In this page, you can find the complete solutions of the fourth exercise of Logic, Sets, and Real Number System chapter from Basic Mathematics Grade XI book published and distributed by Sukunda Pustak Bhawan.

In the above-mentioned book, logic, sets, and real number system is the 1st chapter and has four exercises only. Out of which, this is the solution of the fourth and final exercise.

Check: Basic Mathematics Grade 11 (Sukunda Publication) Guide:
Grade 11 Basic Mathematics by Sukunda Pustak Vawan Notes and Solutions | Nepal

 

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Answers mentioned here are not solved by teachers. These are the solutions written by a student of Grade 11. Answers are all correct. However, the language or process of solving the questions might be informal and in examinations, you might have to add little more language and show more calculations than what has been done here. So, we highly encourage you to view these solutions as guide rather than just copying everything mentioned here. Few questions have been typed while most of them have been updated as pictures.

Real Number System Exercises

1. Evaluate:

a) $|-2| + 4$

Solution:

$= |-2| +4$

$= 2 + 4$

$= 6$


b) $|-5| + |-2| -3$

Solution:

$= |-5| + |-2| -3$

$= 5 + 2 -3$

$= 4$


c) $2 + |-3| - |-5|$

Solution:

$= 2 + |-3| -|-5|$

$= 2 + 3 - 5$

$= 0$


d) $|3 - |-5||$

Solution:

$= |3 - |-5||$

$= |3 - 5|$

$= |-2|$

$= 2$


2. Let (i) x = 2, y = 3 (ii) x = 2, y = -3, verify each of the followings:

Solving for case (i) x = 2, y =3

a) $|x + y | \leq |x| + |y|$

Solution:

$|x + y| = |2 + 3| = |5|$

$\therefore |x+ y| = 5$


$|x| + |y| = |2| + |3| = 2+3$

$\therefore |x| + |y| = 5$


$\implies |x+y| = |x| + |y| \implies \ \text{above relation holds true.}$



b) $|x - y| \geq |x| - |y|$

Solution:

$|x- y| = |2-3| = |-1|$

$\therefore |x-y| = 1$


$|x| - |y| = |2| - |3| = 2-3$

$\therefore |x| - |y| = -1$


$\implies |x-y| > |x| - |y|$



c) $|xy| = |x|.|y|$

Solution:

$|xy| = |2*3| = |6|$

$\therefore |xy| = 6$


$|x|.|y| = |2|*|3| = 2*3$

$\therefore |x|.|y| = 6$


$\implies |xy| = |x|.|y|$



d) $\left | \dfrac{x}{y} \right | = \dfrac[|x|}{|y|$

Solution:

$\left | \dfrac{x}{y} \right | = \left | \dfrac{2}{3] \right |$

$\therefore \left | \dfrac{x}{y} \right | = \dfrac{2}{3}$


$\dfrac{|x|}{|y|} = \dfrac{|2|}{|3|}$

$\therefore \dfrac{|x|}{|y|} = \dfrac{2}{3}$


$\implies \left | \dfrac{x}{y} \right | = \dfrac{|x|}{|y|}$


Solving for case (ii) x = 2, y =-3

a) $|x + y | \leq |x| + |y|$

Solution:

$|x + y| = |2 + (-3)| = |2-3|$

$\therefore |x+ y| = 1$


$|x| + |y| = |2| + |-3| = 2+3$

$\therefore |x| + |y| = 5$


$\implies |x+y| < |x| + |y| \implies \ \text{above relation holds true.}$



b) $|x - y| \geq |x| - |y|$

Solution:

$|x- y| = |2-(-3)| = |2+3|$

$\therefore |x-y| = 5$


$|x| - |y| = |2| - |-3| = 2-3$

$\therefore |x| - |y| = -1$


$\implies |x-y| > |x| - |y|$



c) $|xy| = |x|.|y|$

Solution:

$|xy| = |2*(-3)| = |-6|$

$\therefore |xy| = 6$


$|x|.|y| = |2|*|-3| = 2*3$

$\therefore |x|.|y| = 6$


$\implies |xy| = |x|.|y|$



d) $\left | \dfrac{x}{y} \right | = \dfrac[|x|}{|y|$

Solution:

$\left | \dfrac{x}{y} \right | = \left | \dfrac{2}{(-3)} \right |$

$\therefore \left | \dfrac{x}{y} \right | = \dfrac{2}{3}$


$\dfrac{|x|}{|y|} = \dfrac{|2|}{|-3|}$

$\therefore \dfrac{|x|}{|y|} = \dfrac{2}{3}$


$\implies \left | \dfrac{x}{y} \right | = \dfrac{|x|}{|y|}$



3. Solve the following inequalities

a) $x-1>2$

Solution:

$\implies x - 1>2$

$\implies x -1 +1 >2 + 1$

$\implies x >3$



b) $x- 3 \leq 5$

Solution:

$\implies x - 3 \leq 5$

$\implies x - 3  +3 \leq 5+3$

$\implies x \leq 8$


c) $-1 <x -2 <3$

Solution:

$\implies -1 <x -2<3$

$\implies -1 + 2 < x - 2 +2 <3 + 2$

$\implies 1 <x <5$


d) $-3 \leq 2x - 1 \leq 5$

Solution:

$\implies -3 \leq 2x - 1 \leq 5$

$\implies - 3 +1 \leq 2x - 1 + 1 \leq 5+1$

$\implies -2 \leq 2x \leq 6$

$\implies \dfrac{-2}{2} \leq \dfrac{2x}{2} \leq \dfrac{6}{2}$

$\implies -1 \leq x leq 3$


e) $x^2 - 2x > 0$

Solution:

Given,

$x^2 - 2x > 0$

The equation form of the given inequality is

$x^2 - 2x = 0$

$\implies x(x-2) = 0$

$\implies x = 0, 2$

Let us see the following possible intervals and signs of x(x-2) in those.

Note: Since, x(x-2)>0, we get to know that the result must always be positive. Also, it is greater than zero. So, the solution cannot have 0 and 2 in it.

Intervalsxx-2x(x-2)
$- \infty to 0$--+
$0 to 2$+--
$2 to \infty$+++

Thus, the required intervals are $(- \infty, 0) and (2 ,\infty)$.

$\implies x \in (-\infty , 0) \cup (2 , \infty)$




f) $6 + 5x - x^2 \geq 0$

Solution:

Given,

$6 + 5x - x^2 \geq 0$

The equation form of the given inequality is
$6 +5x  - x^2 = 0$

$\implies - (x^2 -5x -6) = 0$

$\implies -(x-6)(x+1) = 0$

$\implies x = -1, 6$

Let us see the following possible intervals and the signs of $-(x-6)(x+1)$ in these intervals.

Note: Since, $-(x-6)(x+1) \geq 0$, we get to know that the result must always be positive. It can include zero.

Intervalsx-6x+1-(x-6)(x+1)
$- \infty to -1$---
$-1 to 6$-++
$6 to \infty$++-

Hence, the required interval is $[-1,6]$

$\implies x \in [-1,6]$

About the Textbook:

Name: Basic Mathematics Grade XI
Author(s): D.R. Bajracharya | R.M. Shresththa | M.B. Singh | Y.R. Sthapit | B.C. Bajracharya
Publisher: Sukunda Pustak Bhawan (Bhotahity, Kathmandu)
Telephone: 5320379, 5353537
Price: 695 /- (2078 BS)