Medium INTEGER +4 / -1 PYQ · JEE Mains 2025

For $\mathrm{t}>-1$, let $\alpha_{\mathrm{t}}$ and $\beta_{\mathrm{t}}$ be the roots of the equation

$$ \left((\mathrm{t}+2)^{1 / 7}-1\right) x^2+\left((\mathrm{t}+2)^{1 / 6}-1\right) x+\left((\mathrm{t}+2)^{1 / 21}-1\right)=0 \text {. If } \lim \limits_{\mathrm{t} \rightarrow-1^{+}} \alpha_{\mathrm{t}}=\mathrm{a} \text { and } \lim \limits_{\mathrm{t} \rightarrow-1^{+}} \beta_{\mathrm{t}}=\mathrm{b} \text {, } $$

then $72(a+b)^2$ is equal to ___________.

Answer (integer) 98

Solution

<p>$$\begin{aligned} &a+b=\lim _{t \rightarrow-1^{+}}(\alpha+\beta)=\lim _{t \rightarrow-1^{+}}-\frac{(t+2)^{\frac{1}{6}}-1}{(t+2)^{\frac{1}{7}}-1}\\ &\begin{aligned} & \text { let } t+2=y \\ & a+b=\lim _{y \rightarrow 1^{+}} \frac{y^{1 / 6}-1}{y^{1 / 7}-1}=\frac{7}{6} \\ & 72(a+b)^2=72 \frac{49}{36}=98 \end{aligned} \end{aligned}$$</p>

About this question

Subject: Mathematics · Chapter: Limits, Continuity and Differentiability · Topic: Limits and Standard Results

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