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Solve the inequality

What are possible values of x?

\sqrt{x^2} < x+1 ?

a. x>0.5
b. x>0
c. All x
d. x>-0.5
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SathyaNIT
Solve the inequality

x-17>=60/x ?

a. x<-3
b. x<20
c. -3<=x<0 or x>=20
d. -3<x<=0 or x>=20

\(x-17\geq{60/x}\) = \(x^2(x-17)\geq{60x}\)
or \(x[x^2-17x-60]\geq{0}\)
or \(x(x-20)(x+3)\geq{0}\)
Thus, either \(x>={20} OR -3<=x<0\)[As x cannont equal 0].
C
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Solve the inequality

What are possible values of x?

|x-6|>x^2 - 5x+9 ?

a. 1<= x < 3
b. 1<x<3
c. 2<x<5
d. -3<x<1

\(|x-6|>x^2-6x+9+x =| x-6|>(x-3)^2+x\)

For x>6, we have\(x-6>(x-3)^2+x\) -->\(-6>(x-3)^2\) = This is not possible as for all real x, \((x-3)^2\geq{0}\)

Thus, for x<6, we have \(6-x>(x-3)^2+x\) --> \((x-3)^2+2x-6<0\)--> \((x-3)^2+2(x-3)<0\) =\((x-3)(x-3+2)<0\) --> (x-3)(x-1)<0

Thus, 1<x<3.

B.
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Just solve for the case when X is +ve.
You will end up getting x>= 20 or x<= -3.
So finally search for answer that states the following
C and are the ones ..but notice that d does not mentions equal to in the inequality for -3, hence cannot be the correct answer.
So the answer must be C

Consider kudos if my post helps!!!!!!!!!!

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I used the attached graph to solve this inequality

Its clear from the graph that \(\sqrt{x^2}\) is greater than (x+1) only when x is greater than -0.5

So ans is D
Attachments

untitled1.JPG
untitled1.JPG [ 10.31 KiB | Viewed 2958 times ]

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I used the attached graph to solve this inequality

Its clear from the graph that \(\sqrt{x^2}\) is greater than (x+1) only when x is greater than -0.5

So ans is D


This graph doesn't seem right. The mod function a curve that intersects the line y=x somewhere between the ranges x=1 and x=3.
Therefore the range of values for f(x) < x or y< x is 1<x < 3

Thanks
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SathyaNIT
Solve the inequality.

What are possible values of x?

|x^2 - 2x| 3
Hi,

The best approach is to put different value of x and validate the inquality.
For ex, x=0, LHS=RHS (thus, inequality doesn't hold true.)
So, answer choice that has got 0 in the range will be eliminated.
a. 13

Put x= 1, LHS=RHS
a. 13

Put x=4, LHS = RHS
a. 13[/s]
Thus, the only choice which remains is (A)

Check, put x=2, LHS < RHS
Therefore, Answer (A).



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