Let f(x) = x^2 + frac{1}{x^2} and g(x) = x - | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given $h(x) = \frac{x^2 + \frac{1}{x^2}}{x - \frac{1}{x}}$.We can rewrite the numerator as $(x - \frac{1}{x})^2 + 2$.So,$h(x) = \frac{(x - \frac{1

Vedclass · Accepted Answer

$2\sqrt{2}$

Vedclass · Answer

(C) Given $h(x) = \frac{x^2 + \frac{1}{x^2}}{x - \frac{1}{x}}$.We can rewrite the numerator as $(x - \frac{1}{x})^2 + 2$.So,$h(x) = \frac{(x - \frac{1}{x})^2 + 2}{x - \frac{1}{x}} = (x - \frac{1}{x}) + \frac{2}{x - \frac{1}{x}}$.Let $t = x - \frac{1}{x}$. Since $x \in R - \{-1, 1, 0\}$,$t$ can take any real value except $0$.Then $h(t) = t + \frac{2}{t}$.For $t > 0$,by the $AM$-$GM$ inequality,$t + \frac{2}{t} \ge 2\sqrt{t \cdot \frac{2}{t}} = 2\sqrt{2}$.The equality holds when $t = \frac{2}{t}$,i.e.,$t^2 = 2$,so $t = \sqrt{2}$ (since $t > 0$).For $t  0$. Then $h(t) = -u - \frac{2}{u} = -(u + \frac{2}{u}) \le -2\sqrt{2}$.Thus,the local minimum value of $h(x)$ is $2\sqrt{2}$.

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

Similar Questions

The maximum volume (in cubic units) of the cylinder which can be inscribed in a sphere of radius $12$ units is

Suppose the cubic $x^3 - px + q$ has three distinct real roots where $p > 0$ and $q > 0$. Then which one of the following holds?

The maximum volume (in $m^3$) of the right circular cone having slant height $3 \, m$ is

The height of a right circular cylinder of maximum volume inscribed in a sphere of radius $3$ is

If $y = a \log x + bx^2 + x$ has extreme values at $x = 1$ and $x = 2$,then $(a, b) = \dots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module