f(x)=1 + 2 sin x + 2 cos2x, 0≤x≤pi/2
A. f(x) is greatest at pi/6
B. f(x) is least at 0,pi/2
C. increasing in (0,pi/6) and decreasing in(pi/6,pi/2)
D. f(x) is continuous in [0,pi/2]
11First
dy/dx =2cosx - 4cos x sinx
dy/dx = 2[cosx - sin2x]
Critical points
dy/dx = 0
cosx = sin2x
sinx = 1/2
x = pi/6
Now draw the number line
On drawing the number line we get that it is increasing before pi/6 and decreasing atfer pi/6 Therefore at pi/6 the F(x) is greatest.
Since it is differentiable at all points hence it is continuous at all points form 0 to pi/2
hence the ans is A,C,D
621 + 2 sin x + 2 cos2x
=3+2 sin x - 2 sin2x
=-2(sin2x-sinx +1/4)+7/2
=-2(sinx - 1/2)2+7/2
the maximum value is 7/2 taken at sin x = 1/2
and nimimum value is taken when sin x=-1
Ronald.. think of how the last part can be wrong..
D is absolutely correct.. :)
62what do you mean by rhl not equl..
The function is sum of 2 functions which are all continuous..!!!
Am i sleeping?
11Sorry nishant bhaiyya
Made a silly mistake
Really sorry
Actually i am a bit sleepy
