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JEE Main 20257 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
7 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Quadratic Equation

JEE Main 2025 · 7 April · Evening Shift · Mathematics · Quadratic Equation

For t -1 , let α_t and β_t be the roots of the equation ((t+2)^...

For t>−1t \gt -1, let αt\alpha_t and βt\beta_t be the roots of the equation ((t+2)17−1)x2+((t+2)16−1)x+((t+2)121−1)=0\left((t+2)^{\frac{1}{7}}-1\right) x^2+\left((t+2)^{\frac{1}{6}}-1\right) x+\left((t+2)^{\frac{1}{21}}-1\right)=0
If lim⁡t→−1+αt=a\lim _{t \rightarrow-1^{+}} \alpha_t=a and lim⁡t→−1+βt=b\lim _{t \rightarrow-1^{+}} \beta_t=b, then 72(a+b)272(a+b)^2 is equal to ________.

Correct answer

0

Step-by-step solution

a+b=lim⁡t→−1+(α+β)=lim⁡t→−1+−(t+2)16−1(t+2)17−1 let t+2=ya+b=lim⁡y→1+y1/6−1y1/7−1=7672(a+b)2=724936=98\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} \\ & \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}
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