if `(6x^2-5x-3)/(x^2-2x+6)<=4`, then the least and the highest values of `4x^2` are`:` (a) 0 and 81 (b) 9 and 81 (c) 36 and 81 (d) None of these

Options

A. 0 and 81
B. 9 and 81
C. 36 and 81
D. None of these

Answer

To solve the inequality \( \frac{6x^2 - 5x - 3}{x^2 - 2x + 6} \leq 4 \), we will follow these steps:

### Step 1: Rearranging the Inequality
We start by moving 4 to the left side:

\[ \frac{6x^2 - 5x - 3}{x^2 - 2x + 6} - 4 \leq 0 \]

This can be rewritten as:

\[ \frac{6x^2 - 5x - 3 - 4(x^2 - 2x + 6)}{x^2 - 2x + 6} \leq 0 \]

### Step 2: Simplifying the Numerator
Now, we simplify the numerator:

\[ 6x^2 - 5x - 3 - 4(x^2 - 2x + 6) = 6x^2 - 5x - 3 - 4x^2 + 8x - 24 \]

Combining like terms gives:

\[ (6x^2 - 4x^2) + (-5x + 8x) + (-3 - 24) = 2x^2 + 3x - 27 \]

Thus, we have:

\[ \frac{2x^2 + 3x - 27}{x^2 - 2x + 6} \leq 0 \]

### Step 3: Finding Critical Points
Next, we find the critical points by setting the numerator equal to zero:

\[ 2x^2 + 3x - 27 = 0 \]

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\[ x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot (-27)}}{2 \cdot 2} \]
\[ x = \frac{-3 \pm \sqrt{9 + 216}}{4} \]
\[ x = \frac{-3 \pm \sqrt{225}}{4} \]
\[ x = \frac{-3 \pm 15}{4} \]

Calculating the two possible values:

1. \( x = \frac{12}{4} = 3 \)
2. \( x = \frac{-18}{4} = -\frac{9}{2} \)

### Step 4: Analyzing the Sign of the Expression
We need to analyze the sign of \( \frac{2x^2 + 3x - 27}{x^2 - 2x + 6} \) in the intervals determined by the critical points \( x = -\frac{9}{2} \) and \( x = 3 \).

1. **For \( x < -\frac{9}{2} \)**: Choose \( x = -5 \)
- Numerator: \( 2(-5)^2 + 3(-5) - 27 = 50 - 15 - 27 = 8 \) (positive)
- Denominator: \( (-5)^2 - 2(-5) + 6 = 25 + 10 + 6 = 41 \) (positive)

2. **For \( -\frac{9}{2} < x < 3 \)**: Choose \( x = 0 \)
- Numerator: \( 2(0)^2 + 3(0) - 27 = -27 \) (negative)
- Denominator: \( (0)^2 - 2(0) + 6 = 6 \) (positive)

3. **For \( x > 3 \)**: Choose \( x = 4 \)
- Numerator: \( 2(4)^2 + 3(4) - 27 = 32 + 12 - 27 = 17 \) (positive)
- Denominator: \( (4)^2 - 2(4) + 6 = 16 - 8 + 6 = 14 \) (positive)

### Step 5: Conclusion on Intervals
The expression \( \frac{2x^2 + 3x - 27}{x^2 - 2x + 6} \leq 0 \) holds in the interval \( x \in \left[-\frac{9}{2}, 3\right] \).

### Step 6: Finding Values of \( 4x^2 \)
Now, we need to find the minimum and maximum values of \( 4x^2 \) in this interval.

1. **At \( x = 0 \)**:
\[ 4(0)^2 = 0 \]

2. **At \( x = -\frac{9}{2} \)**:
\[ 4\left(-\frac{9}{2}\right)^2 = 4 \cdot \frac{81}{4} = 81 \]

3. **At \( x = 3 \)**:
\[ 4(3)^2 = 36 \]

### Final Values
The least value of \( 4x^2 \) is \( 0 \) and the highest value is \( 81 \).

### Answer
Thus, the least and highest values of \( 4x^2 \) are \( 0 \) and \( 81 \), respectively. The correct option is (a) 0 and 81.