Calculus – Functions & Integration – JEE Main 28 Jan 2025 Shift 2

Solution:

Given $g(x) = \int_{0}^{x} tf(t) dt$.

Using Newton-Leibniz rule to differentiate with respect to $x$:
$$g'(x) = xf(x) \Rightarrow f(x) = \frac{g'(x)}{x}$$

We are given $g(x^3) = x^6 + x^7$.

Let $x^3 = u \Rightarrow x = u^{1/3}$.

Substituting in the given equation:
$$g(u) = (u^{1/3})^6 + (u^{1/3})^7$$
$$g(u) = u^2 + u^{7/3}$$

Now, differentiate $g(u)$ with respect to $u$:
$$g'(u) = 2u + \frac{7}{3}u^{4/3}$$

We know that $f(u) = \frac{g'(u)}{u}$.
$$f(u) = \frac{2u + \frac{7}{3}u^{4/3}}{u} = 2 + \frac{7}{3}u^{1/3}$$

We need to find $\sum_{r=1}^{15} f(r^3)$.

Substitute $u = r^3$:
$$f(r^3) = 2 + \frac{7}{3}(r^3)^{1/3} = 2 + \frac{7}{3}r$$

Now calculate the sum:
$$S = \sum_{r=1}^{15} \left( 2 + \frac{7}{3}r \right)$$
$$S = \sum_{r=1}^{15} 2 + \frac{7}{3} \sum_{r=1}^{15} r$$
$$S = (15 \times 2) + \frac{7}{3} \times \frac{15(15+1)}{2}$$
$$S = 30 + \frac{7}{3} \times \frac{15 \times 16}{2}$$
$$S = 30 + \frac{7}{3} \times 120$$
$$S = 30 + 7 \times 40$$
$$S = 30 + 280 = 310$$

Ans. (4)