Let `f(a) = g(a) =k` and their nth derivatives `f^(n)(a), g^(n)(a)` exist and are not equal for some n. Further if
`"lt"_(x to a) (f(a) g(x) -f(a) -g(a) f(x) + g(a))/(g(x)-f(x)) =4`, then the value of k =

Options

A. 0
B. 1
C. 2
D. 4

Answer

To solve the problem, we start with the given limit expression:

\[ \lim_{x \to a} \frac{f(a) g(x) - f(a) - g(a) f(x) + g(a)}{g(x) - f(x)} = 4 \]

Given that \( f(a) = g(a) = k \), we can substitute these values into the limit expression:

\[ \lim_{x \to a} \frac{k g(x) - k - k f(x) + k}{g(x) - f(x)} = 4 \]

This simplifies to:

\[ \lim_{x \to a} \frac{k (g(x) - f(x))}{g(x) - f(x)} = 4 \]

Now, we notice that both the numerator and denominator approach 0 as \( x \to a \), which gives us the indeterminate form \( \frac{0}{0} \). To resolve this, we can apply L'Hôpital's Rule, which states that if we have an indeterminate form \( \frac{0}{0} \), we can take the derivative of the numerator and the derivative of the denominator.

Taking the derivative of the numerator:

\[ \text{Numerator: } \frac{d}{dx}(k g(x) - k - k f(x) + k) = k g'(x) - k f'(x) \]

Taking the derivative of the denominator:

\[ \text{Denominator: } \frac{d}{dx}(g(x) - f(x)) = g'(x) - f'(x) \]

Now, we apply L'Hôpital's Rule:

\[ \lim_{x \to a} \frac{k(g'(x) - f'(x))}{g'(x) - f'(x)} = 4 \]

Assuming \( g'(x) \neq f'(x) \) at \( x = a \), we can simplify this limit:

\[ k = 4 \]

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

\[ \boxed{4} \]