Inequalities - Solved examples

1. Solve 1x>15${\displaystyle \frac{1}{x}>\frac{1}{5}}$
(a)   We cannot cross multiply since we don’t know whether x is positive or negative.
Case 1: If x > 0
⇒$\Rightarrow$ 5 > x
⇒$\Rightarrow$ x < 5
So  0 < x < 5

Case 2: If x < 0
⇒$\Rightarrow$ 5 < x
⇒$\Rightarrow$ x > 5
No solution.
So Solution set for given inequation is 0 < x < 5.

2. Solve 1x<15${\displaystyle \frac{1}{x}<\frac{1}{5}}$
We cannot cross multiply since we don’t know whether x is positive or negative.

Case 1: If x > 0
5 < x
⇒$\Rightarrow$ x > 5

Case 2: If x < 0
5 > x
⇒$\Rightarrow$ x < 5
But we know that x < 0
Solution set for given inequation is x < 0 or x > 5.

3. Solve: x+3x+4≥1${\displaystyle \frac{x+3}{x+4}\ge 1}$

x+3x+4−1≥0${\displaystyle \frac{x+3}{x+4}-1\ge 0}$
x+3−x−4x+4≥0${\displaystyle \frac{x+3-x-4}{x+4}\ge 0}$
−1x+4>0${\displaystyle \frac{-1}{x+4}>0}$
[ab>0anda<0⇒b<0]$\left[{\displaystyle \frac{a}{b}>0anda<0\Rightarrow b<0}\right]$
x (– ∞$\mathrm{\infty }$ , 4]

4. If x2${\mathrm{x}}^2$ – 7x + 12 < 0. find the range of x.
x2${\mathrm{x}}^2$ – 7x + 12 < 0
(x – 3) (x – 4) < 0
3 < x < 4
This is of type a – b < 0. This is possible when the product of two is negative. Rather than trying possibilities, a simpler way also exists. (x – 3) will be negative when if x < 3 and positive of x > 3.
Similarly, (x – 4) will be negative if x < 4 and positive if x > 4
Combining above possibilities, we have-
If x > 4, both terms will be positive and hence the product will be positive.
If 3 < x < 4, (x – 3) remain positive and by (x – 4) will be negative. Hence the product will be negative. since, we need the product to be negative, the solution will be 3 < x < 4.
Rather writing above all, we can simply plot the points where the terms will change the signs, on a number line as follows.

The number line represents all values from – ∞$\mathrm{\infty }$ to + ∞$\mathrm{\infty }$. However, it is broken in various regions, three regions for this inequality. Now we have to identify those values of x that satisfy the given inequality,

For the right most region, x > 4, all terms will be positive and hence the product will be positive. For the region from the right, one term will turn negative and thus the product will be negative in this range. For the third region from right side, two terms will turn negative making product positive for this range of x.

5. x2${\mathrm{x}}^2$ – x –  30 > 0, find the range of x.
(x + 5) (x – 6) > 0
Drawing and representing this on number line.

i, e. X doesn’t lie between – 5  and 6. So, x > 6 or x < – 5

6. x2${\mathrm{x}}^2$ – 4x + 3 ≤$\le$ 0, find the range of x.
(x – 1) (x – 3) ≤$\le$ 0
Representing this on number line-

 

i, e x lies between 1 ≤$\le$  x ≤$\le$ 3.

7. x2${\mathrm{x}}^2$+ 8x – 33 ≥$\ge$  0
(x + 11) (x – 3) ≥$\ge$ 0
Representing this on number line-

Hence, x3 ≥$\ge$ or x ≤$\le$  – 11.


8. Solve for the following conditions x2${\mathrm{x}}^2$ + 8x – 33 ≥$\ge$ 0, x2${\mathrm{x}}^2$ ≥$\ge$ 36

Solution: First we  have to find the solution set for separate equations.
(a)  x2${\mathrm{x}}^2$ + 8x – 33 ≥$\ge$ 0
(x + 11) (x – 3) ≥$\ge$ 0
Hence, x ≥$\ge$ 3 or x ≤$\le$  – 11

(b) x2${\mathrm{x}}^2$ ≥$\ge$ 36
x2${\mathrm{x}}^2$ – (6)2 ≥$\ge$ 0
(x + 6) (x – 6) ≥$\ge$ 0
Hence, x ≤$\le$ – 6 or x ≥$\ge$ 6
Now combining the solutions from (a) and (b).

 


 

From above two number line we can conclude that the region from x ≤$\le$ – 11 or x ≥$\ge$ 6 is possible for both. Hence, required solution is x ≤$\le$ – 11 or x ≥$\ge$ 6. 

