If `f(x)={{:(5x-4",",0ltxle1),(4x^(2)+3ax",",1ltxlt2):}`

Options

A. 1
B. 0
C. `-1`
D. `(1)/(3)`

Answer

To solve the problem, we need to ensure that the function \( f(x) \) is continuous at \( x = 1 \). The function is defined piecewise as follows:

\[ f(x) = \begin{cases} 5x - 4 & \text{for } 0 < x \leq 1 \\ 4x^2 + 3ax & \text{for } 1 < x < 2 \end{cases} \]

### Step 1: Find the left-hand limit (LHL) as \( x \) approaches 1
The left-hand limit as \( x \) approaches 1 is given by the first piece of the function:

\[ \text{LHL} = \lim_{x \to 1^-} f(x) = 5(1) - 4 = 5 - 4 = 1 \]

### Step 2: Find the right-hand limit (RHL) as \( x \) approaches 1
The right-hand limit as \( x \) approaches 1 is given by the second piece of the function:

\[ \text{RHL} = \lim_{x \to 1^+} f(x) = 4(1)^2 + 3a(1) = 4 + 3a \]

### Step 3: Set LHL equal to RHL for continuity
For the function to be continuous at \( x = 1 \), we need:

\[ \text{LHL} = \text{RHL} \]

Substituting the values we found:

\[ 1 = 4 + 3a \]

### Step 4: Solve for \( a \)
Rearranging the equation gives:

\[ 3a = 1 - 4 \]
\[ 3a = -3 \]
\[ a = -1 \]

### Conclusion
The value of \( a \) that makes the function continuous at \( x = 1 \) is:

\[ \boxed{-1} \]