Question

For which pair of functions is (g circle f) (a) = StartAbsoluteValue a EndAbsoluteValue minus 2?
f(a) = a2 – 4 and g (a) = StartRoot a EndRoot
f (a) = one-half a minus 1 and g(a) = 2a – 2
f(a) = 5 + a2 and g (a) = StartRoot a minus 5 EndRoot minus 2
f(a) = 3 – 3a and g(a) = 4a – 5

Please explain in detail I am very confused about the whole thing thx for ur help in advance :)

Answer 1

Given:

$(g\circ f)(a)=|a|-2$

To find:

The functions f(x) and g(x).

Solution:

We know that,

$(g\circ f)(a)=g[f(a)]$

If $f(a)=a^2-4$ and $g(a)=\sqrt{a}$, then

$(g\circ f)(a)=g[a^2-4]$

$(g\circ f)(a)=\sqrt{a^2-4}\neq |a|-2$

Option A is incorrect.

If $f(a)=\dfrac{1}{2}a-1$ and $g(a)=2a-2$, then

$(g\circ f)(a)=g[\dfrac{1}{2}a-1]$

$(g\circ f)(a)=2(\dfrac{1}{2}a-1)-2$

$(g\circ f)(a)=a-2-2$

$(g\circ f)(a)=a-4\neq |a|-2$

Option B is incorrect.

If $f(a)=5+a^2$ and $g(a)=\sqrt{a-5}-2$, then

$(g\circ f)(a)=g[5+a^2]$

$(g\circ f)(a)=\sqrt{5+a^2-5}-2$

$(g\circ f)(a)=\sqrt{a^2}-2$

$(g\circ f)(a)=|a|-2$

Option C is correct.

If $f(a)=3-3a$ and $g(a)=4a-5$, then

$(g\circ f)(a)=g[3-3a]$

$(g\circ f)(a)=4(3-3a)-5$

$(g\circ f)(a)=12-12a-5$

$(g\circ f)(a)=7-12x\neq |a|-2$

Option D is incorrect.

Therefore, the correct option is C.