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JEE Mains · Maths · STD 11 - 12. limits

For \(t \gt -1\), let \(\alpha_t\) and \(\beta_t\) be the roots of the equation
\(((t+2)^{\frac{1}{7}}-1) x^2+((t+2)^{\frac{1}{6}}-1) x~+\) \(((t+2)^{\frac{1}{21}}\) \(-~1)=0\)
If \(\lim _{t \rightarrow-1^{+}} \alpha_t=a\) and \(\lim _{t \rightarrow-1^{+}} \beta_t=b\), then \(72(a+b)^2\) is equal to ________.

  1. A 92
  2. B 94
  3. C 96
  4. D 98
Verified Solution

Answer & Solution

Correct Answer

(D) 98

Step-by-step Solution

Detailed 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} \\ & \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…