The number of points, where the curve f ( x ) = e ^ 8x - e ^ 6x - 3...

Given,

${e}^{8x} - {e}^{6x} - 3{e}^{4x} - {e}^{2x} + 1 = 0$

$⇒ {e}^{4x} - {e}^{2x} - 3 - \dfrac{1}{{e}^{2x}} + \dfrac{1}{{e}^{4x}} = 0$

$⇒ {e}^{4x} + \dfrac{1}{{e}^{4x}} + 2 - \left({e}^{2x} + \dfrac{1}{{e}^{2x}}\right) = 5$

$⇒ {\left({e}^{2x} + \dfrac{1}{{e}^{2x}}\right)}^{2} - \left({e}^{2x} + \dfrac{1}{{e}^{2x}}\right) = 3$

Now let \left({e}^{2x} + \dfrac{1}{{e}^{2x}}\right) = t$</p><p>So, the equation becomes</p><p>${t}^{2} - t - 5 = 0$</p><p>$⇒ t = \dfrac{1 + \sqrt{21}}{2}$, ignoring negative sign as exponential function are positive,</p><p>Now${e}^{2x} + \dfrac{1}{{e}^{2x}} = \dfrac{1 + \sqrt{21}}{2}$</p><p>$⇒ {e}^{4x} - \dfrac{1 + \sqrt{21}}{2}{e}^{2x} + 1 = 0$which is quadratic equation in${e}^{2x} with upward parabola,

Now by $A . M \ge G . M$ we get, ${e}^{2x} + \dfrac{1}{{e}^{2x}} \ge 2$

So, $y = \dfrac{1 + \sqrt{21}}{2} > 2$, hence it will cut at two distinct point,

Hence, there will be two solution.