(i) Consider `f(x) = sin 2x " in " [0, (pi)/(2)]`
Since the sine functions is continouous at each `x in R`, it follows that `f(x) = sin 2x` is continuous on `[0, (pi)/(2)]`
Alos, `f'(x) = 2 cos 2x`, which clearly exists for all `x in [0, (pi)/(2)]`
So, `f(x)` is differentiable on `[0, (pi)/(2)]`
Alos, `f(0) = f((pi)/(2)) = 0`
Thus, all the conditions of Rolle's theorem are satisfied.
So, there must exist a real number `c in [0, (pi)/(2)]` such that `f'(x) = 0`
Now, `f'(c) = 0 hArr 2 co s2c = 0 hArr cos 2c = 0`
`hArr 2c = (pi)/(2)" i.e., " c = (pi)/(4)`
Thus, `c = (pi)/(4) in [0, (pi)/(2)]"such that" f'(c) = 0`
Hence, Rolle's theorem is verified
(ii) Consider `f(x) = (sin x + cos x) " in " [0,(pi)/(2)]`
By the continuity of the sine function, the cosine function and the sum of continuous function, it follows that `f(x)` is continuous on `[0, (pi)/(2)]`
Also, `f(x) = (cos x - sin x)`, which clearly exists for all values of `x in [0, (pi)/(2)]`
So, f(x) is differentiable on `[0, (pi)/(2)]` Also, `f(0) = f ((pi)/(2))= 1`
Thus, all the conditions of Rolle's theorem are satisfied. So, there must exist some `c in [0, (pi)/(2)]`such that `f'(c) = 0`
Now, `f'(c) = 0 hArr cos c - sin c = 0 hArr cos c = sin c hArr c = (pi)/(4)`
Thus, `c = (pi)/(4) in [0, (pi)/(2)]` such that `f'(c) = 0`
Hence, Rolle's theorem is verified.
(iii) Consider `f(x) = cos 2 (x - (pi)/(4)) " in " [0, (pi)/(2)]`
SInce the cosine function is continuous everywhere, it follows that `f(x) = cos 2 (x , (pi)/(4))` is continuous on `[0, (pi)/(2)]`.
Also, `f'(x) = -2 sin (2x - (pi)/(2)) = 2 cos 2x`, which clearly exists
for all `x in [0, (pi)/(2)]`
`:. f(x)` is differentiable on `[0, (pi)/(2)]`
Further, `f(0) = cos 2 (-(pi)/(4)) = cos (pi)/(2) = 0`
And, `f((pi)/(2)) = cos 2 ((pi)/(2) (pi)/(4)) = cos (pi)/(2) = 0`
`:. f(0) = f ((pi)/(2)) = 0`
Thus, all the conditions of Rolle's theorem are satisfied. So, there must exist `c in [0, (pi)/(2)]` such that `f'(c) = 0`
Now, `f'(c) = 0 hArr 2 co s2c = 0 hArr 2c = (pi)/(2) rArr c = (pi)/(4)`
Thus, `c = (pi)/(4) in [0, (pi)/(2)]` such that `f'(c) = 0`
Hence, Rolle's theorem is verified.
(iv) Consider `f'(x) = (sin x - sin 2x) " in " [0, pi]`
Since the sine function is continuous, it follows that `g(x) = sin x and h (x) = sin 2x` are both continuous and so their difference is also continuous
Consequently, `f(x) = g(x) - h(x)` is difference is also continuous.
Consequently, `f(x) = g(x) - g(x)` is differentiable on `[0, pi]`.
Also, `f'(x) = (cos x - 2 cos 2x)`, which clearly exists for all `x in [0, pi]`.
`:. f(x)` is differentiable on `[0, pi]`
And, `f(0) = f(pi) = 0`
Thus, all the conditions of Rolle's theorem are satisifed.
So, there must exist a real number `c in [0, pi]`such that `f'(c) = 0`
Now, `f'(c) = 0 hArr cos c - 2 cos 2c = 0`
`hArr cos c - 2 (2 cos ^(2) c -1) = 0`
`hArr 4 cos^(2) c - cos c - 2 = 0`
`hArr cos c = (1 +- sqrt33)/(8) = 0.8431 or -0.5931`
`hArr cos c = 0.8431 or cos (180^(@) - c) = 0.5931`
`hArr c = 32^(@) .32' or c = 126^(@)23'`
Thus, `c in [0, pi]` such that `f'(c) = 0`
Hence, Rolle's theorem is satisfied.
