1. For continuity of the piecewise function, we must check the junction points: $x=1,\; x=3,\; x=5.$

   The function is

   5, & x\le 1 \\ a+bx, & 1<x<3 \\ b+5x, & 3\le x<5 \\ 30, & x\ge 5 \end{cases}$$

2. Continuity at $x=1$:

   Since for $x\le 1$, $f(1)=5$.

   Right-hand limit at $x=1$ is $\lim_{x\to 1^+}f(x)=a+b.$

   For continuity, a+b=5. \tag{1}

3. Continuity at $x=3$:

   Left-hand limit at $x=3$ from the second piece: $\lim_{x\to 3^-}f(x)=a+3b.$

   Value at $x=3$ from the third piece: $f(3)=b+5(3)=b+15.$

   For continuity, $a+3b=b+15$ a+2b=15. \tag{2}

4. Continuity at $x=5$:

   Left-hand limit at $x=5$ from the third piece: $\lim_{x\to 5^-}f(x)=b+25.$

   Value at $x=5$ from the fourth piece: $f(5)=30.$

   For continuity, $b+25=30$ b=5. \tag{3}

5. Substitute $b=5$ into (1): $a+5=5 \implies a=0.$

   Check in (2): $a+2b=0+2(5)=10 \ne 15.$

   So the condition at $x=3$ fails.

6. Therefore, no values of $a$ and $b$ can make the function continuous at all three points simultaneously.

7. Checking options:

   • A: $a=0, b=5$ fails at $x=3$.
   • B: $a=-5, b=10$ fails at $x=1$ and $x=5$.
   • C: $a=5, b=5$ fails at $x=1$ and $x=3$.
   • D: true.

Hence, the function is not continuous for any values of $a$ and $b$.