$\begin{array}{l} I f A, B, C a r e t h e s e t s o f a l l v a l u e s o f x, f o r w h i c h x^{2} - 5 x - 14 i s p o s i t i v e, - 6 x^{2} + 2 x - 3 i s n e g a t i v e a n d 4 x - 5 x^{2} + 2 i s p o s i t i v e r e s p e c t i v e l y, t h e n A \cap B \cap C = \end{array}$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$(\frac{2 - \sqrt{14}}{5}, \frac{2 + \sqrt{14}}{5})$

$R$

$ϕ$

$(- 2, 7)$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} L e t x^{2} - 5 x - 14 > 0 \Rightarrow x^{2} - 7 x + 2 x - 14 > 0 \\ \Rightarrow (x - 7) (x + 2) > 0 \\ A = (- \infty, - 2) \cup (7, \infty) \\ L e t 6 x^{2} + 2 x - 3 < 0 \Rightarrow 6 x^{2} - 2 x + 3 > 0 \\ \Rightarrow 6 [x^{2} - \frac{2}{6} x + \frac{1}{2}] > 0 \\ \Rightarrow 6 [x^{2} - \frac{2}{6} x + \frac{1}{36} - \frac{1}{36} + \frac{1}{2}] > 0 \\ \Rightarrow 6 ({(x - \frac{1}{6})}^{2} + \frac{17}{36}) > 0 \forall x \in R \\ ∴ B = (- \infty, \infty) \\ L e t 4 x - 5 x^{2} + 2 > 0 \Rightarrow 5 x^{2} - 4 x - 2 < 0 \\ x = \frac{4 \pm \sqrt{16 + 40}}{2 (5)} = \frac{2 \pm \sqrt{14}}{5} \\ ∴ C = (\frac{2 - \sqrt{14}}{5}, \frac{2 + \sqrt{14}}{5}) \\ ∴ A \cap B \cap C = ϕ \end{array}$