JEE Mains · Maths · STD 12 - 5. continuity and differentiation
The function \(f : R \rightarrow R\) defined by \(f(x)=\lim _{n \rightarrow \infty} \frac{\cos (2 \pi x)-x^{2 n} \sin (x-1)}{1+x^{2 n+1}-x^{2 n}}\) is continuous for all \(x\) in.
- A \(R -\{-1\}\)
- B \(R -\{-1,1\}\)
- C \(R -\{1\}\)
- D \(R -\{0\}\)
Answer & Solution
Correct Answer
(B) \(R -\{-1,1\}\)
Step-by-step Solution
Detailed explanation
in should be given as a natural number. \(f\left(x=\left\{\begin{array}{cc} \frac{-\sin (x-1)}{x-1} & x<-1 \\ -(\sin 2+1) & x=-1 \\\cos 2 \pi x & -11\end{array}\right.\right.\) \(f ( x )\) is discontinuous at \(x =-1\) and \(x =1\)
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