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(B) For the function $f(x)$ to be continuous at $x = 0$,the limit of $f(x)$ as $x \rightarrow 0$ must exist and be equal to $f(0)$. $f(0) = k$. Now,we

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$5$

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(B) For the function $f(x)$ to be continuous at $x = 0$,the limit of $f(x)$ as $x \rightarrow 0$ must exist and be equal to $f(0)$. $f(0) = k$. Now,we calculate the limit: $\lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0} \frac{1 + 3 x^2 - \cos 2 x}{x^2}$ Using the trigonometric identity $\cos 2x = 1 - 2 \sin^2 x$: $\lim_{x \rightarrow 0} \frac{1 + 3 x^2 - (1 - 2 \sin^2 x)}{x^2}$ $= \lim_{x \rightarrow 0} \frac{1 + 3 x^2 - 1 + 2 \sin^2 x}{x^2}$ $= \lim_{x \rightarrow 0} \frac{3 x^2 + 2 \sin^2 x}{x^2}$ $= \lim_{x \rightarrow 0} \left( \frac{3 x^2}{x^2} + 2 \frac{\sin^2 x}{x^2} \right)$ $= \lim_{x \rightarrow 0} \left( 3 + 2 \left( \frac{\sin x}{x} \right)^2 \right)$ Since $\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$,we have: $= 3 + 2(1)^2 = 3 + 2 = 5$. Since the function is continuous at $x = 0$,$f(0) = \lim_{x \rightarrow 0} f(x)$,therefore $k = 5$.

If $f: R \rightarrow R$ defined by $f(x) = \begin{cases} \frac{1 + 3 x^2 - \cos 2 x}{x^2}, & x \neq 0 \\ k, & x = 0 \end{cases}$ is continuous at $x = 0$,then $k$ is equal to

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