The number of points, where the curve f ( x ) = e 8 x - e 6 x - 3 e 4 x - e 2 x + 1 , x cuts the x -axis, is equal to ______ . [2023]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
The number of points, where the curve $f (x) = e^{8 x} - e^{6 x} - 3 e^{4 x} - e^{2 x} + 1, x \in ℝ$ cuts the $x$ -axis, is equal to ______ . [2023]

Ans.
(2)

$Given: f (x) = e^{8 x} - e^{6 x} - 3 e^{4 x} - e^{2 x} + 1$

$= (e^{4 x} + \frac{1}{e^{4 x}}) - (e^{2 x} + \frac{1}{e^{2 x}}) = 3$

$= {(e^{2 x} + \frac{1}{e^{2 x}})}^{2} - (e^{2 x} + \frac{1}{e^{2 x}}) = 5$

$Let e^{2 x} + \frac{1}{e^{2 x}} = t$

$\Rightarrow t^{2} - t - 5 = 0 \Rightarrow t = \frac{1 \pm \sqrt{1 + 20}}{2} = \frac{1 \pm \sqrt{21}}{2}$

$\Rightarrow t = \frac{1 + \sqrt{21}}{2} and t = \frac{1 - \sqrt{21}}{2} (rejected)$

$\Rightarrow e^{2 x} + \frac{1}{e^{2 x}} = \frac{1 + \sqrt{21}}{2}$

$Let e^{2 x} = t, then t + \frac{1}{t} = \frac{1 + \sqrt{21}}{2}$

$\Rightarrow t^{2} + 1 = (\frac{1 + \sqrt{21}}{2}) t$

$It is a quadratic equation.$

$∴$ $2 values of e^{2 x} are possible.$

$Hence, 2 real solutions exist.$

Want Us to Email you About Special Offers & Updates?