The number of points, where the curve f(x)=e^ | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

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

Vedclass · Accepted Answer

$2$

Vedclass · Answer

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 $t$.

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

Similar Questions

If $f(x)=\frac{2^x}{2^x+\sqrt{2}}, x \in R$, then $\sum_{k=1}^{81} f\left(\frac{k}{82}\right)$ is equal to :

The graph of the function $y = f(x)$ is symmetrical about the line $x = 2$, then

Consider a function $f : N \rightarrow R$, satisfying $f(1)+2 f(2)+3 f(3)+\ldots+x f(x)=x(x+1) f(x) ; x \geq 2$ with $f(1)=1$. Then $\frac{1}{f(2022)}+\frac{1}{f(2028)}$ is equal to

If in greatest integer function, the domain is a set of real numbers, then range will be set of

Show that function $f : R \rightarrow\{ x \in R :-1< x <1\}$ defined by $f ( x )=\frac{x}{1+|x|^{\prime}} x \in R$ is one-one and onto function.