**Question 3:**Given that

*f*:

*x*→

*hx*+

*k*and

*f*

^{2}:

*x*→ 4

*x*+ 15.

(a) Find the value of

*h*and of

*k*.

(b) Take the value of

*h*> 0, find the values of

*x*for which

*f*(

*x*

^{2}) = 7

*x*

*Solution*:**(a)**

Given

*f*(

*x*) =

*hx*+

*k*

*f*

^{2}(

*x*) =

*ff*(

*x*) =

*f*(

*hx*+

*k*)

=

*h*(

*hx*+

*k*) +

*k*

=

*h*

^{2}

*x*+

*hk*+

*k*

*f*

^{2}(

*x*) = 4

*x*+ 15

*h*

^{2}

*x*+

*hk*+

*k*= 4

*x*+ 15

*h*

^{2}= 4

*h*= ± 2

when,

*h*= 2

*hk*+

*k*= 15

2

*k*+

*k*= 15

*k*= 5

When,

*h*= –2

*hk*+

*k*= 15

–2

*k*+

*k*= 15

*k*= –15

**(b)**

*h*> 0,

*h*= 2,

*k*= 5

Given

*f*(

*x*) =

*hx*+

*k*

*f*(

*x*) = 2

*x*+ 5

*f*(

*x*

^{2}) = 7

*x*

2 (

*x*

^{2}) + 5 = 7

*x*

2

*x*

^{2}– 7

*x*+ 5 = 0

(2

*x*– 5)(

*x*–1) = 0

2

*x*– 5 = 0 or

*x*–1= 0

*x***= 5/2**

*x***= 1**