The point in the interval -0- 2- where fx-ex sin x has maximum slope- is

The correct option is A π2 f'(x)=e^x(sin x+cos x) f'(x)=e^x(sin x+cos x+cos x sin x) ⇒ f'(x)=e^x cos x=0 ⇒ x=π2 ...

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Global Maxima

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Question

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

\frac{π}{4}

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\frac{π}{2}

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π

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\frac{3 π}{2}

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(0, 2 π)

f (x) = e^{x} sin x

^{'} a^{'}

^{'} b^{'}

f (x)

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} x + a \sqrt{2} sin x, & 0 \leq x < \frac{π}{4} 2 x cot x + b, & \frac{π}{4} \leq x \leq \frac{π}{2} a cos 2 x - b sin x, & \frac{π}{2} < x \leq π \end{matrix}

x = \frac{π}{4}, \frac{π}{2}

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

f (x) = e^{x} sin x

[0, π]

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

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