Question

Let g be continuous function on R such that $\int g\left(x\right)dx=f\left(x\right)+C$, where C is constant of integration. If f(x) is an odd function, $f\left(1\right)=3$ and ${\int }_{-1}^1{f}^2\left(x\right)g\left(x\right)dx=\lambda$, then $\frac{\lambda }{2}$ is equal to

• A. 9
• B. 10
• C. 11
• D. 12

Answer 1

The correct option is A 9

$\int g\left(x\right)dx=f\left(x\right)+C$
$\Rightarrow g\left(x\right)=f^{\prime }\left(x\right)$
Now,${\int }_{-1}^1{f}^2\left(x\right)g\left(x\right)dx=\lambda$
Applying integration by parts,
$\Rightarrow {\left[f^2\right(x\left)f\right(x\left)\right]}_{-1}^1-{\int }_{-1}^{1}2f\left(x\right)f^{\prime }\left(x\right)f\left(x\right)dx=\lambda$
$\Rightarrow {\left[f^3\right(x\left)\right]}_{-1}^1-{\int }_{-1}^{1}2f^2\left(x\right)g\left(x\right)dx=\lambda$
$\Rightarrow {\left[f\right(1\left)\right]}^3-{\left[f\right(-1\left)\right]}^3=3\lambda$
$\Rightarrow 27+27=3\lambda$ ( $\because$ f(x) is an odd function)
$\Rightarrow \frac{\lambda }{2}=9$