Question

The value of '$a$' for which exactly one root of the equation $e^a{x}^2-e^{2a}x+e^{a-1}=0$ lies between $1$ and $2$ are given by,

• A. $\ln \left[\frac{5-\sqrt{17}}{4}\right]<a<\ln \left[\frac{5+\sqrt{17}}{4}\right]$
• B. $0<a<100$
• C. $\ln \frac{5}{4}$
• D. None of these

Answer 1

The correct option is A $\ln \left[\frac{5-\sqrt{17}}{4}\right]<a<\ln \left[\frac{5+\sqrt{17}}{4}\right]$

Let $f\left(x\right)=e^a{x}^2-e^{2a}x+e^a-1$

Clearly this represents a parabola opening upwards.

If exactly one root lies between $1$ and $2$, then $f\left(1\right)f\left(2\right)<0$.

$\Rightarrow \{e^a{\left(1\right)}^2-e^{2a}(1)+e^a-1\}\{e^a{\left(2\right)}^2-e^{2a}(2)+e^a-1\}<0\Rightarrow (e^a-e^{2a}+e^a-1)({4e}^a-{2e}^{2a}+e^a-1)<0\Rightarrow ({2e}^a-e^{2a}-1)({5e}^a-{2e}^{2a}-1)<0\Rightarrow (e^{2a}-{2e}^a+1)({2e}^{2a}-{5e}^a+1)<0\Rightarrow {(e^a-1)}^2({2e}^{2a}-{5e}^a+1)<0\Rightarrow ({2e}^{2a}-{5e}^a+1)<0$

Now let we solve for $({2e}^{2a}-{5e}^a+1)=0$

Using quadratic formula, we have

$e^a=\frac{5\pm \sqrt{25-4\left(2\right)\left(1\right)}}{4}=\frac{5\pm \sqrt{17}}{4}$

$\Rightarrow \frac{5-\sqrt{17}}{4}<e^a<\frac{5+\sqrt{17}}{4}\Rightarrow \ln \left[\frac{5-\sqrt{17}}{4}\right]<a<\ln \left[\frac{5+\sqrt{17}}{4}\right]$

So, option A is correct.