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

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

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(D) For the function $f(x)$ to be continuous,it must be continuous at the points $x=1$,$x=3$,and $x=5$. At $x=1$: $\lim_{x 	o 1^-} f(x) = 5$ and $\lim_{x 	o 1^+} f(x) = a+b$. Thus,$a+b=5$ (Eq. $i$). At $x=3$: $\lim_{x 	o 3^-} f(x) = a+3b$ and $\lim_{x 	o 3^+} f(x) = b+15$. Thus,$a+3b = b+15 \Rightarrow a+2b=15$ (Eq. $ii$). At $x=5$: $\lim_{x 	o 5^-} f(x) = b+25$ and $\lim_{x 	o 5^+} f(x) = 30$. Thus,$b+25=30 \Rightarrow b=5$ (Eq. $iii$). Substituting $b=5$ into Eq. $ii$: $a+2(5)=15 \Rightarrow a=5$. Now,check if $a=5$ and $b=5$ satisfy Eq. $i$: $5+5 = 10 
eq 5$. Since the conditions for continuity at $x=1, 3, 5$ cannot be satisfied simultaneously,$f$ is not continuous for any values of $a$ and $b$.

Let $f: R \rightarrow R$ be the function defined by $f(x) = \begin{cases} 5, & \text{if } x \leq 1 \\ a+bx, & \text{if } 1 < x < 3 \\ b+5x, & \text{if } 3 \leq x < 5 \\ 30, & \text{if } x \geq 5 \end{cases}$. Then $f$ is:

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