Mathematics - Definite Integration Question with Solution | TestHub

Statement-1: $f(x)={\int }_0^1(xf(t)+1)dt=x{\int }_0^{1}f(t)dt+{\int }_0^{1}1dt$. Let $A={\int }_0^{1}f(t)dt$. Then $f(x)=Ax+1$.

Substitute $f(t)$ back into $A$: $A={\int }_0^1(At+1)dt=A[\frac{t^2}{2}{]}_0^1+[t{]}_0^1=\frac{A}{2}+1$.

So, $A-\frac{A}{2}=1⟹\frac{A}{2}=1⟹A=2$.

Thus, $f(x)=2x+1$.

Now, evaluate ${\int }_0^{3}f(x)dx={\int }_0^3(2x+1)dx=[x^2+x{]}_0^3=(3^2+3)-(0^2+0)=9+3=12$.

So, Statement-1 is TRUE.

Statement-2: $f(x)=3x+1$. This contradicts our finding that $f(x)=2x+1$.

So, Statement-2 is FALSE.

Therefore, Statement-1 is true and Statement-2 is false.

The final answer is C