Let `f:RtoR` be a function defined as `f(x)={(5,"if", xle1),(a+bx,"if", 1ltxlt3),(b+5x,"if",3lexlt5),(30,"if",xge5):}` Then f is :

Options

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

Explanation

To determine the nature of the function \( f: \mathbb{R} \to \mathbb{R} \) 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 of the function at the points where the definition of the function changes, specifically at \( x = 1 \), \( x = 3 \), and \( x = 5 \).

### Step 1: Continuity at \( x = 1 \)

For \( f(x) \) to be continuous at \( x = 1 \), we need:

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

Calculating these limits:

- \( f(1) = 5 \)
- \( \lim_{x \to 1^-} f(x) = 5 \)
- \( \lim_{x \to 1^+} f(x) = a + b(1) = a + b \)

Setting these equal gives us the first equation:

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

### Step 2: Continuity at \( x = 3 \)

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

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

Calculating these limits:

- \( f(3) = b + 5(3) = b + 15 \)
- \( \lim_{x \to 3^-} f(x) = a + b(3) = a + 3b \)
- \( \lim_{x \to 3^+} f(x) = b + 15 \)

Setting these equal gives us the second equation:

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

### Step 3: Solve the equations

Now we have two equations:

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

Subtract Equation 1 from Equation 2:

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

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

\[ a + 10 = 5 \quad \Rightarrow \quad a = 5 - 10 = -5 \]

### Step 4: Continuity at \( x = 5 \)

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

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

Calculating these limits:

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

Setting these equal gives us:

\[ 35 \neq 30 \]

### Conclusion

Since the limit from the left at \( x = 5 \) does not equal the value of the function at \( x = 5 \), \( f(x) \) is not continuous at \( x = 5 \).

Thus, the function \( f \) is not continuous for any values of \( a \) and \( b \).

### Final Answer

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

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