Question

If we apply the Rolle's theorem to f(x)=eˣsinx x ∈[0,π ],then c=.......,Select Proper option from the given options.
(a) 3π/4
(b) 5π/4
(c) π/4
(d) 7π/4

Answer 1

Rolle's theorem is application of mean value theorem. in case of mean value theorem, 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 $f'(c)=\frac{f(b)-f(a)}{b-a}$
if f'(c) = 0, it is said as Rolle's theorem,

here, f(x) = e^xsinx , x belongs to [0, π]
now, f'(x) = e^xsinx + e^xcosx
take a point c in interval [0, π] such that 0 < c < π.
then, from. Rolle's theorem, f'(c) =0
e^c [sinc + cosc] = 0
we know, e^c ≠ 0 so, [sinc + cosc] =0
sinc = -cosc
tanc = -1 = tan(π - π/4) = tan3π/4
c = 3π/4

hence, option (a) is correct.

Answer 2

Hello,

Answer:The answer is option a, c= 3π/4.

Solution:

Rolle's theorem is applicable,if

1) function is continuous in given closed interval.

2) function is differentiable in given closed interval.
3) f(a)=f(b)

we need not to check these conditions, here since function is Rolle's applicable.

$f(x) = {e}^{x} sinx \\ f'(x) = {e}^{x}sinx + {e}^{x} cosx \\ \\ f'(c) = {e}^{c}sin \: c + {e}^{c} cosc = 0 \\ {e}^{c} (sin \: c + cos \: c) = 0 \\ \\ sin \: c = - cos \: c \\ \frac{sin \: c}{cos \: c} = - 1 \\ \\ tan \: c = - 1 \\ \\ tan \: c = tan \: (\pi - \frac{\pi}{4} ) = tan \: ( \frac{3\pi}{4} ) \\ \\ c = \frac{3\pi}{4}$
So, the answer is option a, c= 3π/4.

Hope it helps you.