The number of real roots of the equation e6 x-e4 x-2 e3 x-12 e2 x+ex+1=0 is:

Question

The number of real roots of the equation e6 x-e4 x-2 e3 x-12 e2 x+ex+1=0 is: (A) 2 (B) 4 (C) 6 (D) 1

Tardigrade Admin · Accepted Answer

e6 x-e4 x-2 e3 x-12 e2 x+ex+1=0 ⇒(e3 x-1)2-ex(e3 x-1)=12 e2 x (e3 x-1)2(ex-e-x-e-2 x)=12 ⇒ ex-e-x-e-2 x=(12/e3 x-1) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/545f7660fe1a993c275d93d88cb4df21-.png" /> ⇒ No. of real roots =2

Q. The number of real roots of the equation $e^{6 x} - e^{4 x} - 2 e^{3 x} - 12 e^{2 x} + e^{x} + 1 = 0$ is:

2

4

6

1

$e^{6 x} - e^{4 x} - 2 e^{3 x} - 12 e^{2 x} + e^{x} + 1 = 0$
$\Rightarrow (e^{3 x} - 1)^{2} - e^{x} (e^{3 x} - 1) = 12 e^{2 x}$
$(e^{3 x} - 1)^{2} (e^{x} - e^{- x} - e^{- 2 x}) = 12$
$\Rightarrow e^{x} - e^{- x} - e^{- 2 x} = \frac{12}{e ^{3 x} - 1}$

$\Rightarrow$ No. of real roots $= 2$

Q. The number of real roots of the equation e6x−e4x−2e3x−12e2x+ex+1=0 is:

2

4

6

1

e6x−e4x−2e3x−12e2x+ex+1=0 ⇒(e3x−1)2−ex(e3x−1)=12e2x (e3x−1)2(ex−e−x−e−2x)=12 ⇒ex−e−x−e−2x=e3x−112​ ⇒ No. of real roots =2

Q. The number of real roots of the equation $e^{6 x} - e^{4 x} - 2 e^{3 x} - 12 e^{2 x} + e^{x} + 1 = 0$ is:

$e^{6 x} - e^{4 x} - 2 e^{3 x} - 12 e^{2 x} + e^{x} + 1 = 0$
$\Rightarrow (e^{3 x} - 1)^{2} - e^{x} (e^{3 x} - 1) = 12 e^{2 x}$
$(e^{3 x} - 1)^{2} (e^{x} - e^{- x} - e^{- 2 x}) = 12$
$\Rightarrow e^{x} - e^{- x} - e^{- 2 x} = \frac{12}{e ^{3 x} - 1}$

$\Rightarrow$ No. of real roots $= 2$