If f′(x)=g(x) and g′(x)=−f(x) for all x and f(2)=4=f′(2),then (f(24))2+(g(24))2 is

# If  and

1. A

32

2. B

24

3. C

64

4. D

48

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### Solution:

We have,$\frac{\mathrm{d}}{\mathrm{dx}}\left\{\left(\mathrm{f}\left(\mathrm{x}\right){\right)}^{2}+\left(\mathrm{g}\left(\mathrm{x}\right){\right)}^{2}\right\}$

$\begin{array}{l}=2\mathrm{f}\left(\mathrm{x}\right)\cdot {\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)+2\mathrm{g}\left(\mathrm{x}\right)\cdot {\mathrm{g}}^{\mathrm{\prime }}\left(\mathrm{x}\right)\\ =2\mathrm{f}\left(\mathrm{x}\right)\mathrm{g}\left(\mathrm{x}\right)+2\mathrm{g}\left(\mathrm{x}\right)\left(-\mathrm{f}\left(\mathrm{x}\right)\right)\\ =2\mathrm{f}\left(\mathrm{x}\right)\mathrm{g}\left(\mathrm{x}\right)-2\mathrm{f}\left(\mathrm{x}\right)\mathrm{g}\left(\mathrm{x}\right)\\ =0\end{array}$

$\therefore \left(\mathrm{f}\left(\mathrm{x}\right){\right)}^{2}+\left(\mathrm{g}\left(\mathrm{x}\right){\right)}^{2}$ is constant.

Again, $\begin{array}{l}\mathrm{g}\left(\mathrm{x}\right)={\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)\\ \mathrm{g}\left(2\right)={\mathrm{f}}^{\mathrm{\prime }}\left(2\right)=4\\ \left(\mathrm{f}\left(24\right){\right)}^{2}+\left(\mathrm{g}\left(24\right){\right)}^{2}\end{array}$

$\begin{array}{l}=\left(\mathrm{f}\left(2\right){\right)}^{2}+\left(\mathrm{g}\left(2\right){\right)}^{2}\\ =\left(4{\right)}^{2}+\left(4{\right)}^{2}\\ =16+16=32\end{array}$

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