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Algebra
Just five ( simple) Questions im not getting :
Q.1. What is the total number of integral solutions to x1. x2. x3. x4. x5 = 840 ??
(ans = 3125)
Q. 2. a, b are positive reals such that a + b = 1.
Prove that (a + 1/a)2 + (b + 1/b)2 >= 25/2 .
[ This Qsn I have solved by a method, I want 2 know if there r other methods ]
Q. 3. a, b, c, p, q, r are real nos. such that
ax2 + bx + c >= 0, and px2 + qx + r >= 0
for all real x. Prove for all real x that
apx2 + bqx + cr >= 0 .
[ I know D <= 0 but not getting d calculation part :( ]
Q. 4. Let p(x) be a quadratic polynomial
p(x) = ax2 + bx + c
such that | p(x) | <= 1 for | x | <= 1. Prove that
| cx2 + bx + a | <= 2 for | x | <= 1.
Q. 5. Find real roots of
[ After squaring twice, this is what I get :
x4 – 2ax2 – x + a2 – a = 0
How 2 do this ?? ]
Comments (9)
1.
Referred to this solution by Brad sir : http://www.goiit.com/posts/list/algebra-find-the-number-of-positive-integral-solutions-of-the-54893.htm
840 = 7*3*5*2*2*2
Now we have been given x1 , x2 , x3 , x4 , x5
now 5 or 3 or 7 , so number of ways = 5 3 ways
So we have 2 to deal with
So number of ways of arranging the 2 s are 5 C1 + 5 C2 +5C3= 25
So total number of ways = 5 3 * 25 = 125 * 25 = 3125
+4
quadratic polynomial question.
are u sure it is right?
-1<=x<=1 -1<=f<=1
further g(1)=f(1)
g(-1)=f(-1)
implies mod(g(1))<=1 and
and modg(-1)<=1
that means g attains both extrema values in the interval [-1,1](2 and -2)
that is impossible for a 2 degree polynomial which has only one local extremum.
so the problem seems incorrect to me...
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