9. (x+2x2−16)−(5x+4)<(6x4x−x2)$\left({\displaystyle \frac{x+2}{x^2-16}}\right)-\left({\displaystyle \frac{5}{x+4}}\right)<\left({\displaystyle \frac{6x}{4x-x^2}}\right)$
(1)  4
(2)  3
(3)  2
(4)  1
(5)  -1
Ans:
(x+2x2−16)−(5x+4)<(6x4x−x2)$\left({\displaystyle \frac{x+2}{x^2-16}}\right)-\left({\displaystyle \frac{5}{x+4}}\right)<\left({\displaystyle \frac{6x}{4x-x^2}}\right)$

Cancelling x, in the right hand term

(x+2x2−16)−(5x+4)<(64−x)$\left({\displaystyle \frac{x+2}{x^2-16}}\right)-\left({\displaystyle \frac{5}{x+4}}\right)<\left({\displaystyle \frac{6}{4-x}}\right)$

Multiplying the numerator and denominator by - 1 in right hand term

(x+2x2−16)−(5x+4)<−(6x−4)$\left({\displaystyle \frac{x+2}{x^2-16}}\right)-\left({\displaystyle \frac{5}{x+4}}\right)<-\left({\displaystyle \frac{6}{x-4}}\right)$

x+2(x−4)(x+4)−5x+4+6x−4<0${\displaystyle \frac{x+2}{(x-4)(x+4)}-{\displaystyle \frac{5}{x+4}+{\displaystyle \frac{6}{x-4}<0}}}$

2x+46(x−4)(x+4)<0${\displaystyle \frac{2x+46}{(x-4)(x+4)}<0}$

x+23(x−4)(x+4)<0${\displaystyle \frac{x+23}{(x-4)(x+4)}<0}$

So the above inequality is valid for, numerator is negative and denominator is positive. So x < -23 and x < -4, x > 4. No number satisfy this range
or the above inequality is valid for, numerator is positive and denominator is negative. So x > -23 and -4 < x < 4.
Hence 3 is the largest integer which satisfies the inequality.
Hence option 2

10. (x−10x−9)−(30x2−81)>(15xx+9)$\left({\displaystyle \frac{x-10}{x^{}-9}}\right)-\left({\displaystyle \frac{30}{x^2-81}}\right)>\left({\displaystyle \frac{15x}{x+9}}\right)$
(1) 12  
(2) 15  
(3) 9
(4) 0  
(5) 8
Ans:
x−10x−9−30x2−81>15xx+9${\displaystyle \frac{x-10}{x-9}-{\displaystyle \frac{30}{x^2-81}>{\displaystyle \frac{15x}{x+9}}}}$

x−10x−9−30(x−9)(x+9)>15xx+9${\displaystyle \frac{x-10}{x-9}-{\displaystyle \frac{30}{(x-9)(x+9)}>{\displaystyle \frac{15x}{x+9}}}}$

(x−10)(x+9)−30−15x(x−9)(x−9)(x+9)>0${\displaystyle \frac{(x-10)(x+9)-30-15x(x-9)}{(x-9)(x+9)}>0}$

x2−x−90−30−15x2+135x(x−9)(x+9)>0${\displaystyle \frac{x^2-x-90-30-15x^2+135x}{(x-9)(x+9)}>0}$

−120−14x2+134x(x−9)(x+9)>0${\displaystyle \frac{-120-14x^2+134x}{(x-9)(x+9)}>0}$

−7x2+67x−60(x−9)(x+9)>0${\displaystyle \frac{-7x^2+67x-60}{(x-9)(x+9)}>0}$

7x2−67x+60(x+9)(x−9)<0${\displaystyle \frac{7x^2-67x+60}{(x+9)(x-9)}<0}$

(7x−60)(x−1)(x+9)(x−9)<0${\displaystyle \frac{(7x-60)(x-1)}{(x+9)(x-9)}<0}$

x∈(−9,1)∪(60/7,9)$x\in (-9,1)\cup (60/7,9)$

So the largest integer that satisfies the inequality is 0.
Hence option 4

 11. (x−3)(6−x)(x−8)2>0$(x-3)(6-x)(x-8{)}^2>0$
(1) 3<x>10 and x ≠$\ne$8
(2) 3<x<9
(3) 6<x<8 and x≠$\ne$3
(4) 3<x<6
(5) 6<x<8
Ans:
(x−3)(6−x)(x−8)2>0$(x-3)(6-x)(x-8{)}^2>0$
(x−3)(x−6)(x−8)2<0$(x-3)(x-6)(x-8{)}^2<0$
As (x−8)2${(\mathrm{x}-8)}^2$ is always positive, the sign of the above inequality depends on (x−3)(x−6)<0$(x-3)(x-6)<0$
So x should lie in between 3 and 6
Hence option 4

12. x2−7x+12x2+x+1<0${\displaystyle \frac{x^2-7x+12}{x^2+x+1}<0}$
(1) 3<x<4 
(2) x<3 and x>4
(3)  x<3 
(4) 4<x
(5) x<4
Ans:
x2−7x+12x2+x+1<0${\displaystyle \frac{x^2-7x+12}{x^2+x+1}<0}$
(x−3)(x−4)x2+x+1<0${\displaystyle \frac{(x-3)(x-4)}{x^2+x+1}<0}$
For denominator 2x2+2x+1$2x^2+2x+1$ we have D<0.  Hence is positive for all real values of x.
The inequality reduces to (x+8)(x-2)<0
So we have , -8<x<2
Hence option 3