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प्रश्न
Which of the following functions are strictly decreasing on `(0, pi/2)`?
- cos x
- cos 2x
- cos 3x
- tan x
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उत्तर
(A) Let (x) = cos x, then
f' (x) = - sin x < 0 for all `x in (0, pi/2)` ....`[∴ sin x > 0 AA x in (0, pi/2)]`
(B) Let f(x) = cos 2x, then
f'(x) = -2 sin 2x < 0 for all `x in (0, pi/2)` ...(`∵ sin x > 0 (0, pi) = sin 2x > 0 (0, pi/2)`
= f is strictly decreasing on `(0, pi/2)`
(c) Let f(x) = cos 3x, then f'(x) = -3 sin 3x,
which assume +ve as well as -ve values in`(0, pi/2)`
`[0 < x < pi/2 = 0 <3x < (3pi)/2]`and `sin 3x > 0 (0, pi/2), sin 3x < o (pi, (3pi)/2)`
∴ f is neither increasing nor decreasing on `(0, pi/2)`
(D) Let f(x) = tan x, then f'(x) = sec2 x > 0 for all `x in (0, pi/2)` ....`[∵ sec^2 x > 0 AA x in (0, pi/2)]`
= f is strictly increasing on `(0, pi/2)`
Thus, we find that the function in (A) and (B) are strictly decreasing on `(0, pi/2).`
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