Mathematics

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

9

SOLUTION
$\int g(x){ d }x=f(x)+C$
$\Rightarrow g(x)=f'(x)$

Now,$\int _{ -1 }^{ 1 } f^{ 2 }(x)g(x){ d }x=\lambda$
Applying integration by parts,

$\Rightarrow { [f^{ 2 }(x)f(x)] }_{ -1 }^{ 1 }-\int _{ -1 }^{ 1 }{ 2f(x)f'(x)f(x)dx } =\lambda$

$\Rightarrow { [f^{ 3 }(x)] }_{ -1 }^{ 1 }-\int _{ -1 }^{ 1 }{ 2{ f }^{ 2 }(x)g(x)dx } =\lambda$

$\Rightarrow { [f(1)] }^{ 3 }-{ [f(-1)] }^{ 3 }=3\lambda$

$\Rightarrow 27+27=3\lambda$ ( $\because$ f(x) is an odd function)

$\Rightarrow \displaystyle \frac{\lambda}{2}=9$

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Single Correct Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 109

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