A
$14$
B
$7$
C
$14 \sqrt{2}$
D
$7 \sqrt{2}$

Solution

$\lim_{x \to \frac{π}{4}} \frac{8 \sqrt{2} - {(\cos x + \sin x)}^{7}}{\sqrt{2} - \sqrt{2} \sin 2 x} (\frac{0}{0} form)$
$= \lim_{x \to \frac{π}{4}} \frac{- 7 {(\cos x + \sin x)}^{6} (- \sin x + \cos x)}{- 2 \sqrt{2} \cos 2 x}$ (using L'Hospital Rule)
$= \lim_{x \to \frac{π}{4}} \frac{56 (\cos x - \sin x)}{2 \sqrt{2} \cos 2 x} (\frac{0}{0} form)$
$= \lim_{x \to \frac{π}{4}} \frac{- 56 (\sin x + \cos x)}{- 4 \sqrt{2} \sin 2 x}$ (using L'Hospital Rule)
$= \frac{56 \sqrt{2}}{4 \sqrt{2}} = 14$

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

limx→π482-cosx+sinx72-2sin2x is equal to

Solution

Relevant Questions

$\lim_{x \to \frac{π}{4}} \frac{8 \sqrt{2} - {(\cos x + \sin x)}^{7}}{\sqrt{2} - \sqrt{2} \sin 2 x}$ is equal to