Let `f(x)=(20)/(4x^2-9x^2+6x)`
Statement -1 : Range of f=[6,20]
Statement -2 f(x) increases (1/2,1) and decrease on `(1,oo) cup (-oo,0)cup (0,1//2)`

Options

A. Statement-1 True statement -1 is True,Statement -2 is True statement -2 is a correct explanation for Statement-6
B. Statement-1 True statement -1 is True,Statement -2 is True statement -2 is not a correct explanation for Statement-6
C. Statement-1 True statement -1 is True,Statement -2 is False
D. Statement-1 is False ,Statement -2 is True

Explanation

To solve the problem step by step, we will analyze the function \( f(x) = \frac{20}{4x^3 - 9x^2 + 6x} \) and determine the validity of the two statements regarding its range and behavior (increasing/decreasing).

### Step 1: Simplify the Function
We start with the function:
\[ f(x) = \frac{20}{4x^3 - 9x^2 + 6x} \]
We can rewrite the denominator:
\[ f(x) = \frac{20}{x(4x^2 - 9x + 6)} \]

### Step 2: Find Critical Points
To analyze the increasing and decreasing behavior of \( f(x) \), we need to find the derivative \( f'(x) \) and set it to zero to find critical points.

Using the quotient rule:
\[ f'(x) = \frac{(0)(4x^3 - 9x^2 + 6x) - 20(12x^2 - 18x + 6)}{(4x^3 - 9x^2 + 6x)^2} \]
This simplifies to:
\[ f'(x) = \frac{-20(12x^2 - 18x + 6)}{(4x^3 - 9x^2 + 6x)^2} \]
Setting the numerator to zero:
\[ 12x^2 - 18x + 6 = 0 \]
Dividing by 6:
\[ 2x^2 - 3x + 1 = 0 \]
Factoring:
\[ (2x - 1)(x - 1) = 0 \]
Thus, the critical points are:
\[ x = \frac{1}{2}, \quad x = 1 \]

### Step 3: Determine Intervals of Increase and Decrease
We will test the intervals determined by the critical points \( x = \frac{1}{2} \) and \( x = 1 \):
- For \( x < \frac{1}{2} \)
- For \( \frac{1}{2} < x < 1 \)
- For \( x > 1 \)

Choose test points:
1. \( x = 0 \) (for \( x < \frac{1}{2} \))
2. \( x = \frac{3}{4} \) (for \( \frac{1}{2} < x < 1 \))
3. \( x = 2 \) (for \( x > 1 \))

Calculating \( f'(x) \) at these points:
- \( f'(0) < 0 \) (decreasing)
- \( f'(\frac{3}{4}) > 0 \) (increasing)
- \( f'(2) < 0 \) (decreasing)

### Step 4: Analyze the Behavior
From the analysis:
- \( f(x) \) is decreasing on \( (-\infty, 0) \cup (1, \infty) \)
- \( f(x) \) is increasing on \( \left(\frac{1}{2}, 1\right) \)

### Step 5: Determine the Range of \( f(x) \)
To find the range, we evaluate \( f(x) \) at the critical points:
- As \( x \to 0 \), \( f(x) \to \infty \)
- At \( x = \frac{1}{2} \):
\[ f\left(\frac{1}{2}\right) = \frac{20}{4\left(\frac{1}{2}\right)^3 - 9\left(\frac{1}{2}\right)^2 + 6\left(\frac{1}{2}\right)} = \frac{20}{\frac{1}{2} - \frac{9}{4} + 3} = \frac{20}{\frac{1}{2} - \frac{9}{4} + \frac{12}{4}} = \frac{20}{\frac{1}{2} + \frac{3}{4}} = \frac{20}{\frac{5}{4}} = 16 \]
- At \( x = 1 \):
\[ f(1) = \frac{20}{4 - 9 + 6} = \frac{20}{1} = 20 \]

Thus, the range of \( f(x) \) is \( (16, \infty) \).

### Conclusion
- **Statement 1**: False (the range is not \([6, 20]\))
- **Statement 2**: True (the function increases on \((\frac{1}{2}, 1)\) and decreases on the other intervals)