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प्रश्न
In each of the following, find the value of the constant k so that the given function is continuous at the indicated point;
\[f\left( x \right) = \begin{cases}kx + 1, \text{ if } & x \leq \pi \\ \cos x, \text{ if } & x > \pi\end{cases}\] at x = π
योग
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उत्तर
\[f\left( x \right) = \begin{cases}kx + 1, \text{ if } & x \leq \pi \\ \cos x, \text{ if } & x > \pi\end{cases}\]
We have
(LHL at x = \[\pi\] = \[\lim_{x \to \pi^-} f\left( x \right) = \lim_{h \to 0} f\left( \pi - h \right) = \lim_{h \to 0} k\left( \pi - h \right) + 1 = k\pi + 1\]
(LHL at x = \[\pi\] = \[\lim_{x \to \pi^-} f\left( x \right) = \lim_{h \to 0} f\left( \pi - h \right) = \lim_{h \to 0} k\left( \pi - h \right) + 1 = k\pi + 1\]
\[\pi\]= \[\lim_{x \to \pi^+} f\left( x \right) = \lim_{h \to 0} f\left( \pi + h \right) = \lim_{h \to 0} \cos\left( \pi + h \right) = cos\pi = - 1\]
If f(x) is continuous at x =
\[\pi\] , then
\[\lim_{x \to \pi^-} f\left( x \right) = \lim_{x \to \pi^+} f\left( x \right)\]
\[\Rightarrow k\pi + 1 = - 1\]
\[ \Rightarrow k = \frac{- 2}{\pi}\]
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