AP EAMCET · Maths · Continuity and Differentiability
If \(f(x)= \begin{cases}\frac{x-[x]}{x-2}, & x>2 \\ b, & x=2 \\ \frac{\left|x^2-x-2\right|}{a\left(2+x-x^2\right)}, & -1 < x \leq 2 \\ 2 a-b, & x \leq-1\end{cases}\)
is continuous on \(\mathbb{R}\), then \(\lim _{x \rightarrow 0} \frac{\sin ^2 a x+x \tan b x}{x^2}=\)
- A \(0\)
- B \(1\)
- C \(2\)
- D \(3\)
Answer & Solution
Correct Answer
(C) \(2\)
Step-by-step Solution
Detailed explanation
Given \(f(x)=\left\{\begin{array}{cc}\frac{x-[x]}{x-2} & x>2 \\ b & x=2 \\ \frac{\left|x^2-x-2\right|}{a\left(2+x-x^2\right)} & -1 < x \leq 2 \\ 2 a-b & x \leq-1\end{array}\right.\) Since \(f(x)\) is continuous on \(\mathrm{R}\) So,…
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