Question

Show that the function given by f(x) = sin x is (a) strictly increasing in{0,π/2} (b) strictly decreasing in {π/2,π} (c) neither increasing nor decreasing in (0, π)

Answer 1

1. any function is strictly increasing in [a,b]
only when f'(x) > 0 in [a,b]
2. function is strictly decreasing in [a,b]
only when f'(x) < 0 in [a,b]
3. function is neither increasing nor decreasing in [a,b] when f'(x) > 0 and f'(x) < 0 in [a,b]


(a) the function is f(x) = sinx
differentiate with respect to x,
f'(x) = cosx , but we know cox> 0 in (0,π/2)
so,f(x) is strictly increasing in (0,π/2)

(b) f'(x) = cosx < 0 in (π/2, π)
e.g.,f'(x) < 0 in (π/2, π) so, f(x) is strictly decreasing in (π/2, π)

(c) from (a), we see f(x) is strictly increasing in (0,π/2) and from (b), 2e see f(x) is strictly decreasing in (π/2, π) . therefore we csn say that f(x) is firstly increasing and then decreasing in (0,π) . therefore f(x) is neither increasing nor decreasing function in (0,π)

Answer 2

Heya mate,

Plz refer to the attachment for ur answer☺

Hope it works