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Let f be twice differentiable function such that \mathrm{f^{\prime \prime}(x)=-f(x)} and \mathrm{f^{\prime}(x)=g(x), \quad h(x)=|f(x)|^2+|g(x)|^2. h(5)=11}  then h(10) is equal to

Option: 1

22


Option: 2

11


Option: 3

0


Option: 4

none of these


Answers (1)

best_answer

We have  \mathrm{h(x)=(f(x))^2+\left(\left.g(x)\right|^2\right.}
\mathrm{ \Rightarrow h^{\prime}(x)=2 f(x) 2 f^{\prime}(x)+2 g(x) g^{\prime}(x) .}
Now,   \mathrm{ f^{\prime}(x)=g(x)~ and ~f^{\prime \prime}(x)=-f(x) .}
\mathrm{ \Rightarrow f^{\prime \prime}(x)=g^{\prime}(x) \text { and } f^{\prime \prime}=-f(x) }
\mathrm{ \Rightarrow-f(x)=g^{\prime}(x) }

Thus, \mathrm{ f^{\prime}(x)=g(x)~and ~g^{\prime}(x)=-f(x) }
\mathrm{ \therefore h^{\prime}(x)=-2 g(x) g^{\prime}(x)+2 g(x) g^{\prime}(x)=0, \forall x \\ }
\mathrm{ \Rightarrow h(x)=\text { cost for all } x . }
But h(5)=11. Hence, h(x)=11 for all x.

Posted by

Pankaj Sanodiya

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