Question

Find the domain of q(x) = cos^–1 (4x² – 3). Hence, find the value of x for which q(x) = 0. Also, write the range of 3q(x) – π.

Answer 1

–1 ≤ 4x² – 3 ≤ 1

2 ≤ 4x² ≤ 4

`1/2` ≤ x² ≤ 1

x∈ `[-sqrt(1/2),-sqrt1]∪[sqrt1,sqrt(1/2)]`

Domain = `[-1/sqrt2),-1]∪[1,1/sqrt2]`

∈(x) = 0

cos^–1 (4x² – 3) = 0

4x² – 3 = cos0

4x² – 3 = 1

4x² = 4

x² = 1

x = ±1

F(x) = 3q(x) – π

F(x) = 3cos^–1 (4x² – 3) – π

`F((-1)/sqrt2)=3cos^-1(1)-pi`

F(1) = – π

`F(1/sqrt2)=2pi`

3π – π = 2π

3(0) – π = –π

Range = [–π, 2π]