Question

Let $f\left(x\right)=\{\begin{array}{cc}\frac{x^p}{(\sin x{)}^q},& if0<x\le \frac{\pi }{2}\\ 0,& ifx=0\end{array},(p,q,ϵR)$. Then Lagrange's mean value theorem is applicable to $f\left(x\right)$ in closed interval $[0,x]$.

• A. For all $p,q$
• B. Only when $p>q$
• C. Only when $p<q$
• D. For no value of $p,q$

Answer 1

The correct option is B Only when $p>q$

$f\left(x\right)=\{\begin{array}{cc}\frac{x^p}{(\sin x{)}^q},& if0<x\le \frac{\pi }{2}\\ 0,& ifx=0\end{array},(p,q,ϵR)$

Lagrange's Mean value theorem is applicable for any $f$, if it is continuous at $(a,b)$ and differential with in the closed interval $[a,b]$.

i.e. if $\underset{x\to a}{lim}f\left(x\right)=f\left(a\right)$

Here, $f\left(a\right)=f\left(0\right)=0$

$f\left(x\right)$ is differentiable at $(0,x)$

Consider, $\underset{x\to 0}{lim}f\left(x\right)$

$=\underset{x\to 0}{lim}\frac{x^p}{(\sin x{)}^q}$

$=\underset{x\to 0}{lim}\frac{x^p\cdot x^q\cdot x^{-q}}{(\sin x{)}^q}$

$=\underset{x\to 0}{lim}\frac{x^q\cdot x^{p-q}}{(\sin x{)}^q}$

$=\underset{x\to 0}{lim}\frac{x^q}{(\sin x{)}^q}\times \underset{x\to 0}{lim}{x}^{p-q}$

$=\underset{x\to 0}{lim}{\left(\frac{x}{\sin x}\right)}^q\times \underset{x\to 0}{lim}{x}^{p-q}$

$=\underset{x\to 0}{lim}{x}^{p-q}$ ....... $[\because \underset{x\to 0}{lim}\frac{x}{\sin x}=1]$

Now, $\underset{x\to 0}{lim}{x}^{p-q}=\{\begin{array}{cc}0,& ifp>q\\ \infty ,& ifp<q\\ 1,& ifp=q\end{array}$

$\therefore \underset{x\to 0}{lim}f\left(x\right)=0$ only when $p>q$

Hence, Lagrange's Mean value theorem is applicable to $f\left(x\right)$ in $[0,x]$ only when $p>q$.