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(B) For a function $f(x)$ to be continuous at $x = 2$,the left-hand limit,right-hand limit,and the value of the function at $x = 2$ must be equal.$1$.

Vedclass · Accepted Answer

$3$

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(B) For a function $f(x)$ to be continuous at $x = 2$,the left-hand limit,right-hand limit,and the value of the function at $x = 2$ must be equal.$1$. Right-hand limit $(RHL)$: $\lim_{x 	o 2^+} f(x) = \lim_{x 	o 2^+} (2x - 1) = 2(2) - 1 = 3$.$2$. Left-hand limit $(LHL)$: $\lim_{x 	o 2^-} f(x) = \lim_{x 	o 2^-} (x^2 - 1) = (2)^2 - 1 = 4 - 1 = 3$.$3$. Value of the function at $x = 2$: $f(2) = k$.Since the function is continuous,$	ext{LHL} = 	ext{RHL} = f(2)$.Therefore,$3 = 3 = k$,which implies $k = 3$.

The function $f$ is defined by $f(x) = \begin{cases} 2x - 1, & \text{if } x > 2 \\ k, & \text{if } x = 2 \\ x^2 - 1, & \text{if } x < 2 \end{cases}$. If $f$ is continuous at $x = 2$,then the value of $k$ is equal to:

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