Question

The least value of \[a \in R\] for which \[4a{x^2} + \dfrac{1}{x} \geqslant 1\] for all \[x \geqslant 0\]is
A.\[\dfrac{1}{{64}}\]
B. \[\dfrac{1}{{32}}\]
C. \[\dfrac{1}{{27}}\]
D. \[\dfrac{1}{{25}}\]

Answer 1

Hint: Shift the reciprocal to RHS of the inequality and calculate the value of 4a in terms of x. Differentiate the function obtained in RHS of the inequality and substitute back in the inequality. Use the concept that if a constant is greater than the given function, then it will be greater than the differentiation of the function as well.
* Differentiation of \[{x^n}\]with respect to x is given by \[\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}\]

Complete step by step answer:
We are given the inequality \[4a{x^2} + \dfrac{1}{x} \geqslant 1\]
Shift \[\dfrac{1}{x}\] to RHS of the inequality
\[ \Rightarrow 4a{x^2} \geqslant 1 - \dfrac{1}{x}\]
Take LCM in RHS of the inequality
\[ \Rightarrow 4a{x^2} \geqslant \dfrac{{x - 1}}{x}\]
Divide both sides of the equation by \[{x^2}\]
\[ \Rightarrow \dfrac{{4a{x^2}}}{{{x^2}}} \geqslant \dfrac{{x - 1}}{{x.{x^2}}}\]
Cancel same factors from numerator and denominator in LHS of the equation and collect powers in denominator of RHS
\[ \Rightarrow 4a \geqslant \dfrac{{x - 1}}{{{x^3}}}\]
Separate the terms of fraction in RHS of the equation
\[ \Rightarrow 4a \geqslant \dfrac{x}{{{x^3}}} - \dfrac{1}{{{x^3}}}\]
Cancel same factors from numerator and denominator in RHS of the equation
\[ \Rightarrow 4a \geqslant \dfrac{1}{{{x^2}}} - \dfrac{1}{{{x^3}}}\]
Divide both sides by 4,
\[ \Rightarrow \dfrac{{4a}}{4} \geqslant \dfrac{1}{4}\left( {\dfrac{1}{{{x^2}}} - \dfrac{1}{{{x^3}}}} \right)\]
Cancel same factors from numerator and denominator
\[ \Rightarrow a \geqslant \dfrac{1}{4}\left( {\dfrac{1}{{{x^2}}} - \dfrac{1}{{{x^3}}}} \right)\] … (1)
Now we differentiate RHS of the equation
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{1}{{{x^2}}} - \dfrac{1}{{{x^3}}}} \right) = \dfrac{{ - 2}}{{{x^3}}} + \dfrac{3}{{{x^4}}}\]
Equate the differentiation to zero
\[ \Rightarrow \dfrac{{ - 2}}{{{x^3}}} + \dfrac{3}{{{x^4}}} = 0\]
Shift one term to RHS
\[ \Rightarrow - \dfrac{2}{{{x^3}}} = - \dfrac{3}{{{x^4}}}\]
Cancel same terms from both sides of the equation
\[ \Rightarrow \dfrac{2}{1} = \dfrac{3}{x}\]
Cross multiply the values to obtain value of x
\[ \Rightarrow x = \dfrac{3}{2}\]
Substitute the value of x in equation (1)
\[ \Rightarrow a \geqslant \dfrac{1}{4}\left( {\dfrac{1}{{{{\left( {\dfrac{3}{2}} \right)}^2}}} - \dfrac{1}{{{{\left( {\dfrac{3}{2}} \right)}^3}}}} \right)\]
\[ \Rightarrow a \geqslant \dfrac{1}{4}\left( {\dfrac{4}{9} - \dfrac{8}{{27}}} \right)\]
Take LCM in RHS of the equation
\[ \Rightarrow a \geqslant \dfrac{1}{4}\left( {\dfrac{{12 - 8}}{{27}}} \right)\]
\[ \Rightarrow a \geqslant \dfrac{1}{4}\left( {\dfrac{4}{{27}}} \right)\]
Cancel same factors from numerator and denominator of fraction in RHS
\[ \Rightarrow a \geqslant \dfrac{1}{{27}}\]
\[\therefore \]The least value of ‘a’ is \[\dfrac{1}{{27}}\]

\[\therefore \]Option C is correct.

Note: Many students make the mistake of differentiating the terms of fraction as they have variable in denominator, keep in mind we can write the variable in numerator with negative power equal to variable in denominator. Also, while shifting values from one side to another side, always change signs from positive to negative and vice–versa.