Given: f(x) = sinx – sin2x in [0,π]

Now, we have to show that f(x) verify the Mean Value Theorem

First of all, Conditions of Mean Value theorem are:

a) f(x) is continuous at (a,b)

b) f(x) is derivable at (a,b)

If both the conditions are satisfied, then there exist some ‘c’ in (a,b) such that

Condition 1:

f(x) = sinx – sin 2x

Since, f(x) is a trigonometric function and we know that, sine function are defined for all real values and are continuous for all x ∈ R

⇒ f(x) = sinx – sin 2x is continuous at x ∈ [0,π]

Hence, condition 1 is satisfied.

Condition 2:

f(x) = sinx – sin 2x

f’(x) = cosx – 2 cos2x

⇒ f(x) is differentiable at [0,π]

Hence, condition 2 is satisfied.

Thus, Mean Value Theorem is applicable to the given function

Now,

f(x) = sinx – sin2x x ∈ [0,π]

f(a) = f(0) = sin(0) – sin2(0) = 0 [∵ sin(0°) = 0]

f(b) = f(π) = sin(π) – sin2(π) = 0 – 0 = 0

[∵ sin π = 0 & sin 2π = 0]

Now, let us show that there exist c ∈ (0,1) such that

f(x) = sinx – sin2x

On differentiating above with respect to x, we get

f’(x) = cosx – 2cos2x

Put x = c in above equation, we get

f’(c) = cos(c) – 2cos2c …(i)

By Mean Value Theorem,

[from (i)]

⇒ cos c – 2cos2c = 0

⇒ cos c – 2(2cos² c – 1) = 0 [∵ cos 2x = 2cos²x –1]

⇒ cos c – 4cos² c + 2 = 0

⇒ 4 cos² c – cos c – 2 = 0

Now, let cos c = x

⇒ 4x² – x – 2 = 0

Now, to find the factors of the above equation, we use

[above we let cos c = x]

So, value of

Thus, Mean Value Theorem is verified.