Mathematics Question on Applications of Derivatives

Question

Which of the following functions are strictly decreasing on (0,π/2)?

Answer 1

(A) Let f1(x) = cos x.

=f1(x) = -sin x

In interval (0, $\frac \pi2$ ), f'1(x) = -sin x<0.

$\implies$ f1(x) = cos x is strictly decreasing in interval (0, $\frac \pi2$ ).

(B) Let f2(x) = cos2x.

f'2(x) = -2sin 2x

Now 0<x< $\frac \pi2$ $\implies$ 0<2x<π $\implies$ sin 2x>0 $\implies$ -2 sin 2x<0

f'1(x) = -2sin 2x < 0 on (0, $\frac \pi2$ )

$\implies$ f'2(x) = cos 2x is strictly decreasing in interval (0, $\frac \pi2$ ).

(C) Let f3(x) = cos3x.

f'3(x) = -sin3x

Now, f'3(x) = 0

$\implies$ sin3x=0 $\implies$ 3x=π, as x ε (0, $\frac \pi2$ )

$\implies$ x = $\frac \pi2$

The point x= $\frac \pi3$ divides the interval (0, $\frac \pi2$ ) into two disjoint intervals

i.e., 0 (0, $\frac \pi3$ ) and ( $\frac \pi3$ , $\frac \pi2$ ).

Now, in interval(0, $\frac \pi3$ ), f3(x) =-3 sin3x<0 [as 0<x< $\frac \pi3$ =0<3x< $\pi$ ].

∴ f3 is strictly decreasing in interval (0, $\frac \pi3$ )

However, in interval ( $\frac \pi3$ , $\frac \pi2$ ), f3(x)=-3sin 3x>0 [as $\frac \pi3$ <x< $\frac \pi2$$\implies$ π<3x< $\frac {3\pi}{2}$ ]

∴ f3 is strictly increasing in interval ( $\frac \pi3$ , $\frac \pi2$ ).

Hence, f3 is neither increasing nor decreasing in interval (0, $\frac \pi2$ )

(D) Let f4(x) = tan x.

$\implies$ f4(x) = sec2 x

In interval (0, $\frac \pi2$ ), f'4(x) = sec2x>0.

f4 is strictly increasing in interval (0, $\frac \pi2$ ).

Therefore, functions cos x and cos 2x are strictly decreasing in (0, $\frac \pi2$ ).

Hence, the correct answers are (A) and (B).