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Question: Let f(0) = f ' (0) = 0 and f " (x) = sec<sup>4</sup>x + 4, then f(x) =...

Let f(0) = f ' (0) = 0 and f " (x) = sec4x + 4, then f(x) =

A

23\frac{2}{3}log sec x + 16\frac{1}{6} tan2 x + 2x2

B

23\frac{2}{3}log cos x + 16\frac{1}{6}tan2 x + 2x2

C

23\frac{2}{3}log sec x + 16\frac{1}{6} tan2 x

D

None of these

Answer

23\frac{2}{3}log sec x + 16\frac{1}{6} tan2 x + 2x2

Explanation

Solution

Integrate we have

f ' (x) = sec2x.(1+tan2x)dx\int_{}^{}{\sec^{2}x.(1 + \tan^{2}x)dx}+ 4x + C1

f ' (x) = tan x + tan3x3\frac{\tan^{3}x}{3} + 4x + c1

put x = 0

a = 0 \end{matrix}$$ again integrate f(x) = log sec x + $\frac{1}{3}$ $\int_{}^{}{\tan^{3}xdx}$ + 2x<sup>2</sup> + c<sub>2</sub> = log secx + $\frac{1}{3}$ $\frac{\tan^{2}x}{2}$ – $\frac{1}{3}$ log sec x + c<sub>2</sub> + 2n<sup>2</sup>put x = 0 c<sub>2</sub> = 0 \\ $\begin{matrix} f(x) = 2x^{2} + \frac{1}{6}\tan^{2}x + \frac{2}{3}{logsec}x \end{matrix}$