Let `x=2` be a root of the equation `x^(2)+px+q=0` and `f(x)={((1-cos(x^(2)-4px+q^(2)+8q+16))/((x-2p)^(4)),,xne2p),(0,,x=2p):}` Then `lim_(x rarr2p)[f(x)]`,where [.] denotes greatest integer function, is

Options

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

Answer

To solve the problem step by step, we need to analyze the given information and compute the limit of the function \( f(x) \) as \( x \) approaches \( 2p \).

### Step 1: Understanding the Function
We have the function defined as:
\[ f(x) = \frac{1 - \cos(x^2 - 4px + q^2 + 8q + 16)}{(x - 2p)^4}, \quad x \neq 2p \]
and \( f(2p) = 0 \).

### Step 2: Finding the Limit
We need to compute:
\[ \lim_{x \to 2p} f(x) \]
This involves substituting \( x = 2p + h \) where \( h \to 0 \):
\[ f(2p + h) = \frac{1 - \cos((2p + h)^2 - 4p(2p + h) + q^2 + 8q + 16)}{(h)^4} \]

### Step 3: Simplifying the Argument of Cosine
Calculating the expression inside the cosine:
\[ (2p + h)^2 = 4p^2 + 4ph + h^2 \]
\[ -4p(2p + h) = -8p^2 - 4ph \]
Combining these:
\[ (2p + h)^2 - 4p(2p + h) = 4p^2 + 4ph + h^2 - 8p^2 - 4ph = -4p^2 + h^2 \]
Thus, we have:
\[ f(2p + h) = \frac{1 - \cos(-4p^2 + h^2 + q^2 + 8q + 16)}{h^4} \]

### Step 4: Applying the Limit
As \( h \to 0 \), the expression \( -4p^2 + h^2 + q^2 + 8q + 16 \) approaches \( -4p^2 + q^2 + 8q + 16 \). We can denote this limit as \( L \):
\[ L = -4p^2 + q^2 + 8q + 16 \]
Thus:
\[ f(2p + h) = \frac{1 - \cos(L)}{h^4} \]

### Step 5: Using Taylor Expansion
Using the Taylor expansion for \( \cos \) around 0:
\[ 1 - \cos(L) \approx \frac{L^2}{2} \quad \text{for small } L \]
So:
\[ f(2p + h) \approx \frac{\frac{L^2}{2}}{h^4} \]

### Step 6: Evaluating the Limit
As \( h \to 0 \):
\[ \lim_{h \to 0} f(2p + h) = \lim_{h \to 0} \frac{\frac{L^2}{2}}{h^4} \]
This limit diverges unless \( L = 0 \). If \( L = 0 \), then:
\[ L = -4p^2 + q^2 + 8q + 16 = 0 \]

### Step 7: Final Limit Calculation
If \( L \neq 0 \), the limit diverges to infinity. If \( L = 0 \), then:
\[ \lim_{x \to 2p} f(x) = \infty \]
Thus, we need to evaluate:
\[ \lim_{x \to 2p} [f(x)] \]

### Step 8: Greatest Integer Function
Since \( f(x) \) approaches infinity, the greatest integer function \( [f(x)] \) will also approach infinity.

### Conclusion
Thus, the limit we are looking for is:
\[ \lim_{x \to 2p} [f(x)] = \infty \]