Number of points, where curve f(x)=e^{8 x}-e^{6 x}-3 e^{4 x}-e^{2 x}+1, x ∈ R cuts x -axis, is equ

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1.Relation and Function

hard

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 $x$-axis, is equal to

A

$2$

B

$4$

C

$6$

D

$8$

(JEE MAIN-2023)

Solution

Let $e ^{2 x }= t$

$\Rightarrow t^4-t^3-3 t^2-t+1=0$

$\Rightarrow t^2+\frac{1}{t^2}-\left(t+\frac{1}{t}\right)-3=0$

$\Rightarrow\left(t+\frac{1}{t}\right)^2-\left(t+\frac{1}{t}\right)-5=0$

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

Two real values of $t$.

Standard 12

Mathematics

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