JEE Advanced · Mathematics · 16. Limits
Let \(\alpha(a)\) and \(\beta(a)\) be the roots of the equation \((\sqrt[3]{1+a}-1) x^{2}+(\sqrt{1+a}-1) x+(\sqrt[6]{1+a}-1)=0\) where \(a\) \(>-1\). Then \(\lim _{a \rightarrow 0^{+}} \alpha(a)\) and \(\lim _{x \rightarrow 0^{+}} \beta(a)\) are
- A \(-\frac{5}{2}\) and 1
- B \(-\frac{1}{2}\) and \(-1\)
- C \(-\frac{7}{2}\) and 2
- D \(-\frac{9}{2}\) and 3
Answer & Solution
Correct Answer
(B) \(-\frac{1}{2}\) and \(-1\)
Step-by-step Solution
Detailed explanation
\((\sqrt[3]{1+a}-1) x^{2}+(\sqrt{1+a}-1) x+(\sqrt[6]{1+a}-1)=0\)
Let \(a+1=y\), then equation reduces to \(\left(y^{1 / 3}-1\right) x^{2}+\left(y^{1 / 2}-1\right) x+\left(y^{1 / 6}-1\right)=0\)
On dividing both sides by \(y-1\), we get \(\left(\frac{y^{1 / 3}-1}{y-1}\right) x^{2}+\left(\frac{y^{1 / 2}-1}{y-1}\right) x+\left(\frac{y^{1 / 6}-1}{y-1}\right)=0\)
On taking limit as \(y \rightarrow 1\) i.e. \(a \rightarrow 0\) on both sides, we get \(\frac{1}{3} x^{2}+\frac{1}{2} x+\frac{1}{6}=0 \Rightarrow 2 x^{2}+3 x+1=0\)
\(\Rightarrow x=-1,-\frac{1}{2}\) (roots of the equation)
\(\therefore \quad \lim _{a \rightarrow 0^{+}} \alpha(a)=-1, \lim _{a \rightarrow 0^{+}} \beta(a)=-\frac{1}{2}\)
Let \(a+1=y\), then equation reduces to \(\left(y^{1 / 3}-1\right) x^{2}+\left(y^{1 / 2}-1\right) x+\left(y^{1 / 6}-1\right)=0\)
On dividing both sides by \(y-1\), we get \(\left(\frac{y^{1 / 3}-1}{y-1}\right) x^{2}+\left(\frac{y^{1 / 2}-1}{y-1}\right) x+\left(\frac{y^{1 / 6}-1}{y-1}\right)=0\)
On taking limit as \(y \rightarrow 1\) i.e. \(a \rightarrow 0\) on both sides, we get \(\frac{1}{3} x^{2}+\frac{1}{2} x+\frac{1}{6}=0 \Rightarrow 2 x^{2}+3 x+1=0\)
\(\Rightarrow x=-1,-\frac{1}{2}\) (roots of the equation)
\(\therefore \quad \lim _{a \rightarrow 0^{+}} \alpha(a)=-1, \lim _{a \rightarrow 0^{+}} \beta(a)=-\frac{1}{2}\)
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