Question

Possible value of '$a$' such that $f\left(x\right)=0$ has exactly one real root lying in the interval $(2,3)$ are

• A. $\left(\frac{1}{2},\infty \right)$
• B. $\left(\frac{1}{3},\frac{1}{2}\right)$
• C. $\left(\frac{1}{4},\frac{1}{3}\right)$
• D. $\left(\frac{1}{4},\frac{1}{2}\right)$

Answer 1

The correct option is B $\left(\frac{1}{3},\frac{1}{2}\right)$

For the quadratic function $f\left(x\right)=a{x}^2+bx+c$, it's given that $a,b,c$ are non-zero.
Hence, $f\left(0\right)=c\ne 0$
It's given that $f:[0,2]\to [0,2]$ is bijective.
Since $f\left(0\right)\ne 0$, $f\left(0\right)=2$ for it to be bijective.
$\Rightarrow f\left(2\right)=0$
So, three cases possible as shown in the figure, which will satisfy the second condition as well.
$f\left(0\right)=2\Rightarrow c=2$
$f\left(2\right)=0\Rightarrow 4a+2b+2=0\Rightarrow 2a+b=-1$
From the figure, $f^{\prime }\left(0\right)<0\Rightarrow b<0$
$f^{\prime }\left(2\right)<0\Rightarrow 2a\left(2\right)+b<0\Rightarrow 4a+(-1-2a)<0$
$\Rightarrow a<\frac{1}{2}$
Since it's given that exactly one root lies between $2$ and $3$, only case of figure 2 is possible.
$f\left(3\right)>0$
$\Rightarrow 9a+3b+2>0$
$\Rightarrow 9a+3(-1-2a)+2>0$
$\Rightarrow 3a-1>0$
$\Rightarrow a>\frac{1}{3}$
Hence, 'a' lies in the interval $(\frac{1}{3},\frac{1}{2})$