MCQ
If $|\text{x} + 3|\geq10,$ then:
  • A
    $\text{x}\in(-13,7\big)$
  • B
    $\text{x}\in(-13,7\big]$
  • C
    $\text{x}\in(-\infty -13\big]\cup \big[7,\infty)$
  • $\text{x}\in\big[-\infty -13\big]\cup \big[7,\infty)$

Answer

Correct option: D.
$\text{x}\in\big[-\infty -13\big]\cup \big[7,\infty)$
$|\text{x} + 3|\geq10$
$\Rightarrow\text{x} + 3\leq-10$ or $\text{x}+3\geq10$
$\Rightarrow\text{x}\leq -13 $ or $\text{x}\geq7$
$\Rightarrow\text{x}\in\big[-\infty -13\big]\cup \big[7,\infty)$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

The equation of a common tangent to the conics $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$ and $\frac{{{y^2}}}{{{a^2}}} - \frac{{{x^2}}}{{{b^2}}} = 1$ is
If two points $P$ and $Q$ on the hyperbola $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$ whose centre is $C$, are such that $CP$ is perpendicular to $CQ, ( a < b )$ , then value of, $\frac{1}{{{{(CP)}^2}}} + \frac{1}{{{{(CQ)}^2}}} = $
The solution set of $6 x-1 > 5$ is :
If the first term of a $G.P. a_1, a_2, a_3......$ is unity such that $4a_2 + 5a_3$ is least, then the common ratio of $G.P.$ is
Choose the correct answer. If $\text{f}(\text{x})=\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}$ then $\frac{\text{dy}}{\text{dx}}$ at $x = 1$ is equal to:
The middle term in the expansion of $\Big(\frac{2\text{x}^{2}}{3}+\frac{3}{2\text{x}^{2}}\Big)^{10}$ is:
A beam of light travels along the line $y = -4$ from right to left and strikes a parabola at point $P.$ If focus and directrix of parabola are $(2, 0)$ and $x = -2$ respectively, then the coordinates of the point where the reflected beam contact the parabola again
If $f(x)=\left\{\begin{array}{ll}\sin x, & x \neq n \pi, \quad n \in I \\ 2, & \text { otherwise }\end{array}\right.$ and $g(x)=\left\{\begin{array}{ll}x^{2}+1, & x \neq 0,2 \\ 2, & x=0 \\ 4, & x=2\end{array}\right.$ then $\lim _{x \rightarrow 0}$$g[f(x)]$ is
If $\text{f(z)}=\frac{7-\text{z}}{1-\text{z}^2},$ where $\text{z}=1+2\text{i},$ then $|\text{f(z)}|$ is:
If $n \in N \Big(\frac{\text{n}+1}{2}\Big)^\text{n}\geq\text{n!}$ is true when: