Inequalities with modulus function

You must commit to memory these thumb rules before you solve the questions.  If |f(x) | <  k, we get two equations f(x) < k and f(x) > - k
Take an example |x2−8x|$|x^2-8x|$<9
If x2−8x$x^2-8x$ is positive, then |x2−8x|=x2−8x$|x^2-8x|=x^2-8x$
We get x2−8x$x^2-8x$<9
If  |x2−8x|$|x^2-8x|$ is negative then |x2−8x|=−(x2−8x)$|x^2-8x|=-(x^2-8x)$
We get −(x2−8x)<9⇒x2−8x>−9$-(x^2-8x)<9\Rightarrow x^2-8x>-9$
Therefore we get two equations x2−8x<9$x^2-8x<9$ and x2−8x$x^2-8x$ > - 9

1. |x2−8x|$|x^2-8x|$<9
1. -1 ≤$\le$ x < 9
2. x∈(−1,4−7)−−√∪(4+7√,9)$x\in (-1,4-\sqrt{7)}\cup (4+\sqrt{7},9)$
3. x∈(1−4−7√)∪(4,9)$x\in (1-4-\sqrt{7})\cup (4,9)$
4. x∈(−1,4−7√)∪(4+7√,9)$x\in (-1,4-\sqrt{7})\cup (4+\sqrt{7},9)$
5. 1≤x≤−9$1\le x\le -9$
Ans:
From the above discussion,
If x2−8x$x^2-8x$ is positive, then |x2−8x|=x2−8x$|x^2-8x|=x^2-8x$
and |x2−8x|$|x^2-8x|$ is negative then |x2−8x|=−(x2−8x)$|x^2-8x|=-(x^2-8x)$
x2−8x<9$x^2-8x<9$ ------(1)
and −(x2−8x)<9⇒x2−8x>−9$-({\mathrm{x}}^2-8x)<9\Rightarrow {\mathrm{x}}^2-8x>-9$ -------(2)
From (1) (x+1)(x−9)<0⇒−1<x<9$(x+1)(x-9)<0\Rightarrow -1<x<9$ ------(3)
From (2) x2−8x+9>0${\mathrm{x}}^2-8x+9>0$
Using, −b±b2−4ac−−−−−−−√2a=8±(−8)2−4.1.9−−−−−−−−−−−√2.1=4±7√${\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}={\displaystyle \frac{8\pm \sqrt{{(-8)}^2-\mathrm{4.1.9}}}{2.1}=4\pm \sqrt{7}}}$
We know that for f(x)> 0, x should not lie in the roots.
x<4−7√$x<4-\sqrt{7}$ or x>4+7√$x>4+\sqrt{7}$ ---------(4)
From (3) and (4), we get
x∈(−1,4−7√)∪(4+7√,9)$x\in (-1,4-\sqrt{7})\cup (4+\sqrt{7},9)$
Hence option 2

2. |x2−12x|<45$|x^2-12x|<45$
1. −15≤x≤3$-15\le x\le 3$
2. 15≤x≤−3$15\le x\le -3$
3. 15<x<−3$15<x<-3$
4. −15<x<3$-15<x<3$
5. −15≤x<−3$-15\le x<-3$
Ans:
|x2−12x|<45$|x^2-12x|<45$
We have x2−12x<45$x^2-12x<45$ and x2−12x>−45$x^2-12x>-45$
 x2−12x−45<0$x^2-12x-45<0$  --------- (1)and x2−12x+45>0$x^2-12x+45>0$ ------(2)
Solving the first equation,
(x−15)(x+3)<0$(x-15)(x+3)<0$
−3<x<15$-3<\mathrm{x}<15$ ---------(3)
For second equation, b2−4ac=(−12)2−4.1.45=144−180=−36<0$b^2-4ac=(-12{)}^2-\mathrm{4.1.45}=144-180=-36<0$
So no real solution exist.
So only −3<x<15$-3<\mathrm{x}<15$ holds good.
Option 4.

3. |x2−5x|<4x$|x^2-5x|<4x$
(1) x < 9
(2) 15<x-3
(3)  1 < x < 9
(4) x>1
(5) -9<x<-1
Ans: |x2−5x|<4x$|x^2-5x|<4x$
x2−5x<4x$x^2-5x<4x$ and x2−5x>−4x$x^2-5x>-4x$
x2−5x−4x<0$x^2-5x-4x<0$ and x2−5x+4x>0$x^2-5x+4x>0$
x(x−9)<0$x(x-9)<0$ and x(x−1)>0$\mathrm{x}(\mathrm{x}-1)>0$
For x(x−9)<0$\mathrm{x}(\mathrm{x}-9)<0$ we have x∈(0,9)$\mathrm{x}\in (0,9)$
For x(x−1)>0$\mathrm{x}(\mathrm{x}-1)>0$ we have x < 0 and x > 1
Taking the common region from above we get 1<x<9
Hence Option 3

4. |x2+12x|<16x$|x^2+12x|<16x$
(1) -28<x<4
(2) 0<x<4
(3) 4<x<28
(4) -28<x<-4
ANS :
|x2+12x|<16x$|x^2+12x|<16x$
x2+12x<16x$x^2+12x<16x$ and x2+12x>−16x$x^2+12x>-16x$
x2+12x−16x<0$x^2+12x-16x<0$ and x2+12x+16x>0$x^2+12x+16x>0$
x(x−4)<0$x(x-4)<0$ and x(x+28)>0$\mathrm{x}(\mathrm{x}+28)>0$
For x(x−4)<0$\mathrm{x}(\mathrm{x}-4)<0$ we have x∈(0,4)$\mathrm{x}\in (0,4)$
For x(x+28)>0$\mathrm{x}(\mathrm{x}+28)>0$ we have x∈(−∞,−28)∪(0,∞)$\mathrm{x}\in (-\mathrm{\infty },-28)\cup (0,\mathrm{\infty })$
Taking the common region from above we get 0<x<4.
Hence option 2


5.  |x2+7x+14|<4x+14$|x^2+7x+14|<4x+14$
(1) -3<x<7
(2) -7<X<11
(3) -3<x<0
(4) -4<x<3
(5) -7<x<-3
Ans:
|x2+7x+14|<4x+14$|x^2+7x+14|<4x+14$
x2+7x+14<4x+14${\mathrm{x}}^2+7x+14<4x+14$ and x2+7x+14>−(4x+14)${\mathrm{x}}^2+7x+14>-(4x+14)$
x2+7x−4x<0${\mathrm{x}}^2+7x-4x<0$ and x2+7x+14>−4x−14${\mathrm{x}}^2+7x+14>-4x-14$
x(x+3)<0$x(x+3)<0$ and x2+11x+28>0${\mathrm{x}}^2+11\mathrm{x}+28>0$
x(x+3)<0$x(x+3)<0$ and (x+7)(x+4)>0$(\mathrm{x}+7)(\mathrm{x}+4)>0$
For x(x+3)<0$\mathrm{x}(\mathrm{x}+3)<0$ we have x∈(−3,0)$\mathrm{x}\in (-3,0)$
For (x+7)(x+4)>0$(\mathrm{x}+7)(\mathrm{x}+4)>0$ we have x∈(−∞,−7)∪(−4,∞)$\mathrm{x}\in (-\mathrm{\infty },-7)\cup (-4,\mathrm{\infty })$
So, we have, -3<x<0
Hence option 3

6. x2−14x+42<|2x−22|$x^2-14x+42<|2x-22|$
(1) x<10
(2) x<8
(3) 2<x<10
(4) 2<x<8
(5) 0<x<8
Ans:
If (2x - 22) > 0 then |2x−22|$|2x-22|$ = 2x - 22
If (2x - 22) < 0 then |2x−22|$|2x-22|$ = - (2x - 22)
So we have x2−14x+42<2x−22${\mathrm{x}}^2-14x+42<2x-22$ and x2−14x+42>−(2x−22)${\mathrm{x}}^2-14x+42>-(2x-22)$
x2−14x−2x+42+22<0${\mathrm{x}}^2-14x-2x+42+22<0$ and x2−14x+42+2x−22<0${\mathrm{x}}^2-14x+42+2x-22<0$
x2−16x+64<0${\mathrm{x}}^2-16x+64<0$ and x2−12x+20<0${\mathrm{x}}^2-12x+20<0$
(x−8)(x−8)<0$(\mathrm{x}-8)(x-8)<0$ and (x−2)(x−10)<0$(\mathrm{x}-2)(\mathrm{x}-10)<0$
(x−8)2$(\mathrm{x}-8{)}^2$can’t be less than 0 for real values of x.
For (x-2)(x-10)<0 we havex∈(2,10)$\mathrm{x}\in (2,10)$
Note that here we have to take union of the two sets of solutions and not intersection.
Hence option 3