Q. 755.0( 1 Vote )

Verify mean value theorem for each of the functions given

f(x) = sinx – sin2x in

Answer :

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.


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