JEE Main 2022 · 25 Jul · Shift 2 · Mathematics Q6

MathematicsSingle correct+4 / -1Question 6 of 90

$\displaystyle\lim_{x\to\frac\pi4}\frac{8\sqrt2-(\cos x+\sin x)^7}{\sqrt2-\sqrt2\sin2x}$ is equal to

14\sqrt2

7\sqrt2

Answer and worked solution

Official answer: A

Substitute $t=\cos x+\sin x$ : $t^2=1+\sin2x$ , so the denominator is $\sqrt2(2-t^2)$ and the numerator $(\sqrt2)^7-t^7$ .

Limit as $t\to\sqrt2$ : $\dfrac{7(\sqrt2)^6}{\sqrt2\cdot2\sqrt2}=\dfrac{56}4=14$ .

(A)

Topics:Limits

JEE Main 2022 · 25 Jul · Shift 2, Mathematics question 6