If $f\left(x\right)=a{x}^2+bx+c,\mathrm{}f\left(-1\right)>\frac{1}{2},f\left(1\right)<-1$ and $f\left(-3\right)<-\frac{1}{2}$, then

• A

  $a=0$
• Correct answer
  B

  $a<0$
• C

  $a>0$
• D

  Sign of $a$ can not be determined

Solution

Clearly, parabola is opening downwards $\Rightarrow a<0.$

Hence, the correct answer is option: 'B'.