JEE MAIN - Mathematics (2025 - 7th April Evening Shift - No. 25)
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
98
Explanation
$$\begin{aligned}
&a+b=\lim _{t \rightarrow-1^{+}}(\alpha+\beta)=\lim _{t \rightarrow-1^{+}}-\frac{(t+2)^{\frac{1}{6}}-1}{(t+2)^{\frac{1}{7}}-1}\\
&\begin{aligned}
& \text { let } t+2=y \\
& a+b=\lim _{y \rightarrow 1^{+}} \frac{y^{1 / 6}-1}{y^{1 / 7}-1}=\frac{7}{6} \\
& 72(a+b)^2=72 \frac{49}{36}=98
\end{aligned}
\end{aligned}$$
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