Question

The least value of $\alpha \in R$ for which $4\alpha {{x}^{2}}+\dfrac{1}{x}\ge 1$ for all x>0 is

& A.\dfrac{1}{64} \\\ & B.\dfrac{1}{32} \\\ & C.\dfrac{1}{27} \\\ & D.\dfrac{1}{25} \\\ \end{aligned}$$

Answer 1

For finding the least value of $\alpha \in R$ in the given function, we will first suppose the given function as f(x). Then we will use the first derivative test to calculate the value of x at which function changes its nature. After that, we will substitute the value of x in the function $f\left( x \right)\ge 1$ and find the range of $\alpha$. From this range, we will be able to find the minimum (least) value of $\alpha$. For the first derivative test, we will find f'(x) and then put it equal to 0 to find the value of x.

Complete step-by-step solution
Here we are given the function as $4\alpha {{x}^{2}}+\dfrac{1}{x}\ge 1$ where x>0. We need to find the least value of $\alpha \in R$. For this, let us first suppose that the function $f\left( x \right)=4\alpha {{x}^{2}}+\dfrac{1}{x}$.
Let us now use the first derivative test to find the value of x where this function changes its nature (changes from going higher to satisfy going lower or vice versa).
For this, let us find the derivative of f(x): $f\left( x \right)=4\alpha {{x}^{2}}+\dfrac{1}{x}$.
Differentiating both sides with respect to x, we get:

& \dfrac{d}{dx}f\left( x \right)=\dfrac{d}{dx}\left( 4\alpha {{x}^{2}}+\dfrac{1}{x} \right) \\\ & \Rightarrow f'\left( x \right)=\dfrac{d}{dx}\left( 4\alpha {{x}^{2}}+\dfrac{1}{x} \right)+\dfrac{d}{dx}\left( \dfrac{1}{x} \right) \\\ & \Rightarrow f'\left( x \right)=4\alpha \dfrac{d{{x}^{2}}}{dx}+\dfrac{d}{dx}\left( \dfrac{1}{x} \right) \\\ \end{aligned}$$ As we know, $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ so we get: $$\begin{aligned} & \Rightarrow f'\left( x \right)=4\alpha \left( 2{{x}^{2-1}} \right)+\left( -1 \right){{x}^{-1-1}} \\\ & \Rightarrow f'\left( x \right)=8\alpha x-{{x}^{-2}} \\\ & \Rightarrow f'\left( x \right)=8\alpha x-\dfrac{1}{{{x}^{2}}} \\\ \end{aligned}$$ Now putting f'(x) = 0 will give us value of x, so we get: $$\Rightarrow 8\alpha x-\dfrac{1}{{{x}^{2}}}=0$$ Taking $\dfrac{1}{{{x}^{2}}}$ to other side we get: $$\begin{aligned} & \Rightarrow 8\alpha x=\dfrac{1}{{{x}^{2}}} \\\ & \Rightarrow 8\alpha =\dfrac{1}{{{x}^{3}}} \\\ \end{aligned}$$ Taking reciprocal on both sides we get: $$\Rightarrow {{x}^{3}}=\dfrac{1}{8\alpha }$$ Taking cube root on both sides we get: $$\begin{aligned} & \Rightarrow {{\left( {{x}^{3}} \right)}^{\dfrac{1}{3}}}={{\left( \dfrac{1}{8\alpha } \right)}^{\dfrac{1}{3}}} \\\ & \Rightarrow x=\dfrac{1}{{{\left( 8\alpha \right)}^{\dfrac{1}{3}}}} \\\ \end{aligned}$$ We know that ${{2}^{3}}=8$ so ${{8}^{\dfrac{1}{3}}}=2$ we get: $$\Rightarrow x=\dfrac{1}{2{{\left( \alpha \right)}^{\dfrac{1}{3}}}}$$ Hence our function changes its nature at the point $x=\dfrac{1}{2{{\left( \alpha \right)}^{\dfrac{1}{3}}}}$. We are given $f\left( x \right)\ge 1$ so we get: $4\alpha {{x}^{2}}+\dfrac{1}{x}\ge 1$. Let us put the value of x in f(x) we will get a range of $\alpha $. Hence we get: $$\begin{aligned} & \Rightarrow 4\alpha {{\left( \dfrac{1}{2{{\left( \alpha \right)}^{\dfrac{1}{3}}}} \right)}^{2}}+\dfrac{1}{\left( \dfrac{1}{2{{\left( \alpha \right)}^{\dfrac{1}{3}}}} \right)}\ge 1 \\\ & \Rightarrow 4\alpha \left( \dfrac{1}{{{2}^{2}}{{\left( \alpha \right)}^{\dfrac{2}{3}}}} \right)+2{{\left( \alpha \right)}^{\dfrac{1}{3}}}\ge 1 \\\ & \Rightarrow \dfrac{4\alpha }{4{{\left( \alpha \right)}^{\dfrac{2}{3}}}}+2{{\left( \alpha \right)}^{\dfrac{1}{3}}}\ge 1 \\\ & \Rightarrow \dfrac{\alpha }{{{\alpha }^{\dfrac{2}{3}}}}+2{{\alpha }^{\dfrac{1}{3}}}\ge 1 \\\ \end{aligned}$$ We know that, $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$ so we get: $$\begin{aligned} & \Rightarrow {{\alpha }^{1-\dfrac{2}{3}}}+2{{\alpha }^{\dfrac{1}{3}}}\ge 1 \\\ & \Rightarrow {{\alpha }^{\dfrac{1}{3}}}+2{{\alpha }^{\dfrac{1}{3}}}\ge 1 \\\ & \Rightarrow 3{{\alpha }^{\dfrac{1}{3}}}\ge 1 \\\ \end{aligned}$$ Taking 3 to the other side, we get: $$\Rightarrow {{\alpha }^{\dfrac{1}{3}}}\ge \dfrac{1}{3}$$ Taking cube on both sides we get: $$\Rightarrow {{\left( {{\alpha }^{\dfrac{1}{3}}} \right)}^{3}}\ge {{\left( \dfrac{1}{3} \right)}^{3}}$$ We know $3\times 3\times 3=27$ so we get: $$\Rightarrow \alpha \ge \dfrac{1}{27}$$ So we can see that $\alpha $ is always greater than or equal to $\dfrac{1}{27}$. We can say that $\dfrac{1}{27}$ will be the least value of $\alpha $. Hence the least value of $\alpha $ is $\dfrac{1}{27}$. **Hence option C is the correct answer.** **Note:** Students should take care of the signs while solving inequalities. Take care while taking cubes or cube roots. f'(x) = 0 always tells us the value of x when f(x) changes its nature. This nature could be going from higher to lower or vice versa.