Question

The number of integral values of $\alpha$ for which the quadratic expression $\alpha x^2+|2\alpha -3|x-6$ is positive for exactly two integral values of $x$ is equal to

• A. 3
• B. 2
• C. 1
• D. 0

Answer 1

Answer: (D)

Note that $f(0)=-6$.
Hence the curve cuts the $y$-axis at $(0,-6)$.
If $\alpha >0$ then the graph will as shown and this case there will be infinite values of $x$ for which $f(x)$ will be positive, hence $\alpha$ will be negative.

Now, for $\alpha <0$
$2\alpha -3<0$
Hence, $\alpha x^2+(3-2\alpha )x-6=0$
$x(\alpha x+3)-2(\alpha x+3)=0$
$(\alpha x+3)(x-2)=0$
$\therefore x=\frac{-3}{\alpha }\text{ or }x=2$
for exactly 2 integers, we must have $-4<\frac{-3}{\alpha }\le 5\Rightarrow \alpha \in \left(\frac{-3}{4},\frac{-3}{5}\right]$
Hence number of integral values is zero.