Question

If the derivative of f(x) w.r.t x is $\frac{\frac{1}{2}–\sin^{2}x}{f(x)}$ then period of f(x) is-

• A. 2p
• B. p
• C. p/2
• D. None

Answer 1

Answer: (B)

p

Answer 2

f ¢(x) = $\frac{\frac{1}{2}–\sin^{2}x}{f(x)}$̃ f(x) f ¢ (x) = $\frac{1}{2}$ – sin²x

̃ $\Rightarrow - \operatorname { cosec } ^ { 2 } x d x = d t$

̃ $= \int \frac { - d t } { t ^ { 2 } } = \frac { 1 } { t } + c$ = $\frac{x}{2}$– $\left( \frac{x}{2}–\frac{\sin 2x}{4} \right) + C$

(f(x))² = $\frac{\sin 2x}{2}$ + C, f(x) = $\sqrt{\frac{|\sin 2x|}{2} + C}$ period of f(x) is $\frac{2\pi}{2}$ = p