The number of elements in the set {x in R : ( | vedclass

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

Question

(B) To find the number of elements in the set,we solve the equation $(|x|-3)|x+4|=6$.Let $f(x) = |x|-3$ and $g(x) = \frac{6}{|x+4|}$.We look for the n

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) To find the number of elements in the set,we solve the equation $(|x|-3)|x+4|=6$.Let $f(x) = |x|-3$ and $g(x) = \frac{6}{|x+4|}$.We look for the number of intersection points of the graphs of $y = |x|-3$ and $y = \frac{6}{|x+4|}$.The graph of $y = |x|-3$ is a $V$-shaped curve with its vertex at $(0, -3)$ and x-intercepts at $x = 3$ and $x = -3$.The graph of $y = \frac{6}{|x+4|}$ is a hyperbola-like curve with a vertical asymptote at $x = -4$ and it is always positive.By observing the graphs,the curve $y = |x|-3$ intersects $y = \frac{6}{|x+4|}$ at two distinct points: one in the interval $(3, \infty)$ and one in the interval $(-\infty, -4)$.Therefore,the number of elements in the set is $2$.

The number of elements in the set $\{x \in R : (|x|-3)|x+4|=6\}$ is equal to

Similar Questions

Which of the following functions are odd?
$I. f(x)=x\left(\frac{e^x-1}{e^x+1}\right)$
$II. f(x)=k^x+k^{-x}+\cos x$
$III. f(x)=\log \left(x+\sqrt{x^2+1}\right)$

For a real number $r$,we denote by $[r]$ the largest integer less than or equal to $r$. If $x, y$ are real numbers with $x, y \geq 1$,then which of the following statements is always true?

If $f(x) = \log \left( \frac{1 + x}{1 - x} \right)$,then $f(x)$ is

Let $f$ be an odd function defined on the set of real numbers such that for $x \geq 0$,$f(x) = 3 \sin x + 4 \cos x$. Then $f(x)$ at $x = -\frac{11\pi}{6}$ is equal to:

Assertion $(A)$: $\coth x = \frac{1-k}{1+k}$ where $0 < k < 2$.
Reason $(R)$: The graph of $y = \tanh x$ always lies between the lines $y = -1$ and $y = 1$.
Choose the correct option:

Vedclass Test Series

Exam Paper Generator

Online Exam Module