Question

If $f\left(x\right)=p\left|\sin x\right|+q{e}^{\left|x\right|}+r{\left|x\right|}^3$ and if $f\left(x\right)$ is differentiable at $x=0$, then

• A. $p=q=r=0$
• B. $p+q=0$, $r$ is any real number
• C. $q=0,r=0,p$ is any real number
• D. $r=0,p=0,q$ is any real number

Answer 1

Answer: (A), (B)

$p+q=0$, $r$ is any real number

Explanation for the correct options

Step 1. Find the left hand derivative.

The function given is $f\left(x\right)=p\left|\sin x\right|+q{e}^{\left|x\right|}+r{\left|x\right|}^3$.

As $x\to 0^{-}$, the function can be written as $f\left(x\right)=-p\sin x+q{e}^{-x}-r{x}^3$.

Now the left hand derivative at $x=0$ is given as:

$\begin{array}{rcl}\mathrm{LHD}& =& \underset{h\to 0}{\lim }\frac{f\left(0-h\right)-f\left(0\right)}{-h}\\ & =& \underset{h\to 0}{\lim }\frac{-p\sin \left(-h\right)+q{e}^{-\left(-h\right)}-r{\left(-h\right)}^3-\left(-p\sin 0+q{e}^{-0}-r\times 0^3\right)}{-h}\\ & =& \underset{h\to 0}{\lim }\frac{p\sin h+q{e}^h+r{h}^3-\left(0+q-0\right)}{-h}\\ & =& \underset{h\to 0}{\lim }\frac{p\sin h+q{e}^h+r{h}^3-q}{-h}\end{array}$

Differentiating the numerator and denominator,

$\begin{array}{rcl}\underset{h\to 0}{\lim }\frac{p\sin h+q{e}^h+r{h}^3-q}{-h}& =& \underset{h\to 0}{\lim }\frac{p\cos h+q{e}^h+r3h^2-0}{-1}\\ & =& -p\cos 0-q{e}^0-3r\times 0^2\\ & =& -p-q\end{array}$

Thus the left hand derivative at $x=0$ is given as $\mathrm{LHD}=-p-q$.

Step 2. Find the right hand derivative.

The function given is $f\left(x\right)=p\left|\sin x\right|+q{e}^{\left|x\right|}+r{\left|x\right|}^3$.

As $x\to 0^{+}$, the function can be written as $f\left(x\right)=p\sin x+q{e}^x+r{x}^3$.

Now the right hand derivative at $x=0$ is given as:

$\begin{array}{rcl}\mathrm{RHD}& =& \underset{h\to 0}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{h}\\ & =& \underset{h\to 0}{\lim }\frac{p\sin h+q{e}^h+r{h}^3-\left(p\sin 0+q{e}^0+r\times 0^3\right)}{h}\\ & =& \underset{h\to 0}{\lim }\frac{p\sin h+q{e}^h+r{h}^3-\left(0+q+0\right)}{h}\\ & =& \underset{h\to 0}{\lim }\frac{p\sin h+q{e}^h+r{h}^3-q}{h}\end{array}$

Differentiating the numerator and denominator:

$\begin{array}{rcl}\underset{h\to 0}{\lim }\frac{p\sin h+q{e}^h+r{h}^3-q}{h}& =& \underset{h\to 0}{\lim }\frac{p\cos h+q{e}^h+r3h^2-0}{1}\\ & =& p\cos 0+q{e}^0+3r\times 0^2\\ & =& p+q\end{array}$

Thus the right hand derivative at $x=0$ is given as $\mathrm{RHD}=p+q$.

Step 3. Find the required condition.

As the function $f\left(x\right)$ is differentiable at $x=0$, so the left hand derivative must be equal to the right hand derivative, so $\mathrm{LHD}=\mathrm{RHD}$. Now

$$\begin{gathered} -p-q=p+q \\ \Rightarrow -2\left(p+q\right)=0 \\ \Rightarrow p+q=0 \end{gathered}$$

So, the required conditions are $p+q=0$, and $r$ is any real number.

Hence, the correct options are A and B.