MCQ
Which of the following is not a convex set?
  • A
    {(x, y) ; 2x + 5y ≤ 7}
  • B
    {(x, y) : x2 + y2 ≤ 4}
  • C
    {x : |x| = 5}
  • D
    {(x, y) : 3x2 + 2y2 ≤ 6}

Answer

  1. {x : |x| = 5}

Solution:

|x| = 5 is not a convex set as any two points from negative and positive x-axis if are joined will not lie in set.

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

If $\int \operatorname{cosec}^5 x d x=\alpha \cot x \operatorname{cosec} x\left(\operatorname{cosec}^2 x+\frac{3}{2}\right)+\beta \log _e\left|\tan \frac{x}{2}\right|+C$ where $\alpha, \beta \in \mathbb{R}$ and $\mathrm{C}$ is constant of integration , then the value of $8(\alpha+\beta)$ equals....................
Number of value of $'a'$ for which the system of equations,$A^2 x + (2 - a) y = 4 + a^2$ $a x + (2 a - 1) y = a^5 - 2$ possess no solution is
If $tan^{-1} (x^2 + 3|x|-4 )= tan^{-1} (4 \pi + sin^{-1}(sin14))$, then the value of $cos^{-1}(cos3|x|)$ is equal to
If the direction ratios of two lines are given by 3lm - 4ln + mn = 0 and l + 2m + 3n = 0, then the angle between the lines is:

  1. $\frac{\pi}{6}$

  2. $\frac{\pi}{4}$

  3. $\frac{\pi}{3}$

  4. $\frac{\pi}{2}$

Suppose $f ( x )$ is a polynomial of degree four, having critical points at $-1,0,1$ . If $T =\{ x \in R \mid f ( x )= f (0)\},$ then the sum of squares of all the elements of $T$ is
If $\cos^{-1}\text{x}+\sin^{-1}\text{x}=\pi,$ then the value of x is:
  1. $\frac{3}{2}$
  2. $\frac{1}{\sqrt{2}}$
  3. $\frac{\sqrt{3}}{2}$
  4. $\frac{2}{\sqrt{3}}$
$\int_0^{\pi /2} {{{\cos }^2}x\,dx = } $
If $A =$ $\left[ {\begin{array}{*{20}{c}}0&1&2\\1&2&3\\3&a&1\end{array}} \right]$ ,$A^{-1} =$$\left[ {\begin{array}{*{20}{c}}{1/2}&{ - 1/2}&{1/2}\\{ - 4}&3&c\\{5/2}&{ - 3/2}&{1/2}\end{array}} \right]$, then
The maximum value of function $f(x) = \int\limits_0^1 {t\,\sin \,\left( {x + \pi t} \right)} dt,\,x \in \,R$ is
If $\left| {\,\begin{array}{*{20}{c}}{ - {a^2}}&{ab}&{ac}\\{ab}&{ - {b^2}}&{bc}\\{ac}&{bc}&{ - {c^2}}\end{array}\,} \right| = K{a^2}{b^2}{c^2},$ then $K = $