The least value of α∈R for which, 4αx^2+1/x ≥1, for all x>0, is

Correct answer: (C) 1/27. fx=4αx^2+1/x;x>0 ⇒f^(')x=8αx-1x^2=(8αx^3-1x^2⇒fx attains its minimum at x=1^(8α13)f18α^13=1⇒ 4α1^(8α23)+8α^13=1⇒ 3α^13=1 ⇒ α=1)/27

Q20 / 54 · JEE Advanced 2016 (Paper 1) · Mathematics

The least value of $α \in R$ for which, $4 α x^{2} + \frac{1}{x} \geq 1,$ for all $x > 0,$ is

A
$\frac{1}{64}$
B
$\frac{1}{32}$
Correct answer
C
$\frac{1}{27}$
D
$\frac{1}{25}$

Solution

f (x) = 4 α x^{2} + \frac{1}{x}; x > 0

\Rightarrow f^{'} (x) = 8 α x - \frac{1}{x^{2}} = \frac{8 α x^{3} - 1}{x^{2}}

\Rightarrow f (x)

attains its minimum at

x = {(\frac{1}{8 α})}^{\frac{1}{3}}

f \{{(\frac{1}{8 α})}^{\frac{1}{3}}\} = 1

\Rightarrow

4 α {(\frac{1}{8 α})}^{\frac{2}{3}} + {(8 α)}^{\frac{1}{3}} = 1

\Rightarrow

3 α^{\frac{1}{3}} = 1 \Rightarrow α = \frac{1}{27}

Hence, the correct answer is option: 'C'.

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The least value of α∈R for which, 4αx2+1x ≥1, for all x>0, is

Solution

Relevant Questions

The least value of $α \in R$ for which, $4 α x^{2} + \frac{1}{x} \geq 1,$ for all $x > 0,$ is