हिंदी

In the following, determine the value of constant involved in the definition so that the given function is continuou: \[f\left( x \right) = \begin{cases}k( x^2 + 3x), & \text{ if }  x < 0 \\ \cos 2x ,

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प्रश्न

In the following, determine the value of constant involved in the definition so that the given function is continuou:  \[f\left( x \right) = \begin{cases}k( x^2 + 3x), & \text{ if }  x < 0 \\ \cos 2x , & \text{ if }  x \geq 0\end{cases}\]

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उत्तर

Given:

 \[f\left( x \right) = \begin{cases}k( x^2 + 3x), & \text{ if }  x < 0 \\ \cos 2x , & \text{ if }  x \geq 0\end{cases}\]
If  \[f\left( x \right)\]  is continuous at x = 0, then 
\[\lim_{x \to 0^-} f\left( x \right) = \lim_{x \to 0^+} f\left( x \right)\]
\[\Rightarrow \lim_{h \to 0} f\left( - h \right) = \lim_{h \to 0} f\left( h \right)\]
\[ \Rightarrow \lim_{h \to 0} \left( k\left( \left( - h \right)^2 - 3h \right) \right) = \lim_{h \to 0} \left( \cos 2h \right)\]
\[ \Rightarrow 0 = 1 \left[ \text{It is not possible} \right]\]
Hence, there does not exist any value of k, which can make the given function continuous.
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अध्याय 8: Continuity - Exercise 9.2 [पृष्ठ ३५]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 8 Continuity
Exercise 9.2 | Q 4.3 | पृष्ठ ३५

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