Question

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

• A. $(-\infty ,2)\cup (3,\infty )$
• B. $(-\infty ,2]\cup [3,\infty )$
• C. $(2,3)$
• D. None of the above

Answer 1

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

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

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: $D>0$
$\Rightarrow (a-2{)}^2-4.1\left(\frac{a-3}{4}\right)>0$
$\Rightarrow (a-2)(a-3)>0$
$\Rightarrow (a-2)<0$ or $(a-3)>0$
$\Rightarrow a<2$ or $a>3$
$\Rightarrow a\in (-\infty ,2)\cup (3,\infty )=A$

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 (i): $D\ge 0$
$\Rightarrow (a-2)(a-3)\ge 0$
$\therefore a\in (-\infty ,2]\cup [3,\infty )=B$

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

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

Condition (iv) : $k_1<-\frac{b}{2a}<k_2$
$\Rightarrow 0<\frac{a-2}{2}<\frac{1}{2}$
$\Rightarrow 0<a-2<1$
$\Rightarrow 2<a<3$
$\Rightarrow a\in (2,3)=E$
If we look carefully, nothing is common among B,C,D,E
$\therefore$ Solution set of $a$ for case 2 is, $B\cap C\cap D\cap E=\varphi$

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