Question

Find the range of values of $a$ for which at least one root lies in $(0,\frac{1}{2})$ for the quadratic equation $4x^2-4(a-2)x+(a-2)=0\forall a\in R.$

• A. $[2,3]$
• B. $(2,3)$
• C. $(-\infty ,2)\cup (3,\infty )$
• D. $(-\infty ,2]\cup [3,\infty )$

Answer 1

The correct option is C $(-\infty ,2)\cup (3,\infty )$

Given: $4x^2-4(a-2)x+(a-2)=0\forall a\in R$
To find: Values of $a$ for which at least one root lies in $(0,\frac{1}{2})$

Step-1: Draw graph for the given condition.
Step-2: Write the applicable conditions.
Step-3: Solve all the conditions and combine the results for a.

$4x^2-4(a-2)x+(a-2)=0$
$\Rightarrow x^2-(a-2)x+\frac{a-2}{4}=0$

Case-1: Exactly one root lies in $(0,\frac{1}{2})$


Condition $1:D>0$
$\Rightarrow (a-2{)}^2-4\cdot 1\left(\frac{a-2}{4}\right)>0$
$\Rightarrow (a-2)(a-3)>0$
$\Rightarrow (a-2)<0$ or $(a-3)>0$
$\Rightarrow a<2$ or $a>3$
$\Rightarrow a\in (-\infty ,2)\cup (3,\infty )\cdots \left(A\right)$

Condition $2:f\left(k_1\right).f\left(k_2\right)<0$
$\Rightarrow \frac{a-2}{4}.\frac{3-a}{4}<0$
$\Rightarrow (a-2)(a-3)>0$ (Same as previous interval)

Case 2 : Both the root lies in $(0,\frac{1}{2})$


Condition $1:D\ge 0$
$\Rightarrow (a-2)(a-3)\ge 0$
$\therefore a\in (-\infty ,2]\cup [3,\infty )\cdots \left(B\right)$

Condition $2:f\left(k_1\right)>0$
$\Rightarrow \frac{a-2}{4}>0$
$\Rightarrow a>2$
$\Rightarrow a\in (2,\infty )\cdots \left(C\right)$

Condition $3:f\left(k_2\right)>0$
$\Rightarrow \frac{3-a}{4}>0$
$\Rightarrow (a-3)<0$
$\Rightarrow a\in (-\infty ,3)\cdots \left(D\right)$

Condition $4:k_1<-\frac{b}{2a}<k_2$
$\Rightarrow 0<\frac{a-2}{2}<\frac{1}{2}$
$\Rightarrow 0<a-2<1$
$\Rightarrow 2<a<3$
$\Rightarrow a\in (2,3)\cdots \left(E\right)$
Now, No region is common among the four solutions of different conditions
$\therefore$ Solution set of $a$ for case 2 is, $B\cap C\cap D\cap E=\varphi$

Thus, general solution set of $a$ is $A\cup (B\cap C\cap D\cap E)=A\cup \varphi$
$\therefore$ Required range of $a$ is $(-\infty ,2)\cup (3,\infty )$