# Let f(x) = $\alpha {{x}^{2}}-2+\dfrac{1}{x}$ where $\alpha $ is a real constant. The smallest $\alpha $ for which f(x)$\ge $0 for all x>0 is ?

(a) $\dfrac{{{2}^{2}}}{{{3}^{3}}}$

(b) $\dfrac{{{2}^{3}}}{{{3}^{3}}}$

(c) $\dfrac{{{2}^{4}}}{{{3}^{3}}}$

(d) $\dfrac{{{2}^{5}}}{{{3}^{3}}}$

Answer

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Hint: To solve the above problem, we need to be aware about the basic concepts of the extremum (that is, minimum and maximum) of a function. We will use the principle of derivatives to solve this function. We will use the property that for minimum, we have,

f’(x) = 0 and f’’(x) > 0 [Here, f’(x) is same as $\dfrac{d}{dx}\left( f(x) \right)$ and f’’(x) is same as $\dfrac{{{d}^{2}}}{d{{x}^{2}}}\left( f(x) \right)$.

Complete step by step answer:

Now, before solving we try to understand the basics about extremum points. Basically, an extremum is a point of a function at which it has the highest (maximum) or lowest (minimum) value. In general, these points occur where there is a change in sign of the slope of the graph at that point. For example, we take a simple case of f(x) = ${{x}^{2}}$. Here, we see that the sign of the slope changes when we cross x=0. Thus, x = 0 is an extremum point. Now, we try to solve the problem in hand keeping above points in mind.

We have, f(x) = $\alpha {{x}^{2}}-2+\dfrac{1}{x}$. Now, to find extremum point, in general, we perform f’(x) = 0 [f’(x) is same as $\dfrac{d}{dx}\left( f(x) \right)$]

We know that, $\dfrac{d({{x}^{2}})}{dx}=2x,\text{ }\dfrac{d(\text{constant})}{dx}=0,\text{ }\dfrac{d}{dx}\left( \dfrac{1}{x} \right)=-\dfrac{1}{{{x}^{2}}}$, thus, we get,

f’(x) = $2\alpha x-\dfrac{1}{{{x}^{2}}}$ -- (1)

We equate this to zero, we get,

$2\alpha x-\dfrac{1}{{{x}^{2}}}$=0

$\begin{align}

& 2\alpha x=\dfrac{1}{{{x}^{2}}} \\

& {{x}^{3}}=\dfrac{1}{2\alpha } \\

& x={{\left( \dfrac{1}{2\alpha } \right)}^{\dfrac{1}{3}}} \\

\end{align}$

Now, we want this point to be minimum since, we have to find the smallest $\alpha $, we have another condition,

f’’(x) > 0

Using $\dfrac{d}{dx}\left( \dfrac{1}{{{x}^{2}}} \right)=-\dfrac{2}{{{x}^{3}}}$, thus doing derivative of (1), we have,

f’’(x) = 2$\alpha $- ($-\dfrac{2}{{{x}^{3}}}$)

f’’(x) = 2$\alpha $+$\dfrac{2}{{{x}^{3}}}$

Now, in the question, for f’’(x) to be greater than zero, we have,

2$\alpha $+$\dfrac{2}{{{x}^{3}}}$> 0

$\alpha $> $-\dfrac{1}{{{x}^{3}}}$ - (2)

Now, we try to satisfy, f(x)$\ge $0 for $x={{\left( \dfrac{1}{2\alpha } \right)}^{\dfrac{1}{3}}}$,

thus, we have,

f${{\left( \dfrac{1}{2\alpha } \right)}^{\dfrac{1}{3}}}$$\ge $0

\[\begin{align}

& \alpha {{\left( \dfrac{1}{2\alpha } \right)}^{\dfrac{2}{3}}}-2+{{\left( \dfrac{1}{2\alpha }

\right)}^{-\dfrac{1}{3}}}\ge 0 \\

& {{\alpha }^{\dfrac{1}{3}}}{{2}^{-\dfrac{2}{3}}}-2+{{2}^{\dfrac{1}{3}}}{{\alpha }^{\dfrac{1}{3}}}\ge 0 \\

& {{\alpha }^{\dfrac{1}{3}}}\left( \dfrac{1+2}{{{2}^{\dfrac{2}{3}}}} \right)-2\ge 0 \\

\end{align}\]

\[\begin{align}

& {{\alpha }^{\dfrac{1}{3}}}\left( \dfrac{3}{{{2}^{\dfrac{2}{3}}}} \right)\ge 2 \\

& {{\alpha }^{\dfrac{1}{3}}}\ge \dfrac{{{2}^{\dfrac{5}{3}}}}{3} \\

& \alpha \ge \dfrac{{{2}^{5}}}{{{3}^{3}}} \\

\end{align}\]

Thus, the minimum value of \[\alpha =\dfrac{{{2}^{5}}}{{{3}^{3}}}\].

