Let f be a continuous function satisfying \int_0^t^2(f(x)+x^2) d x=\frac43 t^3, \forall t \gt 0 . Then f(\frac\pi^24) is equal to :

Definite Integration PYQ — JEE Main 2023 · 10 Apr · Shift 2 · Q26 with Solution

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Correct answer: B

We are given

\int_{0}^{t^{2}} \left(f(x)+x^{2}\right)\,dx=\frac{4}{3}t^{3}, \qquad \forall t>0.

We need to find

f\left(\frac{\pi^2}{4}\right).

Differentiate both sides with respect to $t$ using the Fundamental Theorem of Calculus and chain rule.

Let

F(t)=\int_{0}^{t^{2}} \left(f(x)+x^{2}\right)\,dx.

Then

F'(t)=\left(f(t^2)+(t^2)^2\right)\cdot \frac{d}{dt}(t^2).

So,

F'(t)=\left(f(t^2)+t^4\right)\cdot 2t.

Differentiate the right-hand side:

\frac{d}{dt}\left(\frac{4}{3}t^3\right)=4t^2.

Hence,

2t\left(f(t^2)+t^4\right)=4t^2.

Since $t>0$ , divide by $2t$ :

f(t^2)+t^4=2t.

Therefore,

f(t^2)=2t-t^4.

To find $f\left(\frac{\pi^2}{4}\right)$ , set

t^2=\frac{\pi^2}{4} \implies t=\frac{\pi}{2}

because $t>0$ .

Then

f\left(\frac{\pi^2}{4}\right)=2\left(\frac{\pi}{2}\right)-\left(\frac{\pi}{2}\right)^4.

Compute:

2\cdot \frac{\pi}{2}=\pi,

and

\left(\frac{\pi}{2}\right)^4=\frac{\pi^4}{16}.

So,

f\left(\frac{\pi^2}{4}\right)=\pi-\frac{\pi^4}{16}.

Factor out $\pi$ :

f\left(\frac{\pi^2}{4}\right)=\pi\left(1-\frac{\pi^3}{16}\right).

Compare with the options:

A: $-\pi\left(1+\frac{\pi^3}{16}\right)$
B: $\pi\left(1-\frac{\pi^3}{16}\right)$
C: $-\pi^2\left(1+\frac{\pi^2}{16}\right)$
D: $\pi^2\left(1-\frac{\pi^2}{16}\right)$

Thus the correct option is

\boxed{\text{B}}

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View written solutionFree

Correct answer: B

We are given

\int_{0}^{t^{2}} \left(f(x)+x^{2}\right)\,dx=\frac{4}{3}t^{3}, \qquad \forall t>0.

We need to find

f\left(\frac{\pi^2}{4}\right).

Differentiate both sides with respect to $t$ using the Fundamental Theorem of Calculus and chain rule.

Let

F(t)=\int_{0}^{t^{2}} \left(f(x)+x^{2}\right)\,dx.

Then

F'(t)=\left(f(t^2)+(t^2)^2\right)\cdot \frac{d}{dt}(t^2).

So,

F'(t)=\left(f(t^2)+t^4\right)\cdot 2t.

Differentiate the right-hand side:

\frac{d}{dt}\left(\frac{4}{3}t^3\right)=4t^2.

Hence,

2t\left(f(t^2)+t^4\right)=4t^2.

Since $t>0$ , divide by $2t$ :

f(t^2)+t^4=2t.

Therefore,

f(t^2)=2t-t^4.

To find $f\left(\frac{\pi^2}{4}\right)$ , set

t^2=\frac{\pi^2}{4} \implies t=\frac{\pi}{2}

because $t>0$ .

Then

f\left(\frac{\pi^2}{4}\right)=2\left(\frac{\pi}{2}\right)-\left(\frac{\pi}{2}\right)^4.

Compute:

2\cdot \frac{\pi}{2}=\pi,

and

\left(\frac{\pi}{2}\right)^4=\frac{\pi^4}{16}.

So,

f\left(\frac{\pi^2}{4}\right)=\pi-\frac{\pi^4}{16}.

Factor out $\pi$ :

f\left(\frac{\pi^2}{4}\right)=\pi\left(1-\frac{\pi^3}{16}\right).

Compare with the options:

A: $-\pi\left(1+\frac{\pi^3}{16}\right)$
B: $\pi\left(1-\frac{\pi^3}{16}\right)$
C: $-\pi^2\left(1+\frac{\pi^2}{16}\right)$
D: $\pi^2\left(1-\frac{\pi^2}{16}\right)$

Thus the correct option is

\boxed{\text{B}}

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