For t –1, let and be the roots of the equation . If , then is equal to __________. [2025]

Question

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 $\lim_{t \to - 1^{+}} α_{t} = a and \lim_{t \to - 1^{+}} β_{t} = b$ , then $72 {(a + b)}^{2}$ is equal to __________. [2025]

Smart Achievers · Accepted Answer

(98)

$α_{t}$

$= \lim_{t \to - 1^{+}} \frac{- [{(t + 2)}^{\frac{1}{6}} - 1]}{{(t + 2)}^{\frac{1}{7}} - 1}$ [Sum of roots]

Let t + 2 = y, we get

$a + b = \lim_{y \to 1^{+}} - \frac{y^{1 / 6} - 1}{y^{1 / 7} - 1} = - \frac{7}{6}$

So, $72 {(a + b)}^{2} = 72 (\frac{49}{36}) = 98$ .

Solution

Q.
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 $\lim_{t \to - 1^{+}} α_{t} = a and \lim_{t \to - 1^{+}} β_{t} = b$ , then $72 {(a + b)}^{2}$ is equal to __________. [2025]

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Solution

Q. For t > –1, let αt and βt be the roots of the equation ((t+2)1/7–1)x2+((t+2)1/6–1)x+((t+2)1/21–1)=0. If limt→–1+αt=a and limt→–1+βt=b, then 72(a+b)2 is equal to __________. [2025]

Ans. (98) a+b=limt→–1+(αt+βt) =limt→–1+–[(t+2)16–1](t+2)17–1 [Sum of roots] Let t + 2 = y, we get a+b=limy→1+–y1/6–1y1/7–1=-76 So, 72(a+b)2=72(4936)=98.

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Q.
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 $\lim_{t \to - 1^{+}} α_{t} = a and \lim_{t \to - 1^{+}} β_{t} = b$ , then $72 {(a + b)}^{2}$ is equal to __________. [2025]