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04 Sep 2013, 23:52
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05 Sep 2013, 00:10
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kumar83
Is x^3-6x^2+11x-6<0
(a) 1<x<=2
(b) 2<=x<3
Please Explain too.
(a) 1<x<=2
(b) 2<=x<3
Please Explain too.
Factorise the given expression : \(x^3-6x^2+5x+6x-6 = x(x^2-6x+5) +6(x-1) = x(x-5)(x-1)+6(x-1) = (x-1)(x^2-5x+6) = (x-1)(x-2)(x-3)\)
Thus, the question boils down to : Is (x-1)(x-2)(x-3)<0. Now, the given expression when plotted for negative and positive ranges , would yield :
---(-ve)---1----(+ve)----2---(-ve)-----3----(+ve)----
Now from F.S 1, for 1<x<2, we have the given expression as always positive. Also, for the boundary condition of x=2, the given expression is zero. Thus, in both the instances, the expression is NOT negative. Sufficient.
From F.S 2, for 2<x<3, the expression is negative,however, for the boundary condition, i.e. x=2, the expression is zero(and not negative). Thus, Insufficient.
A.
06 Sep 2013, 00:15
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kumar83
Is x^3 - 6x^2 + 11x - 6 < 0
(1) 1 < x <= 2
(2) 2 <= x < 3
(1) 1 < x <= 2
(2) 2 <= x < 3
We know how to deal with quadratics but we usually do not deal with third degree equations. If you do get a third degree equation, it will have one very easy root i.e. 0 or 1 or -1 or 2 or -2. Try a few of these values to get the first root. Here x = 1 works. It is easy to see since you have two 6s and an 11 as the co-efficients.
So you know that (x-1) is a factor. Now figure out the quadratic which when multiplied by (x-1) gives the third degree expression
\((x - 1)(x^2 - 5x + 6) = x^3 - 6x^2 + 11x - 6\)
How do you get this quadratic? Let's see.
\((x - 1)(ax^2 + bx + c) = x^3 - 6x^2 + 11x - 6\)
Coefficient of x^3 on right hand side is 1. So you know that all you need is x^2 so that it multiplies with x to give x^3 on left hand side too. So a must be 1.
\((x - 1)(x^2 + bx + c) = x^3 - 6x^2 + 11x - 6\)
The constant term is easy to figure out too. It should multiply with -1 to give -6 on right hand side. Hence c must be 6.
\((x - 1)(x^2 + bx + 6) = x^3 - 6x^2 + 11x - 6\)
The middle term is slightly more complex. bx multiplies with x to give x^2 term on right hand side (i.e. -6x^2) but you also get the x^2 term by multiplying -1 with x^2. You need -6x^2 and you have -x^2 so you need another -5x^2 from bx^2. So b must be -5.
\((x - 1)(x^2 - 5x + 6) = x^3 - 6x^2 + 11x - 6\)
Now you just factorize the quadratic in the usual way and use the wavy curve to solve the inequality.
\((x - 1)(x - 2)(x - 3) < 0\)
2 < x < 3 or x < 1
(1) 1 < x <= 2
Between 1 to 2 (inclusive), the expression is not negative. It is 0 or positive.
Sufficient.
(2) 2 <= x < 3
Between 2 and 3, the expression is negative but it is 0 when x = 2. Hence insufficient.
Answer (A)
