Answer :

Given: x = e^{cos2t} and y = e^{sin2t}

Taking log both the sides, we get

log x = log(e^{cos2t}) and log y = log(e^{sin2t})

⇒ logx = cos2t and logy = sin2t …(i)

Differentiate x = e^{cos2t} with respect to t, we get

…(ii)

Now, differentiate y = e^{sin2t} with respect to t, we get

…(iii)

Now,

[from eq(ii) & (iii)]

[from eq. (i)]

[given]

Hence Proved

**OR**

Given: f(x) = 2sinx + 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) = 2sinx + 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) = 2sinx + sin 2x is continuous at x ∈ [0,π]

Hence, condition 1 is satisfied.

Condition 2:

f(x) = 2sinx + sin 2x

f’(x) = 2cosx + 2cos2x

⇒ 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) = 2sinx + sin2x x ∈ [0,π]

f(a) = f(0) = 2sin(0) + sin2(0) = 0 [∵ sin(0°) = 0]

f(b) = f(π) = 2sin(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) = 2sinx + sin2x

On differentiating above with respect to x, we get

f’(x) = 2cosx + 2cos2x

Put x = c in above equation, we get

f’(c) = 2cos(c) + 2cos2c …(i)

By Mean Value Theorem,

[from (i)]

⇒ 2cos c + 2cos2c = 0

⇒ 2cos c + 2(2cos^{2} c – 1) = 0 [∵ cos 2x = 2cos^{2}x –1]

⇒ 2cos c + 4cos^{2} c – 2 = 0

⇒ 4 cos^{2} c + 2cos c – 2 = 0

Now, let cos c = x

⇒ 4x^{2} + 2x – 2 = 0

⇒ 2x^{2} + x – 1 = 0

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

So, the value of

Thus, Mean Value Theorem is verified.

Rate this question :