If f (x )= 1+ sec ⁡x2 ( sec ⁡x-1 ) for 0

Question

If f (x )= 1+ sec ⁡x2 ( sec ⁡x-1 ) for 0<x< 2 and f^ ' (x )=f (x ) g (x ), then g (x )=

Accepted Answer

Correct answer: D. (d) Given ,f (x )= 1+ sec ⁡x2 ( sec ⁡x-1 ) Nowf^ ' (x )= 12 (1+ cos ⁡x ) (1- cos ⁡x ) ( - sin ⁡x1+ cos ⁡x- sin ⁡x1- cos ⁡x)= 12 1+ sec ⁡ (x ) ( sec ⁡ (x )-1 ) -2 cosec ⁡ (x ) ⇒f^ ' (x )=f (x ) g (x ) hence, g (x )=-2 cosec ⁡ (x )

$If f (x) = \frac{1 + sec x}{2 ( sec x - 1 )} for 0 < x < \frac{π}{2} and f^{’} (x) = f (x) \cdot g (x), then g (x) =$

Solution

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