Question

Let the quadratic equation $4x^2-2x+a=0$ has both the roots in the interval $(-1,1)$. Then which of the following is/are true?

• A. Minimum value of $a^2$ is $\frac{1}{16}$
• B. Number of integral values of $a$ is $2$
• C. One of the possible values of $a$ is $\frac{17}{13}$
• D. The values of $\left[a\right]$ are the roots of the equation $x^3+3x^2+2x=0$, where $[.]$ is the greatest integer function

Answer 1

Answer: (B), (D)

The correct options are
B Number of integral values of $a$ is $2$
D The values of $\left[a\right]$ are the roots of the equation $x^3+3x^2+2x=0$, where $[.]$ is the greatest integer function

$4x^2-2x+a=0$ has both the roots in $(-1,1)$
Then
$\left(i\right)D\ge 0\Rightarrow 4-16a\ge 0\Rightarrow a\le \frac{1}{4}\cdots \left(1\right)$

$\left(ii\right)f(-1)>0\Rightarrow 6+a>0\Rightarrow a>-6\cdots \left(2\right)$

$\left(iii\right)f\left(1\right)>0\Rightarrow 2+a>0\Rightarrow a>-2\cdots \left(3\right)$

$\left(iv\right)-1<-\frac{b}{2a}<1\Rightarrow -1<\frac{1}{4}<1$

From equation $\left(1\right),\left(2\right)$ and $\left(3\right)$, we get
$a\in (-2,\frac{1}{4}]$
Therefore, $a^2\in [0,4)$
Minimum value of $a^2$ is $0.$
The number of integral values of $a$ is $2.$
$\left[a\right]=-2,-1,0$

Now, $x^3+3x^2+2x=0$
$\Rightarrow x(x^2+3x+2)=0\Rightarrow x(x+2)(x+1)=0\Rightarrow x=-2,-1,0$
Hence, the values of $\left[a\right]$ are roots of $x^3+3x^2+2x=0$