Question

The minimum value of $a\sec x+b\operatorname{cosec}x$, $0 < a,b < 1$, $0 < x < \dfrac{\pi }{2}$.
(a) $a+b$,
(b) ${{a}^{\dfrac{2}{3}}}+{{b}^{\dfrac{2}{3}}}$,
(c) ${{\left( {{a}^{\dfrac{2}{3}}}+{{b}^{\dfrac{2}{3}}} \right)}^{\dfrac{3}{2}}}$,
(d) None of these.

Answer 1

Hint: We start solving the problem by assigning a variable to the function $a\sec x+b\operatorname{cosec}x$. We then differentiate this function with respect to x and equate it to 0 to get the value of x. We differentiate the function again and substitute the obtained value of x to check whether the function is minimum or maximum at that point. We then substitute this value in function to get the minimum value of function.

Complete step-by-step answer:
According to the problem, we have to find the minimum value of $a\sec x+b\operatorname{cosec}x$, $0 < a < b$, $0 < x < \dfrac{\pi }{2}$.
Let us assume $a\sec x+b\operatorname{cosec}x$, $0 < a < b$, $0 We know that to find the minimum value of $f\left( x \right)$, we differentiate it with respect to x and equates it to 0 to get the value of x. We then find the function ${{f}^{''}}\left( x \right)$ and substitute the obtained value to check whether the value is positive to get maxima at that value of x.
So, let us differentiate $f\left( x \right)=a\sec x+b\operatorname{cosec}x$ with respect to x on both sides.
$\Rightarrow \dfrac{d}{dx}\left( f\left( x \right) \right)=\dfrac{d}{dx}\left( a\sec x+b\operatorname{cosec}x \right)$ ---(1).
We know that $\dfrac{d}{dx}\left( ag\left( x \right)+bh\left( x \right) \right)=a\dfrac{d}{dx}\left( g\left( x \right) \right)+b\dfrac{d}{dx}\left( h\left( x \right) \right)$. We use this in equation (1).
$\Rightarrow \dfrac{d}{dx}\left( f\left( x \right) \right)=\dfrac{d}{dx}\left( a\sec x \right)+\dfrac{d}{dx}\left( b\operatorname{cosec}x \right)$.
\[\Rightarrow \dfrac{d}{dx}\left( f\left( x \right) \right)=a\dfrac{d}{dx}\left( \sec x \right)+b\dfrac{d}{dx}\left( \operatorname{cosec}x \right)\] ---(2).
We know that $\dfrac{d}{dx}\left( \sec x \right)=\sec x\tan x$ and $\dfrac{d}{dx}\left( \operatorname{cosec}x \right)=-\operatorname{cosec}x\cot x$. We use these results in equation (2).
\[\Rightarrow \dfrac{d}{dx}\left( f\left( x \right) \right)=a\left( \sec x.\tan x \right)+b\left( -\operatorname{cosec}x.\cot x \right)\].
$\Rightarrow {{f}^{'}}\left( x \right)=a\sec x.\tan x-b\operatorname{cosec}x\cot x$ ---(3).
To find the value of x at which the minimum occurs, we take ${{f}^{'}}\left( x \right)=0$.
$\Rightarrow a\sec x.\tan x-b\operatorname{cosec}x\cot x=0$.
$\Rightarrow a\dfrac{1}{\cos x}.\dfrac{\sin x}{\cos x}-b\dfrac{1}{\sin x}\dfrac{\cos x}{\sin x}=0$.
$\Rightarrow a\dfrac{\sin x}{{{\cos }^{2}}x}-b\dfrac{\cos x}{{{\sin }^{2}}x}=0$.
$\Rightarrow a\dfrac{\sin x}{{{\cos }^{2}}x}=b\dfrac{\cos x}{{{\sin }^{2}}x}$.
$\Rightarrow \dfrac{\sin x}{{{\cos }^{2}}x}\times \dfrac{{{\sin }^{2}}x}{\cos x}=\dfrac{b}{a}$.
$\Rightarrow \dfrac{{{\sin }^{3}}x}{{{\cos }^{3}}x}=\dfrac{b}{a}$.
\[\Rightarrow {{\left( \dfrac{\sin x}{\cos x} \right)}^{3}}=\dfrac{b}{a}\].
\[\Rightarrow {{\left( \tan x \right)}^{3}}=\dfrac{b}{a}\].
\[\Rightarrow \tan x={{\left( \dfrac{b}{a} \right)}^{\dfrac{1}{3}}}\] ---(4).
