The point in the interval [ 0 , 2 π ] where f ( x ) = e x s i n x has maximum slope, is Options: A . &nbsp π 4 B . &nbsp π 2 C . &nbsp π D . &nbsp 3 π 2

Answer: Option B : B We have, f ′ ( x ) = e x + c o s x + s i n x e x A n d f ′ ( x ) = − s i n x e x + c o s x e x + c o s x e x + s i n x c o s x e x . N o w , f ′ ( x ) = 2 c o s x c o s x e x = 0 ⇒ c o s x = 0 ⇒ x = π 2 . A l s o , f ′ ( x ) = − 2 s i n x e x + 2 c o s x e x = − v e ∴ Slope is maximum at x = π 2 . Hence (b) is the correct answer.

The point in the interval [0,2π] where f(x)=ex sin x has

Discussion Forum : Application Of Derivatives

Question -

The point in the interval $[0, 2 π]$ where $f (x) = e^{x} s i n x$ has maximum slope, is

Options:

A .

\frac{π}{4}

B .

\frac{π}{2}

C .

π

D .

3 \frac{π}{2}

Answer: Option B
:
B
We have,

f^{'} (x) = e^{x} + c o s x + s i n x e^{x} A n d f^{'} (x) = - s i n x e^{x} + c o s x e^{x} + c o s x e^{x} + s i n x c o s x e^{x} . N o w, f^{'} (x) = 2 c o s x c o s x e^{x} = 0 \Rightarrow c o s x = 0 \Rightarrow x = \frac{π}{2} . A l s o, f^{'} (x) = - 2 s i n x e^{x} + 2 c o s x e^{x} = - v e

∴

Slope is maximum at

x = \frac{π}{2} .

Hence (b) is the correct answer.

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Discussion Forum : Application Of Derivatives

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