मराठी

If F ( X ) = { X 2 2 , If 0 ≤ X ≤ 1 2 X 2 − 3 X + 3 2 , P If 1 < X ≤ 2 . Show that F is Continuous at X = 1.

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प्रश्न

If \[f\left( x \right) = \begin{cases}\frac{x^2}{2}, & \text{ if } 0 \leq x \leq 1 \\ 2 x^2 - 3x + \frac{3}{2}, & \text P{ \text{ if }  }  1 < x \leq 2\end{cases}\]. Show that f is continuous at x = 1.

 
बेरीज
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उत्तर

Given: 

\[f\left( x \right) = \binom{\frac{x^2}{2}, \text{ if } 0 \leq x \leq 1}{2 x^2 - 3x + \frac{3}{2}, if 1 < x \leq 2}\] 
We have
(LHL at x = 1) = 
\[\lim_{x \to 1^-} f\left( x \right) = \lim_{h \to 0} f\left( 1 - h \right)\]
\[= \lim_{h \to 0} \frac{\left( 1 - h \right)^2}{2} = \frac{1}{2}\]
(RHL at x = 1) =  
\[\lim_{x \to 1^+} f\left( x \right) = \lim_{h \to 0} f\left( 1 + h \right)\]
\[= \lim_{h \to 0} \left[ 2 \left( 1 + h \right)^2 - 3\left( 1 + h \right) + \frac{3}{2} \right] = 2 - 3 + \frac{3}{2} = \frac{1}{2}\]
Also,
\[f\left( 1 \right) = \frac{\left( 1 \right)^2}{2} = \frac{1}{2}\]
\[\lim_{x \to 1^-} f\left( x \right) = \lim_{x \to 1^+} f\left( x \right) = f\left( 1 \right)\]
Hence, the given function is continuous at
\[x = 1\]
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पाठ 8: Continuity - Exercise 9.1 [पृष्ठ २१]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 8 Continuity
Exercise 9.1 | Q 38 | पृष्ठ २१

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