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Given: x = ecos2t and y = esin2t

Taking log both the sides, we get

log x = log(ecos2t) and log y = log(esin2t)

logx = cos2t and logy = sin2t …(i)

Differentiate x = ecos2t with respect to t, we get

…(ii)

Now, differentiate y = esin2t 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(2cos2 c – 1) = 0 [ cos 2x = 2cos2x –1]

2cos c + 4cos2 c – 2 = 0

4 cos2 c + 2cos c – 2 = 0

Now, let cos c = x

4x2 + 2x – 2 = 0

2x2 + x – 1 = 0

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

So, the value of

Thus, Mean Value Theorem is verified.

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