मराठी

Extend the Definition of the Following by Continuity F ( X ) = 1 − Cos 7 ( X − π ) 5 ( X − π ) 2 at the Point X = π.

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प्रश्न

Extend the definition of the following by continuity 

\[f\left( x \right) = \frac{1 - \cos7 (x - \pi)}{5 (x - \pi )^2}\]  at the point x = π.
बेरीज
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उत्तर

Given: 

\[f\left( x \right) = \frac{1 - \cos7\left( x - \pi \right)}{5 \left( x - \pi \right)^2}, x = \pi\]

If f(x) is continuous at x =

\[\pi\]then
\[\lim_{x \to \pi} f\left( x \right) = f\left( \pi \right)\]
\[ \Rightarrow \lim_{x \to \pi} \frac{1 - \cos7\left( x - \pi \right)}{5 \left( x - \pi \right)^2} = f\left( \pi \right)\]
\[ \Rightarrow \frac{2}{5} \lim_{x \to \pi} \frac{\sin^2 \left( \frac{7\left( x - \pi \right)}{2} \right)}{\left( x - \pi \right)^2} = f\left( \pi \right)\]
\[ \Rightarrow \frac{2}{5} \times \frac{49}{4} \lim_{x \to \pi} \frac{\sin^2 \left( \frac{7\left( x - \pi \right)}{2} \right)}{\frac{49}{4} \left( x - \pi \right)^2} = f\left( \pi \right)\]
\[ \Rightarrow \frac{2}{5} \times \frac{49}{4} \lim_{x \to \pi} \frac{\sin^2 \left( \frac{7\left( x - \pi \right)}{2} \right)}{\left( \frac{7}{2}\left( x - \pi \right) \right)^2} = f\left( \pi \right)\]
\[ \Rightarrow \frac{2}{5} \times \frac{49}{4} \lim_{x \to \pi} \left[ \frac{\sin\left( \frac{7\left( x - \pi \right)}{2} \right)}{\left( \frac{7}{2}\left( x - \pi \right) \right)} \right]^2 = f\left( \pi \right)\]
\[ \Rightarrow \frac{2}{5} \times \frac{49}{4} \times 1 = f\left( \pi \right)\]
\[ \Rightarrow \frac{1}{5} \times \frac{49}{2} \times 1 = f\left( \pi \right)\]
\[ \Rightarrow \frac{49}{10} = f\left( \pi \right)\]
Hence, the given function will be continuous at
\[x = \pi\]
\[f\left( \pi \right) = \frac{49}{10}\] .
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पाठ 8: Continuity - Exercise 9.1 [पृष्ठ २०]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 8 Continuity
Exercise 9.1 | Q 33 | पृष्ठ २०

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