First slide

Differentiability

Question

$If f^{''} (x) = - f (x) and g (x) = f^{'} (x) and F (x) = {(f (\frac{x}{2}))}^{2} + {(g (\frac{x}{2}))}^{2} and given that F (5) = 5, then F (10) is$

Moderate

A

5
B

10
C

0
D

15

Solution

$F^{'} (x) = [f (\frac{x}{2}) f^{'} (\frac{x}{2}) + g (\frac{x}{2}) g^{'} (\frac{x}{2})]$
$Here, g (x) = f^{'} (x)$
$and g^{'} (x) = f^{''} (x) = - f (x)$
$So, F^{'} (x) = f (\frac{x}{2}) g (\frac{x}{2}) - f (\frac{x}{2}) g (\frac{x}{2}) = 0$
$Hence, F (x) is a constant function.$
$Therefore, F (10) = 5 (\sin c e F (5) = 5)$

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