Question

The number of real roots of the equation $e^{6x} - e^{4x} - 2e^{3x} - 12e^{2x} + e^x + 1 = 0$ is :

• A. 1
• B. 2
• C. 4
• D. 6

Hint:

For transcendental equations like $P(e^x)=0$, the number of real roots for $x$ is equal to the number of *positive* roots of the polynomial $P(t)=0$. Descartes' rule of signs is a very powerful tool to find an upper bound on the number of roots quickly.

Answer 1

The Correct Option is B

To find the number of real roots of the equation \(e^{6x} - e^{4x} - 2e^{3x} - 12e^{2x} + e^x + 1 = 0\), we can use a substitution method. Let:

\(y = e^x\)

This changes the equation to:

\(y^6 - y^4 - 2y^3 - 12y^2 + y + 1 = 0\) 

Now, we need to determine the number of real positive solutions for this polynomial equation since \(y = e^x > 0\) for all real \(x\).

1. The polynomial \(y^6 - y^4 - 2y^3 - 12y^2 + y + 1\) is continuous and differentiable for all real numbers.
2. To find critical points, we calculate the derivative and set it to zero:
3. \(\frac{d}{dy}(y^6 - y^4 - 2y^3 - 12y^2 + y + 1) = 6y^5 - 4y^3 - 6y^2 - 24y + 1\)
4. Solve \(6y^5 - 4y^3 - 6y^2 - 24y + 1 = 0\) for \(y\) to find critical points.
5. A thorough analysis or graphing can be employed to determine sign changes and verify the count of roots.
6. Due to the complexity of performing algebraic manipulations directly by hand, graphing calculators or computer algebra systems (CAS) like Desmos, GeoGebra, Wolfram Alpha, etc., are typically used to visualize the function over the interval where \(y > 0\).

An analysis or plotting shows that there are two changes of sign for real positive \(y\), indicating two real roots in this context.

Therefore, the number of real roots for the equation is:

2