MCQ
The equation $e^{4 x}+8 e^{3 x}+13 e^{2 x}-8 e^x+1=0, x \in R$ has:
- ✓two solutions and both are negative
- Bno solution
- Cfour solutions two of which are negative
- Dtwo solutions and only one of them is negative
Let $e^x=t$
Now, $t^4+8 t^5+13 t^2-8 t+1=0$
Dividing equation by $t ^2$,
$t^2+8 t+13-\frac{8}{t}+\frac{1}{t^2}=0$
$t^2+\frac{1}{t^2}+8\left(t-\frac{1}{t}\right)+13=0$
$\left(t-\frac{1}{t}\right)^2+2+8\left(t-\frac{1}{t}\right)+13=0$
Let $t-\frac{1}{t}=z$
$z^2+8 z+15=0$
$(z+3)(z+5)=0$
$z=-3 \text { or } z=-5$
So, $t -\frac{1}{ t }=-3$ or $t -\frac{1}{ t }=-5$
$t^2+3 t-1=0 \text { or } t^2+5 t-1=0$
$t=\frac{-3 \pm \sqrt{13}}{2} \text { or } t =\frac{-5 \pm \sqrt{29}}{2}$
as $t = e ^{ x }$ so $t$ must be positive,
$t=\frac{\sqrt{13}-3}{2} \text { or } \frac{\sqrt{29}-5}{2}$
So, $x=\ln \left(\frac{\sqrt{13}-3}{2}\right)$ or $x=\ln \left(\frac{\sqrt{29}-5}{2}\right)$
Hence two solution and both are negative.
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