Suppose `|(f'(x),f(x)),(f''(x),f'(x))|=0` where `f(x)` is continuous differentiable function with `f'(x) !=0` and satisfies `f(0)=1` and `f'(0)=2`, then `f(x)=e^(lambda x)+k`, then `lambda+k` is equal to ..........

Options

A. a) 2
B. b) 4
C. c) 0
D. d) -2

Answer

To solve the problem, we start with the given condition involving the determinant of a matrix formed by the derivatives of the function \( f(x) \). The condition is:

\[ |(f'(x), f(x)), (f''(x), f'(x))| = 0 \]

This implies that the determinant of the matrix is zero. We can express this determinant as follows:

\[ f'(x) \cdot f'(x) - f(x) \cdot f''(x) = 0 \]

This simplifies to:

\[ (f'(x))^2 - f(x) f''(x) = 0 \]

Rearranging gives us:

\[ (f')^2 = f f'' \]

Next, we can rewrite this in a more manageable form:

\[ \frac{f'}{f} = \frac{f''}{f'} \]

Now, we integrate both sides. Let \( c \) be an arbitrary constant:

\[ \int \frac{f'}{f} \, dx = \int \frac{f''}{f'} \, dx \]

This leads to:

\[ \ln |f| = \ln |f'| + c \]

Exponentiating both sides gives:

\[ f = e^{c} f' \]

Now, we can express \( f \) in terms of \( f' \):

\[ f = k f' \]

where \( k = e^{c} \) is a constant.

Next, we substitute the values of \( f(0) \) and \( f'(0) \) given in the problem. We know:

\[ f(0) = 1 \quad \text{and} \quad f'(0) = 2 \]

Substituting \( x = 0 \):

\[ f(0) = k f'(0) \implies 1 = k \cdot 2 \implies k = \frac{1}{2} \]

Now we have:

\[ f = \frac{1}{2} f' \]

From here, we can express \( f' \) in terms of \( f \):

\[ f' = 2f \]

Next, we integrate \( f' = 2f \):

\[ \frac{df}{dx} = 2f \]

Separating variables gives:

\[ \frac{df}{f} = 2dx \]

Integrating both sides results in:

\[ \ln |f| = 2x + C \]

Exponentiating yields:

\[ f = e^{2x + C} = e^{C} e^{2x} \]

Let \( A = e^{C} \), so:

\[ f(x) = A e^{2x} \]

Now using the initial condition \( f(0) = 1 \):

\[ f(0) = A e^{0} = A = 1 \]

Thus, we have:

\[ f(x) = e^{2x} \]

Now, we need to find \( \lambda + k \) where \( f(x) = e^{\lambda x} + k \). Here, \( \lambda = 2 \) and \( k = 0 \):

\[ \lambda + k = 2 + 0 = 2 \]

Thus, the final answer is:

\[ \lambda + k = 2 \]