JEE Mains · Maths · STD 11 - 4.2 Quadratic equations and inequations
Let a and β be the roots of equation \( x^{2}+2ax+(3a+10) = 0 \) such that \( \alpha<1<\beta \) Then the set of all possible values of a is :
- A \( (-\infty,\frac{-11}{5})\cup(5,\infty) \)
- B \( (-\infty,-2)\cup(5,\infty) \)
- C \( (-\infty,-3) \)
- D \( (-\infty,\frac{-11}{5}) \)
Answer & Solution
Correct Answer
(D) \( (-\infty,\frac{-11}{5}) \)
Step-by-step Solution
Detailed explanation
\(\because \alpha<1<\beta\) \( f(1)<0 \) \( \Rightarrow1+2a+(3a+10)<0 \) \( \Rightarrow5a+11<0 \) \( a<\frac{-11}{5} \) \(\therefore a \in\left(-\infty, \frac{-11}{5}\right)\)
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