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Solution for Verify Rolle'S Theorem for the Following Function on the Indicated Interval F(X) = Sin4 X + Cos4 X on [ 0 , π 2 ] ? - CBSE (Science) Class 12 - Mathematics

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Question

Verify Rolle's theorem for the following function on the indicated interval f(x) = sin4 x + cos4 x on \[\left[ 0, \frac{\pi}{2} \right]\] ?

Solution

The given function is \[f\left( x \right) = \sin^4 x + \cos^4 x\] . 

Since 

\[\sin x \text { and } \cos x\] are everywhere continuous and differentiable,

\[f\left( x \right) = \sin^4 x + \cos^4 x\] is continuous on \[\left[ 0, \frac{\pi}{2} \right]\] and differentiable on \[\left( 0, \frac{\pi}{2} \right)\] .
Also,
\[f\left( \frac{\pi}{2} \right) = f\left( 0 \right) = 1\]
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

\[f\left( x \right) = \sin^4 x + \cos^4 x\]

\[ \Rightarrow f'\left( x \right) = 4 \sin^3 x\cos x - 4 \cos^3 x\sin x\]

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

\[ \Rightarrow 4 \sin^3 x\cos x - 4 \cos^3 x\sin x = 0\]

\[ \Rightarrow \sin^3 x\cos x - \cos^3 x\sin x = 0\]

\[ \Rightarrow \tan^3 x - \tan x = 0\]

\[ \Rightarrow \tan x\left( \tan^2 x - 1 \right) = 0\]

\[ \Rightarrow \tan x = 0, \tan^2 x = 1\]

\[ \Rightarrow \tan x = 0, \tan x = \pm 1\]

\[ \Rightarrow x = 0, x = \frac{\pi}{4}, \frac{3\pi}{4}\]

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.
  Is there an error in this question or solution?
Solution for question: Verify Rolle'S Theorem for the Following Function on the Indicated Interval F(X) = Sin4 X + Cos4 X on [ 0 , π 2 ] ? concept: Maximum and Minimum Values of a Function in a Closed Interval. For the courses CBSE (Science), CBSE (Commerce), CBSE (Arts), PUC Karnataka Science
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