Let f be a twice differentiable function on \mathbbR . If f^\prime(0)=4 and f(x) + \int\limits_0^x (x - t)f'(t)dt = ( e^2x + e^ - 2x )\cos 2x + 2 \over ax , then (2 a+1)^5\, a^2 is equal to \underline\hspace2cm .

Definite Integration PYQ — JEE Main 2022 · 25 Jul · Shift 2 · Q42 with Solution

View written solutionFree

Correct answer: 8

Given integral equation

We have

f(x)+\int_0^x (x-t)f'(t)\,dt=\left(e^{2x}+e^{-2x}\right)\cos 2x+\frac{2}{a}x.

Let

R(x)=\left(e^{2x}+e^{-2x}\right)\cos 2x+\frac{2}{a}x.

So,

f(x)+\int_0^x (x-t)f'(t)\,dt=R(x).

Differentiate once

Using Leibniz rule,

\frac{d}{dx}\int_0^x (x-t)f'(t)\,dt=\int_0^x f'(t)\,dt=f(x)-f(0).

Hence,

f'(x)+f(x)-f(0)=R'(x). \tag{1}

Differentiate again

Differentiating (1),

f''(x)+f'(x)=R''(x). \tag{2}

Compute derivatives of the RHS

First note

e^{2x}+e^{-2x}=2\cosh 2x.

Let

g(x)=\left(e^{2x}+e^{-2x}\right)\cos 2x.

We compute derivatives directly.

g'(x)=\left(2e^{2x}-2e^{-2x}\right)\cos 2x-2\left(e^{2x}+e^{-2x}\right)\sin 2x.

Now at $x=0$ ,

g(0)=2, \qquad g'(0)=0.

Also,

R'(x)=g'(x)+\frac{2}{a},

so at $x=0$ ,

R'(0)=\frac{2}{a}.

From (1) with $x=0$ ,

f'(0)+f(0)-f(0)=R'(0)

which gives

f'(0)=R'(0)=\frac{2}{a}.

Given $f'(0)=4$ , therefore

4=\frac{2}{a}\implies a=\frac12.

Compute the required expression

We need

(2a+1)^5 a^2.

Substitute $a=\frac12$ :

2a+1=2\cdot \frac12+1=2,

(2a+1)^5 a^2=2^5\left(\frac12\right)^2=32\cdot \frac14=8.

Final answer

\boxed{8}

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View written solutionFree

Correct answer: 8

Given integral equation

We have

f(x)+\int_0^x (x-t)f'(t)\,dt=\left(e^{2x}+e^{-2x}\right)\cos 2x+\frac{2}{a}x.

Let

R(x)=\left(e^{2x}+e^{-2x}\right)\cos 2x+\frac{2}{a}x.

So,

f(x)+\int_0^x (x-t)f'(t)\,dt=R(x).

Differentiate once

Using Leibniz rule,

\frac{d}{dx}\int_0^x (x-t)f'(t)\,dt=\int_0^x f'(t)\,dt=f(x)-f(0).

Hence,

f'(x)+f(x)-f(0)=R'(x). \tag{1}

Differentiate again

Differentiating (1),

f''(x)+f'(x)=R''(x). \tag{2}

Compute derivatives of the RHS

First note

e^{2x}+e^{-2x}=2\cosh 2x.

Let

g(x)=\left(e^{2x}+e^{-2x}\right)\cos 2x.

We compute derivatives directly.

g'(x)=\left(2e^{2x}-2e^{-2x}\right)\cos 2x-2\left(e^{2x}+e^{-2x}\right)\sin 2x.

Now at $x=0$ ,

g(0)=2, \qquad g'(0)=0.

Also,

R'(x)=g'(x)+\frac{2}{a},

so at $x=0$ ,

R'(0)=\frac{2}{a}.

From (1) with $x=0$ ,

f'(0)+f(0)-f(0)=R'(0)

which gives

f'(0)=R'(0)=\frac{2}{a}.

Given $f'(0)=4$ , therefore

4=\frac{2}{a}\implies a=\frac12.

Compute the required expression

We need

(2a+1)^5 a^2.

Substitute $a=\frac12$ :

2a+1=2\cdot \frac12+1=2,

(2a+1)^5 a^2=2^5\left(\frac12\right)^2=32\cdot \frac14=8.

Final answer

\boxed{8}

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