Jee Main 2024 01 02 2024 Shift 2 - Question12

Answer: (1)

Solution:

$F(x)=\int _0^{x} t f(t) d t$

and $F\left(x^{2}\right)=x^{4}+x^{5}=\left(x^{2}\right)^{2}+\left(x^{2}\right)^{5 / 2}$

$F(x)=x+x^{5 / 2}$

$F^{\prime}(x)=1+\frac{5}{2} x^{3 / 2}=x f(x)$

$\Rightarrow f(x)=\frac{1}{x}+\frac{5}{2} \sqrt{x}$

$f\left(x^{2}\right)=\frac{1}{x^{2}}+\frac{5 x}{2}$

$x^{2} f\left(x^{2}\right)=1+\frac{5}{2} x^{3}$

$\sum _{x=1}^{12} x^{2} f\left(x^{2}\right)=\sum _{x=1}^{12}\left(1+\frac{5}{2} x^{3}\right)$

$=12+\frac{5}{2} \times \frac{12 \times 13 \times 25}{6}$

$=12+5 \times 13 \times 25=1637$