[sumitbce]

Originally Posted by sumitbce

Some question from my side...


(1) Solve each inequality.
(a) (x + 5)2(x + 3)(x - 1) > 0

(b) x(1 - x2)3 > 7(1 - x2)3

(c) 4x4 - 25x2 + 36 = 0

(d)

Any update on this....

[shreyans_nitt]

Originally Posted by sumitbce

Any update on this....

hey i have solved all ...plz check the threads....last one by me is wrong....as i know....tell about others

[aayush_badmash]

Originally Posted by sumitbce

Some question from my side...


(1) Solve each inequality.
(a) (x + 5)2(x + 3)(x - 1) > 0

(b) x(1 - x2)3 > 7(1 - x2)3

(c) 4x4 - 25x2 + 36 = 0

(d)

a . (-infinity , -3) U (1 , infinity)
b . (-infinity , -1) U (1 , 7)
c . [-2,-3/2]U [3/2,2] --- inclusive of the range
d . (0,1] --- inclusive of 1 and excluding 0

are they correct ???

[shreyans_nitt]

Originally Posted by aayush_badmash

a . (-infinity , -3) U (1 , infinity)
b . (-infinity , -1) U (1 , 7)
c . [-2,-3/2]U [3/2,2] --- inclusive of the range
d . (0,1] --- inclusive of 1 and excluding 0

are they correct ???

First one shud exclude x= -5...........

[aayush_badmash]

Originally Posted by sumitbce

One more inequalities -- I will post the solution in the evening
(1/x) + (1/1+x) < (2/x+2)

my take -

(-infinity , -2/3)

[aayush_badmash]

Originally Posted by shreyans_nitt

First one shud exclude x= -5...........

ohh yes...
dont know how i missed it ...!

thanks

[shreyans_nitt]

Originally Posted by shreyans_nitt

Last Ones:

7. If ax^2 + bx + c, where a,b,c are real, has no real zeroes, and if c<0 then :

a. a<0
b. a+b+c>0
c. a>0
d. none of these

8. If r is the root of f(x) = x^4 + ax^3 - 6x^2 - ax + 1 = 0, then which among the following is also a root of f(x) = 0? (1) -1/r (2) (1+r)/(1-r) (3) (r-1)/(r+1) (4) All of these (5) Exactly two of these





9. If x, y, z are any real numbers then the minimum possible value of x^2 + 2y^2 + z^2 + 2yz subject of x + 2y + z = –6 is

(a) –6 (b) 6 (c) 12 (d) none of the foregoing


10. The integer solutions (a,b) to a^4 + 2(a^3) + 2(a^2) +2a + 1 = b^2 is/are (a) 0 (b) 1 (c) more than 2 but finite (d) infinitely many

anybody solving????!!!!!!!!!

[MaskedMenace]

Originally Posted by shreyans_nitt

anybody solving????!!!!!!!!!

If x, y, z are any real numbers then the minimum possible value of x^2 + 2y^2 + z^2 + 2yz subject of x + 2y + z = –6 is

(a) –6 (b) 6 (c) 12 (d) none of the foregoing

k = y+z

x+y+k = -6

x^2 + y^2 + k^2

Now, Using Cauchy-schwarz inequality

(1.x + 1.y + 1.k)^2 <= (1^2 + 1^2 +1^2)(x^2 + y^2 + k^2)

36 <= 3(x^2 +y^2 + k^2)

so, x^2 + y^2 + k^2 >= 12

[shashank3012]

Originally Posted by shreyans_nitt

anybody solving????!!!!!!!!!

7th one,i couldn't comprehend,9th one,see post 713

[sumitbce]

Originally Posted by slam

=> (x-7) (1-x^2)^3 > 0
=> (x-7) (1+x)^2 (1-x)^3 > 0
=> (x-7) (x+1)^3 (x-1)^3 < 0
=> the critical points are -1,1,7
Draw the number line to get:
exp < 0 for x in (-inf, -1) U (1,7). Is this correct?

ya this is right answer. .. :-)