Let f be a real valued continuous function defined on the positive real axis such that g ( x )=int 0 x t f( t ) dt . If g (x 3 )=x 6 +x 7 , then value of sum r=1 15 f (r 3 ) is :

Mathematics · Definite Integration

JEE Main 2025 — 28 January, Evening Shift — Question 7

Let $f$ be a real valued continuous function defined on the positive real axis such that $\mathrm{g}(\mathrm{x})=\int_{0}^{\mathrm{x}} \mathrm{t} f(\mathrm{t}) \mathrm{dt}$ . If $g\left(x^{3}\right)=x^{6}+x^{7}$ , then value of $\sum_{r=1}^{15} f\left(r^{3}\right)$ is :

Option A:
320
Option B:
340
Option C:
270
Option D:
310
Correct

Answer: D

Step-by-step solution

$g(x)=x^2+x^{\frac{7}{3}}$

$g^{\prime}(x)=2 x+\frac{7}{3} x^{\frac{4}{3}}$

$\mathrm{f}(\mathrm{x})=\frac{\mathrm{g}^{\prime}(\mathrm{x})}{\mathrm{x}}$

$f(x)=2+\frac{7}{3} x^{\frac{1}{3}}$

$f\left(r^{3}\right)=2+\frac{7 r}{3}$

$\sum_{r=1}^{15}\left(2+\frac{7}{3} r\right)=310$

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Exam: JEE Main 2025
Paper: 28 January, Evening Shift
Subject: Mathematics
Chapter: Definite Integration
Topic: Leibnitz rule & its application in limits

Let f: R arrow R be a twice differentiable function such that f(2)=1 . If F( x)= x f( x) for all x in R , int 0^2 x F^prime(x) d x=6 and…

Answer: D

Step-by-step solution

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