Question:
The number of real roots of the equation
$e^{t x}+2 e^{3 x}-e^{x}-6=0$ is :
Correct Option: , 3
Solution:
Let $\mathrm{e}^{\mathrm{x}}=\mathrm{t}>0$
$f(t)=t^{4}+2 t^{3}-t-6=0$
$f^{\prime}(t)=4 t^{3}+6 t^{2}-1$
$f^{\prime \prime}(\mathrm{t})=12 \mathrm{t}^{2}+12 \mathrm{t}>0$
$f(0)=-6, f(1)=-4, f(2)=24$
$\Rightarrow$ Number of real roots $=1$