मराठी

F ( X ) = { − 4 X + 5 , 0 ≤ X ≤ 1 2 X − 3 , 1 < X ≤ 2 Discuss the Applicability of Rolle'S Theorem for the Following Function on the Indicated Intervals ?

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प्रश्न

\[f\left( x \right) = \begin{cases}- 4x + 5, & 0 \leq x \leq 1 \\ 2x - 3, & 1 < x \leq 2\end{cases}\] Discuss the applicability of Rolle's theorem for the following function on the indicated intervals ?

बेरीज
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उत्तर

The given function is 

\[f\left( x \right) = \begin{cases}{ - 4x + 5, 0 \leq x \leq 1\\2x - 3, 1 < x \leq 2}\end{cases}\]

At = 0, we have

\[\lim_{x \to 1^-} f\left( x \right) = \lim_{h \to 0} f\left( 1 - h \right) = \lim_{h \to 0} \left[ - 4\left( 1 - h \right) + 5 \right] = 1\]
And
\[\lim_{x \to 1^+} f\left( x \right) = \lim_{h \to 0} f\left( 1 + h \right) = \lim_{h \to 0} \left[ 2\left( 1 + h \right) - 3 \right] = - 1\]
\[\therefore\] \[\lim_{x \to 1^-} f\left( x \right) \neq \lim_{x \to 1^+} f\left( x \right)\]
Thus,  \[f\left( x \right)\]  is discontinuous at  \[x = 1\] 

Hence, Rolle's theorem is not applicable for the given function.

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पाठ 14: Mean Value Theorems - Exercise 15.1 [पृष्ठ ८]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 14 Mean Value Theorems
Exercise 15.1 | Q 1.6 | पृष्ठ ८

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