SATHEE: Application of Derivatives 4 Question 57

Application of Derivatives 4 Question 57

####60. If $a x^{2} + b / x \geq c$ for all positive $x$ where $a > 0$ and $b > 0$ , then show that $27 a b^{2} \geq 4 c^{3}$ .

$(1982, 2 M)$

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Answer:

Correct Answer: 60. (2)

Solution:

Given, $a x^{2} + \frac{b}{x} \geq c, \forall x > 0; a, b > 0$

Let

$f (x) = a x^{2} + \frac{b}{x} - c$

$∴ f^{'} (x) = 2 a x - \frac{b}{x^{2}} = \frac{2 a x^{3} - b}{x^{2}}$

$\Rightarrow f^{''} (x) = 2 a + \frac{2 b}{x^{3}} > 0$ [since, $a, b$ are all positive]

Now, put $f^{'} (x) = 0 \Rightarrow x = {\frac{b}{2 a}}^{1 / 3} > 0 [∵ a, b > 0]$

At $x = {\frac{b}{2 a}}^{1 / 3}, f^{''} (x) = + ve$

$\Rightarrow f (x)$ has minimum at $x = {\frac{b}{2 a}}^{1 / 3}$ .

$\begin{aligned} and f {\frac{b}{2 a}}^{1 / 3} = & a {\frac{b}{2 a}}^{2 / 3} + \frac{b}{(b / 2 a)^{1 / 3}} - c \geq 0 \\ = \frac{2 a}{b}^{1 / 3} \cdot \frac{3 b}{2} - c \geq 0 \\ \Rightarrow & \frac{2 a}{b}^{1 / 3} \cdot \frac{3 b}{2} \geq c \end{aligned}$

On cubing both sides, we get

$\begin{aligned} \frac{2 a}{b} \cdot \frac{27 b^{3}}{8} \geq c^{3} \\ \Rightarrow & 27 a b^{2} \geq 4 c^{3} \end{aligned}$

Application of Derivatives 4 Question 57

Answer:

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Application of Derivatives 4 Question 57

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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