Question

One maximum point of ${{\sin }^{p}}x.{{\cos }^{q}}x$ is
(a) $x={{\tan }^{-1}}\left( \sqrt{\dfrac{p}{q}} \right)$
(b) $x={{\tan }^{-1}}\left( \sqrt{\dfrac{q}{p}} \right)$
(c) $x={{\tan }^{-1}}\left( \dfrac{p}{q} \right)$
(d) $x={{\tan }^{-1}}\left( \dfrac{q}{p} \right)$

Answer 1

For solving this problem we consider the given function as $f\left( x \right)$. For finding the maximum or minimum points of the function we need to take ${f}'\left( x \right)=0$ and solve for $'x'$. These values of $'x'$ will give the maximum points of the given function. For finding the derivative of function we use product rule that is $\dfrac{d}{dx}\left( u.v \right)=u.\dfrac{dv}{dx}+v.\dfrac{du}{dx}$.

Complete step-by-step solution
Let us assume that the given function as
$f\left( x \right)={{\sin }^{p}}x.{{\cos }^{q}}x$
We know that for finding maximum or minimum points of the function we need to take ${f}'\left( x \right)=0$ and solve for $'x'$.
Let us find ${f}'\left( x \right)$.
Let us assume $u={{\sin }^{p}}x$ and $v={{\cos }^{q}}x$.
Then we can write
${f}'\left( x \right)=\dfrac{d}{dx}\left( u.v \right)$
We know that when finding the derivative of product of two different functions we need to use product rule that is $\dfrac{d}{dx}\left( u.v \right)=u.\dfrac{dv}{dx}+v.\dfrac{du}{dx}$
By using product rule we get

& \Rightarrow {f}'\left( x \right)=\dfrac{d}{dx}\left( u.v \right)=u.\dfrac{dv}{dx}+v.\dfrac{du}{dx} \\\ & \Rightarrow {f}'\left( x \right)={{\sin }^{p}}x.\dfrac{d}{dx}\left( {{\cos }^{q}}x \right)+{{\cos }^{q}}x.\dfrac{d}{dx}\left( {{\sin }^{p}}x \right)....equation(i) \\\ \end{aligned}$$ We know that when there is some power for a function in derivative we need to use chain rule as $$\dfrac{d}{dx}\left( {{u}^{n}} \right)=n.{{u}^{n-1}}.\dfrac{du}{dx}$$ By using the chain rule we can write $$\Rightarrow \dfrac{d}{dx}\left( {{\cos }^{q}}x \right)=q.{{\cos }^{q-1}}x.\dfrac{d}{dx}\left( \cos x \right)$$ We know that $$\dfrac{d}{dx}\left( \cos x \right)=-\sin x$$, by substituting in above equation we get $$\Rightarrow \dfrac{d}{dx}\left( {{\cos }^{q}}x \right)=-q.{{\cos }^{q-1}}x.\sin x$$ Similarly by using the chain rule we can write $$\Rightarrow \dfrac{d}{dx}\left( {{\sin }^{p}}x \right)=p.{{\sin }^{p-1}}x.\dfrac{d}{dx}\left( \sin x \right)$$ We know that $$\dfrac{d}{dx}\left( \sin x \right)=\cos x$$, by substituting in above equation we get $$\Rightarrow \dfrac{d}{dx}\left( {{\sin }^{p}}x \right)=p.{{\sin }^{p-1}}x.\cos x$$ Now, by substituting the required values in equation (i) we get $$\begin{aligned} & \Rightarrow {f}'\left( x \right)={{\sin }^{p}}x.\dfrac{d}{dx}\left( {{\cos }^{q}}x \right)+{{\cos }^{q}}x.\dfrac{d}{dx}\left( {{\sin }^{p}}x \right) \\\ & \Rightarrow {f}'\left( x \right)={{\sin }^{p}}x\left( -q.{{\cos }^{q-1}}x.\sin x \right)+{{\cos }^{q}}x\left( p.{{\sin }^{p-1}}x.\cos x \right) \\\ & \Rightarrow {f}'\left( x \right)=-q.{{\sin }^{p+1}}x.{{\cos }^{q-1}}x+p.{{\sin }^{p-1}}x.{{\cos }^{q+1}}x \\\ \end{aligned}$$ Since we need to find the maximum point let us take $${f}'\left( x \right)=0$$. By taking $${f}'\left( x \right)=0$$ we get $$\begin{aligned} & \Rightarrow {f}'\left( x \right)=0 \\\ & \Rightarrow -q.{{\sin }^{p+1}}x.{{\cos }^{q-1}}x+p.{{\sin }^{p-1}}x.{{\cos }^{q+1}}x=0 \\\ & \Rightarrow p.{{\sin }^{p-1}}x.{{\cos }^{q+1}}x=q.{{\sin }^{p+1}}x.{{\cos }^{q-1}}x \\\ \end{aligned}$$ By taking similar terms on same side we get $$\begin{aligned} & \Rightarrow p.{{\cos }^{2}}x=q.{{\sin }^{2}}x \\\ & \Rightarrow {{\left( \dfrac{\sin x}{\cos x} \right)}^{2}}=\dfrac{p}{q} \\\ \end{aligned}$$ We know that $$\tan x=\dfrac{\sin x}{\cos x}$$. By substituting in above equation we get $$\begin{aligned} & \Rightarrow \tan x=\pm \sqrt{\dfrac{p}{q}} \\\ & \Rightarrow x={{\tan }^{-1}}\left( \pm \sqrt{\dfrac{p}{q}} \right) \\\ \end{aligned}$$ We know that$${{\tan }^{-1}}\theta $$ is an increasing function in its domain we can take $$'\pm '$$ outside, then we get $$\Rightarrow x=\pm {{\tan }^{-1}}\left( \sqrt{\dfrac{p}{q}} \right)$$ Here, we have two points of maximum that is $$\Rightarrow x=+{{\tan }^{-1}}\left( \sqrt{\dfrac{p}{q}} \right)$$ or $$x=-{{\tan }^{-1}}\left( \sqrt{\dfrac{p}{q}} \right)$$. **One answer is given in options, that is an option (a).** **Note:** Students will make mistakes without considering $$'\pm '$$. It will be wrong if we don’t take $$'\pm '$$. This symbol tells us that there will be two values satisfying that equation. If we didn’t consider all the possible answers we may get wrong answers in the end. The symbol $$'\pm '$$ plays an important role in multiple-choice questions.