JEE Mains · Maths · STD 12 - 5. continuity and differentiation
Let \(\mathrm{a}, \mathrm{b} \in R, \mathrm{b} \neq 0\), Define a function \(f(x)= \begin{cases}\operatorname{a} \sin \frac{\pi}{2}(x-1), & \text { for } x \leq 0 \\ \frac{\tan 2 x-\sin 2 x}{b x^{3}}, & \text { for } x>0\end{cases}\) If \(f\) is continuous at \(x=0\), then \(10-a b\) is equal to ...... .
- A \(10\)
- B \(14\)
- C \(8\)
- D \(3\)
Answer & Solution
Correct Answer
(B) \(14\)
Step-by-step Solution
Detailed explanation
\(f(x)= \begin{cases}a \sin \frac{\pi}{2}(x-1), & x \leq 0 \\ \frac{\tan 2 x-\sin 2 x}{b x^{3}}, & x>0\end{cases}\) For continuity at \(' 0 '\) \(\lim _{x \rightarrow 0^{+}} f(x)=f(0)\) \(\Rightarrow \lim _{x \rightarrow 0^{+}} \frac{\tan 2 x-\sin 2 x}{b x^{3}}=-a\)…
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