Question

lf $a\in [-10,0]$ then the probability that the graph of the function $y=x^2+2(a+3)x-(2a+3)$ is strictly above x-axis is

• A. $\frac{3}{5}$
• B. $\frac{2}{5}$
• C. $\frac{1}{5}$
• D. $\frac{4}{5}$

Answer 1

The correct option is B $\frac{2}{5}$

The graph is above x-axis means that the roots are imaginary, i.e. the discriminant is negative.
Thus, ${\left\{2\right(a+3\left)\right\}}^2+4\ast 1\ast (2a+3)<0$
Or, $-6<a<-2$
But, $a\in [-10,0]$
Hence, the overlap is $a\in (-6,-2)$
Hence, the required probability = $\frac{\text{length of favourable line}}{\text{total length of line}}=\frac{4}{10}=\frac{2}{5}$