Question

10 A function f is defined by f : x  → 5 − 2 sin 2x for 0 ≤ x ≤ 0.
(i) Find the range of f. [2]
(ii) Sketch the graph of y = fx. [2]
(iii) Solve the equation fx = 6, giving answers in terms of 0. [3]
The function g is defined by g : x  → 5 − 2 sin 2x for 0 ≤ x ≤ k, where k is a constant. (iv) State the largest value of k for which g has an inverse. [1] (v) For this value of k, find an expression for g−1 x. [3]

PLEASE HELP WITH PART 10i ASAP.

Answer 1

1)we know, that, -1≤sin2x≤1
=2≤-2sin2x≤-2
now, Adding+5 on both the sides,
7≤5-2sinx≤3=now, we can see that, 5-2sin2x can take all the real values between 7 and 3.
so its range is [3,7].
3)given f(x)=6
=>5-2sin2x=6
=>-1=2sin2x=>sin2x=-1/2
4)given, x->5-2sin2x
=>g(x)=5-2sin2x
so, it is said that the value of k for which the function g(x) has an inverse (1) so ,
the inverse function is sin2y=5-x/2
and thus sin(2)=5-x/2=>y=5-2sin(2).
now, k=5-2sin2.
and the expression is g^-1(x)=sin^-1(5-x/4).