We are given function: [tex]M(x) = 6x^2-12.[/tex]
Let us find it's inverse.
In order to find it's inverse, we need to get function equal to y.
[tex]y= 6x^2-12[/tex]
Switching x and y's .
[tex]x= 6y^2-12[/tex]
Now, solving it for y.
[tex]x+12 = 6y^2[/tex]
Dividing both sides by 6, we get
[tex]y^2=\frac{x}{6} +\frac{12}{6}[/tex]
[tex]y^2=\frac{x}{6} +2[/tex]
Taking square root on both sides, we get
[tex]\sqrt{y^2} =\sqrt{\frac{x}{6} +2}[/tex]
[tex]y=\sqrt{\frac{x}{6} +2}[/tex]
Re-writing in the form of given inverse.
[tex]m^{-1}(x)=\sqrt{\frac{x}{3\times2} +2}[/tex]
On comparing with given M^-1 = √((x/3k)+k).
[tex]\sqrt{\frac{x}{3k} +k} = \sqrt{\frac{x}{3\times2} +2}[/tex]
k=2.
Answer:
Option 2 - k=2
Step-by-step explanation:
Given : [tex]M(x)=6x^2-12[/tex] and [tex]M^{-1}= \sqrt{\frac{x}{3k}+k}[/tex]
To find : What value of k would make these inverses?
Solution :
First we find the inverse of M(x)
Let, [tex]y=6x^2-12[/tex]
Interchange x and y,
[tex]x=6y^2-12[/tex]
Now, find the value of y
[tex]x+12=6y^2[/tex]
[tex]\frac{x+12}{6}=y^2[/tex]
Taking root both side,
[tex]y=\sqrt{\frac{x+12}{6}}[/tex]
[tex]y=\sqrt{\frac{x}{6}+\frac{12}{6}}[/tex]
[tex]y=\sqrt{\frac{x}{6}+2}[/tex]
[tex]M^{-1}=\sqrt{\frac{x}{6}+2}[/tex]
Now, we compare or equate both the inverse function.
[tex]\sqrt{\frac{x}{6}+2}= \sqrt{\frac{x}{3k}+k}[/tex]
[tex]\sqrt{\frac{x}{3(2)}+2}= \sqrt{\frac{x}{3k}+k}[/tex]
On caparison the value of k=2.
Therefore, Option C is correct.
