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

Correct answer: C. (1)/(27) f(x) = 4α x ^ 2 + (1)/(x); x > 0 ⇒ f ^ ' (x) = 8 - 1 x ^ 2 = 8α x ^ 3 - 1 x ^ 2 ⇒ f(x) attains its minimum at x = ((1)/(8α)) ^ (1)/(3) f\ ((1)/(8α))(1)/(3)\ = 1 ⇒ 4α ((1)/(8α)) ^ (2)/(3) + (8α) ^ (1)/(3) = 1 ⇒ 3 α ^ (1)/(3) = 1 ⇒ α = (1)/(27)

JEE Advanced 2016Paper 1

Exam: JEE Advanced
Year: 2016
Paper: Paper 1
Subject: Mathematics
Chapter: Application of Derivatives

JEE Advanced 2016 · Paper 1 · Mathematics · Application of Derivatives

The least value of α ∈ R for which, 4α x ^ 2 + (1)/(x) ≥ 1, for all...

The least value of

\alpha \in R

for which,

4\alpha{x}^{2} + \dfrac{1}{x} \ge 1,

for all

x > 0,

Options

A $\dfrac{1}{64}$
B $\dfrac{1}{32}$
C $\dfrac{1}{27}$
D $\dfrac{1}{25}$

Correct answer

C. $\dfrac{1}{27}$

Step-by-step solution

f\left(x\right) = 4\alpha{x}^{2} + \dfrac{1}{x}; x > 0

⇒ {f}^{'}\left(x\right) = 8\alphax - \dfrac{1}{{x}^{2}} = \dfrac{8\alpha{x}^{3} - 1}{{x}^{2}}

⇒ f\left(x\right)

attains its minimum at

x = {\left(\dfrac{1}{8\alpha}\right)}^{\dfrac{1}{3}}

f\left\{\left(\dfrac{1}{8\alpha}\right)\dfrac{1}{3}\right\} = 1

⇒

4\alpha{\left(\dfrac{1}{8\alpha}\right)}^{\dfrac{2}{3}} + {\left(8\alpha\right)}^{\dfrac{1}{3}} = 1

⇒

3{\alpha}^{\dfrac{1}{3}} = 1 ⇒ \alpha = \dfrac{1}{27}

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