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Verify Rolle'S Theorem for the Following Function on the Indicated Interval F (X) = Cos 2 (X − π/4) on [0, π/2] ? - Mathematics

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प्रश्न

Verify Rolle's theorem for each of the following function on the indicated interval f (x) = cos 2 (x − π/4) on [0, π/2] ?

बेरीज
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उत्तर

The given function is \[f\left( x \right) = \cos2\left( x - \frac{\pi}{4} \right) = \cos\left( 2x - \frac{\pi}{2} \right) = \sin2x\] .

Since \[\sin2 \ x \] is everywhere continuous and differentiable.

Therefore, \[\sin2x\] is continuous on \[\left[ 0, \frac{\pi}{2} \right]\] and differentiable on

\[\left( 0, \frac{\pi}{2} \right)\] .
Also,
\[f\left( \frac{\pi}{2} \right) = f\left( 0 \right) = 0\]
Thus \[f\left( x \right)\] satisfies all the conditions of Rolle's theorem.
Now, we have to show that there exists
\[c \in \left( 0, \frac{\pi}{2} \right)\] such that 
\[f'\left( c \right) = 0\] .
We have
\[f\left( x \right) = \sin2x\]
\[ \Rightarrow f'\left( x \right) = 2\cos2x\]
\[\therefore f'\left( x \right) = 0\]
\[ \Rightarrow 2\cos2x = 0\]
\[ \Rightarrow \cos2x = 0\]
\[ \Rightarrow x = \frac{\pi}{4}\]
Thus,
\[c = \frac{\pi}{4} \in \left( 0, \frac{\pi}{2} \right)\] such that \[f'\left( c \right) = 0\] .
Hence, Rolle's theorem is verified.
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पाठ 15: Mean Value Theorems - Exercise 15.1 [पृष्ठ ९]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 15 Mean Value Theorems
Exercise 15.1 | Q 3.01 | पृष्ठ ९

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