सिद्ध कीजिये कि `(0, 2pi)` में फलन `f(x) = x - sin x` वर्धमान है।

f(x) = sinx - sin2x in [0,pi]

In [0,pi/2] the function f(x) = log sin x is

Verify Rolle's theorem for each of the following functions: (i) f(x) = sin 2x " in " [0, (pi)/(2)] (ii) f(x) = (sin x + cos x) " in " [0, (pi)/(2)] (iii) f(x) = cos 2 (x - (pi)/(4)) " in " [0, (pi)/(2)] (iv) f(x) = (sin x - sin 2x) " in " [0, pi]

Let f'(sin x) lt 0 and f''(sin x) gt 0 AA x in (0, pi/2) and g (x) = f (sin x) + f (cos x) , then

Verify Rolle's theorem for each of the following functions : f(x) = sin x - sin 2x " in " [0, 2pi]

If f(x)=sin x/ e^(x) in [0,pi] then f(x)

Consider a function f defined by f(x)=sin^(-1) sin ((x+sinx)/2) AA x in [0,pi] which satisfies f(x)+f(2pi-x)=pi AA x in [pi, 2pi] and f(x)=f( 4pi-x) for all x in [2pi,4pi] then If alpha is the length of the largest interval on which f(x) is increasing, then alpha =