Question:

The number of points where $f(x)=e^{8x}-e^{6x}-3e^{4x}-e^{2x}+1=0$ cuts the $x$-axis is equal to............

The number of points where $f(x)=e^{8x}-e^{6x}-3e^{4x}-e^{2x}+1=0$ cuts the $x$-axis is equal to 0.

The number of points where $f(x)=e^{8x}-e^{6x}-3e^{4x}-e^{2x}+1=0$ cuts the $x$-axis is equal to 1.

The number of points where $f(x)=e^{8x}-e^{6x}-3e^{4x}-e^{2x}+1=0$ cuts the $x$-axis is equal to 2.

The number of points where $f(x)=e^{8x}-e^{6x}-3e^{4x}-e^{2x}+1=0$ cuts the $x$-axis is equal to 3.

Step-by-Step Solution

Key Concept: Let $t=e^{2x}+\frac{1}{e^{2x}}\geq2$. Equation becomes $t^2-t-5=0\Rightarrow t=\frac{1+\sqrt{21}}{2}>2$. Two solutions for $x$.

2 real roots.

Correct Answer: 2