NCERT Solutions
Class 12 Maths
Application of Derivatives

Ex.6.2 Q.12
Which of the following functions are strictly decreasing on (0, π/2)?
(A) cos x
(B) cos 2x
(C) cos 3x
(D)tan x
(A) Let f(x) = cos x
So, f’(x) = -sin x
In interval (0, π/2), f’(x) = -sin x < 0
So, f(x) = cos x is strictly decreasing in interval (0, π/2)
(B) Let f(x) = cos 2x
So, f’(x) = -2 sin 2x
Now, 0 < x < π/2
⟹ 0 < 2x < π
⟹ sin 2x > 0
⟹ -2 sin 2x < 0
So, f(x) = cos 2x is strictly decreasing in interval (0, π/2).
(C) Let f(x) = cos 3x
So, f’(x) = -3 sin 3x
Now, f’(x) = 0
-3 sin 3x = 0
sin 3x = 0
3x = π as x є (0, π/2)
x =
The point x = π3 divides the interval (0,
) into two disjoint intervals i.e. (0,
) and (
,
).
Now, in interval (0, π/3), f’(x) = -3 sin 3x < 0 [As 0 < x < ⟹ 0 < 3x < π]
So, f(x) = cos 3x is strictly decreasing in interval (0, )
However in interval ( ,
), f’(x) = -3 sin 3x > 0 [As
< x <
π < 3x <
]
So, f(x) = cos 3x is strictly increasing in interval ( ,
)
So, f(x) = cos 3x is neither increasing nor decreasing in the interval (0, π/2).
(A) Let f(x) = tan x
So, f’(x) = sec2 x
In interval (0, ), f’(x) = sec2 x > 0
So, f(x) = tan x is strictly increasing in interval (0, ).
Therefore, functions cos x and cos 2x are strictly decreasing in (0, ).
Hence, the correct answers are A and B.