For t -1 , let α t and β t be the roots of the equation ((t+2) 1/7 -1 ) x 2 + ((t+2) 1/6 -1 ) x+ ((t+2) 1/21 -1 )=0 . If lim t rightarrow-1 + α t =a and lim t rightarrow-1 + β t =b , then 72(a+b) 2 is equal to \ \ \ \ .

Mathematics · Limits, Continuity and Differentiability

JEE Main 2025 — 7 April, Evening Shift — Question 41

For $t>-1$ , let $\alpha_{t}$ and $\beta_{t}$ be the roots of the equation $\left((t+2)^{\frac{1}{7}}-1\right) x^{2}+\left((t+2)^{\frac{1}{6}}-1\right) x+$

$\left((t+2)^{\frac{1}{21}}-1\right)=0$ . If $\lim _{t \rightarrow-1^{+}} \alpha_{t}=a$ and $\lim _{t \rightarrow-1^{+}} \beta_{t}=b$ , then $72(a+b)^{2}$ is equal

to $\_\_\_\_$ .

Answer: 98

Numerical answer — enter this value.

Step-by-step solution

$\alpha+\beta=\frac{(t+2)^{1 / 6}-1}{(t+2)^{1 / 7}-1}$

$\alpha \beta=\frac{(t+2)^{1 / 21}-1}{(t+2)^{1 / 7}-1}$

$\lim _{t \rightarrow-1}(\alpha+\beta)=\frac{-\frac{1}{6}}{\frac{1}{7}}=\frac{-7}{6}=a+b$

$\lim _{t \rightarrow-1}(\alpha \beta)=\frac{\frac{1}{21}}{\frac{1}{7}}=\frac{7}{21}=\frac{1}{3}=a b$

$\Rightarrow \quad(a+b)^{2}=\frac{49}{36}$

$\Rightarrow \quad 72(a+b)^{2}=98$

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Exam: JEE Main 2025
Paper: 7 April, Evening Shift
Subject: Mathematics
Chapter: Limits, Continuity and Differentiability
Topic: Evaluation of Limit of Functions

If the function f(x)=tan (tan x)-sin (sin x)/tan x-sin x is continuous at x=0 , then f(0) is equal to \ \ \ \ .

Answer: 98

Step-by-step solution

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