CAT 2008: Quantitative Questions a Day 1 to 50 - The discussions

[v-factor]

|2x-1|-3|x+1| = a
There are 3 intervals
Case 1:
x > 1/2, we get x = -a-4
=> a< -9/2

Case 2:
-1< x < 1/2, we get x = -(a+2)/5
=> -9/2 < a < 3

Case 3:
x < -1, we get x = a-4
=>a < 3

Case 1 and 2: No value falls in the given range
Case 1 and 3: a < -9/2
=> p-q = -2a
=> -5 < a < -1
Case 2 and 3: -9/2 < a < 3
=> p-q = (-6a+1/5
=> -16/3 < a < 4/3

So the max(a) satisfying the condition is 4/3
min(a) is -5
Thus, 4/3-(-5) = 19/3

[Daniel Ocean]

there are three ranges:
in order to have roots,
(-inf, -1) => a<3 .....(I)
(-1,1/2) => 3> a >-4.5 .....(II)
(1/2,inf) => a<-4.5 .....(III)

we can see that II & III are mutually exclusive.

so the two roots lie under conditions (I & II) or (I & III)

(I & II) => -4.5 < a < 3
or
(I & III) => a > -4.5

under the second condition, we will not have a finite difference for Max(a)-Min(a)
so we have to take first condition, (I & II) => ans = (b) 15/2

ps: there is definitely a better solution using the range for |p-q|. but i dont have it

[kill_cat]

Hello everyone,

This is my first post on PG. .
kudos to al the maths masters here on this thread.
here goes my solution.

defintion of modulus:
For any real number a the absolute value or modulus of a is denoted by |a|,
and is defined as
|a| = a, if a>=0;
-a, if a<0;
given equation is |2x-1|-3|x+1|=a;
case 1:2x-1>=0 and x+1>=0

=> x>=1/2 and x>=-1 ==> x>=1/2 (a common domain for both values of x)

coming to the equation we have from definition of modulus,

2x-1-3(x+1)=a;
==> x=-a-4;
hence,
-a-4>=1/2
==> a<=-9/2 ------------------(1);

case 2: 2x-1>=0 and x+1 <0
==> x>=1/2 and x<-1
==> there is no domain for these values of x (together) and hence this case is not applicable.

case 3:2x-1 < 0 and x+1 < 0
==> x<1/2 and x<-1
==>x<-1 (combining the domains)
hence we have from the equation,
-(2x-1)+3(x+1)=a;
=> x=a-4;
as we have x<-1,

a-4<-1
==> a<3.---------------------(2)

case 4: 2x-1<0 and x+1>=0;
==> x<1/2 and x>=-1
here the domain is an interval i.e -1<=x<1/2.
from the expression -(2x-1)-3(x+1)=a
==> -5x-2=a;
x=-(a+2)/5.
as x lies in an interval,
-1<= -(a+2)/5 < 1/2
==> -5<= -a-2 < 5/2

==> -3 <= -a < 9/2
==> -9/2 < a < 3--------------(3).

from values of (1) and (3) we can see that there is no value which satisfies the range of a (mutually exclusive).hence this case is ignored.
llY from values of (1) and (2) ==> there is no finite minima value.
only possible option is case (2) and (3).
here range of a is -9/2 to to 3.
hence max(a)- min(a) =3+9/2 = 15/2.

I go with option (b).

ps: I am unable to figure out why the question contains something related to p and q.


Thanks,
Anil.

[prade]

|2x-1| - 3|x+1| = a

when x <-1
x+4 = a and a<3 .......(1)
when x>=-1 and x< 1/2
x = -(a+2)/5 => -1/2 < a <= 3 .......(2)
when x>=1/2
x = -(a+4) => a<=-9/2 .......(3)

from (1) and (2)

|p-q| = |a-4 + (a+2)/5| -1/2 < = a < 3
now 2<=|p-q| <=10
10<= |6a-18| <= 50
10<=-6a + 18 <= 50
-16/3 <= a <= 4/3
hence -1/2 < = a <= 4/3 => possible a

from (2) and (3)
2<= |-(a+4) + (a+2)/5| <= 10 where -9/2 <= a
1/2<=a<=8 => no possible solution

from (1) and (3)
2<= |a-4+a+4| <= 10 and a<=-9/2
1<=|a| <= 5 HENCE min a=-5

=> max(a) = 4/3 and min(a) = -5

max(a) - min(a) = 19/3 option (a)

[zion1729]

x = -(a+4), x > 1/2 (Interval - I)

x = -(a+2)/5, -1 <= x <= 1/2 (Interval - II)

x = a-4, x< -1 (Interval - III)

-(a+4) > 1/2 => a < -9/2 => -(a+2)/5 > 1/2

Hence if a root lies in interval I, it doesn't lie in interval II

Also,

a-4<-1 => a < 3 Hence for -9/2 < a < 3, there might exist two roots p,q in interval II and III respectively such that 2 <= |p-q| <= 10. We see that in the above interval, all a satisfy 2 <= 2/5|2a-11| <= 10.

We also see that if a <-9/2, then the roots p,q lie in intervals I and III respectively, in which case |p-q| = |2a|, which implies -5<a<-1 or 1<a<5 - the second result is ruled out as a <-9/2. The intersection between the first result and a <-9/2 is -5<a<-9/2.

