Question:
If $\int_{0}^{x} f(\mathrm{t}) \mathrm{d} \mathrm{t}=x^{2}+\int_{x}^{1} \mathrm{t}^{2} f(\mathrm{t}) \mathrm{dt}$, then $f^{\prime}(1 / 2)$ is:
Correct Option: 1
Solution:
$\int_{0}^{x} f(t) d t=x^{2}+\int_{x}^{1} t^{2} f(t) d t$
$\Rightarrow \quad f(x)=2 x-x^{2} f(x)$
$\Rightarrow \quad f(x)=\frac{2 x}{1+x^{2}}$
$\Rightarrow \quad f^{\prime}(x)=\frac{2\left(1-x^{2}\right)}{\left(1+x^{2}\right)^{2}}$
Then,
$f^{\prime}(1 / 2)=\frac{2\left(1-\frac{1}{4}\right)}{\left(1+\frac{1}{4}\right)^{2}}=\frac{3}{2} \times \frac{16}{25}=\frac{24}{25}$