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← JEE Main 2022 · 28 Jun · Shift 2
MathematicsSingle correct+4 / -1Question 7 of 90

JEE Main 2022 · 28 Jun · Shift 2, Mathematics question 7

Let f:R→Rf:\mathbf R\to\mathbf R be a differentiable function such that f(π4)=2f\left(\dfrac\pi4\right)=\sqrt2, f(π2)=0f\left(\dfrac\pi2\right)=0 and f′(π2)=1f'\left(\dfrac\pi2\right)=1 and let g(x)=∫xπ/4(f′(t)sec⁡t+tan⁡tsec⁡t f(t))dtg(x)=\displaystyle\int_x^{\pi/4}\left(f'(t)\sec t+\tan t\sec t\,f(t)\right)dt for x∈[π4,π2)x\in\left[\dfrac\pi4,\dfrac\pi2\right). Then lim⁡x→(π2)−g(x)\displaystyle\lim_{x\to\left(\frac\pi2\right)^-}g(x) is equal to :

Choose one option
Answer and worked solution

Official answer: B

Integrand is ddt(f(t)sec⁡t)\frac d{dt}\left(f(t)\sec t\right), so g(x)=f(π4)2−f(x)cos⁡x=2−f(x)cos⁡xg(x)=f\left(\frac\pi4\right)\sqrt2-\dfrac{f(x)}{\cos x}=2-\dfrac{f(x)}{\cos x}.

Limit. By L'Hospital, f(x)cos⁡x→f′(π/2)−1=−1\dfrac{f(x)}{\cos x}\to\dfrac{f'(\pi/2)}{-1}=-1, so g→3g\to3.

(B)