If `g(x)=x^2+x-2a n d1/2gof(x)=2x^2-5x+2,` then which is not a possible `f(x)?` ``(a)2x-3` (b) `-2x+2` (c)`x-3` (d) None of these

Options

A. `2x-3`
B. `-2x+2`
C. `x-3`
D. None of these

Explanation

To solve the problem, we need to find which function \( f(x) \) is not possible given the equations \( g(x) = x^2 + x - 2 \) and \( \frac{1}{2} g(f(x)) = 2x^2 - 5x + 2 \).

### Step-by-Step Solution:

1. **Understanding the given functions**:
We have:
\[ g(x) = x^2 + x - 2 \]
and
\[ \frac{1}{2} g(f(x)) = 2x^2 - 5x + 2. \]
To eliminate the fraction, we multiply both sides by 2:
\[ g(f(x)) = 4x^2 - 10x + 4. \]

2. **Substituting \( f(x) \) into \( g(x) \)**:
We know that:
\[ g(f(x)) = f(x)^2 + f(x) - 2. \]
Therefore, we can set the two expressions for \( g(f(x)) \) equal to each other:
\[ f(x)^2 + f(x) - 2 = 4x^2 - 10x + 4. \]

3. **Rearranging the equation**:
Rearranging gives us:
\[ f(x)^2 + f(x) - 4x^2 + 10x - 6 = 0. \]

4. **Identifying the quadratic form**:
This is a quadratic equation in terms of \( f(x) \):
\[ f(x)^2 + f(x) + (-4x^2 + 10x - 6) = 0. \]
Here, \( a = 1 \), \( b = 1 \), and \( c = -4x^2 + 10x - 6 \).

5. **Finding the discriminant**:
For \( f(x) \) to have real values, the discriminant must be non-negative:
\[ D = b^2 - 4ac = 1^2 - 4(1)(-4x^2 + 10x - 6). \]
Simplifying this:
\[ D = 1 + 16x^2 - 40x + 24 = 16x^2 - 40x + 25. \]

6. **Setting the discriminant to be non-negative**:
We need:
\[ 16x^2 - 40x + 25 \geq 0. \]
This is a quadratic equation. We can find its roots using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{40 \pm \sqrt{0}}{32} = \frac{40}{32} = \frac{5}{4}. \]
Since the discriminant is zero, the quadratic touches the x-axis at \( x = \frac{5}{4} \) and is non-negative everywhere else.

7. **Finding possible \( f(x) \)**:
The roots of the quadratic equation give us possible forms for \( f(x) \):
\[ f(x) = \frac{-1 \pm \sqrt{16x^2 - 40x + 25}}{2}. \]
This leads to two potential functions:
\[ f(x) = 2x - 3 \quad \text{and} \quad f(x) = -2x + 2. \]
We can also check for \( f(x) = x - 3 \).

8. **Checking the options**:
- \( f(x) = 2x - 3 \) is valid.
- \( f(x) = -2x + 2 \) is valid.
- \( f(x) = x - 3 \) needs to be checked.

Substituting \( f(x) = x - 3 \) into the quadratic equation:
\[ (x - 3)^2 + (x - 3) - 4x^2 + 10x - 6 = 0. \]
This simplifies to:
\[ x^2 - 6x + 9 + x - 3 - 4x^2 + 10x - 6 = 0, \]
which leads to:
\[ -3x^2 + 5x = 0. \]
This does not yield a valid quadratic form, hence \( f(x) = x - 3 \) is not a possible function.

### Conclusion:
The function that is not a possible \( f(x) \) is:
**(c) \( x - 3 \)**.