Limits, Continuity & Differentiability
Limit using 1-cos approximation and L'Hopital
nta_pyq_2023_jan
Grade 12
Question:
Let x = 2 be a root of the equation x^2 + px + q = 0 and f(x) = \begin{cases} \dfrac{1-\cos(x^2-4px+q^2+8q+16)}{(x-2p)^4}, & x \neq 2p \\ 0, & x = 2p \end{cases}. Then \lim_{x \to 2p^+} [f(x)], where [.] denotes greatest integer function, is
Step-by-Step Solution
Key Concept: Use 1-\cos\theta \approx \theta^2/2 for \theta \to 0. Substitute the root condition to simplify the numerator expression.
Since x=2 is a root: 4+2p+q=0. Near x=2p, the numerator \approx \frac{1}{2}(\text{expression})^2 / (x-2p)^4 \to 1/2. So [f(x)] = [1/2] = 0.
Correct Answer: 3