PUC Karnataka Science Class 12Department of Pre-University Education, Karnataka
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Solution for If F (X) = Ex Sin X in [0, π], Then C in Rolle'S Theorem is (A) π/6 (B) π/4 (C) π/2 (D) 3π/4 - PUC Karnataka Science Class 12 - Mathematics

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Question

If f (x) = ex sin x in [0, π], then c in Rolle's theorem is

(a) π/6

(b) π/4

(c) π/2

(d) 3π/4

Solution

(d) 3π/4
The given function is 

\[f\left( x \right) = e^x \sin x\].
Differentiating the given function with respect to x, we get 

\[f'\left( x \right) = e^x \cos n x + \sin x e^x \]

\[ \Rightarrow f'\left( c \right) = e^c \cos c + \sin c e^c \]

\[\text{Now }, e^x cos x \text { is continuous and derivable in R } . \]

\[\text { Therefore, it is continuous on } \left[ 0, \pi \right] \text { and derivable on} \left( 0, \pi \right) . \]

\[ \therefore f'\left( c \right) = 0 \]

\[ \Rightarrow e^c \left( \cos c + \sin c \right) = 0\]

\[ \Rightarrow \left( \cos c + \sin c \right) = 0 \left( \because e^c \neq 0 \right)\]

\[ \Rightarrow \tan c = - 1\]

\[ \Rightarrow c = \frac{3\pi}{4}, \frac{7\pi}{4}, . . . \]

\[ \therefore c = \frac{3\pi}{4} \in \left( 0, \pi \right)\]

Thus, 

\[c = \frac{3\pi}{4} \in \left( 0, \pi \right)\]

for which Rolle's theorem holds.
Hence, the required value of c is 3π/4.

  Is there an error in this question or solution?
Solution If F (X) = Ex Sin X in [0, π], Then C in Rolle'S Theorem is (A) π/6 (B) π/4 (C) π/2 (D) 3π/4 Concept: Maximum and Minimum Values of a Function in a Closed Interval.
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