Question

Find the values of ${}^{\prime }{t}^{\prime }$ in the equation $x^2-2tx+t^2-1=0$ such that exactly one root lies in between the numbers $2$ and $4$, and no root of the equation is either $2$ (or) $4$.

• A. $1<t<5$
• B. $1\le t\le 5$
• C. $1\le t\le 3$ and $4<t<5$
• D. $1<t<3$ and $3<t<5$

Answer 1

The correct option is D

$1<t<3$ and $3<t<5$

Comparing $f\left(x\right)=x^2-2tx+t^2-1$ to $a{x}^2+bx+c$,

$a=1,b=-2t,c=t^2-1$

As $a>0$, graph of $f\left(x\right)$ is upward (cup shaped) parabola

For roots to be real, $b^2-4ac>0$

$\Rightarrow 4t^2-4(t^2-1)>0$

$\Rightarrow 4>0$ [Always true]

Observe from the graph that $f\left(2\right)>0$ and $f\left(4\right)<0$.

Hence $f\left(2\right).f\left(4\right)<0$ . . . $\left(1\right)$

Also, for root $\ne +2$ (or) $4,f\left(2\right)\ne 0$ and $f\left(4\right)\ne 0$ . . . $\left(2\right)$

$\left(1\right)\Rightarrow f\left(2\right)f\left(4\right)<0$

$[4-4t+t^2-1][16-8t+t^2-1]<0$

$\Rightarrow (t^2-4t+3)(t^2-8t+15)<0$

$\Rightarrow (t-1)(t-3)(t-3)(t-5)<0$

$\Rightarrow (t-3{)}^2(t-1\left)\right(t-5)<0$

$\Rightarrow t\in (1,5)$

$\left(2\right)\Rightarrow f\left(2\right)\ne 0$

$\Rightarrow t\ne 1,3$

$\left(3\right)\Rightarrow f\left(4\right)\ne 0\Rightarrow t\ne 3,5$

So, $t\in (1,5)-3$.