Question
If $$f\left( 0 \right) = f'\left( 0 \right) = 0$$ and $$f''\left( x \right) = {\tan ^2}x$$ then $$f\left( x \right)$$ is :
A.
$$\log \,\sec \,x - \frac{1}{2}{x^2}$$
B.
$$\log \,\cos \,x + \frac{1}{2}{x^2}$$
C.
$$\log \,\sec \,x + \frac{1}{2}{x^2}$$
D.
none of these
Answer :
$$\log \,\sec \,x - \frac{1}{2}{x^2}$$
Solution :
$$\eqalign{
& f'\left( x \right) = \int {{{\tan }^2}x\,dx = \int {\left( {{{\sec }^2}x - 1} \right)dx} = \tan \,x - x + k} \cr
& \therefore f'\left( 0 \right) = k = 0 \cr
& \therefore f'\left( x \right) = \tan \,x - x \cr
& \therefore f\left( x \right) = \int {\left( {\tan \,x - x} \right)dx = \log \,\sec \,x - \frac{{{x^2}}}{2} + c} \cr
& \therefore f\left( 0 \right) = c = 0 \cr} $$