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JEE Mains · Maths · STD 12 - 5. continuity and differentiation

અહી \(\mathrm{a}, \mathrm{b} \in R, \mathrm{b} \neq 0\), વિધેય નીચે મુજબ વ્યાખ્યાયિત છે \(f(x)= \begin{cases}\operatorname{a} \sin \frac{\pi}{2}(x-1), & \text { for } x \leq 0 \\ \frac{\tan 2 x-\sin 2 x}{b x^{3}}, & \text { for } x>0\end{cases}\) જો \(f\) એ \(x=0\) આગળ સતત હોય તો \(10-a b\) ની કિમંત મેળવો.

  1. A \(10\)
  2. B \(14\)
  3. C \(8\)
  4. D \(3\)
Verified Solution

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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