QJEEM23Ma9511: JEE Main 2023 Mathematics - Class 11 JEEM23Ma9511 Solution: The number of points, where the curve cuts -axis, is equal to…

Firstly, we know that the given function

$f(x)=e^{8x}-e^{6x}-3e^{4x}-e^{2x}+1$ intersects the x-axis

where $f(x) = 0$ . Setting $f(x)$ equal to zero gives us :

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

Let $t = e^{2x}$ . The equation now becomes :

$t^4 - t^3 - 3t^2 - t + 1 = 0.$

Dividing by $t^2$ and rearranging the equation gives :

$t^2 - t - 3 - \frac{1}{t} + \frac{1}{t^2} = 0.$

If we let $y = t + \frac{1}{t}$ , we get :

$y^2 - y - 5 = 0.$

This quadratic equation in y can be solved using the quadratic formula to give two roots :

$y = \frac{1 \pm \sqrt{21}}{2}.$

Since $y = t + \frac{1}{t}$ , and t > 0 (as $t = e^{2x}$ ), we must choose the root where y > 2 . Thus, we take $y = \frac{1 + \sqrt{21}}{2}$ .

So, we have :

$t + \frac{1}{t} = \frac{1 + \sqrt{21}}{2}.$

Solving for $t$ gives us :

$t^2 - yt + 1 = 0,$

or

$t = \frac{y \pm \sqrt{y^2 - 4}}{2}.$

Substituting $y = \frac{1 + \sqrt{21}}{2}$ into the formula gives :

$t = \frac{\frac{1 + \sqrt{21}}{2} \pm \sqrt{\left(\frac{1 + \sqrt{21}}{2}\right)^2 - 4}}{2}.$

This quadratic formula for $t$ yields two possible values (t = 2.37 and t = 0.42). Both of them are positive, thus both are acceptable values for $t = e^{2x}$ .

Finally, to get $x$ , take the natural log of both sides and divide by 2 :

$x = \frac{1}{2}\ln(t).$

This gives us two solutions for $x$ , corresponding to the two positive solutions for $t$ . So, the number of points where the curve $f(x) = e^{8x} - e^{6x} - 3e^{4x} - e^{2x} + 1$ intersects the x-axis is 2.