Question

If we apply the mean value theorem to f(x)=2sinx+sin2x, then c=.......,Select Proper option from the given options.
(a) π
(b) π/4
(c) π/2
(d) π/3

Answer 1

I think question is said to apply Rolle's theorem, because interval is not given in question.

Mean value theorem : It states that if f(x) is defined and continuous on the interval [a,b] and differentiable on (a,b), then there is at least one number c in the interval (a,b) (that is a< c < b) such that $f'(c)=\frac{f(b)-f(a)}{b-a}$
when f'(c) = 0, it is known as Rolle's theorem.

so, first of all we have to find f'(x)
f'(x) = 2cosx + 2cos2x
now, f'(c) = 2cosc + 2cos2c = 0
cosc =- cos2c
cosc = -2cos²c + 1 [ as you know, cos2A = cos²A-sin²A = 2cos²A - 1]
2cos²c + cosc - 1 = 0
2cos²c + 2cosc - cosc - 1 = 0
cosc = -1 , 1/2
so, c = π , or π/3

Answer 2

Hello,



Solution:

If Mean value theorem is applicable on the given function then

1) f(x) is continuous in the given interval

2) f(x) is differentiable in the given interval

3) there must be a value c ,f'(c) = f(b)-f(a)/b-a

f(x)=2sinx+sin2x

f'(x) = 2cos x+ 2cos 2x

2cos c+ 2cos 2c=0
$cos \: c + cos \: 2c = 0 \\ cos \: c + 2 {cos}^{2} c - 1 = 0 \\ 2 {cos}^{2} c + cos \: c - 1 = 0 \\ 2 {cos}^{2} c +2 cos \: c \: - cos \: c - 1 = 0 \\ 2cos \: c(cos \: c + 1) - 1(cos \: c + 1) = 0 \\ \\ (cos \: c + 1)(2cos \: c - 1) = 0 \\ \\ cos \: c = - 1 \\ c = {cos}^{ - 1}( - 1) \\ c = \pi \\ 2cos \: c = 1 \\ \\ cos \: c = \frac{1}{2} \\ c = \frac{\pi}{3}$
There is two values of c,since there is not shown any close interval,so we can take both if them.

Option a and d both are correct.