JEE Main 2021 — Mathematics Calculus

$${e}^{6x}-{e}^{4x}-2{e}^{3x}-12{e}^{2x}+{e}^{x}+1=0$$e6x−e4x−2e3x−12e2x+ex+1=0

$${({e}^{3x})}^{2}-2{e}^{3x}+1-{e}^{4x}+{e}^{x}=12{e}^{2x}$$(e3x)2−2e3x+1−e4x+ex=12e2x

$$\Rightarrow {({e}^{3x}-1)}^{2}-{e}^{x}({e}^{3x}-1)=12{e}^{2x}$$⇒(e3x−1)2−ex(e3x−1)=12e2x

$$\Rightarrow ({e}^{3x}-1)({e}^{3x}-1-{e}^{x})=12{e}^{2x}$$⇒(e3x−1)(e3x−1−ex)=12e2x

$$\Rightarrow \frac{({e}^{3x}-1-{e}^{x})}{{e}^{2x}}=\frac{12}{({e}^{3x}-1)}$$⇒e2x(e3x−1−ex)​=(e3x−1)12​

$$\Rightarrow {e}^{x}-{e}^{-x}-{e}^{-2x}=\frac{12}{{e}^{3x}-1}$$⇒ex−e−x−e−2x=e3x−112​

Assume, f(x)={e}^{x}-{e}^{-x}-{e}^{-2x}&g(x)=\frac{12}{{e}^{3x}-1}

$$f(x)={e}^{x}-{e}^{-x}-{e}^{-2x}$$f(x)=ex−e−x−e−2x

$${f}^{'}(x)={e}^{x}+{e}^{-x}+2{e}^{-2x}$$f′(x)=ex+e−x+2e−2x

Clearly, {f}^{'}(x)>0\forall x\in R(\because {e}^{x}&{e}^{-x}>0\forall x\in R)

So, here $f(x)$f(x) is increasing and $f(0)=-1$f(0)=−1and clearly $g(x)$g(x) is decreasing function because it is reciprocal function.

From the graph, we can say that number of real roots $=2$=2