Question 13 Marks
Write the following set of real numbers in a set - builder form :
(a) $\left\{x: x^2-9 x+20=0\right\}$
(b) $\left\{x: x^3+1=0\right\}$
(c) $\left\{x: x^2-2=10\right\}$
(d) $\{x: x$ is a prime number, $x \leq 11\}$
(a) $\left\{x: x^2-9 x+20=0\right\}$
(b) $\left\{x: x^3+1=0\right\}$
(c) $\left\{x: x^2-2=10\right\}$
(d) $\{x: x$ is a prime number, $x \leq 11\}$
Answer
View full question & answer→(a) $\left\{x: x^2-9 x+20=0\right\}$
The given set is a collection of roots of equation $x^2-9 x+20=0$.
Hence,
$\begin{aligned}& x^2-9 x+20=0 \\\Rightarrow\quad & x^2-5 x-4 x+20=0 \\\Rightarrow\quad & x(x-5)-4(x-5)=0 \\\Rightarrow\quad & (x-5)(x-4)=0 \\\Rightarrow\quad & x=5,4\end{aligned}$
$\therefore \quad$ Required set $=\{4,5\}$
(b) $\left\{x: x^3+1=0\right\}$
Solving the equation $x^3+1=0$
$(x+1)\left(x^2-x+1\right)=0$
or $x=-1, \frac{1 \pm i \sqrt{3}}{2}$
But $\frac{1 \pm i \sqrt{3}}{2}$ are not real numbers.
$\therefore \quad$ Required set $=\{-1\} \quad$
(c) $\left\{x: x^2-2=10\right\}$
Solving equation $x^2-2=10$
or $x^2=12 \quad \therefore \quad x=\sqrt{12}$
or $x= \pm 2 \sqrt{3}$
$\therefore$ Given set $=\{2 \sqrt{3},-2 \sqrt{3}\}$
(d) $\{x: x$ is a prime number, $x \leq 11\}$
Prime number upto 11 are $2,3,5,7,11$
Hence the required set $=\{2,3,5,7,11\}$
The given set is a collection of roots of equation $x^2-9 x+20=0$.
Hence,
$\begin{aligned}& x^2-9 x+20=0 \\\Rightarrow\quad & x^2-5 x-4 x+20=0 \\\Rightarrow\quad & x(x-5)-4(x-5)=0 \\\Rightarrow\quad & (x-5)(x-4)=0 \\\Rightarrow\quad & x=5,4\end{aligned}$
$\therefore \quad$ Required set $=\{4,5\}$
(b) $\left\{x: x^3+1=0\right\}$
Solving the equation $x^3+1=0$
$(x+1)\left(x^2-x+1\right)=0$
or $x=-1, \frac{1 \pm i \sqrt{3}}{2}$
But $\frac{1 \pm i \sqrt{3}}{2}$ are not real numbers.
$\therefore \quad$ Required set $=\{-1\} \quad$
(c) $\left\{x: x^2-2=10\right\}$
Solving equation $x^2-2=10$
or $x^2=12 \quad \therefore \quad x=\sqrt{12}$
or $x= \pm 2 \sqrt{3}$
$\therefore$ Given set $=\{2 \sqrt{3},-2 \sqrt{3}\}$
(d) $\{x: x$ is a prime number, $x \leq 11\}$
Prime number upto 11 are $2,3,5,7,11$
Hence the required set $=\{2,3,5,7,11\}$