Question

Let $f$ be a function defined by $f\left(x\right)=\left\{\begin{array}{ll}\frac{\tan x}{x},& x\ne 0\\ 1,& x=0\end{array}\right.$

Statement $\text{I}$: $x=0$ is a point of minima of $f$.

Statement $\text{II}$ : $f\text{'}\left(0\right)=0$

• A. Statement $\text{I}$ is correct, Statement $\text{II}$ is correct
• B. Statement $\text{I}$ is correct, statement $\text{II}$ is correct; statement $\text{II}$ is the correct explanation for Statement $\text{I}$
• C. Statement $\text{I}$ is correct, statement $\text{II}$ is correct; Statement $\text{II}$ is not a correct explanation for Statement $\text{I}$.
• D. Statement $\text{I}$ is correct, statement $\text{II}$ is incorrect

Answer 1

The correct option is C

Statement $\text{I}$ is correct, statement $\text{II}$ is correct; Statement $\text{II}$ is not a correct explanation for Statement $\text{I}$.

Explanation for correct option:

Using L' hospital's rule to determine if the statements are correct:

Consider the given equation,

$f\left(x\right)=\left\{\begin{array}{ll}\frac{\tan x}{x},& x\ne 0\\ 1,& x=0\end{array}\right.$

Explanation for Statement $\text{II}$ : $f\text{'}\left(0\right)=0$

We know that,

$f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$

Then,

$$\begin{gathered} f\text{'}\left(0\right)=\underset{h\to 0}{\lim }\frac{f\left(h\right)-f\left(0\right)}{h} \\ f\text{'}\left(0\right)=\underset{x\to 0}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x} \\ f\text{'}\left(0\right)=\underset{x\to 0}{\lim }\frac{{\displaystyle \frac{\tan x}{x}}-1}{x} \\ f\text{'}\left(0\right)=\underset{x\to 0}{\lim }\frac{\tan x-x}{x} \end{gathered}$$

According to the L' hospital's rule

$f\text{'}\left(0\right)=\underset{x\to 0}{\lim }\frac{se{c}^{2}x-1}{2x}$

We know that

$$\begin{gathered} 1+{\tan }^{2}x=se{c}^{2}x \\ {\tan }^{2}x=se{c}^{2}x-1 \end{gathered}$$

$$\begin{gathered} f\text{'}\left(0\right)=\underset{x\to 0}{\lim }\frac{{\tan }^{2}x}{2x} \\ f\text{'}\left(0\right)=\underset{x\to 0}{\lim }\frac{\tan x}{x}\times \frac{\tan x}{2} \end{gathered}$$

We know that,$\frac{\tan x}{x}=1$

$$\begin{gathered} f\text{'}\left(0\right)=\underset{x\to 0}{\lim }\frac{\tan x}{2} \\ f\text{'}\left(0\right)=0 \end{gathered}$$

Hence, Statement $\text{II}$ is correct

Explanation for the Statement $\text{I}$ : $x=0$ is a point of minima of $f$.

$f\left(x\right)=\frac{\tan x}{x}$

Differentiate the above Equation

$f\text{'}\left(x\right)=\frac{d}{dx}\left(\frac{\tan x}{x}\right)......\left(1\right)$

We know that

$\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v{\displaystyle \frac{du}{dx}}-u{\displaystyle \frac{dv}{dx}}}{v^2}$

Differentiating the Equation $\left(1\right)$

$$\begin{gathered} f\text{'}\left(x\right)=\frac{xse{c}^{2}x-\tan x}{x^2} \\ f\text{'}\left(x\right)=\frac{{\displaystyle \frac{x}{{\cos }^{2}x}}-{\displaystyle \frac{\sin x}{\cos x}}}{x^2} \\ f\text{'}\left(x\right)=\frac{x-\left(\sin x\right)\left(\cos x\right)}{x^2{\cos }^{2}x} \end{gathered}$$

Multiply and divide by $2$

$f\text{'}\left(x\right)=\frac{2x-2\left(\sin x\right)\left(\cos x\right)}{2x^2{\cos }^{2}x}$

We know that

$$\begin{gathered} 2\left(\sin x\right)\left(\cos x\right)=\sin 2x \\ {x}^2{\cos }^{2}x>0 \end{gathered}$$

Then,

$f\text{'}\left(x\right)=2x-\sin 2x$

Let, $2x\to t$

$f\text{'}\left(x\right)=t-\sin t$

$f\left(x\right)=t-\sin t$

$\left\{\begin{array}{ll}\text{for}t>0& f\left(t\right)>0t>\sin t\\ \text{for}t<0& f\left(t\right)<0t<\sin t\end{array}\right.$

$f\left(x\right)$ changes its sign from negative to positive as $x$ passes through $0$.

Hence, $x=0$ is a point of minima of $f$ so that Statement $\text{I}$ is also correct.

and it clears that Statement $\text{I}$ is correct, statement $\text{II}$ is correct; Statement $\text{II}$ is not a correct explanation for Statement $\text{I}$

Therefore, the correct answer is Option (C).