The number of solutions of the equation 2^x = | vedclass

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

Question

(C) To find the number of solutions of the equation $2^x = x^2$,we analyze the intersection points of the functions $f(x) = 2^x$ and $g(x) = x^2$.$1$.

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) To find the number of solutions of the equation $2^x = x^2$,we analyze the intersection points of the functions $f(x) = 2^x$ and $g(x) = x^2$.$1$. For $x $2$. For $x > 0$: - At $x = 2$,$2^2 = 4$ and $2^2 = 4$. Thus,$x = 2$ is a solution.- At $x = 4$,$2^4 = 16$ and $4^2 = 16$. Thus,$x = 4$ is a solution.- Between $x = 2$ and $x = 4$,the function $x^2$ grows faster than $2^x$ initially,but $2^x$ eventually overtakes $x^2$. Since $2^x$ is convex and $x^2$ is convex,there are no other solutions for $x > 0$ besides $x = 2$ and $x = 4$.Thus,the solutions are $x \approx -0.766$,$x = 2$,and $x = 4$.There are a total of $3$ solutions.

The number of solutions of the equation $2^x = x^2$ is

Similar Questions

If $f:[0,2) \rightarrow R$ is defined by $f(x)=\begin{cases} 1+\frac{2x}{k} & \text{for } 0 \leq x < 1 \\ kx & \text{for } 1 \leq x < 2 \end{cases}$ where $k>0$,and $f$ is such that $\lim_{x \rightarrow 1^{-}} f(x)=\lim_{x \rightarrow 1^{+}} f(x)$,then find the value of $k^2$.

For some $a, b, c \in N$,let $f(x)=ax-3$ and $g(x)=x^b+c$,$x \in R$. If $(fog)^{-1}(x)=\left(\frac{x-7}{2}\right)^{1/3}$,then $(fog)(ac) + (gof)(b)$ is equal to $..........$

Let $f$ and $g$ be increasing and decreasing functions,respectively,from $[0, \infty)$ to $[0, \infty)$. Let $h(x) = f(g(x))$. If $h(0) = 0$,then $h(x) - h(1)$ is:

If $f(x)=3[x]+\{x+1\}$,where $[x]$ is the greatest integer function of $x$ and $\{x\}$ is the fractional part function of $x$,then $f(-1.32)=$

Let $f(x) = x^{12} - x^9 + x^4 - x + 1$. Which of the following is true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module