`f(x)={:{(5, x le 1) ,(a+bx,1 lt x lt 3),(b+5x,3 le x lt 5),(30,xge5):}` then (a) `f(x)` is discontiuous `AA` a `in R, b in R` (b) `f(x)` is discontiuous if `a=0` &`b=5` (c) `f(x)` is discontiuous if `a=5` &`b=0` (d) `f(x)` is discontiuous if `a=-5` &`b=10`

Options

A. continuous if a =5 and b = 5
B. continuous if a =-5 and b = 10
C. continuous if a = 0 and b = 5
D. not continuous for any values of a and b

Answer

To determine the continuity of the piecewise function \( f(x) \), we need to check the continuity at the points where the definition of the function changes, specifically at \( x = 1 \), \( x = 3 \), and \( x = 5 \).

The function is defined as follows:

\[ f(x) = \begin{cases} 5 & \text{if } x \leq 1 \\ a + bx & \text{if } 1 < x < 3 \\ b + 5x & \text{if } 3 \leq x < 5 \\ 30 & \text{if } x \geq 5 \end{cases} \]

### Step 1: Check continuity at \( x = 1 \)

To check continuity at \( x = 1 \), we need to ensure that:

\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) \]

From the definition of \( f(x) \):
- \( f(1) = 5 \)
- \( \lim_{x \to 1^-} f(x) = 5 \)
- \( \lim_{x \to 1^+} f(x) = a + b(1) = a + b \)

Setting these equal gives us the equation:

\[ a + b = 5 \quad \text{(Equation 1)} \]

### Step 2: Check continuity at \( x = 3 \)

Next, we check continuity at \( x = 3 \):

\[ \lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x) = f(3) \]

From the definition of \( f(x) \):
- \( f(3) = b + 5(3) = b + 15 \)
- \( \lim_{x \to 3^-} f(x) = a + b(3) = a + 3b \)

Setting these equal gives us the equation:

\[ a + 3b = b + 15 \quad \text{(Equation 2)} \]

Rearranging Equation 2:

\[ a + 2b = 15 \]

### Step 3: Solve the system of equations

Now we have the system of equations:

1. \( a + b = 5 \)
2. \( a + 2b = 15 \)

Subtract Equation 1 from Equation 2:

\[ (a + 2b) - (a + b) = 15 - 5 \]
\[ b = 10 \]

Substituting \( b = 10 \) back into Equation 1:

\[ a + 10 = 5 \implies a = 5 - 10 = -5 \]

### Step 4: Check continuity at \( x = 5 \)

Now we check continuity at \( x = 5 \):

\[ \lim_{x \to 5^-} f(x) = b + 5(5) = 10 + 25 = 35 \]
\[ f(5) = 30 \]

Since \( \lim_{x \to 5^-} f(x) \neq f(5) \), the function is discontinuous at \( x = 5 \).

### Conclusion

The function \( f(x) \) is discontinuous for \( a = -5 \) and \( b = 10 \).