Q. Let $f(x) = ax^{2} + bx + c\$and$\ f( - 1) < 1,f(1) > - 1,f(3) < - 4\$and$\ a \neq 0,$then
A) $a > 0$
B) $a < 0$
C) Sign of a cannot be determined
D) None of the above
Explanation
(b)
$\because f( - 1) < 1 \Rightarrow a - b + c < 1$ …(i)
and $f(1) > - 1,f(3) < - 4,\$then
$a + b + c > - 1$ …(ii)
$9a + 3b + c < - 4$ …(iii)
From Eq. (ii),
$- a - b - c < 1\ \ \ \ \ \$….(iv)
On solving Eqs. (i), (iii) and (iv), we get $a < - \frac{1}{8} \Rightarrow a\$is negative
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.