JEE Mains · Maths · STD 11 - 12. limits
Let \(\{x\}\) denote the fractional part of \(\mathrm{x}\) and \(f(x)=\frac{\cos ^{-1}\left(1-\{x\}^2\right) \sin ^{-1}(1-\{x\})}{\{x\}-\{x\}^3}, x \neq 0\). If \(L\) and \(\mathrm{R}\) respectively denotes the left hand limit and the right hand limit of \(f(x)\) at \(x=0\), then \(\frac{32}{\pi^2}\left(L^2+R^2\right)\) is equal to ...........
- A \(18\)
- B \(20\)
- C \(22\)
- D \(30\)
Answer & Solution
Correct Answer
(A) \(18\)
Step-by-step Solution
Detailed explanation
Finding right hand limit \(\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)\) \(=\lim _{h \rightarrow 0} f(h)\) \(=\lim _{h \rightarrow 0} \frac{\cos ^{-1}\left(1-h^2\right) \sin ^{-1}(1-h)}{h\left(1-h^2\right)}\)…
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