Question

$f$ and $g$ be two positive real function defined on $[-1,1]$ such that $f(-x)=\frac{1}{f\left(x\right)}$ and $g$ is an even function with${\int }_1^{1}g\left(x\right)dx=1$ then ${\int }_{-1}^{1}f\left(x\right)g\left(x\right)dx$ satisfies

• A. $I\ge 1$
• B. $I<1$
• C. $I>\frac{1}{2}$
• D. $I=1$

Answer 1

The correct option is B $I<1$

$\begin{array}{cc}\Rightarrow I={\int }_{-1}^{1}f\left(x\right)g\left(x\right)& \\ \Rightarrow I={\int }_{-1}^{1}f\left(-x\right)g\left(-x\right)& \\ \Rightarrow I={\int }_{-1}^1\frac{1}{f\left(x\right)}\cdot g\left(x\right)& \left[g\left(-x\right)=g\left(x\right)\right]\\ \Rightarrow I=\left[\frac{1}{f\left(x\right)}{\int }_{-1}^{1}g\left(x\right)-{\int }_{-1}^1\frac{d\left(\frac{1}{f\left(x\right)}\right)}{dx}\cdot {\int }_{-1}^{1}g\left(x\right)dx\right]dx& \\ \Rightarrow I=\frac{1}{f\left(x\right)}\times 1-{\int }_{-1}^1\frac{d\left(\frac{1}{f\left(x\right)}\right)}{dx}\times 1dx& \\ \Rightarrow I=\frac{1}{f\left(x\right)}-{\int }_{-1}^1\left(\frac{1}{f\left(x\right)}\right)& \\ \Rightarrow I=\frac{1}{f\left(x\right)}-\frac{1}{f\left(x\right)}=0& \\ \therefore I<1& \end{array}$