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CONTINUITY AND DIFFERENTIABILITY — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSCONTINUITY AND DIFFERENTIABILITY2 Marks
Question
Is the function defined by $f(x) = x^2 – \sin x + 5$ continuous at $\text{x} = \pi$ ?
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Answer

$\text{f(x)}=\text{x}^{2} - \sin\text{x}+ 5$
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\pi}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\pi}\text{(x}^{2} - \sin \text{x} + 5)$
$= \pi^{2} - \sin\pi + 5 = \pi^{2} - 0 + 5 = \pi^{2} + 5$
Also $\text{f}(\pi)= \pi^{2} - \sin\pi + 5 = \pi^{2} - 0 + 5 = \pi^{2} + 5$
$\therefore \ ^{\ \ \text{Lt}}_{\text{x}\rightarrow{\pi}}\text{f(x)} = \text{f}(\pi)$
$\therefore f$ is continous at $\text{x} = \pi.$

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