JEE Main 28 June 2022 Shift 2 Solved Paper

Section: Mathematics

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Question : 7 of 90

Marks: +1, -0

Let f:R→R be a differentiable function such that f(‌

π

4

)=√2,f(‌

π

2

)=0 and f′(‌

π

2

)=1 and
let g(x)=

∫

x

π∕4(f′(t)‌sec‌t+tan‌t‌sec‌t‌f(t))dt for x∈[‌

π

4

,‌

π

2

). Then

lim

x→(‌

π

2

)−

g(x) is equal to :

[28-Jun-2022-Shift-2]

2
3
4
−3

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