CAT 2008: Quantitative Questions a Day 134 -Till end -> The Discussions

[fiction]

x + |x^2 - 1| = k

k = 6/5

x + |x^2 - 1| = 6/5
for x^2 > 1, we get 5x^2 + 5x - 11 = 0 (2 roots satisfying x^2>1)
for x^2 < 1, we get 5x^2 - 5x - 1 = 0 (2 roots satisfying x^2<1)

k = 5/4
x + |x^2 - 1| = 5/4
for x^2 > 1, we get 4x^2 + 4x - 5 = 0 (2 roots satisfying x^2>1)
for x^2 < 1, we get 4x^2 - 4x + 1 = 0 (1 root satisfying x^2>1)


-1 < k < 1

-1 < x + |x^2 - 1| < 1

means, x < 1 (since |x^2 - 1| > 0)

so, for x E (-1, 1), we get x^2 - x - (1+k) = 0
D = 1+4(1-k) > 1, so one root will be greater than 1. Only one root from here satisfies the condition on x

for x < -1, we get x^2 + x -1 - k = 0
D = 1+ 4 (1+k) > 1, one root will be positive, so only one x from here satisfies the condition on x.
Total 2 roots.


Hence (4) All of the foregoing

[fiction]

Originally Posted by slam

case 1: (x^2-1) > 0

=> x^2 + x - (K+1) = 0 => x = (-1 + rt(4k+5) )/2 ; (-1 - rt(4k+5) )/2

case 2: (x^2-1) > 0

=> x^2 - x + (k-1) = 0 => x = (1 + rt(5-4k) )/2 ; (1 - rt(5-4k) )/2

Now, for k = 6/5
we get 2 roots that satisfy x^2 -1 > 0 from case 1 and similarily from case 2. => 4 roots.

For k = 5/4
we get 2 roots from case 1 and 1 root from case 2 (since D = 0 for case 2 now)

For -1 < k < 1
case 1:
(4k+5) lies in (1,9) (exclusive)
=> x lies in (0,1) and (-1,0)
None of these satisfy x^2 -1 > 0 => no roots from here.

case 2:
(5-4k) lies in (1,9) (exclusive)
=> x lies in (1,2) and (-1,0)
=> only one root from here that satisfies x^2 -1 < 0 (the one that lies in (-1,0)

=> we have only options (1) and (2) correct. => exactly two of the foregoing.

Put k =1, we get x = -2.

[implex]

The question can be done by graph easily

let us plot y=x+|x^2-1|
gives Y=(x+1/2)^2 -5/4 for |x|>=1
and Y=5/4 -(x-1/2)^2 for |x|<=1

if Y=5/4 first equation gives two roots second one
if -1<Y<1 each equation will give two roots , one of which will go outside the boundary
and Y=6/5, first equation gives two roots and second gives two roots

all the three conditions are true

option 4 all of the foregoing

[implex]

the other way to solve this is rotate the coordinate axes 45 degrees anti clockwise, it will spill the results in a jiffy we can easily that in the new coordinate system a line parallel to x axis at a distance of (1/sqrt(2)) gives 3 or more solution

[ankaj]

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Quantitative Question # 150
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The number of roots of the equation x + | x^2 - 1 | = k, where k is a real constant is/are



(1) 4 for k = 6/5 (2) 3 for k = 5/4 (3) 2 for -1 < k < 1 (4) All of the foregoing (5) Exactly two of the foregoing


Case 1 when the values of x is >1 and less<-1...value of x^2>1
x+x^2-1=K

Case 2 when the value of x2<1
x-x^2+1=K

K= 6/5
a) 5x^2 + 5x -11 = 0...D= 245.....2 real roots
b) 5x-5x^2-1=0......D= 5...... 2 real roots

K= 5/4
a) 4x^2 + 4x -9 = 0 D= 160....2 real roots
b) 4x-4x^2-1=0......D= 0.....equal roots....1 value

-1<K<1...let K=0
a) x^2 +x -1 = 0 ....D= 5....2 real roots
b) x-x^2+1=0......D= 5..2 real roots
but two if these will not fall in the boundries...so
for this case only two values....

All true....
so 4) All of these..

[nbangalorekar]

Originally Posted by implex

The question can be done by graph easily

let us plot y=x+|x^2-1|
gives Y=(x+1/2)^2 -5/4 for |x|>=1
and Y=5/4 -(x-1/2)^2 for |x|<=1

if Y=5/4 first equation gives two roots second one
if -1<Y<1 each equation will give two roots , one of which will go outside the boundary
and Y=6/5, first equation gives two roots and second gives two roots

all the three conditions are true

option 4 all of the foregoing

the graphical method can be further simplified if we just plot the graph of
y=x+|x^2-1| & just check how many intersection pts we will get for the various k's ( which will be lines parallel to 'x axis'. The no of roots will be = no of intersection pts of that particular graph.
the ans is option 4
some one also post the graph (moi feeling lazy rite now)

[implex]

Originally Posted by nbangalorekar

the graphical method can be further simplified if we just plot the graph of
y=x+|x^2-1| & just check how many intersection pts we will get for the various k's ( which will be lines parallel to 'x axis'. The no of roots will be = no of intersection pts of that particular graph.
the ans is option 4
some one also post the graph (moi feeling lazy rite now)

There is no need just rotate the axis by 45 degress
its 15 secs from there, no math required !!

[implex]

Originally Posted by ankaj

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Quantitative Question # 150
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The number of roots of the equation x + | x^2 - 1 | = k, where k is a real constant is/are



(1) 4 for k = 6/5 (2) 3 for k = 5/4 (3) 2 for -1 < k < 1 (4) All of the foregoing (5) Exactly two of the foregoing


Case 1 when the values of x is >1 and less<-1...value of x^2>1
x+x^2-1=K

Case 2 when the value of x2<1
x-x^2+1=K

K= 6/5
a) 5x^2 + 5x -11 = 0...D= 245.....2 real roots
b) 5x-5x^2-1=0......D= 5...... 2 real roots

K= 5/4
a) 4x^2 + 4x -9 = 0 D= 160....2 real roots
b) 4x-4x^2-1=0......D= 0.....equal roots....1 value

-1<K<1...let K=0
a) x^2 +x -1 = 0 ....D= 5....2 real roots
b) x-x^2+1=0......D= 5..2 real roots

only 1 and 2 options are true...
so 5) exactly two of foregoing..

in your case 3 your boundary values are beaten, just check out of those 4, two do not fall in the boundries

[tuco]

x + |x^2 - 1| = k
this can be x+x^2-1=k when x^2-1 is positive
and x-x^2+1 = k when x^2-1 is negative ie when -1<x<1

when k=6/5 we get two different equations & none of them is a perfect square. solving we 4 roots n all of them hold true to the supposed value range of x. so (a) is true.

when k=5/4 we get two different equations & one of them is a perfect square. solving we get 3 roots n again all of them hold true. so (b) is true.

when -1<k<1. taking k=0
we again get 4 different roots. so (c) is not true.

hence ans-(5) exactly two of the foregoing

[utsavbhatt]

In the answer to Q. 149
the first equation needs to be
3a + 4b = 5 and not 4
Kindly correct me !