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Question
The value of c in Rolle's theorem for the function f (x) = x3 − 3x in the interval [0,\[\sqrt{3}\]] is
Options
1
−1
3/2
1/3
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Solution
1
The given function is \[f\left( x \right) = x^3 - 3x\] .
This is a polynomial function, which is continuous and derivable in R.
Therefore, the function is continuous on [0,\[\sqrt{3}\]] and derivable on (0,\[\sqrt{3}\])
Differentiating the given function with respect to x, we get
\[f'\left( x \right) = 3 x^2 - 3\]
\[ \Rightarrow f'\left( c \right) = 3 c^2 - 3\]
\[ \therefore f'\left( c \right) = 0 \]
\[ \Rightarrow 3 c^2 - 3 = 0\]
\[ \Rightarrow c^2 = 1\]
\[ \Rightarrow c = \pm 1\]
Thus, \[c = 1 \in \left[ 0, \sqrt{3} \right]\] for which Rolle's theorem holds.
Hence, the required value of c is 1.
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