Question

The maximum value of the function $y = 2\tan x - {\tan ^2}x$ over $\left[ {0,\dfrac{\pi }{2}} \right]$ is
A. $\infty$
B. 1
C. 3
D. 2

Answer 1

Hint: Here, we will first calculate the first order derivative $\dfrac{{dy}}{{dx}}$ of the given function $y$ and then take it equal to 0 to find the critical point. Then we will find the sign of the second order derivative $\dfrac{{{d^2}y}}{{d{x^2}}}$ at the obtained critical point. If the sign of second differentiation is negative, then the function has a point of maxima.

Complete step-by-step solution:
Given that the function
$y = 2\tan x - {\tan ^2}x{\text{ ......}}\left( 1 \right)$

We know that the point of maxima is calculated by taking the differentiation $\dfrac{{dy}}{{dx}}$ of the given function $y$ equals to 0 and then finding the sign of the second differentiation $\dfrac{{{d^2}y}}{{d{x^2}}}$. If the second differentiation is negative, the function has a point of maxima.

Differentiating the above function, we get

$$\dfrac{{dy}}{{dx}} = 2{\sec ^2}x - 2\tan x \cdot {\sec ^2}x \\\ = 2{\sec ^2}x\left( {1 - \tan x} \right){\text{ ......}}\left( 2 \right) \\\$$

Taking $\dfrac{{dy}}{{dx}} = 0$ to find the critical point in the above equation, we get

$\Rightarrow 2{\sec ^2}x\left( {1 - \tan x} \right) = 0$
$\Rightarrow {\sec ^2}x = 0$ or $1 - \tan x = 0$

Simplifying the above equations, we get

$\Rightarrow x = \dfrac{\pi }{2}$ or $x = \dfrac{\pi }{4}$

Here $x = \dfrac{\pi }{2}$ is not possible because it is not in the domain of the function.
Therefore, $x = \dfrac{\pi }{4}$ is the critical point.
Differentiating the equation $\left( 2 \right)$ using the product rule, we get

$$\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{d}{{dx}}\left[ {2{{\sec }^2}x\left( {1 - \tan x} \right)} \right] \\\ = 4\sec x \cdot \sec x \cdot \tan x\left( {1 - \tan x} \right) + 2{\sec ^2}x\left( {0 - {{\sec }^2}x} \right) \\\ = 4{\sec ^2}x\tan x - 4{\sec ^2}x{\tan ^2}x - 2{\sec ^4}x \\\$$

Replacing $\dfrac{\pi }{4}$ for $x$ in the above equation, we get

${\left. {\dfrac{{{d^2}y}}{{d{x^2}}}} \right|_{x = 4}} = 4{\sec ^2}\dfrac{\pi }{4}\tan \dfrac{\pi }{4} - 4{\sec ^2}\dfrac{\pi }{4}\tan^2 \dfrac{\pi }{4} - 2{\sec ^4}\dfrac{\pi }{4}$

Using the trigonometric values $\tan \dfrac{\pi }{4} = 1$ and $\sec \dfrac{\pi }{4} = \sqrt 2$ in the above equation, we get

$$\Rightarrow 4{\left( {\sqrt 2 } \right)^2} \cdot 1 - 4{\left( {\sqrt 2 } \right)^2} \cdot {\left( 1 \right)^2} - 2 \cdot {\left( {\sqrt 2 } \right)^4} \\\ \Rightarrow 8 - 8 - 8 \\\ \Rightarrow - 8 \\\$$

Since $- 8$ is negative, the given function has a point of maxima at $x = \dfrac{\pi }{4}$.

We will now find the maximum value of function at point $x = \dfrac{\pi }{4}$ by substituting it in the given equation.

$${\left. y \right|_{x = \dfrac{\pi }{4}}} = 2\tan \dfrac{\pi }{4} - {\tan ^2}\dfrac{\pi }{4} \\\ = 2 \cdot 1 - 1 \\\ = 2 - 1 \\\ = 1 \\\$$

Thus, the maximum value of the given function is 1.

Hence, the option B is correct.

Note: In solving these types of questions, you should be familiar with the steps to find the point of maxima. Then use the given conditions and values given in the question, and substitute in the steps to find the point of maxima of the function. Also, we are supposed to write the values properly to avoid any miscalculation.