Question

VERIFY..............↑↑

Answer 1

Answer:

Hii mate,

f(x)=eˣ  sin x,x∈(0,π)

For Rolle's Theorem,

f(0)=f(π) & f(x) must be continuous & differentiable over [0,π]

let us check if f(0)=f(π)  

⇒f(0)=e⁰sin(0)=0

⇒f(π)=eⁿ  sin(π)=0

therefore, f(0)=f(π)  

the function eˣ  & sin x are both continuous & differentiable over [0,π]

therefore, Rolle's Theorem can be applied for the function given.

There exists  a b∈[0,π] such that f(b)=0

f(x)= eˣ  sin x + e ˣ cos x

f(b)= eᵇ  sin(b) + eᵇ  cos(b) = 0

⇒eᵇ [cos b−sin b]=0

⇒cos b=sin b⇒tan b=1

⇒b= π/4 which ∈[0,π]

Hence, Rolle's Theorem is verified.

HOPE IT HELPS,

PLEASE THANK ,FOLLOW AND MARK AS BRAINLIEST.

Answer 2

$\huge\red{\underline{{\boxed{\textbf{QUESTION}}}}}$

Verify Rolle's Theorem for f(x) = e^xsinx on [0,pi]

$\huge\red{\underline{{\boxed{\textbf{ANSWER}}}}}$

The given function f(x)=e^xsinx

Since sinx and e^x are everywhere continous and differentiable.

Therefore, being a product of these two, f(x) is continuous on [0,π] and differentiable on (0,π).

Also f(π)=f(0)=0

Thus, f(x) satisfies all the conditions of Rolle's Theorem.Now, we have to show that there exists c∈(0,π) such that f'(c)=0

We have,

f(x)=e^xsinx

⇒f'(x)=e^x(sinx+cosx)

∴f'(x)=0

⇒ex(sinx+cosx)=0

⇒sinx+cosx=0

⇒tanx=−1⇒x=π−π/4=3π/4

Since c=3π/4 ∈ (0,π)

such that f'(c)=0

Hence, Rolle's theorem is verified.

__________________

HARSH PRATAP SINGH