`L e tf(x)={|x^2-3x|+a ,0lt=x<3/2 -2x+3,xgeq3/2` If `f(x)` has a local maxima at `x=3/2` , then greatest value of `|4a|` is _________

Explanation

To solve the problem, we need to analyze the function \( f(x) \) given by:

\[ f(x) = \begin{cases} |x^2 - 3x + a| & \text{for } 0 \leq x < \frac{3}{2} \\ -2x + 3 & \text{for } x \geq \frac{3}{2} \end{cases} \]

We are tasked with finding the greatest value of \( |4a| \) given that \( f(x) \) has a local maxima at \( x = \frac{3}{2} \).

### Step 1: Evaluate \( f\left(\frac{3}{2}\right) \)

First, we need to calculate \( f\left(\frac{3}{2}\right) \) using the second piece of the function since \( x = \frac{3}{2} \) falls in the range \( x \geq \frac{3}{2} \):

\[ f\left(\frac{3}{2}\right) = -2\left(\frac{3}{2}\right) + 3 = -3 + 3 = 0 \]

### Step 2: Find the left-hand limit as \( x \) approaches \( \frac{3}{2} \)

Next, we need to find the value of \( f(x) \) as \( x \) approaches \( \frac{3}{2} \) from the left:

\[ f\left(\frac{3}{2}^-\right) = \left|\left(\frac{3}{2}\right)^2 - 3\left(\frac{3}{2}\right) + a\right| \]

Calculating the quadratic expression:

\[ \left(\frac{3}{2}\right)^2 = \frac{9}{4}, \quad 3\left(\frac{3}{2}\right) = \frac{9}{2} = \frac{18}{4} \]

Thus,

\[ f\left(\frac{3}{2}^-\right) = \left|\frac{9}{4} - \frac{18}{4} + a\right| = \left|a - \frac{9}{4}\right| \]

### Step 3: Set up the condition for local maxima

For \( f(x) \) to have a local maximum at \( x = \frac{3}{2} \), the value from the left must be greater than or equal to the value from the right:

\[ \left|a - \frac{9}{4}\right| \geq 0 \]

This condition simplifies to:

\[ \left|a - \frac{9}{4}\right| \geq 0 \]

Since this is always true, we need to ensure that the left-hand limit is equal to the right-hand limit for a local maximum:

\[ \left|a - \frac{9}{4}\right| = 0 \]

This implies:

\[ a - \frac{9}{4} = 0 \quad \Rightarrow \quad a = \frac{9}{4} \]

### Step 4: Find the value of \( 4a \)

Now, we can find \( 4a \):

\[ 4a = 4 \times \frac{9}{4} = 9 \]

### Step 5: Determine the greatest value of \( |4a| \)

Since \( a \) can only take the value \( \frac{9}{4} \) for the local maximum condition to hold, we find:

\[ |4a| = |9| = 9 \]

Thus, the greatest value of \( |4a| \) is:

\[ \boxed{9} \]