Question

Let \(x = 2\) be a root of the equation \(x^2 + px + q = 0\) and \(f(x)=\begin{cases}\frac{1-cos(x^2-4px+q^2+8q+16)}{(x-2p)^4} & x≠2p\\0 & x=2p\end{cases}\).
Then \(lim_{x \rightarrow 2p}[f(x)]\), where [·] denotes greatest integer function, is

• A. 1
• B. 2
• C. 0
• D. -1

Answer 1

The Correct Option is C

To solve the given problem, we first need to understand the function \( f(x) \) and evaluate its limit as \( x \) approaches \( 2p \).

The problem states that \( x = 2 \) is a root of the quadratic equation \( x^2 + px + q = 0 \). Using this information, let's first find out the implications.

1. Substitute \( x = 2 \) into the equation:
   \( 2^2 + 2p + q = 0 \)
   \( 4 + 2p + q = 0 \)
2. Rearrange to find an expression:
   \( q = -4 - 2p \)

Next, substitute the value of \( q \) into \( f(x) \) and evaluate the limit as \( x \to 2p \).

The function \( f(x) \) is defined as:
\( f(x) = \frac{1-\cos(x^2 - 4px + q^2 + 8q + 16)}{(x - 2p)^4} \) for \( x \neq 2p \) and 0 for \( x = 2p \).

1. Substitute \( q = -4 - 2p \) into the expression inside the cosine:
   \( q^2 = (-4 - 2p)^2 = 16 + 16p + 4p^2 \)
2. Evaluate \( q^2 + 8q + 16 \):
   \( 16 + 16p + 4p^2 + 8(-4 - 2p) + 16 = 4p^2 \)
3. This implies:
   \( x^2 - 4px + q^2 + 8q + 16 = x^2 - 4px + 4p^2 = (x - 2p)^2 \)
4. Thus, as \( x \to 2p \), the expression becomes:
   \( \frac{1-\cos((x - 2p)^2)}{(x - 2p)^4} \)

For small values of \( t \) (where \( t = (x - 2p) \)), the approximation \(\cos t \approx 1 - \frac{t^2}{2}\) holds. Therefore:
\( 1 - \cos((x-2p)^2) \approx \frac{((x-2p)^2)^2}{2}\)

1. Substitute the approximation:
   \(\frac{((x-2p)^2)^2/2}{(x-2p)^4} = \frac{1}{2}\)
2. Therefore, \( \lim_{x \to 2p} f(x) = \frac{1}{2}\)

Since the greatest integer function is denoted by [·], we have:
\( [\lim_{x \rightarrow 2p} f(x)] = [\frac{1}{2}] = 0 \)

Therefore, the answer is 0.