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


ANSWER

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
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