Question

If $f\text{'}\left(x\right)={\tan }^{-1}\left(secx+\tan x\right)$, $\left(\frac{-\mathrm{\pi }}{2}\right)<x<\left(\frac{\mathrm{\pi }}{2}\right)$, and $f\left(0\right)=0$, then $f\left(1\right)$ is equal to

• A. $\frac{\left(\mathrm{\pi }+1\right)}{4}$
• B. $\frac{\left(\mathrm{\pi }+2\right)}{4}$
• C. $\frac{1}{4}$
• D. $\frac{\left(\mathrm{\pi }-1\right)}{4}$

Answer 1

The correct option is A

$\frac{\left(\mathrm{\pi }+1\right)}{4}$

Explanation for the correct option:

Step 1: Simplifying the given equation:

Given that,

$\begin{array}{rcl}f\text{'}\left(x\right)& =& {\tan }^{-1}\left(secx+\tan x\right)\\ & =& {\tan }^{-1}\left(\frac{1}{\cos x}+\frac{\sin x}{\cos x}\right)\\ & =& {\tan }^{-1}\left(\frac{1+\sin x}{\cos x}\right)\\ & =& {\tan }^{-1}\left(\frac{{\sin }^2\left({\displaystyle \frac{x}{2}}\right)+{\cos }^2\left({\displaystyle \frac{x}{2}}\right)+2\sin {\displaystyle \frac{x}{2}}\cos {\displaystyle \frac{x}{2}}}{{\cos }^2{\displaystyle \frac{x}{2}}-{\sin }^2{\displaystyle \frac{x}{2}}}\right)\mathbf{[}\mathbf{\because }{\mathbf{sin}}^{\mathbf{2}}\mathbf{A}\mathbf{+}{\mathbf{cos}}^{\mathbf{2}}\mathbf{A}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{}\mathbf{sin}\mathbf{2}\mathbf{A}\mathbf{=}\mathbf{2}\mathbf{cos}\mathbf{A}\mathbf{sin}\mathbf{A}\mathbf{}\mathbf{]}\\ & =& {\tan }^{-1}\left(\frac{{\left(\sin {\displaystyle \frac{x}{2}}+\cos {\displaystyle \frac{x}{2}}\right)}^2}{\left(\cos {\displaystyle \frac{x}{2}}+\sin {\displaystyle \frac{x}{2}}\right)\left(\cos {\displaystyle \frac{x}{2}}-\sin {\displaystyle \frac{x}{2}}\right)}\right)\mathbf{[}\mathbf{\because }\mathbf{}\left(a^2-b^2\right)\mathbf{=}\left(a+b\right)\left(a-b\right)\mathbf{]}\\ & =& {\tan }^{-1}\left(\frac{\sin {\displaystyle \frac{x}{2}}+\cos {\displaystyle \frac{x}{2}}}{\cos {\displaystyle \frac{x}{2}}-\sin {\displaystyle \frac{x}{2}}}\right)\\ & =& {\tan }^{-1}\left(\frac{1+\tan {\displaystyle \frac{x}{2}}}{1-\tan {\displaystyle \frac{x}{2}}}\right)\\ & =& {\tan }^{-1}\left(\tan \left(\frac{\mathrm{\pi }}{4}+\frac{x}{2}\right)\right)\left[\because \tan \left(\frac{\pi }{4}+A\right)=\frac{1+\tan A}{1-\tan A}\right]\\ & =& \frac{\mathrm{\pi }}{4}+\frac{x}{2}\end{array}$

Step 2: Finding the value of $f\left(1\right)$:

Now, integrate the simplified equation,

$$\begin{gathered} \int f\text{'}\left(x\right)=\int \left(\frac{\mathrm{\pi }}{4}+\frac{x}{2}\right) \\ \Rightarrow f\left(x\right)=\frac{\mathrm{\pi x}}{4}+\frac{x^2}{4}+c...\left(1\right) \end{gathered}$$

Subsitute $x=0$ in the equation $\left(1\right)$ to get $c$,

$$\begin{gathered} f\left(0\right)=0+0+c \\ \Rightarrow c=0 \end{gathered}$$

Now, substitute $x=1$and $c=0$ in the equation $\left(1\right)$.

$\begin{array}{rcl}f\left(1\right)& =& \frac{\mathrm{\pi }}{4}+\frac{1}{4}+0\\ & =& \frac{\left(\mathrm{\pi }+1\right)}{4}\end{array}$

Hence, the correct option is A.