JEE Main 2022 · 26 Jul · Shift 1 · Mathematics Q10

MathematicsSingle correct+4 / -1Question 10 of 90

If $a=\displaystyle\lim_{n\to\infty}\sum_{k=1}^n\frac{2n}{n^2+k^2}$ and $f(x)=\sqrt{\dfrac{1-\cos x}{1+\cos x}}$ , $x\in(0,1)$ , then :

2\sqrt2f\left(\dfrac a2\right)=f'\left(\dfrac a2\right)

f\left(\dfrac a2\right)f'\left(\dfrac a2\right)=\sqrt2

\sqrt2f\left(\dfrac a2\right)=f'\left(\dfrac a2\right)

f\left(\dfrac a2\right)=\sqrt2f'\left(\dfrac a2\right)

Answer and worked solution

Official answer: C

a $=2\int_0^1\frac{dx}{1+x^2}=\frac\pi2$ . f $=\tan\frac x2$ , $f'=\frac12\sec^2\frac x2$ .

At $\frac\pi4$ . $f=\tan\frac\pi8=\sqrt2-1$ , $f'=\frac12(4-2\sqrt2)=2-\sqrt2=\sqrt2f$ .

(C)

Topics:Limit of a sum Derivatives of standard functions

JEE Main 2022 · 26 Jul · Shift 1, Mathematics question 10