# Let f be the twice differentiable function such that $f''\left( x \right)=-f\left( x \right)\text{ and }f'\left( x \right)=g\left( x \right).$If $h'\left( x \right)=\left[ f{{\left( x \right)}^{2}}+g{{\left( x \right)}^{2}} \right],h\left( 1 \right)=8,h\left( 0 \right)=2$, then h(2) equals to

A. 1

B. 2

C. 3

D. None of these

Answer

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Hint: Differentiate the function $h'\left( x \right)=f{{\left( x \right)}^{2}}+g{{\left( x \right)}^{2}}$. Then substitute the values and given conditions to the differentiate function.

Complete step-by-step answer:

Given that $h'\left( x \right)=f{{\left( x \right)}^{2}}+g{{\left( x \right)}^{2}}..............\left( 1 \right)$

Differentiate equation (1) with respect to x.

We get;

$h''.\left( x \right)=2f\left( x \right)f'\left( x \right)+2g\left( x \right)g'\left( x \right)................\left( 2 \right)$

It’s given that f ’(x) = g(x)

$\therefore h''\left( x \right)=2f\left( x \right)g\left( x \right)+2g\left( x \right)g'\left( x \right)$

Given f ”(x) = -f(x)

f ’(x) = g’(x), take the differentiation

then f ” (x) = g’(x)

Substitute, the values for f ’(x) and f ”(x)

$\begin{align}

& \therefore h''\left( x \right)=2f\left( x \right)g\left( x \right)+2g\left( x \right)f''\left( x \right) \\

& =2f\left( x \right)g\left( x \right)-2g\left( x \right)f\left( x \right)=0 \\

\end{align}$

Thus h’(x) = k, a constant for all $x\in R$

Let h(x) = kx + m, forms the equation of line

i.e. h(0) = 2, from question

$\Rightarrow $ We get a possible equation as \[2=kx+m\]

$2=k\times 0+m\Rightarrow m=2,\text{ where }x=0$

Then, h(1) = 8

$\Rightarrow $8 = k x 1 + m

8 = k x 1 +2

$\Rightarrow $k = 8 – 2 = 6

Hence, we got the values of m and k

m = 2, k = 6

h(2) , where x = 2

Substitute these to get value of h(2);

$\therefore $ h(2) = 6 x 2 + 2 = 14

Hence, value of h(2) = 14.

Note: f ’(x) = g(x), is differentiated again to form f ”(x) = g’(x). As we got h”(x) = 0, h’(x) = k is constant for all $x\in R$.

Complete step-by-step answer:

Given that $h'\left( x \right)=f{{\left( x \right)}^{2}}+g{{\left( x \right)}^{2}}..............\left( 1 \right)$

Differentiate equation (1) with respect to x.

We get;

$h''.\left( x \right)=2f\left( x \right)f'\left( x \right)+2g\left( x \right)g'\left( x \right)................\left( 2 \right)$

It’s given that f ’(x) = g(x)

$\therefore h''\left( x \right)=2f\left( x \right)g\left( x \right)+2g\left( x \right)g'\left( x \right)$

Given f ”(x) = -f(x)

f ’(x) = g’(x), take the differentiation

then f ” (x) = g’(x)

Substitute, the values for f ’(x) and f ”(x)

$\begin{align}

& \therefore h''\left( x \right)=2f\left( x \right)g\left( x \right)+2g\left( x \right)f''\left( x \right) \\

& =2f\left( x \right)g\left( x \right)-2g\left( x \right)f\left( x \right)=0 \\

\end{align}$

Thus h’(x) = k, a constant for all $x\in R$

Let h(x) = kx + m, forms the equation of line

i.e. h(0) = 2, from question

$\Rightarrow $ We get a possible equation as \[2=kx+m\]

$2=k\times 0+m\Rightarrow m=2,\text{ where }x=0$

Then, h(1) = 8

$\Rightarrow $8 = k x 1 + m

8 = k x 1 +2

$\Rightarrow $k = 8 – 2 = 6

Hence, we got the values of m and k

m = 2, k = 6

h(2) , where x = 2

Substitute these to get value of h(2);

$\therefore $ h(2) = 6 x 2 + 2 = 14

Hence, value of h(2) = 14.

Note: f ’(x) = g(x), is differentiated again to form f ”(x) = g’(x). As we got h”(x) = 0, h’(x) = k is constant for all $x\in R$.

Last updated date: 26th Sep 2023

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