Question

$If{\int }_0^{x}f\left(t\right)dt=x+{\int }_x^{1}tf\left(t\right)dt,$ then the value of f (1) is

• A. $\frac{1}{2}$
• B. $0$
• C. $1$
• D. $-\frac{1}{2}$

Answer 1

The correct option is A $\frac{1}{2}$

${\int }_0^{x}f\left(t\right)dt=x+{\int }_x^{1}tf\left(t\right)dt,$ differentiating on both sides, f(x) = 1 + 0 –x ∙ f(x)
$\Rightarrow f\left(1\right)=1-f\left(1\right)\Rightarrow 2.f\left(1\right)=1\Rightarrow f\left(1\right)=\frac{1}{2}$