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Write full proofs with your answers
1) Find all real numbers a such that there are four different real numbers x satisfying the equation
x(x+1)3=(2x+a)(x+a−1).
2) Find all functions f:R→R such that for x,y∈R we have
f(xf(y)+x)=xy+f(x).
3) Suppose that x, y and z are real numbers satisfying the following three equations.
x+y+zx2+y2+z2x3+y3+z3=2=6=8
Find all possible values of x2015+y2015+z2015.
4) Suppose that a function f:R→R has the property that
f(x)2≤f(y)
for every x>y.
Prove that 0≤f(x)≤1 for all x∈R.
5) Prove the following inequality for all a,b,c>0.
a2+ab+b2a2(a3+b3)+b2+bc+c2b2(b3+c3)+c2+ca+a2c2(c3+a3)≥2abc
#Algebra
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3). let s1,s2,s3 be the symmetric sums. let pn denote the nth power sum. then by newtons sum: ap1=s1=2p2=s1p1−2s2=4−2s2=6⟹s2=−1p3=s1p2−s2p1+3s3=14+3s3=8⟹s3=−2 so, x,y,z are roots of x3−2x2−x+2=0 x3−x−2x2+2=x(x2−1)−2(x2−1)=(x−2)(x−1)(x+1)=0 since the expression is symmetric, order doesnt matter. we have (x,y,z)=(2,1,−1). put this in to get x2015+y2015+z2015=22015 this is the only value possible. are we suppose to expand 22015?
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Do you think we do? Of course not! Nice solution. 7 out of 7.
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Is the answer to the second question is f(x)=0
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a∈(241(27−3),241(27+3))
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For roots to be real and distinct its derivative must have three distinct roots
4x3+6x2+2x+3(a−1).....(1) And for that to happen , we again take its derivative, we get
12x2+12x+2=0
For real and distinct roots
f(x1)f(x2)<0 where x1,2=4−3±3 in (1)
and then i found interval for a
a∈(4321(153−2957),4321(153+2957))
Putting y=0 we get f(xf(0)+x)=f(x)) xf(0)+x=x f(0)=0.........(1) Now we assume that there exists f(y)=−1 Then putting it in original functional equation we get, f(0)=xy+f(x) −xy=f(x)........(2) now put x=y f(y)=−y2 But f(y)=−1 Therefore y=1 or y=−1. Now putting these values in (2) we get f(x)=x or f(x)=−x . since above two functions satisfy the given functional equation and (1) Hence they are the solutions of the given functional equation.
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You would get 3 out of 7 for this proof. In line 2, how do you know f is injective?
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Answer is right I guess (function)
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Proving that the function is injective is very easy. For two reals a and b let f(a)=f(b). We take two cases. In the first case put y=a and in second case put y=b. Now note that in both cases the LHS is the same. Therefore comparing the RHS you obtain that a=b. Hence function is injective
Using AM−GM a2+b2≥2ab 23(a2+b2)≥(a2+ab+b2) This implies a2+ab+b2a2(a3+b3)≥3(a2+b2)2a2(a3+b3) Using Generalized Power mean inequality We get a2+ab+b2a2(a3+b3)≥32a2(2a2+b2) Now by RMS−AM inequality we get a2+ab+b2a2(a3+b3)≥3a2(a+b)≥32a2ab Therefore cyc∑a,b,ca2+ab+b2a2(a3+b3)≥cyc∑a,b,c32a2ab Now using simple AM−GM on the RHS of the above inequality we get cyc∑a,b,ca2+ab+b2a2(a3+b3)≥cyc∑a,b,c32a2ab≥2abc
cyc∑a,b,ca2+ab+b2a2(a3+b3)≥2abc Hence proved .
Could you provide the solutions to all questions please? So that we can reference and improve. Thanks!
@Sharky Kesa : For question 1 my solution is as follows : The given equation simplifies to x^4+3.x^3+x^2+(3-3a)x +(a-a^2)=0. Let the roots be p,q,r,s. Now, sum of roots two at a time=1, sum of roots three at a time =(3a-3) and all four roots at a time=(a-a^2)
Now we use the identity : (pqr+pqs+prs+qrs)^2=(p^2.q^2.r^2 + .........) + 2pqrs(pq+pr+ps+qr+qs+rs) Thus we obtain the inequality : (pqr+pqs+prs+qrs)^2 >= 3pqrs(pq+pr+ps+qr+qs+rs) