(i) f(x) = |x + 2| – 1
f(x) = |x + 2| – 1 ≥ -1
minimum value is – 1 when x + 2 = 0, x = -2
however it has no maximum value.
(ii) g(x) = – | x + 1| + 3 = 3,-1 x + 1|
g(x) < 3 v x + 1 = 0
∴ max. value is 3
when x = – 1
how ever no minimum value.
(iii) h(x) = sin (2x) + 5
maximum value of sin 2x = 1 and minimum value is -1
f(x) = 1 + 5 is 1 + 5
∴ max. h (x) = 6 and
min. h (x) = 4.
(iv) f(x) = |sin 4x + 3|
-1 < sin 4x < 1 ⇒ 3 -1 < sin 4x + 3 < + 1 + 3 + 2′< sin 4 x + 3 < 4
2 < | sin 4x + 3 | < 4
f(x) > 2 and f (x) < 4
min. f(x) = 2 when sin 4x + 3 = 0
max f (x) = 4 when sin 4x + 3 = 0
∴ minimum value is 2 at sin 4x = -1
maximum value is 4 at sin 4x = 1
(v) h(x) = x + 1, x ∈ (-1,1)
given that x ∈ (-1, 1)
i.e. -1 < x < 1
-1 + 1< x + 1 < 1 + 10 < x + 1< 2
∴ x + 1 > 0 or x + 1 < 2
x + 1 > 0 so no minimum value
x + 1 < 2, so no maximum value.