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

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.