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

For a real number $x$,let $[x]$ denote the greatest integer less than or equal to $x$. For $x \in \mathbb{R}$,let $f(x) = [x] \sin(\pi x)$. Then,

Let $R = \{(1, 3), (2, 2), (3, 2)\}$ and $S = \{(2, 1), (3, 2), (2, 3)\}$ be two relations on set $A = \{1, 2, 3\}$. Then $RoS =$

Suppose $f: [-2, 2] \rightarrow R$ is defined by $f(x) = \begin{cases} -1, & -2 \leq x \leq 0 \\ x - 1, & 0 < x \leq 2 \end{cases}$. Then the set $\{x \in [-2, 2] : x \leq 0 \text{ and } f(|x|) = x\}$ is equal to

Let $f(x) = a^x$ $(a > 0)$ be written as $f(x) = f_1(x) + f_2(x)$,where $f_1(x)$ is an even function and $f_2(x)$ is an odd function. Then $f_1(x + y) + f_1(x - y)$ equals

If $f(x) = \frac{1}{\sqrt{x + 2\sqrt{2x - 4}}} + \frac{1}{\sqrt{x - 2\sqrt{2x - 4}}}$ for $x > 2$,then $f(11) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module