Let f be a real valued continuous function defined on the positive real axis such that g(x)=∫_0^x t f(t) dt . If g(x^3)=x^6+x^7 , then value of ∑_ r=1 ^ 15 f(r^3) is :

Correct answer: D. 310 g(x)=x^2+x^ 7 / 3 aligned & g^ (x)=2 x+(7)/(3) x^ 4 / 3 & f(x)= g^ (x) x & f(x)=2+(7)/(3) x^ 1 / 3 & f(r^3)=2+(7)/(3) r & ∑_ r=1 ^ 15 (2+(7)/(3) r)=2(15)+(7)/(3)((15(16))/(2)) aligned =310

JEE Main 202528 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 28 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Definite Integration

JEE Main 2025 · 28 January · Evening Shift · Mathematics · Definite Integration

Let f be a real valued continuous function defined on the positive...

Let

f

be a real valued continuous function defined on the positive real axis such that

g(x)=\int_0^x \mathrm{t} f(\mathrm{t}) \mathrm{dt}

. If

\mathrm{g}\left(x^3\right)=x^6+x^7

, then value of

\sum_{r=1}^{15} f\left(\mathrm{r}^3\right)

is :

Options

A $270$
B $340$
C $320$
D $310$

Correct answer

D. $310$

Step-by-step solution

g(x)=x^2+x^{7 / 3}

\begin{aligned} & g^{\prime}(x)=2 x+\frac{7}{3} x^{4 / 3} \\ & f(x)=\frac{g^{\prime}(x)}{x} \\ & f(x)=2+\frac{7}{3} x^{1 / 3} \\ & f\left(r^3\right)=2+\frac{7}{3} r \\ & \sum_{r=1}^{15}\left(2+\frac{7}{3} r\right)=2(15)+\frac{7}{3}\left(\frac{15(16)}{2}\right)\end{aligned}

=310

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