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

A
32
B
24
C
64
D
48

Solution:

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

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

$∴ (f (x))^{2} + (g (x))^{2}$ is constant.

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

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

If $f^{'} (x) = g (x) and g^{'} (x) = - f (x) for all x$ and $f (2) = 4 = f^{'} (2), then (f (24))^{2} + (g (24))^{2} is$

Solution:

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