Question

If the expression $\left[ mx-1+\dfrac{1}{x} \right]$ is non negative for all positive x. Find the minimum value of m such that the condition is always true?

Answer 1

Hint: First we will take LCM and form a quadratic expression out of the given expression. Now we know that for a quadratic equation to be always positive Discriminant D of the expression must be < 0. And for the equation to be 0 without being negative we want D = 0. Hence with these two equations we will find the required condition on m.

Complete step-by-step answer:
Now first consider the given expression $\left[ mx-1+\dfrac{1}{x} \right]$
Taking LCM we get
$\left[ mx-1+\dfrac{1}{x} \right]=\left[ \dfrac{m{{x}^{2}}-x+1}{x} \right]$
Now we want $\left[ mx-1+\dfrac{1}{x} \right]$ to be non-negative for all positive x hence we want $\left[ \dfrac{m{{x}^{2}}-x+1}{x} \right]$ to be non-negative for all positive x
Now we will consider the two cases separately first being the equation is positive and second being the equation is 0.
Now since x is positive we have a denominator of $\left[ \dfrac{m{{x}^{2}}-x+1}{x} \right]$ is always positive.
Hence for the expression to be positive we want the numerator also to be positive.
Now consider $m{{x}^{2}}-x+1>0$
Now we know that a quadratic equation $a{{x}^{2}}+bx+c=0$ is always greater than 0 if the discriminant is always less than 0 which means ${{b}^{2}}-4ac<0$ .
Hence we get for equation $m{{x}^{2}}-x+1>0$

\[\begin{align}
  & {{\left( -1 \right)}^{2}}-4\left( m \right)\left( 1 \right)<0 \\
 & \Rightarrow 1-4m < 0 \\
 & \Rightarrow 1 < 4m \\
 & \Rightarrow \dfrac{1}{4} < m \\
\end{align}\]

Now for m greater than $\dfrac{1}{4}$ we will always have positive value for $\left[ \dfrac{m{{x}^{2}}-x+1}{x} \right]..................\left( 2 \right)$ .
Now let us check the condition of m $\left[ \dfrac{m{{x}^{2}}-x+1}{x} \right]$ is 0
Now Note that for $\left[ \dfrac{m{{x}^{2}}-x+1}{x} \right]$ to be 0 we need $m{{x}^{2}}-x-1=0$
Now if we have two roots then the quadratic will also take negative values. Hence the condition on roots is that it can have maximum 1 root. Which means we need conditions for common roots.
Hence we have
$\begin{align}
  & {{\left( -1 \right)}^{2}}-4\left( 1 \right)\left( m \right)=0 \\
 & 1=4m \\
 & m=\dfrac{1}{4}.................................\left( 2 \right) \\
\end{align}$
Hence for non-negative solutions, from equation (1) and (2) we get.
$m\ge \dfrac{1}{4}$ .

Note: Non negative means positive or zero. Now while considering the case for the equation to be 0 we need to take care of few things
Now we know that if there are two roots the equation will not be always positive.
Now even for common roots we will have two conditions. Either the expression lies completely above x-axis or completely below x-axis. This depends on the coefficient of ${{x}^{2}}$ if the coefficient of ${{x}^{2}}$ is positive then the graph is always greater than equal to 0. Similarly if the coefficient of ${{x}^{2}}$ is less than 0 then the graph is downwards facing and is always less than equal to 0. Since here we have m > 0 $\left( m>\dfrac{1}{4} \right)$ in our conditions we could easily proceed.