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QJEEM22Ma461b: JEE Main 2022 Mathematics - Class 12 JEEM22Ma461b Solution: Let be a twice differentiable function on . If and , then is…

Let ff be a twice differentiable function on R\mathbb{R} . If f′(0)=4f^{\prime}(0)=4 and f(x)+∫0x(x−t)f′(t)dt=(e2x+e−2x)cos⁡2x+2axf(x) + \int\limits_0^x {(x - t)f'(t)dt = \left( {{e^{2x}} + {e^{ - 2x}}} \right)\cos 2x + {2 \over a}x} , then (2a+1)5 a2(2 a+1)^{5}\, a^{2} is equal to _______________.

Step-by-Step Verified Solution

∵f(x)+∫0x(x−t)f′(t)dtamp;=(e2x+e−2x)cos⁡2x+2xa  ...(i)\begin{aligned}\because f(x)+\int_0^x(x-t) f^{\prime}(t) d t & =\left(e^{2 x}+e^{-2 x}\right) \cos 2 x+\frac{2 x}{a} ~~...(i)\end{aligned}

Here f(0)=2...(ii)f(0)=2 \hspace{0.5cm} ...(ii)

On differentiating equation (i) w.r.t. xx we get :

amp;f′(x)+∫0xf′(t)dt+xf′(x)−xf′(x)amp;=2(e2x−e−2x)cos⁡2x−2(e2x+e−2x)sin⁡2x+2aamp;⇒f(x)+f(x)−f(0)amp;=2(e2x−e−2x)cos⁡2x−2(e2x+e−2x)sin⁡2x+2aamp; Replace x by 0 we get :amp;⇒4=2a⇒a=12⋅amp;∴(2a+1)5⋅a2=25⋅122=23=8\begin{aligned}& f^{\prime}(x)+\int_0^x f^{\prime}(t) d t+x f^{\prime}(x)-x f^{\prime}(x) \\\\& = 2\left(e^{2 x}-e^{-2 x}\right) \cos 2 x-2\left(e^{2 x}+e^{-2 x}\right) \sin 2 x+\frac{2}{a} \\\\& \Rightarrow \quad f(x)+f(x)-f(0)\\\\&=2\left(e^{2 x}-e^{-2 x}\right) \cos 2 x-2\left(e^{2 x}+e^{-2 x}\right) \sin 2 x+\frac{2}{a} \\\\& \quad \text { Replace } x \text { by } 0 \text { we get }: \\\\& \Rightarrow \quad 4=\frac{2}{a} \Rightarrow a=\frac{1}{2} \cdot \\\\& \therefore \quad(2 a+1)^5 \cdot a^2=2^5 \cdot \frac{1}{2^2}=2^3=8\end{aligned}

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