If $f (x) = \int \frac{2 sin 6 x - 5 sin 4 x + 10 sin 2 x}{2 cos 5 x - 3 cos 3 x + 7 cos x} d x, f (0) = 1,$ then which of the following option(s) is/are correct ?

f (x) + f^{'} (x) + 1 = 0

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f (x) + f^{''} (x) = 3

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f_{m a x} = 1

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f_{m i n} = 1

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Solution

The correct option is D $f_{m i n} = 1$
Let
$I = \int \frac{2 sin 6 x - 5 sin 4 x + 10 sin 2 x}{2 cos 5 x - 3 cos 3 x + 7 cos x} d x = \int \frac{(2 sin 6 x - 2 sin 4 x) - (3 sin 4 x - 3 sin 2 x) + 7 sin 2 x}{2 cos 5 x - 3 cos 3 x + 7 cos x} d x = \int \frac{4 (cos 5 x sin x) - 6 cos 3 x sin x + 7 (2 sin x cos x)}{2 cos 5 x - 3 cos 3 x + 7 cos x} d x = \int \frac{2 sin x (2 cos 5 x - 3 cos 3 x + 7 cos x)}{2 cos 5 x - 3 cos 3 x + 7 cos x} d x = \int 2 sin x d x = - 2 cos x + C$
$\Rightarrow f (x) = - 2 cos x + C$
As, $f (0) = 1, C = 3$
$\Rightarrow f (x) = 3 - 2 cos x$
and $f^{'} (x) = 2 sin x,$
$f^{''} (x) = 2 cos x \Rightarrow f (x) + f^{''} (x) = 3$
Also, $f_{m a x} = 5, f_{m i n} = 1$

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