official quant thread for cat08

[the_hate]

Originally Posted by kaps_lock

Actually i had probs in 6 questions.I had posted 4 earlier now here is other 2:-

Q1.If x+y=4. Find the max/min possible value of x^2 + y^2

1.Min , 8
2.Max , 8
3.Max , 16
4.Min , 16

x+y=4> y=4-x

now, x^2+y^2
= x^2+(4-x)^2
= x^2+16+x^2-8x
= 2(x^2-4x+ 8 )
= 2[(x-2)^2+4]
so, at x=2, value is min,which = 8(placing x=2)

[kewl.bantee]

Originally Posted by implex

for such questions use cauchy inequality and jensons inequality you are done

suppose f(t)=t^2
clealry its a convex function
so [f(x)+(y)]/2>=f([x+y]/2)
x^2+y^2>=[(x+y)^2]/2=16/2=8

so minimum value is 8

if [f(x)+(y)]/2>=f([x+y]/2)
then (x^2+y^2)/2>=[(x+y)^2]/2 which gives (x^2+y^2)>=(x+y)^2
and (x^2+y^2)>=4^2 i.e.16
so min value of (x^2+y^2)is 16 and maximum is infinite

[getneonow]

Originally Posted by kewl.bantee

if [f(x)+(y)]/2>=f([x+y]/2)
then (x^2+y^2)/2>=[(x+y)^2]/2 which gives (x^2+y^2)>=(x+y)^2
and (x^2+y^2)>=4^2 i.e.16
so min value of (x^2+y^2)is 16 and maximum is infinite

dude..u forgot to divide the result by 2..

min is 16/2 = 8

or min occurs at x=y=2

and max can be infinite since x,y E R (x=5,y=-1 ; x= 6, y =-2 and so on)

Neo

[implex]

Problem set # 3
1) given that 5x+12y=60 find the minimum value of (x^2+y^2)^(1/2)

2) find all x in -(pie)<=x<=(pie)

(sinx)cox4x +2(sin2x)^2=1 -4[sin(pie/4-x/2]^2

3 Let r be a positive rational no, but not an integer. Let n be a positive integer such that [nr] is a prime. [.] means greatest integer function( lower floor). Then which is always true

a) There is no such n

b) if n1 and n2 are such n's then n1=n2 always

c) if n1, n2 and n3 are such n's then n3=n1+n2

d) none of these

4) if x+y+xy=75 and x^2-y^2=315 how many (x,y) pairs are possible?
a)0 b) 1 c) 2 d) 4 e) none of these

5) If all digits are assigned an alphabet which replaces it throughout the equation and SLED + SNOW=RIDE and SLED-SNOW=BOB then find each alphabet.


6) given that a clock is such that, its difficult to know which is the minute hand. Ravi saw that the diff between the two hands is 9 minutes. He was not sure of the time. He asked his father. His father asked, is it night, he said no,
his father told the correct time. Whats the time?

[implex]

Originally Posted by kewl.bantee

if [f(x)+(y)]/2>=f([x+y]/2)
then (x^2+y^2)/2>=[(x+y)^2]/2 <----- error

which gives (x^2+y^2)>=(x+y)^2
and (x^2+y^2)>=4^2 i.e.16
so min value of (x^2+y^2)is 16 and maximum is infinite

application of a formula is very easy, but one mistake and u r doomed

[getneonow]

Originally Posted by implex

Problem set # 3
1) given that 5x+12y=60 find the minimum value of (x^2+y^2)^(1/2)

If we plot the line 5x+2y = 60 and make a triangle with the origin and the intercepts as the vertices then the min occurs at one of the two intercepts on axis.

i.e at (12,0) or at (0,5) ..the min here occurs at latter point..

so min value is 5

Neo

[implex]

Originally Posted by getneonow

If we plot the line 5x+2y = 60 and make a triangle with the origin and the intercepts as the vertices then the min occurs at one of the two intercepts on axis.

i.e at (12,0) or at (0,5) ..the min here occurs at latter point..

so min value is 5

Neo

no its wrong, we can get values lower than 5

this problem can be solved by at least 3-4 methods

[getneonow]

Originally Posted by implex

Problem set # 3


3 Let r be a positive rational no, but not an integer. Let n be a positive integer such that [nr] is a prime. [.] means greatest integer function( lower floor). Then which is always true

a) There is no such n

b) if n1 and n2 are such n's then n1=n2 always

c) if n1, n2 and n3 are such n's then n3=n1+n2

d) none of these

Let us take the simplest case where [nr] is prime i.e when [nr]=2

=> nr E [2,3]

We can have many sets of (n,r) such as (4,0.5) , (20,0.1) and so on..

So ans is d) ..none

Neo

[bhavin422]

Originally Posted by getneonow

If we plot the line 5x+2y = 60 and make a triangle with the origin and the intercepts as the vertices then the min occurs at one of the two intercepts on axis.

i.e at (12,0) or at (0,5) ..the min here occurs at latter point..

so min value is 5

Neo

U missed a trick dude..
(x^2+y^2)^1/2 is d distance of point from origin...

so going by your trick, a perpendicular drawn from origin to line 5x + 12y =60 is the shortest distance...
Hence x=2.4 and y=4 ...min distance is 21.76^1/2..

i hope i have not missed a trick here..

[implex]

Originally Posted by bhavin422

U missed a trick dude..
(x^2+y^2)^1/2 is d distance of point from origin...

so going by your trick, a perpendicular drawn from origin to line 5x + 12y =60 is the shortest distance...
Hence x=2.4 and y=4 ...min distance is 21.76^1/2..

i hope i have not missed a trick here..

you hit on one of the methods there are many more!