JEE Mains · Maths · STD 11 - 12. limits
Let \([ t ]\) denote the greatest integer \(\leq t\) and \(\{ t \}\) denote the fractional part of \(t\). Then integral value of \(\alpha\) for which the left hand limit of the function \(f(x)=[1+x]+\frac{\alpha^{2[x]+[x]}+[x]-1}{2[x]+\{x\}}\) at \(x=0\) is equal to \(\alpha-\frac{4}{3}\) is
- A \(1\)
- B \(3\)
- C \(5\)
- D \(7\)
Answer & Solution
Correct Answer
(B) \(3\)
Step-by-step Solution
Detailed explanation
\(f(x)=[1+x]+\frac{\alpha^{2[x]+\{x\}}+[x]-1}{2[x]+\{x\}}\) \(\lim \limits_{x \rightarrow 0^{-}} f(x)=\alpha-\frac{4}{3} \Rightarrow 0+\frac{\alpha^{-1}-2}{-1}=\alpha-\frac{4}{3}\) \(\Rightarrow 2-\frac{1}{\alpha}=\alpha-\frac{4}{3}\)…
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