ExamBro
ExamBro
JEE Mains · Maths · STD 11 - 12. limits

\(t \gt -1\) के लिए, माना \(\alpha_t\) और \(\beta_t\) समीकरण
\(((t+2)^{\frac{1}{7}}-1) x^2+((t+2)^{\frac{1}{6}}-1) x~+\) \(((t+2)^{\frac{1}{21}}\) \(-~1)=0\) के मूल हैं।
यदि \(\lim _{t \rightarrow-1^{+}} \alpha_t=a\) और \(\lim _{t \rightarrow-1^{+}} \beta_t=b\), तो \(72(a+b)^2\) = __________ है।

  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…

Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app