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

Step 1: Using the given equation. We are given that \( x = 2 \) is a root of the equation: \[ x^2 + px + q = 0. \] Substitute \( x = 2 \) into this equation: \[ 2^2 + 2p + q = 0 \quad \Rightarrow \quad 4 + 2p + q = 0 \quad \Rightarrow \quad q = -2p - 4. \quad \cdots (1) \] Step 2: Substituting into the function \( f(x) \). The function is given as: \[ f(x) = \frac{1 - \cos(x^2 - 4px + q^2 + 8q + 16)}{(x - 2p)^4}. \] Substitute \( q = -2p - 4 \) from equation (1): \[ f(x) = \frac{1 - \cos(x^2 - 4px + (-2p - 4)^2 + 8(-2p - 4) + 16)}{(x - 2p)^4}. \] Simplify the expression inside the cosine: \[ q^2 + 8q + 16 = (-2p - 4)^2 + 8(-2p - 4) + 16. \] Expanding this gives: \[ q^2 + 8q + 16 = 4p^2 + 16p + 16 - 16p - 32 + 16 = 4p^2 + 0p + 0 = 4p^2. \] Now the function becomes: \[ f(x) = \frac{1 - \cos(x^2 - 4px + 4p^2)}{(x - 2p)^4}. \] Step 3: Apply L'Hôpital's Rule.
Since the expression is in the form \( \frac{0}{0} \) as \( x \to 2p \), we can apply L'Hôpital's Rule. First, differentiate the numerator and denominator with respect to \( x \).
Numerator: \[ \frac{d}{dx} \left[ 1 - \cos(x^2 - 4px + 4p^2) \right] = \sin(x^2 - 4px + 4p^2) \cdot (2x - 4p). \] Denominator: \[ \frac{d}{dx} \left[ (x - 2p)^4 \right] = 4(x - 2p)^3. \] Thus, the limit becomes: \[ \lim_{x \to 2p} \frac{\sin(x^2 - 4px + 4p^2) \cdot (2x - 4p)}{4(x - 2p)^3}. \] Step 4: Simplifying the limit.
At \( x = 2p \), the numerator becomes \( \sin(0) \cdot 0 = 0 \) and the denominator also becomes \( 0 \), so we apply L'Hôpital's Rule again. The limit simplifies to: \[ \lim_{x \to 2p} f(x) = 0. \] Step 5: Conclusion.
Thus, the correct value of the limit is \( 0 \), and the correct answer is option (3).