Prove that the function f given by f (x) = log |cos x| is decreasing on (0, π/2) and increasing on (3π/2, 2π)

Solution:

Increasing functions are those functions that increase monotonically within a particular domain,

and decreasing functions are those which decrease monotonically within a particular domain.

Logs (or) logarithms are nothing but another way of expressing exponents.

We have

f (x) = log |cos x|

Therefore,

f' (x) = 1/cos x (- sin x)

= - tan x

In interval (0, π/2),

tan x > 0

⇒ - tan x < 0

Hence, f' (x) < 0

Thus, f is strictly decreasing on (0, π/2)

In interval (3π/2, 2π),

tan x < 0

⇒ - tan x > 0

Hence,

f' (x) > 0

Thus, f is strictly increasing on (3π/2, 2π)

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NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.2 Question 17

Prove that the function f given by f (x) = log |cos x| is decreasing on (0, π/2) and increasing on (3π/2, 2π)

Summary:

Hence we have concluded that the function f given by f (x) = log |cos x| is strictly decreasing on (0, π/2) and strictly increasing on (3π/2, 2π)