Answer:
Option (a) is correct.
[tex]\frac{f(x)}{g(x)}=\sqrt{x+1}[/tex]
Step-by-step explanation:
Given : [tex]f(x)=\sqrt{x^2-1}[/tex] and [tex]g(x)=\sqrt{x-1}[/tex]
We have to find the value of [tex](\frac{f}{g})(x)[/tex]
Consider [tex](\frac{f}{g})(x)[/tex]
It is same as [tex](\frac{f}{g})(x)=\frac{f(x)}{g(x)}[/tex]
Substitute the value of f(x) and g(x) , we have,
[tex]\frac{f(x)}{g(x)}=\frac{\sqrt{x^2-1}}{\sqrt{x-1}}[/tex]
Combining same powers , [tex]\quad \frac{\sqrt{x}}{\sqrt{y}}=\sqrt{\frac{x}{y}}[/tex] , We have,
[tex]=\sqrt{\frac{x^2-1}{x-1}}[/tex]
Factorize numerator as , [tex]x^2-1:\quad \left(x+1\right)\left(x-1\right)[/tex]
We have , [tex]=\frac{\left(x+1\right)\left(x-1\right)}{x-1}[/tex]
Cancel out (x-1) , we get, (x + 1)
[tex]\frac{f(x)}{g(x)}=\sqrt{x+1}[/tex]
Thus, Option (a) is correct.
