Let `f(x) = (alpha x^(2))/(x + 1), x ne - 1 `, The value of `alpha` for which f(a) = a , `(a ne 0 )` is

Options

A. `1 - (1)/(a)`
B. `(1)/(a)`
C. `1 + (1)/(a)`
D. `(1)/(a) - 1 `

Answer

To find the value of \(\alpha\) for which \(f(a) = a\), given the function \(f(x) = \frac{\alpha x^2}{x + 1}\) and \(a \neq 0\), we can follow these steps:

### Step 1: Set up the equation
We know that \(f(a) = a\). Therefore, we can write:
\[ f(a) = \frac{\alpha a^2}{a + 1} \]
Setting this equal to \(a\), we have:
\[ \frac{\alpha a^2}{a + 1} = a \]

### Step 2: Cross-multiply to eliminate the fraction
To eliminate the fraction, we can cross-multiply:
\[ \alpha a^2 = a(a + 1) \]

### Step 3: Expand the right side
Now, we expand the right side:
\[ \alpha a^2 = a^2 + a \]

### Step 4: Rearrange the equation
Next, we rearrange the equation to isolate \(\alpha\):
\[ \alpha a^2 - a^2 = a \]
Factoring out \(a^2\) on the left side gives:
\[ (\alpha - 1)a^2 = a \]

### Step 5: Solve for \(\alpha\)
Now, we can solve for \(\alpha\):
\[ \alpha - 1 = \frac{a}{a^2} \]
This simplifies to:
\[ \alpha - 1 = \frac{1}{a} \]
Adding 1 to both sides results in:
\[ \alpha = 1 + \frac{1}{a} \]

### Final Result
Thus, the value of \(\alpha\) for which \(f(a) = a\) is:
\[ \alpha = 1 + \frac{1}{a} \]

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