JEE MAIN - Mathematics (2025 - 28th January Evening Shift - No. 18)
Let $f$ be a real valued continuous function defined on the positive real axis such that $g(x)=\int\limits_0^x t f(t) d t$. If $g\left(x^3\right)=x^6+x^7$, then value of $\sum\limits_{r=1}^{15} f\left(r^3\right)$ is :
270
340
310
320
Explanation
$$\begin{aligned}
& g(x)=x 2+x^{\frac{7}{3}} \\
& g^{\prime}(x)=2 x+\frac{7}{3} x^{\frac{4}{3}} \\
& f(x)=\frac{g^{\prime}(x)}{x} \\
& f(x)=2+\frac{7}{3} x^{\frac{1}{3}} \\
& f\left(r^3\right)=2+\frac{7 r}{3} \\
& \sum_{r=1}^{15}\left(1+\frac{7}{3} r\right)=310
\end{aligned}$$
Comments (0)
