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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 _________________

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

Re: Solve the inequality [#permalink]
18 May 2013, 08:10

aceacharya wrote:

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

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. 1<x<3 b. -1<x<3 c. 0<x<4 d. x>3

Put x= 1, LHS=RHS a. 1<x<3 b. -1<x<3 c. 0<x<4 d. x>3

Put x=4, LHS = RHS a. 1<x<3 b. -1<x<3 c. 0<x<4 d. x>3 Thus, the only choice which remains is (A)

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

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