Question

The real number $x$ when added to its inverse gives the maximum value of sum at $x$ equal to?
a) $1$
b) $- 1$
c) $2$
d) $- 2$

Answer 1

In order to find the maximum value of the sum of $x$ and its inverse, first we need to determine the inverse of $x$ and then we will use maxima minima properties. Letting the sum to be equal to $y$, we need to first evaluate $\dfrac{{dy}}{{dx}} = 0$ and obtain the critical values of $x$. After that, we need to find $\dfrac{{{d^2}y}}{{d{x^2}}}$ and check for what values of $x$, the value of $\dfrac{{{d^2}y}}{{d{x^2}}}$ is less than $0$. The values of $x$ for which $\dfrac{{{d^2}y}}{{d{x^2}}} < 0$ will give us the maximum value of $y$.

Complete step-by-step answer:
We first find the inverse of the real number $x$.
A number $b$ is said to be the inverse of number $a$ if $ab = 1$.
Let the inverse of $x$ be equal to $z$.
According to the definition, $xz = 1$
Reshuffling the terms, we get
$\Rightarrow z = \dfrac{1}{x}$
Hence, the inverse of $x$ is $\dfrac{1}{x}$.
Letting the sum of $x$ and its inverse to be equal to $y$, we get
$\Rightarrow y = x + \dfrac{1}{x}$
Now, we need to evaluate $\dfrac{{dy}}{{dx}} = 0$.
Let us first find $\dfrac{{dy}}{{dx}}$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {x + \dfrac{1}{x}} \right)$
Using $\dfrac{d}{{dx}}\left( {f\left( x \right) + g\left( x \right)} \right) = \dfrac{d}{{dx}}\left( {f\left( x \right)} \right) + \dfrac{d}{{dx}}\left( {g\left( x \right)} \right)$, we get
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{dx}}{{dx}} + \dfrac{d}{{dx}}\left( {\dfrac{1}{x}} \right)$
Writing $\dfrac{1}{x} = {x^{ - 1}}$, we get
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{dx}}{{dx}} + \dfrac{d}{{dx}}\left( {{x^{ - 1}}} \right)$
Now, using $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$, we get
$\Rightarrow \dfrac{{dy}}{{dx}} = 1\left( {{x^{1 - 1}}} \right) + \left( { - 1} \right){x^{ - 1 - 1}}$
Solving the powers of $x$, we get
$\Rightarrow \dfrac{{dy}}{{dx}} = {x^0} - {x^{ - 2}}$
We know, ${x^0} = 1$ and writing ${x^{ - 2}} = \dfrac{1}{{{x^2}}}$, we get
$\Rightarrow \dfrac{{dy}}{{dx}} = 1 - \dfrac{1}{{{x^2}}} - - - - - - (1)$
Now, we need to evaluate $\dfrac{{dy}}{{dx}} = 0$
Hence, calculating this, we get
$\Rightarrow \dfrac{{dy}}{{dx}} = 1 - \dfrac{1}{{{x^2}}} = 0$
$\Rightarrow 1 - \dfrac{1}{{{x^2}}} = 0$
Reshuffling the terms, we get
$\Rightarrow 1 = \dfrac{1}{{{x^2}}}$
Now, By cross multiplying ,we get
$\Rightarrow {x^2} = 1$
Hence, we get
$\Rightarrow x = \pm 1$
So, solving $\dfrac{{dy}}{{dx}} = 0$, we get $x = 1$ or $x = - 1$.
Now, we need to check whether $\dfrac{{{d^2}y}}{{d{x^2}}} > 0$ or $\dfrac{{{d^2}y}}{{d{x^2}}} < 0$ at $x = 1$ and $x = - 1$
So, we now, calculate $\dfrac{{{d^2}y}}{{d{x^2}}}$
We know, $\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{d}{{dx}}\left( {\dfrac{{dy}}{{dx}}} \right)$. Using (1), we get
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{d}{{dx}}\left( {\dfrac{{dy}}{{dx}}} \right) = \dfrac{d}{{dx}}\left( {1 - \dfrac{1}{{{x^2}}}} \right)$
Writing $\dfrac{1}{{{x^2}}} = {x^{ - 2}}$, we get
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{d}{{dx}}\left( {1 - {x^{ - 2}}} \right)$
Now, using $\dfrac{d}{{dx}}\left( {f\left( x \right) - g\left( x \right)} \right) = \dfrac{d}{{dx}}\left( {f\left( x \right)} \right) - \dfrac{d}{{dx}}\left( {g\left( x \right)} \right)$, $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$ and $\dfrac{d}{{dx}}\left( c \right) = 0$, where $c$ is a constant term, we get
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{d}{{dx}}\left( 1 \right) - \dfrac{d}{{dx}}\left( {{x^{ - 2}}} \right)$
Here, $1$ is a constant term. So, we have
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = 0 - \left( { - 2} \right){x^{ - 2 - 1}}$
Solving the powers of $x$, we get
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = 0 - \left( { - 2} \right){x^{ - 3}}$
Now, solving the terms, we get
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = 2{x^{ - 3}}$
Now, writing ${x^{ - 3}} = \dfrac{1}{{{x^3}}}$, we get
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{2}{{{x^3}}}$
Hence, we get $\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{2}{{{x^3}}}$
Now, we need to check the value of $\dfrac{{{d^2}y}}{{d{x^2}}}$ at $x = 1$ and $x = - 1$
First checking $\dfrac{{{d^2}y}}{{d{x^2}}}$ at $x = 1$
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}}{|_{x = 1}} = \dfrac{2}{{{{\left( 1 \right)}^3}}}$
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}}{|_{x = 1}} = \dfrac{2}{1} = 2$
Hence, we get
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}}{|_{x = 1}} = 2 > 0$
So, value of $\dfrac{{{d^2}y}}{{d{x^2}}}$ at $x = 1$ is greater than zero.
Now, checking $\dfrac{{{d^2}y}}{{d{x^2}}}$ at $x = - 1$, we get
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}}{|_{x = - 1}} = \dfrac{2}{{{{\left( { - 1} \right)}^3}}}$
Now, solving the denominator, we get
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}}{|_{x = - 1}} = \dfrac{2}{{ - 1}}$
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}}{|_{x = - 1}} = - 2 < 0$
Hence, we get $\dfrac{{{d^2}y}}{{d{x^2}}}$ at $x = - 1$ is less than zero.
We know, $a$ is said to be the maximum value of $y$ if $\dfrac{{{d^2}y}}{{d{x^2}}} < 0$ at $x = a$.
Hence, we get the maximum value of $y$ at $x = - 1$.
So, the maximum value of the sum of a real number $x$ and its inverse is at $x = - 1$.

So, the correct answer is “Option A”.

Note: Here, to solve this question we have used a second derivative test of maxima and minima. We can also solve this question using the first derivative test of maxima and minima. For that, we need to check whether the sign of the first derivative of the function changes from negative to positive or positive to negative when we take one value smaller than the critical value and one value greater than the critical value. If it changes from positive to negative, then the critical point is the point of maxima and if it changes from negative to positive, then the critical point is the point of minima.