Let f : R→R be a twice differentiable function such that f(2)=1.

Question

Let \(\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}\) be a twice differentiable function such that \(f(2)=1.\) If \(\mathrm{F}(\mathrm{x})=\mathrm{x} f(\mathrm{x})\) for all \(\mathrm{x} \in \mathrm{R},\) \(\int\limits_{0}^{2} x F^{\prime}(x) d x=6\) and \(\int\limits_{0}^{2} x^{2} F^{\prime \prime}(x) d x=40,\) then \(F^{\prime}(2)+\int\limits_{0}^{2} F(x) d x\) is equal to :

(1) 11

(2) 15

(3) 9

(4) 13

Answer 1

Correct option is (1) 11  

\(\int\limits_0^2 x F^{\prime}(x)=6\)

\({\left[F^{\prime}(x) \cdot \frac{x^2}{2}\right]_0^2-\int\limits_0^2 F^{\prime \prime}(x) \cdot \frac{x^2}{2} d x=6} \)

\( 2 F^{\prime}(2)=26 \Rightarrow F^{\prime}(2)=13\)

Given: \(F(x)=x f(x)\)

\(F^{\prime}(x)=x f'(x)+f(x) \)

\(\text { Put } x=2 \)

\(F^{\prime}(2)=2 f(2)+f(2) \)

\( f(2)=6\)

\(\text { As } \int\limits_0^2 x F^{\prime}(x)=6\)

\(\Rightarrow [x F(x)]_0^2-\int\limits_0^2 F(x) d x=6\)

\(\Rightarrow \int_0^2 F(x) d x=-2\)

\(\therefore F^{\prime}(2)+\int\limits_0^2 F(x) d x=13-2=11\)