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Verify Rolle'S Theorem for the Following Function on the Indicated Interval F(X) = Cos 2x on [−π/4, π/4] ? - Mathematics

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Question

Verify Rolle's theorem for the following function on the indicated interval f(x) = cos 2x on [−π/4, π/4] ?

Sum

Solution

The given function is \[f\left( x \right) = \cos2x\].

Since 

\[\cos2x\] is everywhere continuous and differentiable, \[\cos2x\] is continuous on 

\[\left[ \frac{- \pi}{4}, \frac{\pi}{4} \right]\] and differentiable on \[\left( \frac{- \pi}{4}, \frac{\pi}{4} \right)\] .
Also,
\[f\left( \frac{\pi}{4} \right) = f\left( \frac{- \pi}{4} \right) = 0\]
Thus,\[f\left( x \right)\] satisfies all the conditions of Rolle's theorem. 
Now, we have to show that there exists \[c \in \left( \frac{- \pi}{4}, \frac{\pi}{4} \right)\] such that \[f'\left( c \right) = 0\] .
We have
\[f\left( x \right) = \cos2x\]
\[ \Rightarrow f'\left( x \right) = - 2\sin2x\]
\[\therefore f'\left( x \right) = 0\]
\[ \Rightarrow - 2\sin2x = 0\]
\[ \Rightarrow \sin2x = 0\]
\[ \Rightarrow \sin2x = 0\]
\[ \Rightarrow x = 0\]
Since \[c = 0 \in \left( \frac{- \pi}{4}, \frac{\pi}{4} \right)\] such that \[f'\left( c \right) = 0\] .
​Hence, Rolle's theorem is verified.
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Chapter 15: Mean Value Theorems - Exercise 15.1 [Page 9]

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RD Sharma Mathematics [English] Class 12
Chapter 15 Mean Value Theorems
Exercise 15.1 | Q 3.03 | Page 9

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