Answer :

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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