The number of values of `lamda` for which `(lamda^(2)-3 lamda +2) x^(2) +(lamda^(2)-5 lamda +6) x +lamda^(2)-4=0` is an identity in x is

Options

A. 1
B. 2
C. -2
D. 0

Answer

To solve the problem, we need to determine the number of values of \( \lambda \) for which the quadratic equation

\[ (\lambda^2 - 3\lambda + 2)x^2 + (\lambda^2 - 5\lambda + 6)x + (\lambda^2 - 4) = 0 \]

is an identity in \( x \). This means that the coefficients of \( x^2 \), \( x \), and the constant term must all be zero.

### Step 1: Set the coefficient of \( x^2 \) to zero

The coefficient of \( x^2 \) is \( \lambda^2 - 3\lambda + 2 \). We set it to zero:

\[ \lambda^2 - 3\lambda + 2 = 0 \]

### Step 2: Factor the quadratic equation

Factoring the quadratic gives:

\[ (\lambda - 2)(\lambda - 1) = 0 \]

### Step 3: Find the values of \( \lambda \)

From the factors, we find:

\[ \lambda - 2 = 0 \quad \Rightarrow \quad \lambda = 2 \]
\[ \lambda - 1 = 0 \quad \Rightarrow \quad \lambda = 1 \]

So, the values of \( \lambda \) from this equation are \( \lambda = 1 \) and \( \lambda = 2 \).

### Step 4: Set the coefficient of \( x \) to zero

Next, we set the coefficient of \( x \) to zero:

\[ \lambda^2 - 5\lambda + 6 = 0 \]

### Step 5: Factor this quadratic equation

Factoring gives:

\[ (\lambda - 3)(\lambda - 2) = 0 \]

### Step 6: Find the values of \( \lambda \)

From this factorization, we find:

\[ \lambda - 3 = 0 \quad \Rightarrow \quad \lambda = 3 \]
\[ \lambda - 2 = 0 \quad \Rightarrow \quad \lambda = 2 \]

So, the values of \( \lambda \) from this equation are \( \lambda = 2 \) and \( \lambda = 3 \).

### Step 7: Set the constant term to zero

Now, we set the constant term to zero:

\[ \lambda^2 - 4 = 0 \]

### Step 8: Solve for \( \lambda \)

This can be factored as:

\[ (\lambda - 2)(\lambda + 2) = 0 \]

### Step 9: Find the values of \( \lambda \)

From this factorization, we find:

\[ \lambda - 2 = 0 \quad \Rightarrow \quad \lambda = 2 \]
\[ \lambda + 2 = 0 \quad \Rightarrow \quad \lambda = -2 \]

So, the values of \( \lambda \) from this equation are \( \lambda = 2 \) and \( \lambda = -2 \).

### Step 10: Identify common values of \( \lambda \)

Now we have the values of \( \lambda \) from all three conditions:

1. From \( a = 0 \): \( \lambda = 1, 2 \)
2. From \( b = 0 \): \( \lambda = 2, 3 \)
3. From \( c = 0 \): \( \lambda = 2, -2 \)

The common value across all three conditions is \( \lambda = 2 \).

### Conclusion

Thus, there is only **one value of \( \lambda \)** for which the given quadratic equation is an identity in \( x \).

The answer is:

\[ \text{Number of values of } \lambda = 1 \]

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