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Find f(8/27) from ∫₀^{x³} t²f(t)dt = (3/13)x¹³+5 | JEE

JEE Maths question with a full step-by-step solution.

Question

f(x)

is a differentiable function and

\displaystyle\int_{0}^{x^{3}}t^{2}f(t)\,dt=\dfrac{3}{13}x^{13}+5

, then

f\!\left(\dfrac{8}{27}\right)

equals

\dfrac{8}{27}

\dfrac{16}{27}

\dfrac{16}{81}

correct

\dfrac{8}{9}

Solution

Step 1: Differentiate both sides with respect to

x

. By Leibniz's rule on the upper limit

x^{3}

\dfrac{d}{dx}\int_{0}^{x^{3}}t^{2}f(t)\,dt=\big(x^{3}\big)^{2}f\!\big(x^{3}\big)\cdot\dfrac{d}{dx}\big(x^{3}\big)=x^{6}f\!\big(x^{3}\big)\cdot 3x^{2}=3x^{8}f\!\big(x^{3}\big).

Step 2: Differentiate the right-hand side:

\dfrac{d}{dx}\!\left(\dfrac{3}{13}x^{13}+5\right)=3x^{12}.

Step 3: Equate the two:

3x^{8}f\!\big(x^{3}\big)=3x^{12}\ \Rightarrow\ f\!\big(x^{3}\big)=x^{4}.

Step 4: To obtain

f\!\left(\dfrac{8}{27}\right)

, set

x^{3}=\dfrac{8}{27}

, i.e.

x=\dfrac23

f\!\left(\dfrac{8}{27}\right)=\left(\dfrac23\right)^{4}=\dfrac{16}{81}.

Correct answer: (3)

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