Let f : mathbb{R} to mathbb{R} be a function | vedclass

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(D) For $f(x)$ to be continuous,it must be continuous at the transition points $x=1, x=3,$ and $x=5$.At $x=1$:$LHL = \lim_{x 	o 1^-} f(x) = 5$$RHL =

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not continuous for any values of $a$ and $b$

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(D) For $f(x)$ to be continuous,it must be continuous at the transition points $x=1, x=3,$ and $x=5$.At $x=1$:$LHL = \lim_{x 	o 1^-} f(x) = 5$$RHL = \lim_{x 	o 1^+} f(x) = a + b(1) = a + b$For continuity,$a + b = 5$  $(i)$At $x=3$:$LHL = \lim_{x 	o 3^-} f(x) = a + 3b$$RHL = \lim_{x 	o 3^+} f(x) = b + 5(3) = b + 15$For continuity,$a + 3b = b + 15 \implies a + 2b = 15$  $(ii)$At $x=5$:$LHL = \lim_{x 	o 5^-} f(x) = b + 5(5) = b + 25$$RHL = \lim_{x 	o 5^+} f(x) = 30$For continuity,$b + 25 = 30 \implies b = 5$Substituting $b=5$ into $(ii)$:$a + 2(5) = 15 \implies a + 10 = 15 \implies a = 5$Now check these values in $(i)$:$a + b = 5 + 5 = 10 
eq 5$Since the values $a=5$ and $b=5$ do not satisfy the condition at $x=1$,there are no values of $a$ and $b$ for which the function is continuous everywhere.

Let $f : \mathbb{R} \to \mathbb{R}$ be a function defined as $f(x) = \begin{cases} 5, & \text{if } x \le 1 \\ a + bx, & \text{if } 1 < x < 3 \\ b + 5x, & \text{if } 3 \le x < 5 \\ 30, & \text{if } x \ge 5 \end{cases}$. Then $f$ is

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