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

  1. $2, 4$
  2. $2, 14$
  3. $4, 52$
  4. $14, 52$
asked in Numerical Ability by Veteran (115k points) | 975 views

3 Answers

+21 votes
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 $D.$
answered by Boss (43.5k points)
edited by
+1 vote
d )

x=-2 ans is 14

x=4 ans is 52
answered by Loyal (5.3k points)
0
And calculated it how ?? No calculation ??
0 votes
Answer will be none of these..
For x=-2
It will be 2
And for x=4
It will be 52
answered by (495 points)
0

 

for  $x=-2$

The value of $|-x^{3}+x^{2}-x|=|-(2)^{3}+(-2)^{2}-(-2)|=|-(-8)+4+2=|8+4+2|=|14|=14$

Answer:

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