Question

The maximum value of $\left[\frac{\left({\tan }^{2}x-co{t}^{2}x+1\right)}{\left({\tan }^{2}x+co{t}^{2}x+1\right)}\right]$ is

• A. $\frac{3}{2}$
• B. $2$
• C. $\frac{5}{3}$
• D. None of these

Answer 1

The correct option is D

None of these

Explanation for the correct option:

$$\begin{gathered} f\left(x\right)=\left[\frac{\left({\tan }^{2}x-co{t}^{2}x+1\right)}{\left({\tan }^{2}x+co{t}^{2}x+1\right)}\right] \\ =\left[\frac{\left({\tan }^{2}x-{\displaystyle \frac{1}{{\tan }^{2}x}}+1\right)}{\left({\tan }^{2}x+{\displaystyle \frac{1}{{\tan }^{2}x}}+1\right)}\right]\left[\because {\cot }^{2}x=\frac{1}{{\tan }^{2}x}\right] \\ =\left[\frac{\left({\displaystyle \frac{{\tan }^{4}x-1+{\tan }^{2}x}{{\tan }^{2}x}}\right)}{\left({\displaystyle \frac{{\tan }^{4}x+1+{\tan }^{2}x}{{\tan }^{2}x}}\right)}\right] \\ =\left[\frac{\left({\tan }^{4}x+{\tan }^{2}x-1\right)}{\left({\tan }^{4}x+{\tan }^{2}x+1\right)}\right] \end{gathered}$$

Let $t={\tan }^{2}x$ and $f\left(x\right)=y$

$$\begin{gathered} y{t}^2+ty+y-t^2-t+1=0 \\ (y-1)t^2+(y-1)t+(y+1)=0 \\ {(y-1)}^2-4(y^2-1)\ge 0\mathbf{[}\mathbf{\because }\mathbf{t}\mathbf{\in }\mathbf{R}\mathbf{]} \\ -3y^2-2y+5\ge 0 \\ (3y+5)(y-1)\le 0 \\ \left(\frac{-5}{3}\right)\le y\le 1 \end{gathered}$$

But $y=1$is not possible for any $x$

Hence the correct option is option(D)