Let `f : R ->(0,oo)` and `g : R -> R` be twice differentiable functions such that f" and g" are continuous functions on R. suppose `f^(prime)(2)=g(2)=0,f^"(2)!=0` and `g'(2)!=0`, If `lim_(x->2) (f(x)g(x))/(f'(x)g'(x))=1` then

Options

A. f has a local minimum at x=2
B. f has a local maximum at x=2
C. `f''(2)gtf(2)`
D. `f(x)-f''(x)=0` for atleast one `x in R`

Answer

To solve the problem step by step, we will analyze the given conditions and apply L'Hôpital's rule to evaluate the limit.

### Step 1: Understand the Given Conditions
We have two functions, \( f: \mathbb{R} \to (0, \infty) \) and \( g: \mathbb{R} \to \mathbb{R} \), which are twice differentiable. The conditions given are:
- \( f'(2) = 0 \)
- \( g(2) = 0 \)
- \( f''(2) \neq 0 \)
- \( g'(2) \neq 0 \)
- \( \lim_{x \to 2} \frac{f(x)g(x)}{f'(x)g'(x)} = 1 \)

### Step 2: Substitute into the Limit
Substituting \( x = 2 \) into the limit gives us:
\[ \frac{f(2)g(2)}{f'(2)g'(2)} = \frac{f(2) \cdot 0}{0 \cdot g'(2)} = \frac{0}{0} \]
This is an indeterminate form, so we can apply L'Hôpital's rule.

### Step 3: Apply L'Hôpital's Rule
Using L'Hôpital's rule, we differentiate the numerator and the denominator:
\[ \lim_{x \to 2} \frac{f(x)g(x)}{f'(x)g'(x)} = \lim_{x \to 2} \frac{f(x)g'(x) + g(x)f'(x)}{f'(x)g'(x) + g'(x)f''(x)} \]

### Step 4: Substitute \( x = 2 \) Again
Now, substituting \( x = 2 \):
\[ = \frac{f(2)g'(2) + g(2)f'(2)}{f'(2)g'(2) + g'(2)f''(2)} \]
Since \( g(2) = 0 \) and \( f'(2) = 0 \), this simplifies to:
\[ = \frac{f(2)g'(2)}{g'(2)f''(2)} \]

### Step 5: Simplify the Expression
We can cancel \( g'(2) \) from the numerator and denominator (since \( g'(2) \neq 0 \)):
\[ = \frac{f(2)}{f''(2)} \]

### Step 6: Set the Limit Equal to 1
According to the problem, this limit equals 1:
\[ \frac{f(2)}{f''(2)} = 1 \implies f(2) = f''(2) \]

### Step 7: Analyze the Implications
From \( f'(2) = 0 \) and \( f(2) = f''(2) \):
- Since \( f(2) > 0 \) (as \( f \) is always positive), it follows that \( f''(2) > 0 \).
- Therefore, \( f \) has a local minimum at \( x = 2 \).

### Conclusion
The conclusion is that \( f \) has a local minimum at \( x = 2 \).