Advertisements
Advertisements
प्रश्न
In each of the following, find the value of the constant k so that the given function is continuous at the indicated point; \[f\left( x \right) = \begin{cases}k( x^2 + 2), \text{if} & x \leq 0 \\ 3x + 1 , \text{if} & x > 0\end{cases}\]
योग
Advertisements
उत्तर
\[f\left( x \right) = \begin{cases}k( x^2 + 2), \text{if} & x \leq 0 \\ 3x + 1 , \text{if} & x > 0\end{cases}\]
We have
(LHL at x = 0) =
\[\lim_{x \to 0^-} f\left( x \right) = \lim_{h \to 0} f\left( 0 - h \right) = \lim_{h \to 0} k\left( \left( - h \right)^2 + 2 \right) = 2k\]
(RHL at x = 0) =
\[\lim_{x \to 0^+} f\left( x \right) = \lim_{h \to 0} f\left( 0 + h \right) = \lim_{h \to 0} 3h + 1 = 1\]
If f(x) is continuous at x = 0, then
\[\lim_{x \to 0^-} f\left( x \right) = \lim_{x \to 0^+} f\left( x \right)\]
\[ \Rightarrow 2k = 1\]
\[ \Rightarrow k = \frac{1}{2}\]
\[ \Rightarrow 2k = 1\]
\[ \Rightarrow k = \frac{1}{2}\]
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
