`if f(x) ={{:((8^(x)-4^(x)-2^(x)+1)/(x^(2)),xgt0),( x^(2) , x le 0):}`
is continuous at x=0 , then the value of `lamda ` is

Options

A. `4 log _(e)2`
B. `2log_(e)2`
C. `log_(e)2`
D. none of these

Answer

To determine the value of \( \lambda \) for the function

\[ f(x) = \begin{cases} \frac{8^x - 4^x - 2^x + 1}{x^2} & \text{if } x > 0 \\ x^2 + \lambda \ln 4 & \text{if } x \leq 0 \end{cases} \]

to be continuous at \( x = 0 \), we need to ensure that the left-hand limit as \( x \) approaches 0 from the left is equal to the right-hand limit as \( x \) approaches 0 from the right, and also equal to \( f(0) \).

### Step 1: Find \( f(0) \)

For \( x \leq 0 \):

\[ f(0) = 0^2 + \lambda \ln 4 = \lambda \ln 4 \]

### Step 2: Find the right-hand limit \( f(0^+) \)

For \( x > 0 \):

\[ f(0^+) = \lim_{x \to 0^+} \frac{8^x - 4^x - 2^x + 1}{x^2} \]

Substituting \( x = 0 \):

\[ = \frac{8^0 - 4^0 - 2^0 + 1}{0^2} = \frac{1 - 1 - 1 + 1}{0} = \frac{0}{0} \]

This is an indeterminate form, so we apply L'Hôpital's Rule.

### Step 3: Apply L'Hôpital's Rule

Differentiate the numerator and denominator:

- The derivative of the numerator \( 8^x - 4^x - 2^x + 1 \) is:

\[ 8^x \ln 8 - 4^x \ln 4 - 2^x \ln 2 \]

- The derivative of the denominator \( x^2 \) is:

\[ 2x \]

Thus, we have:

\[ f(0^+) = \lim_{x \to 0^+} \frac{8^x \ln 8 - 4^x \ln 4 - 2^x \ln 2}{2x} \]

Substituting \( x = 0 \):

\[ = \frac{8^0 \ln 8 - 4^0 \ln 4 - 2^0 \ln 2}{2 \cdot 0} = \frac{\ln 8 - \ln 4 - \ln 2}{0} = \frac{0}{0} \]

Again, we apply L'Hôpital's Rule.

### Step 4: Apply L'Hôpital's Rule Again

Differentiate again:

- The second derivative of the numerator is:

\[ 8^x (\ln 8)^2 - 4^x (\ln 4)^2 - 2^x (\ln 2)^2 \]

- The second derivative of the denominator is:

\[ 2 \]

Thus, we have:

\[ f(0^+) = \lim_{x \to 0^+} \frac{8^x (\ln 8)^2 - 4^x (\ln 4)^2 - 2^x (\ln 2)^2}{2} \]

Substituting \( x = 0 \):

\[ = \frac{1 \cdot (\ln 8)^2 - 1 \cdot (\ln 4)^2 - 1 \cdot (\ln 2)^2}{2} \]

### Step 5: Simplify the expression

Using the properties of logarithms:

\[ \ln 8 = \ln(2^3) = 3 \ln 2, \quad \ln 4 = \ln(2^2) = 2 \ln 2 \]

Thus,

\[ = \frac{(3 \ln 2)^2 - (2 \ln 2)^2 - (\ln 2)^2}{2} = \frac{9 (\ln 2)^2 - 4 (\ln 2)^2 - (\ln 2)^2}{2} = \frac{4 (\ln 2)^2}{2} = 2 (\ln 2)^2 \]

### Step 6: Set the limits equal for continuity

For continuity at \( x = 0 \):

\[ \lambda \ln 4 = 2 (\ln 2)^2 \]

### Step 7: Solve for \( \lambda \)

Using \( \ln 4 = 2 \ln 2 \):

\[ \lambda (2 \ln 2) = 2 (\ln 2)^2 \]

Dividing both sides by \( 2 \ln 2 \) (assuming \( \ln 2 \neq 0 \)):

\[ \lambda = \frac{2 (\ln 2)^2}{2 \ln 2} = \ln 2 \]

### Final Answer

Thus, the value of \( \lambda \) is:

\[ \lambda = \ln 2 \]