Let f be a continuous function satisfying 0^t^2 ( f(x) + x^2) dx = 43 t^3, t 0. Then f ( ^24) is equal to:

Given: 0^t^2 ( f(x) + x^2) dx = 43 t^3 for t 0. Find: f ( ^24). Differentiate both sides with respect to t using the Leibniz rule: ddt ( 0^t^2 ( f(x) + x^2) dx)= (f(t^2)+t^4) 2t and ddt ( 43t^3)=4t^2 So, 2t (f(t^2)+t^4)=4t^2 Since t0, f(t^2)+t^4=2t Therefore, f(t^2)=2t-t^4 Now put t= 2 so that t^2= ^24 Hence, f ( ^24)=2 2- ( 2)^4= - ^416= (1- ^316) Therefore, the correct option is D. The solution contains a labeling discrepancy because the computed value matches option B, but the page states option D. Given: 0^t^2 ( f(x) + x^2) dx = 43 t^3 Find: which listed option equals f ( ^24). The integrand is f(x)+x^2, so after differentiation the term coming from the upper limit x=t^2 is f(t^2)+(t^2)^2=f(t^2)+t^4 not f(t^2)+t^2. Thus, 2t (f(t^2)+t^4)=4t^2 which gives f(t^2)=2t-t^4 Setting t= 2 gives f ( ^24)= - ^416= (1- ^316) This matches option B exactly. So the working supports B, while the solution headline says D. The computed value is the primary source here.

Let f be a continuous function satisfying 0^t^2 ( f(x) + x^2) dx… — Solution | JEE Mathematics

MCQMediumJEE 2023Definite Integrals

Let $f$ be a continuous function satisfying $\int_0^{t^2} \left( f(x) + x^2 \right) \, dx = \frac{4}{3} t^3$ , $\forall t > 0$ . Then $f \left( \frac{\pi^2}{4} \right)$ is equal to:

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

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given:

\int_0^{t^2} \left( f(x) + x^2 \right) \, dx = \frac{4}{3} t^3

for $t > 0$ .

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

Differentiate both sides with respect to $t$ using the Leibniz rule:

\frac{d}{dt}\left(\int_0^{t^2} \left( f(x) + x^2 \right) \, dx\right)=\left(f(t^2)+t^4\right)\cdot 2t

and

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

So,

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

Since $t>0$ ,

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

Therefore,

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

Now put

t=\frac{\pi}{2}

so that

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

Hence,

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

Therefore, the correct option is D. The solution contains a labeling discrepancy because the computed value matches option B, but the page states option D.

Checking the discrepancy

Given:

\int_0^{t^2} \left( f(x) + x^2 \right) \, dx = \frac{4}{3} t^3

Find: which listed option equals $f\left(\frac{\pi^2}{4}\right)$ .

The integrand is $f(x)+x^2$ , so after differentiation the term coming from the upper limit $x=t^2$ is

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

not $f(t^2)+t^2$ .

Thus,

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

which gives

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

Setting $t=\frac{\pi}{2}$ gives

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

This matches option B exactly. So the working supports B, while the solution headline says D. The computed value is the primary source here.

Common mistakes

While differentiating $\int_0^{t^2}(f(x)+x^2)\,dx$ , replacing $x^2$ by $t^2$ instead of $t^4$ is incorrect because the upper limit is $x=t^2$ , so $x^2=(t^2)^2=t^4$ . Always substitute the upper limit into the entire integrand before multiplying by the derivative of the limit.
Substituting $t=\frac{\pi^2}{4}$ directly into the formula for $f(t^2)$ is wrong because the expression gives values of $f$ at input $t^2$ , not at input $t$ . To find $f\left(\frac{\pi^2}{4}\right)$ , first choose $t$ such that $t^2=\frac{\pi^2}{4}$ .
Trusting the option label printed in the solution without checking the algebra can lead to a wrong choice. Here the computed expression matches option B, even though the solution says D. Match the final value with the options, not only the label.

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Answer

Step-by-step solution

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