Such absolute functions are quite tricky but extremly simple, scoring and take very less time, if you know the them properly.

Worth a watch, if you are struggling...(also have a look at other modulas parts from same guy, if need be)

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in Quantitative Aptitude
Sep 28, 2014

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27 votes

If $x$ is real and $\mid x^2-2x+3 \mid = 11$, then possible values of $\mid -x^3+x^2-x\mid$ include

- $2, 4$
- $2, 14$
- $4, 52$
- $14, 52$

2

Before this lecture if you're new to this type of concept then watch https://youtu.be/XhAb4lTQFTQ . Then after watch about video . It help me to understand the concept rather mugup the properties.

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Best answer

$x^2 - 2x + 3 = 11$ OR $x^2 -2x+3 = -11$

$($Any one of them can be correct because of $\text{mod})$

Lets take first one:

$x^2 - 2x + 3 = 11$

$\implies x^2 - 2x - 8 = 0$

$\implies (x-4) (x+2) =0$

$\implies x= 4 \text{ or } x = -2.$

Now put these values of $x$ in the given equation $\mid -x^3+x^2 -x\mid $

for $x= 4,$ we will get $\mid -64+16-4\mid \quad= 52.$

for $x = -2,$ we will get $\mid 8 + 4 + 2 \mid \quad= 14.$

So, answer is $D.$

$($Any one of them can be correct because of $\text{mod})$

Lets take first one:

$x^2 - 2x + 3 = 11$

$\implies x^2 - 2x - 8 = 0$

$\implies (x-4) (x+2) =0$

$\implies x= 4 \text{ or } x = -2.$

Now put these values of $x$ in the given equation $\mid -x^3+x^2 -x\mid $

for $x= 4,$ we will get $\mid -64+16-4\mid \quad= 52.$

for $x = -2,$ we will get $\mid 8 + 4 + 2 \mid \quad= 14.$

So, answer is $D.$

If we take the second equation $x^{2}-2x+3=-11$

$\implies x^{2}-2x+14 = 0$

Lets find discriminant of a quadratic$:\Delta = b^{2}-4ac$

$\implies \Delta = (-2)^{2}-4\times 1 \times 14 = 4-56 = -52 < 0$

If $\Delta < 0,$ then the expression inside the square root is negative and the roots are both non-real complex roots.

That's by we ignore this equation because it is given that $x$ is real.

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0 votes

edited
Jan 3, 2020
by Lakshman Patel RJIT

1