Let f:R`to` R be a function defined as `f(x)={{:(5,"if" x le 1),(a+bx , "if" 1 lt x lt 3),(b+5x , "if" 3 le x lt 5),(30 , "if" x ge 5):}` then f is

Options

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

Answer

To determine the continuity of the piecewise function \( f(x) \) defined as:

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

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

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

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

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

Calculating these limits:

- From the left (\( x \to 1^- \)):
\[ f(1) = 5 \]

- From the right (\( x \to 1^+ \)):
\[ \lim_{x \to 1^+} f(x) = a + b(1) = a + b \]

Setting these equal for continuity:
\[ 5 = a + b \quad \text{(Equation 1)} \]

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

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

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

Calculating these limits:

- From the left (\( x \to 3^- \)):
\[ \lim_{x \to 3^-} f(x) = a + b(3) = a + 3b \]

- From the right (\( x \to 3^+ \)):
\[ \lim_{x \to 3^+} f(x) = b + 5(3) = b + 15 \]

Setting these equal for continuity:
\[ a + 3b = b + 15 \quad \text{(Equation 2)} \]

### Step 3: Solve the equations

Now we have two equations to solve:

1. \( a + b = 5 \) (Equation 1)
2. \( a + 3b = b + 15 \) (Equation 2)

From Equation 1, we can express \( a \) in terms of \( b \):
\[ a = 5 - b \]

Substituting \( a \) into Equation 2:
\[ (5 - b) + 3b = b + 15 \]
\[ 5 + 2b = b + 15 \]
\[ 2b - b = 15 - 5 \]
\[ b = 10 \]

Now substituting \( b = 10 \) back into Equation 1 to find \( a \):
\[ a + 10 = 5 \]
\[ a = 5 - 10 = -5 \]

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

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

\[ \lim_{x \to 5^-} f(x) = \lim_{x \to 5^+} f(x) = f(5) \]

Calculating these limits:

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

- From the right (\( x \to 5^+ \)):
\[ f(5) = 30 \]

Setting these equal for continuity:
\[ 35 \neq 30 \]

### Conclusion

Since the limits do not match at \( x = 5 \), the function \( f(x) \) is not continuous at this point. Therefore, the function is not continuous for any values of \( a \) and \( b \).

### Final Answer

The function \( f \) is **not continuous** for any values of \( a \) and \( b \).