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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