Hence, the correct answer (d).

Note: We found the minimum of the function, since we had to find the minimum value of $\alpha $ for f(x) to be just greater than or equal to zero. Further, it is important to keep in mind the range of x while arriving at the answer. For example, in the question, it is given that x>0, thus, if we get an extremum point which is negative, we reject that point.

f’(x) = 0 and f’’(x) > 0 [Here, f’(x) is same as $\dfrac{d}{dx}\left( f(x) \right)$ and f’’(x) is same as $\dfrac{{{d}^{2}}}{d{{x}^{2}}}\left( f(x) \right)$.

Complete step by step answer:

Now, before solving we try to understand the basics about extremum points. Basically, an extremum is a point of a function at which it has the highest (maximum) or lowest (minimum) value. In general, these points occur where there is a change in sign of the slope of the graph at that point. For example, we take a simple case of f(x) = ${{x}^{2}}$. Here, we see that the sign of the slope changes when we cross x=0. Thus, x = 0 is an extremum point. Now, we try to solve the problem in hand keeping above points in mind.

We have, f(x) = $\alpha {{x}^{2}}-2+\dfrac{1}{x}$. Now, to find extremum point, in general, we perform f’(x) = 0 [f’(x) is same as $\dfrac{d}{dx}\left( f(x) \right)$]

We know that, $\dfrac{d({{x}^{2}})}{dx}=2x,\text{ }\dfrac{d(\text{constant})}{dx}=0,\text{ }\dfrac{d}{dx}\left( \dfrac{1}{x} \right)=-\dfrac{1}{{{x}^{2}}}$, thus, we get,

f’(x) = $2\alpha x-\dfrac{1}{{{x}^{2}}}$ -- (1)

We equate this to zero, we get,

$2\alpha x-\dfrac{1}{{{x}^{2}}}$=0

$\begin{align}

& 2\alpha x=\dfrac{1}{{{x}^{2}}} \\

& {{x}^{3}}=\dfrac{1}{2\alpha } \\

& x={{\left( \dfrac{1}{2\alpha } \right)}^{\dfrac{1}{3}}} \\

\end{align}$

Now, we want this point to be minimum since, we have to find the smallest $\alpha $, we have another condition,

f’’(x) > 0

Using $\dfrac{d}{dx}\left( \dfrac{1}{{{x}^{2}}} \right)=-\dfrac{2}{{{x}^{3}}}$, thus doing derivative of (1), we have,

f’’(x) = 2$\alpha $- ($-\dfrac{2}{{{x}^{3}}}$)

f’’(x) = 2$\alpha $+$\dfrac{2}{{{x}^{3}}}$

Now, in the question, for f’’(x) to be greater than zero, we have,

2$\alpha $+$\dfrac{2}{{{x}^{3}}}$> 0

$\alpha $> $-\dfrac{1}{{{x}^{3}}}$ - (2)

Now, we try to satisfy, f(x)$\ge $0 for $x={{\left( \dfrac{1}{2\alpha } \right)}^{\dfrac{1}{3}}}$,

thus, we have,

f${{\left( \dfrac{1}{2\alpha } \right)}^{\dfrac{1}{3}}}$$\ge $0

\[\begin{align}

& \alpha {{\left( \dfrac{1}{2\alpha } \right)}^{\dfrac{2}{3}}}-2+{{\left( \dfrac{1}{2\alpha }

\right)}^{-\dfrac{1}{3}}}\ge 0 \\

& {{\alpha }^{\dfrac{1}{3}}}{{2}^{-\dfrac{2}{3}}}-2+{{2}^{\dfrac{1}{3}}}{{\alpha }^{\dfrac{1}{3}}}\ge 0 \\

& {{\alpha }^{\dfrac{1}{3}}}\left( \dfrac{1+2}{{{2}^{\dfrac{2}{3}}}} \right)-2\ge 0 \\

\end{align}\]

\[\begin{align}

& {{\alpha }^{\dfrac{1}{3}}}\left( \dfrac{3}{{{2}^{\dfrac{2}{3}}}} \right)\ge 2 \\

& {{\alpha }^{\dfrac{1}{3}}}\ge \dfrac{{{2}^{\dfrac{5}{3}}}}{3} \\

& \alpha \ge \dfrac{{{2}^{5}}}{{{3}^{3}}} \\

\end{align}\]

Thus, the minimum value of \[\alpha =\dfrac{{{2}^{5}}}{{{3}^{3}}}\].

Hence, the correct answer (d).

Note: We found the minimum of the function, since we had to find the minimum value of $\alpha $ for f(x) to be just greater than or equal to zero. Further, it is important to keep in mind the range of x while arriving at the answer. For example, in the question, it is given that x>0, thus, if we get an extremum point which is negative, we reject that point.

Last updated date: 02nd Oct 2023

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