The number of points,where the curve f(x) = e | vedclass

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(A) To find the points where the curve cuts the $x$-axis,we set $f(x) = 0$.$e^{8x} - e^{6x} - 3e^{4x} - e^{2x} + 1 = 0$.Let $t = e^{2x}$. Since $x \in

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$2$

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(A) To find the points where the curve cuts the $x$-axis,we set $f(x) = 0$.$e^{8x} - e^{6x} - 3e^{4x} - e^{2x} + 1 = 0$.Let $t = e^{2x}$. Since $x \in R$,$t > 0$.The equation becomes $t^4 - t^3 - 3t^2 - t + 1 = 0$.Dividing by $t^2$ (since $t 
eq 0$):$t^2 - t - 3 - \frac{1}{t} + \frac{1}{t^2} = 0$.$(t^2 + \frac{1}{t^2}) - (t + \frac{1}{t}) - 3 = 0$.Let $u = t + \frac{1}{t}$. Then $u^2 = t^2 + 2 + \frac{1}{t^2}$,so $t^2 + \frac{1}{t^2} = u^2 - 2$.Substituting this into the equation:$(u^2 - 2) - u - 3 = 0 \Rightarrow u^2 - u - 5 = 0$.The roots are $u = \frac{1 \pm \sqrt{1 - 4(1)(-5)}}{2} = \frac{1 \pm \sqrt{21}}{2}$.Since $t > 0$,$u = t + \frac{1}{t} \geq 2$.We check the values of $u$:$u_1 = \frac{1 + \sqrt{21}}{2} \approx \frac{1 + 4.58}{2} \approx 2.79 > 2$.$u_2 = \frac{1 - \sqrt{21}}{2} \approx \frac{1 - 4.58}{2} \approx -1.79 For $u_1 = t + \frac{1}{t}$,the equation $t^2 - u_1 t + 1 = 0$ has discriminant $D = u_1^2 - 4 = (2.79)^2 - 4 > 0$,giving two distinct positive values for $t$.For $u_2 = t + \frac{1}{t}$,the equation $t^2 - u_2 t + 1 = 0$ has discriminant $D = u_2^2 - 4 Since $t = e^{2x}$,each positive $t$ gives one real value for $x$.Thus,there are $2$ points where the curve cuts the $x$-axis.

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

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