Q117 / 180 · KCET 2014 · Mathematics
If \( f(x)=f(\Pi+e-x) \) and \( \int_{e}^{I I} f(x) d x=\frac{2}{e+\Pi^{\prime}} \), then \( \int_{e}^{I I} x f(x) d x \) is equal to
- A\( \frac{\Pi+e}{2} \)
- B\( \frac{\Pi-e}{2} \)
- CII \( -e \)
- Correct answerD\( 1 \)
Solution
Given that
\[ \begin{array}{l} f(x)=f(\Pi+e-x) \rightarrow(1) \\ \text { and } \int_{e}^{\Pi} f(x) d x=\frac{2}{e+\Pi} \rightarrow(2) \end{array} \]
Now,
\[ \begin{array}{l} I=\int_{e}^{\Pi} x f(x) d x=\int_{e}^{\Pi}(e+\Pi-x) f(e+\Pi-x) d x \\ =\int_{e}^{\Pi}(e+\Pi-x) f(x) d x=\int_{e}^{\Pi}(e+\Pi) f(x)-I \\ \Rightarrow 2 I=(e+\Pi) \frac{2}{e+\Pi} \Rightarrow I=1 \end{array} \]
\[ \begin{array}{l} f(x)=f(\Pi+e-x) \rightarrow(1) \\ \text { and } \int_{e}^{\Pi} f(x) d x=\frac{2}{e+\Pi} \rightarrow(2) \end{array} \]
Now,
\[ \begin{array}{l} I=\int_{e}^{\Pi} x f(x) d x=\int_{e}^{\Pi}(e+\Pi-x) f(e+\Pi-x) d x \\ =\int_{e}^{\Pi}(e+\Pi-x) f(x) d x=\int_{e}^{\Pi}(e+\Pi) f(x)-I \\ \Rightarrow 2 I=(e+\Pi) \frac{2}{e+\Pi} \Rightarrow I=1 \end{array} \]
Hence, the correct answer is option: 'D'.