Let us differentiate equation (3) with respect to x again.
$\Rightarrow \dfrac{d}{dx}\left( {{f}^{'}}\left( x \right) \right)=\dfrac{d}{dx}\left( a\sec x.\tan x-b\operatorname{cosec}x\cot x \right)$.
$\Rightarrow \dfrac{d}{dx}\left( {{f}^{'}}\left( x \right) \right)=\dfrac{d}{dx}\left( a\sec x.\tan x \right)-\dfrac{d}{dx}\left( b\operatorname{cosec}x\cot x \right)$.
We know that the differentiation of the function $uv$ is defined as $\dfrac{d}{dx}\left( uv \right)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$.
$\Rightarrow \dfrac{d}{dx}\left( {{f}^{'}}\left( x \right) \right)=a\sec x\dfrac{d}{dx}\left( \tan x \right)+a\tan x\dfrac{d}{dx}\left( \sec x \right)-b\operatorname{cosec}x\dfrac{d}{dx}\left( \cot x \right)-b\cot x\dfrac{d}{dx}\left( \operatorname{cosec}x \right)$.
We know that $\dfrac{d}{dx}\left( \tan x \right)={{\sec }^{2}}x$ and $\dfrac{d}{dx}\left( \cot x \right)=-{{\operatorname{cosec}}^{2}}x$.
$\Rightarrow {{f}^{''}}\left( x \right)=a\sec x\left( {{\sec }^{2}}x \right)+a\tan x\left( \sec x.\tan x \right)-b\operatorname{cosec}x\left( -{{\operatorname{cosec}}^{2}}x \right)-b\cot x\left( -\operatorname{cosec}x\cot x \right)$.
$\Rightarrow {{f}^{''}}\left( x \right)=a{{\sec }^{3}}x+a\sec x.{{\tan }^{2}}x+b{{\operatorname{cosec}}^{3}}x+b\operatorname{cosec}x{{\cot }^{2}}x$ ---(5).
From equation (4), we have \[\tan x={{\left( \dfrac{b}{a} \right)}^{\dfrac{1}{3}}}\].
\[\Rightarrow \cot x=\dfrac{1}{\tan x}=\dfrac{1}{{{\left( \dfrac{b}{a} \right)}^{\dfrac{1}{3}}}}\].
\[\Rightarrow \cot x={{\left( \dfrac{a}{b} \right)}^{\dfrac{1}{3}}}\].
We have ${{\sec }^{2}}x=1+{{\tan }^{2}}x$.
$\Rightarrow {{\sec }^{2}}x=1+{{\left( {{\left( \dfrac{b}{a} \right)}^{\dfrac{1}{3}}} \right)}^{2}}$.
$\Rightarrow {{\sec }^{2}}x=1+{{\left( \dfrac{b}{a} \right)}^{\dfrac{2}{3}}}$.
$\Rightarrow {{\sec }^{2}}x=1+\dfrac{{{b}^{\dfrac{2}{3}}}}{{{a}^{\dfrac{2}{3}}}}$.
$\Rightarrow {{\sec }^{2}}x=\dfrac{{{a}^{\dfrac{2}{3}}}+{{b}^{\dfrac{2}{3}}}}{{{a}^{\dfrac{2}{3}}}}$.
$\Rightarrow \sec x=\dfrac{\sqrt{{{a}^{\dfrac{2}{3}}}+{{b}^{\dfrac{2}{3}}}}}{{{a}^{\dfrac{1}{3}}}}$.
We have ${{\operatorname{cosec}}^{2}}x=1+{{\cot }^{2}}x$.
$\Rightarrow {{\operatorname{cosec}}^{2}}x=1+{{\left( {{\left( \dfrac{a}{b} \right)}^{\dfrac{1}{3}}} \right)}^{2}}$.
$\Rightarrow {{\operatorname{cosec}}^{2}}x=1+{{\left( \dfrac{a}{b} \right)}^{\dfrac{2}{3}}}$.
$\Rightarrow {{\operatorname{cosec}}^{2}}x=1+\dfrac{{{a}^{\dfrac{2}{3}}}}{{{b}^{\dfrac{2}{3}}}}$.
$\Rightarrow {{\operatorname{cosec}}^{2}}x=\dfrac{{{b}^{\dfrac{2}{3}}}+{{a}^{\dfrac{2}{3}}}}{{{b}^{\dfrac{2}{3}}}}$.
$\Rightarrow \operatorname{cosec}x=\dfrac{\sqrt{{{a}^{\dfrac{2}{3}}}+{{b}^{\dfrac{2}{3}}}}}{{{b}^{\dfrac{1}{3}}}}$.
We substitute all these values in equation (5).
$\Rightarrow {{f}^{''}}\left( x \right)=a{{\left( \dfrac{\sqrt{{{a}^{\dfrac{2}{3}}}+{{b}^{\dfrac{2}{3}}}}}{{{a}^{\dfrac{1}{3}}}} \right)}^{3}}+a\left( \dfrac{\sqrt{{{b}^{\dfrac{2}{3}}}+{{a}^{\dfrac{2}{3}}}}}{{{a}^{\dfrac{1}{3}}}} \right).{{\left( {{\left( \dfrac{b}{a} \right)}^{\dfrac{1}{3}}} \right)}^{2}}+b{{\left( \dfrac{\sqrt{{{a}^{\dfrac{2}{3}}}+{{b}^{\dfrac{2}{3}}}}}{{{b}^{\dfrac{1}{3}}}} \right)}^{3}}+b\left( \dfrac{\sqrt{{{a}^{\dfrac{2}{3}}}+{{b}^{\dfrac{2}{3}}}}}{{{b}^{\dfrac{1}{3}}}} \right){{\left( {{\left( \dfrac{a}{b} \right)}^{\dfrac{1}{3}}} \right)}^{2}}$.
