Prove that the function f(x) = cos x is :i. strictly decreasing on (0,

1.	Prove that the function f(x) = cos x is :i. strictly decreasing on (0, π)ii. strictly increasing in (π, 2π)iii. neither increasing nor decreasing in (0, 2π)
Answer» Given f(x) = cos x \(\therefore\) f'(x) = -sin x (i) Since for each x ∈ (0, π), sin x > 0 \(\Rightarrow\, \therefore\) f'(x) < 0 So f is strictly decreasing in (0, π) (ii) Since for each x ∈ (π, 2π), sin x < 0 \(\Rightarrow\, \therefore\) f'(x) > 0 So f is strictly increasing in (π, 2π) (iii) Clearly from (1) and (2) above, f is neither increasing nor decreasing in (0, 2π)

Prove that the function f(x) = cos x is :i. strictly decreasing on (0, π)ii. strictly increasing in (π, 2π)iii. neither increasing nor decreasing in (0, 2π)

Answer»

Given f(x) = cos x

\(\therefore\) f'(x) = -sin x

(i) Since for each x ∈ (0, π), sin x > 0

\(\Rightarrow\, \therefore\) f'(x) < 0

So f is strictly decreasing in (0, π)

(ii) Since for each x ∈ (π, 2π), sin x < 0

\(\Rightarrow\, \therefore\) f'(x) > 0

So f is strictly increasing in (π, 2π)

(iii) Clearly from (1) and (2) above, f is neither increasing nor decreasing in (0, 2π)