Hence we have max(a) - min(a) = 3 -(-5) = 8.

Ans. None of these

[ashwin7777]

Puys,
I am not getting these questions after question # 35...Could any one please help me out...


Rgds,
Ashiwin

[Varun Khullar]

i got a few inequalities..
by solving it for various ranges of x
-9/2 < a < 3
a < 3
a< -9/2
then solve |p-q| for each of them ..we get 19/3

[Aarav]

Very well done students This was not an easy problem, but many of you seemed to have cracked this in very 1st attempt.

The official solution to QQAD 038

Solution:

Let f(x) = |2x-1| - 3|x+1|
We have f(x) = x+4 if x <= -1,
-5x-2 if -1 < x < 1/2,
-x-4 if x >= 1/2.

Draw the graph of f(x) now and see that the nodes are (-1, 3) and (1/2, -9/2). The shape of this graph is almost inverted V with V-vertex at (-1, 3) and sligh right shift from (1/2, -9/2).

The equation f(x) = a has no solution if a > 3 because for the line parallel to x-axis doesn't intersect the graph for y > 3.

The equation f(x) = a has 2 solutions when -9/2 <= a <= 3 and the solutions are determined by the intersection of line y = a and the lines y = x+4 and y = -5x-2. The solutions are (a-4, a) and (-(a+2)/5, a).

We are given that solutions p and q are such that 2 <= |p-q| <= 10 => 2 <= |a-4 + (a+2)/5| <= 10 subject to constraint on a => -9/2 <= a <= 4/3
Similarly, the equation f(x) = a has 2 solutions when a < -9/2, and the solutions are determined by the intersection of line y = a and the lines y = x+4 and y = -x-4. The intersection points would be (a-4, a) and (-a-4, a).
We are given that solutions p and q are such that 2 <= |p-q| <= 10 => -5 <= a <= -9/2.

Combining the above we get -5 <= a <= 4/3.

[kill_cat]

Originally Posted by kill_cat

Hello everyone,

This is my first post on PG. .
kudos to al the maths masters here on this thread.
here goes my solution.

defintion of modulus:
For any real number a the absolute value or modulus of a is denoted by |a|,
and is defined as
|a| = a, if a>=0;
-a, if a<0;
given equation is |2x-1|-3|x+1|=a;
case 1:2x-1>=0 and x+1>=0

=> x>=1/2 and x>=-1 ==> x>=1/2 (a common domain for both values of x)

coming to the equation we have from definition of modulus,

2x-1-3(x+1)=a;
==> x=-a-4;
hence,
-a-4>=1/2
==> a<=-9/2 ------------------(1);

case 2: 2x-1>=0 and x+1 <0
==> x>=1/2 and x<-1
==> there is no domain for these values of x (together) and hence this case is not applicable.

case 3:2x-1 < 0 and x+1 < 0
==> x<1/2 and x<-1
==>x<-1 (combining the domains)
hence we have from the equation,
-(2x-1)+3(x+1)=a;
=> x=a-4;
as we have x<-1,

a-4<-1
==> a<3.---------------------(2)

case 4: 2x-1<0 and x+1>=0;
==> x<1/2 and x>=-1
here the domain is an interval i.e -1<=x<1/2.
from the expression -(2x-1)-3(x+1)=a
==> -5x-2=a;
x=-(a+2)/5.
as x lies in an interval,
-1<= -(a+2)/5 < 1/2
==> -5<= -a-2 < 5/2

==> -3 <= -a < 9/2
==> -9/2 < a < 3--------------(3).

from values of (1) and (3) we can see that there is no value which satisfies the range of a (mutually exclusive).hence this case is ignored.
llY from values of (1) and (2) ==> there is no finite minima value.
only possible option is case (2) and (3).
here range of a is -9/2 to to 3.
hence max(a)- min(a) =3+9/2 = 15/2.

I go with option (b).

ps: I am unable to figure out why the question contains something related to p and q.


Thanks,
Anil.

now..i got it right......

2<=|p-q|<=10 ===>take combinations of (1) &(2),(2) & (3),(1) & (3).
here basing on values of a (1) & (3) are mutually exclusive.

(1) (2) (3)
--- --- ----
x -a-4 a-4 -(a+2)/5

range of a a<=-9/2 a<3 -9/2<a<3

(1) & (2):
____________

p-q= a-4+a+4=2a.
common domain of a is a<=-9/2. hence 2a is negative and therfore,
|p-q|=|2a|=-2a;
2<=-2a<=10.
-5<= a <=-1.--------(6) and along with the domain of a<=-9/2,
the final domain for this combination is -5 <= a <= -9/2.----(YYYY)
(2) & (3)
-----------
p-q = a-4+(a+2)/5
=> (6a-1/5.
common domain of a is -9/2 < a < 3, where any value of 6a-18 is not positive.
hence |p-q|= |6a-18/5|= -(6a-1/5.
2<= (-6a+1/5 <=10
10<=-6a+18<=50.
-16/3<= a <= 4/3.--------(7).
final domain for this combination is -9/2 < a< 4/3.------(XXXX)

hence from (XXXX) and (YYYY),
it is max (a)= 4/3 and min (a)=-5.
hence diff= 4/3-(-5)==>4/3+5=19/3.

[kill_cat]

I apologize for such a lengthy post but.....I found that might help some of newbies like me.

Cheers,
Anil.