Define f(x)=(1/2)[| sin x|+ sin x], 0

Question

Define f(x)=(1/2)[| sin x|+ sin x], 0 < x ≤ 2 π. Then, f is (A) increasing in ((π/2), (3 π/2)) (B) decreasing in (0, (π/2)) and increasing in ((π/2), π) (C) increasing in (0, (π/2)) and decreasing in ((π/2), π) (D) increasing in (0, (π/4)) and decreasing in ((π/4), π)

Tardigrade Admin · Accepted Answer

Given, f(x)=(1/2)[| sin x|+ sin x], 0< x ≤ 2 π Case I, when, 0 < x ≤ π f(x)=(1/2)[ sin x+ sin x]= sin x f'(x)= cos x cos x>0, for 0< x< (π/2) (increasing) cos x<0, for (π/2)< x< π (decreasing) Case II When π< x ≤ 2 π f(x)=(1/2)[- sin x+ sin x]=0

Q. Define $f (x) = \frac{1}{2} [∣ sin x ∣ + sin x], 0 < x \leq 2 π$ . Then, $f$ is

increasing in $(\frac{π}{2}, \frac{3 π}{2})$

decreasing in $(0, \frac{π}{2})$ and increasing in $(\frac{π}{2}, π)$

increasing in $(0, \frac{π}{2})$ and decreasing in $(\frac{π}{2}, π)$

increasing in $(0, \frac{π}{4})$ and decreasing in $(\frac{π}{4}, π)$

Q. Define f(x)=21​[∣sinx∣+sinx],0<x≤2π. Then, f is

increasing in (2π​,23π​)

decreasing in (0,2π​) and increasing in (2π​,π)

increasing in (0,2π​) and decreasing in (2π​,π)

increasing in (0,4π​) and decreasing in (4π​,π)

Given, f(x)=21​[∣sinx∣+sinx],0<x≤2π Case I, when, 0<x≤π f(x)=21​[sinx+sinx]=sinx f′(x)=cosx cosx>0, for 0<x<2π​ (increasing) cosx<0, for 2π​<x<π (decreasing) Case II When π<x≤2π f(x)=21​[−sinx+sinx]=0

Q. Define $f (x) = \frac{1}{2} [∣ sin x ∣ + sin x], 0 < x \leq 2 π$ . Then, $f$ is

increasing in $(\frac{π}{2}, \frac{3 π}{2})$

decreasing in $(0, \frac{π}{2})$ and increasing in $(\frac{π}{2}, π)$

increasing in $(0, \frac{π}{2})$ and decreasing in $(\frac{π}{2}, π)$

increasing in $(0, \frac{π}{4})$ and decreasing in $(\frac{π}{4}, π)$