Question

Let $f\left(x\right)$ be a function such that $f\left(0\right)=f^{{}^{\prime }}0$ and $f^{{}^{\prime \prime }}\left(x\right)={\sec }^{4}x+4$ then $f\left(x\right)=$

• A. $\log \left(\sin x\right)+\frac{1}{3}{\tan }^{3}x+6x$
• B. $\frac{2}{3}\log \left(\sec x\right)+\frac{1}{6}{\tan }^{2}x+2x^2$
• C. $\log \left(\cos x\right)+\frac{1}{6}{\cos }^{2}x+\frac{x^2}{5}$
• D. $\log \left(\sec x\right)+\frac{1}{6}{\sin }^{3}x+x^2$

Answer 1

The correct option is B $\frac{2}{3}\log \left(\sec x\right)+\frac{1}{6}{\tan }^{2}x+2x^2$

$f^{{}^{\prime \prime }}\left(x\right)=se{c}^4\left(x\right)+4$

We write above euation as-

$f^{{}^{\prime }}\left(x\right)=\int se{c}^4\left(x\right)dx+4\int dx......\left(1\right)$

Let solve $I=\int se{c}^{4}xdx$

$\int se{c}^{4}xdx=\int se{c}^{2}x.se{c}^{2}xdx$

using Integration by part

$\int udv=uv-\int vdudx$

$u=se{c}^{2}x$ and $dv=se{c}^{2}x$

$\int se{c}^{2}x.se{c}^{2}xdx=tanx.se{c}^{2}x-2\int ta{n}^{2}x.se{c}^{2}xdx$

$\int se{c}^{4}xdx=tanx.se{c}^{2}x-2\int (se{c}^{2}x-1).se{c}^{2}dx\{se{c}^{2}x-1=ta{n}^{2}x\}$

$\int se{c}^{4}xdx=tanx.se{c}^{2}x-2\int (se{c}^{4}x-se{c}^{2}x)dx$

$\int se{c}^{4}xdx=tanx.se{c}^{2}x-2\int se{c}^{4}x+2\int se{c}^{2}xdx$

$3\int se{c}^{4}xdx=tanx.se{c}^{2}x+2\int se{c}^{2}xdx$

$\int se{c}^{4}xdx=\frac{1}{3}tanx.se{c}^{2}x+\frac{2}{3}tanx$

Lets put this result into equation (1)

$f^{{}^{\prime }}\left(x\right)=\frac{1}{3}tanx.se{c}^{2}x+\frac{2}{3}tanx+4x+cwherecisintegrationconstant$

Now let's put boundary condition $f^{\prime }\left(0\right)=0$

$f^{\prime }\left(0\right)=0=\frac{1}{3}tan\left(0\right).sec\left(0\right)+\frac{2}{3}tan\left(0\right)+4.0+c$

$hencec=0$

$f^{{}^{\prime }}\left(x\right)=\frac{1}{3}tanx.se{c}^{2}x+\frac{2}{3}tanx+4x$

$f\left(x\right)=\frac{1}{3}\int tanx.se{c}^{2}xdx+\frac{2}{3}\int tanxdx+4\int xdx$

$lettanx=t,se{c}^{2}xdx=dt$

$f\left(x\right)=\frac{1}{3}\int tdt+\frac{2}{3}log\left(|secx|\right)+2x^2+c_1\{\int tanxdx=log(|secx|\left)\right\}$

$f\left(x\right)=\frac{1}{3}\frac{t^2}{2}+\frac{2}{3}log\left(|secx|\right)+2x^2+c_1$

Now let's again put boundary condition $f\left(0\right)=0$

$f\left(0\right)=0+\frac{2}{3}log\left(sec\right(0\left)\right)+0+c_1=0$

$f\left(0\right)=0+\frac{2}{3}log\left(1\right)+0+c_1=0$

$f\left(0\right)=0+\frac{2}{3}.0+0+c_1=0$

$c_1=0$

$f\left(x\right)=\frac{t^2}{6}+\frac{2}{3}log\left(|secx|\right)+2x^2$

Now back substituting $t$ value-

$f\left(x\right)=\frac{ta{n}^{2}x}{6}+\frac{2}{3}log\left(|secx|\right)+2x^2$

Option $b$ is correct answer.