official quant thread for cat08

[kewl.bantee]

Q.The graph of x^3 - x^2+2x will cut the X axis at how many points? what will be the co-ordinates of the points at which the graph intersects the X-axis??

[satanica]

Originally Posted by implex

Problem Set #2
...
2. For how many positive integer values of a does the equation : sqrt(a+x)+sqrt(a-x)=a have a real solution for x.

1) 0 2) 2 3) 3 4) 4 e) none of these
...

Originally Posted by implex

answer is 3

I'd like to differ here.

Write the equation as,
sqrt(a+x) = a - sqrt(a-x)
Squaring both sides,
a+x = a^2 + (a-x) + 2a(sqrt(a-x))
Simplifying and moving terms around,
a^2 - 2x = 2a(sqrt(a-x))
Squaring both sides again,
a^4 + 4(x^2) - 4x(a^2) = 4(a^2)(a-x)
Simplifying,
a^4 + 4(x^2) = 4(a^3)
x = a[sqrt(a(4-a))]/2

For x to be real, the necessary and sufficient condition is,
a(4-a) >= 0

This will give us folllowing solution-set in integers, a = {0, 1, 2, 3, 4}

As stated in Question, a is positive. Hence, a can take up four values i.e. 1, 2, 3 and 4.

Answer. 4

Note
Also, as can be seen by putting a=4 in the equation, one can solve for x and get x=0 which is a real value.

[kaps_lock]

Originally Posted by thebornattitude

posted by kaps_lock

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

Answer- 2) Max,8
By for max of any sum of any symmetrical function , all terms should be equal...
for eg here
x^2 +y^2 will attain maximum wen x^2=y^2...means either x=y or x=-y

as x+y=4,,, hence x cannot be equal to -y..
so x=y means x=y=2.....
hence x^2 + y^2 attains max at 2^2 +2^2= 8.....hence answer.

I was also thinking the same but answer is min.,8

[thebornattitude]

Ooops sorry, the condition which i told holds true for max/min,,, so check it ...

just put any value ,,,,eg x=1 and y=3,,,so we get x^2 + y^2=10..,,hence 8 is the min value

[implex]

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

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

[jha16june]

Originally Posted by thebornattitude

posted by kaps_lock

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

Answer- 2) Max,8
By for max of any sum of any symmetrical function , all terms should be equal...
for eg here
x^2 +y^2 will attain maximum wen x^2=y^2...means either x=y or x=-y

as x+y=4,,, hence x cannot be equal to -y..
so x=y means x=y=2.....
hence x^2 + y^2 attains max at 2^2 +2^2= 8.....hence answer.

Hey it will be Min 8. Max can be infinity.

[the_hate]

Originally Posted by kewl.bantee

Q.The graph of x^3 - x^2+2x will cut the X axis at how many points? what will be the co-ordinates of the points at which the graph intersects the X-axis??

y=x^3-x^2+2x
=x(x^2-x+2)

now to cut at x-axis-> y=0
x(x^2-x+2) =0

since x^2-x+2>0 always,
x=0,y=0 is the only soln i.e. 1 point

i hope i'm not commitin sm mistake...

[satanica]

Originally Posted by kaps_lock

...
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
...

Solution

x + y = 4
(x+y)^2 = 16
x^2 + y^2 + 2xy = 16
x^2 + y^2 = 16 - 2xy

Given, x+y = 4, max(xy) = 4 which is got when x=y=2.
If xy is maximum, (16 - 2xy) will be minimum. Hence, we will get a minimum value of (x^2 + y^2).

min(x^2 + y^2) = 8

[thebornattitude]

Q2.The roots of the Equation a(x^2)+bx+c=0 are k less than those of the equation p(x^2)+qx+r=0.Find the Equation whose roots are k more than those of p(x^2)+qx+r=0
1.a(x^2)+bx+c=0
2.a((x-2k)^2)+b(x-2k)+c=0
3.a((x+2k)^2)+b(x+2k)+c=0
4.a((x-k)^2)+b(x-k)+c=0

Solution-

answer- 2.a((x-2k)^2)+b(x-2k)+c=0

For any equation ax^2 +bx +c=0... the roots be m,n

then the equation for which the roots are k greater than the roots of the above equation , is

a(x-k)^2 + b(x-k) +c=0 (very easy to prove,, just take sum and product of roots concept)

putting this logic in here, we can get the answer

[implex]

Originally Posted by kewl.bantee

Q.The graph of x^3 - x^2+2x will cut the X axis at how many points? what will be the co-ordinates of the points at which the graph intersects the X-axis??

Y=x(x^2-x+2)=0 clealry x=0 is a solution
the other part of eqn has no real roots

so only one point , i.e x=0 y=0