$\Rightarrow {{f}^{''}}\left( x \right)=\dfrac{a{{\left( {{a}^{\dfrac{2}{3}}}+{{b}^{\dfrac{2}{3}}} \right)}^{\dfrac{3}{2}}}}{a}+\dfrac{a{{\left( {{a}^{\dfrac{2}{3}}}+{{b}^{\dfrac{2}{3}}} \right)}^{\dfrac{1}{2}}}}{{{a}^{\dfrac{1}{3}}}}\times \left( \dfrac{{{b}^{\dfrac{2}{3}}}}{{{a}^{\dfrac{2}{3}}}} \right)+\dfrac{b{{\left( {{a}^{\dfrac{2}{3}}}+{{b}^{\dfrac{2}{3}}} \right)}^{\dfrac{3}{2}}}}{b}+\dfrac{b{{\left( {{a}^{\dfrac{2}{3}}}+{{b}^{\dfrac{2}{3}}} \right)}^{\dfrac{1}{2}}}}{{{b}^{\dfrac{1}{3}}}}\times \left( \dfrac{{{a}^{\dfrac{2}{3}}}}{{{b}^{\dfrac{2}{3}}}} \right)$.
\[\Rightarrow {{f}^{''}}\left( x \right)={{\left( {{a}^{\dfrac{2}{3}}}+{{b}^{\dfrac{2}{3}}} \right)}^{\dfrac{3}{2}}}+{{b}^{\dfrac{2}{3}}}{{\left( {{a}^{\dfrac{2}{3}}}+{{b}^{\dfrac{2}{3}}} \right)}^{\dfrac{1}{2}}}+{{\left( {{a}^{\dfrac{2}{3}}}+{{b}^{\dfrac{2}{3}}} \right)}^{\dfrac{3}{2}}}+{{a}^{\dfrac{2}{3}}}{{\left( {{a}^{\dfrac{2}{3}}}+{{b}^{\dfrac{2}{3}}} \right)}^{\dfrac{1}{2}}}\].
Since the values of a and b are positive and ${{f}^{''}}\left( x \right)$ involves only addition and square roots which makes the value positive.
So, we get ${{f}^{''}}\left( x \right)>0$. Which means that the value has minimum when $\tan x$ is equal to ${{\left( \dfrac{a}{b} \right)}^{\dfrac{1}{3}}}$.
So, let us find the minimum value of $f\left( x \right)=a\sec x+b\operatorname{cosec}x$.
We have minimum value as $f\left( x \right)=a\left( \dfrac{\sqrt{{{a}^{\dfrac{2}{3}}}+{{b}^{\dfrac{2}{3}}}}}{{{a}^{\dfrac{1}{3}}}} \right)+b\left( \dfrac{\sqrt{{{b}^{\dfrac{2}{3}}}+{{a}^{\dfrac{2}{3}}}}}{{{b}^{\dfrac{1}{3}}}} \right)$.
$\Rightarrow f\left( x \right)={{a}^{\dfrac{2}{3}}}\sqrt{{{a}^{\dfrac{2}{3}}}+{{b}^{\dfrac{2}{3}}}}+{{b}^{\dfrac{2}{3}}}\sqrt{{{b}^{\dfrac{2}{3}}}+{{a}^{\dfrac{2}{3}}}}$.
$\Rightarrow f\left( x \right)=\left( {{a}^{\dfrac{2}{3}}}+{{b}^{\dfrac{2}{3}}} \right)\times \left( \sqrt{{{a}^{\dfrac{2}{3}}}+{{b}^{\dfrac{2}{3}}}} \right)$.
$\Rightarrow f\left( x \right)={{\left( {{a}^{\dfrac{2}{3}}}+{{b}^{\dfrac{2}{3}}} \right)}^{1+\dfrac{1}{2}}}$.
$\Rightarrow f\left( x \right)={{\left( {{a}^{\dfrac{2}{3}}}+{{b}^{\dfrac{2}{3}}} \right)}^{\dfrac{3}{2}}}$.
We have found the minimum value of $a\sec x+b\operatorname{cosec}x$, $0∴ The minimum value of $a\sec x+b\operatorname{cosec}x$, $0 < a < b$, $0 < x < \dfrac{\pi }{2}$ is ${{\left( {{a}^{\dfrac{2}{3}}}+{{b}^{\dfrac{2}{3}}} \right)}^{\dfrac{3}{2}}}$.
The correct option for the given problem is (c).

Note: We should know that the value of x we obtained from ${{f}^{'}}\left( x \right)=0$ may not always give a minimum or maxima. So, we need to check the value of ${{f}^{''}}\left( x \right)$ in order to check whether that gives maxima or minima. If we get ${{f}^{''}}\left( x \right)<0$, then x has local maxima and if we get ${{f}^{''}}\left( x \right)>0$, then x has local minima. If ${{f}^{''}}\left( x \right)=0$, then we need to differentiate again and follow the same process. If we get only one value of x while solving ${{f}^{'}}\left( x \right)=0$, it may give us the value of absolute minimum or maximum.