Let f x = x 2 + 1 x 2 and g x = x - 1 x , x ∈ R - - 1 , 0 , 1 . If h x = f x g x , then the local minimum value of h x is:

f x = x 2 + 1 x 2 = x - 1 x 2 + 2 g x = x - 1 x x ∈ R - - 1 , 0 , 1 Let x - 1 x = t h x = f x g x = t 2 + 2 t = t + 2 t t ∈ R - 0 h x = t + 2 t A M ≥ G M Minimum of h x ⇒ t + 2 t 2 ≥ t × 2 t t + 2 t ≥ 2 2 Local minimum = 2 2

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Mathematics - Application of Derivative Question with Solution | TestHub

MathematicsApplication of DerivativeMaxima-MinimaMedium2 minPYQ_2018

MathematicsMediumsingle choice

Let $f (x) = x^{2} + \frac{1}{x^{2}}$ and $g (x) = x - \frac{1}{x}, x \in R - \{- 1, 0, 1\} .$ If $h (x) = \frac{f (x)}{g (x)}$ , then the local minimum value of $h (x)$ is:

Options:

Answer:

Solution:

$f (x) = x^{2} + \frac{1}{x^{2}} = {(x - \frac{1}{x})}^{2} + 2$

$g (x) = x - \frac{1}{x}$

$x \in R - \{- 1, 0, 1\}$

Let $x - \frac{1}{x} = t$

$h (x) = \frac{f (x)}{g (x)} = \frac{t^{2} + 2}{t} = t + \frac{2}{t}$

$t \in R - \{0\}$

$h (x) = t + \frac{2}{t}$

$A M \geq G M$

Minimum of $h (x) \Rightarrow \frac{t + \frac{2}{t}}{2} \geq \sqrt{t \times \frac{2}{t}}$

$t + \frac{2}{t} \geq 2 \sqrt{2}$

Local minimum $= 2 \sqrt{2}$

Stream:JEESubject:MathematicsTopic:Application of DerivativeSubtopic:Maxima-Minima

⏱ 2mℹ️ Source: PYQ_2018

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