JEE Main 2022 · 25 Jul · Shift 2 · Mathematics Q26

← JEE Main 2022 · 25 Jul · Shift 2

MathematicsInteger+4 / -1Question 26 of 90

Let $f$ be a twice differentiable function on $\mathbb R$ . If $f'(0)=4$ and $f(x)+\displaystyle\int_0^x(x-t)f'(t)\,dt=\left(e^{2x}+e^{-2x}\right)\cos2x+\frac2ax$ , then $(2a+1)^5a^2$ is equal to ______.

Your answer

Answer and worked solution

Official answer: 8

Differentiate. $f'(x)+f(x)-f(0)=\dfrac d{dx}\left[(e^{2x}+e^{-2x})\cos2x\right]+\dfrac2a$ . At $x=0$ : $4=0+\frac2a$ , so $a=\frac12$ .

Value $2^5\cdot\frac14=8$ .

Answer: 8

Topics:Leibniz rule Derivatives of standard functions

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JEE Main 2022 · 25 Jul · Shift 2, Mathematics question 26