Question

please solve this question ..........please $<marquee scrollamount=1300>❤ please❤</marquee>$​​

Answer 1

❥____________

As, anything under square root can not be negative,

∴x−2≥0⇒x≥2∴x-2≥0⇒x≥2

So, domain of f(x)f(x) is x∈[2,∞]x∈[2,∞].

( f(x)=1x2−1−−−−−

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

$\displaystyle \: \: f(x) = \sqrt{ \frac{x - 2}{3 - x} }$

Now f(x) is well defined when

$x \geqslant 2 \: \: and \: \: \: 3 - x > 0$

Or

$x \leqslant 2 \: \: \: and \: \: \: 3 - x < 0$

Now

$x \geqslant 2 \: \: and \: \: \: 3 - x > 0 \: \: gives \: 2 \leqslant x < 3$

Again

$x \leqslant 2 \: \: \: and \: \: \: 3 - x < 0 \: \: gives \: \: absurd \: result$

So the required Domain is

$\{ \:x \in \mathbb{R}\: : \: 2 \leqslant x < 3\} = [ \: 2 \: , 3)$

Let

$\displaystyle \: \: </strong><strong>y</strong><strong> </strong><strong>= \sqrt{ \frac{x - 2}{3 - x} }$

Then

y² = (x-2)/(3-x)

=> 3 y² - xy²= x - 2

=> x = (3y²+2)/(y²+1)

Which is Positive for all real values of x

So the required Range is the set of all Positive real numbers $\</strong><strong> \mathbb{</strong><strong>R</strong><strong>+</strong><strong>}\:$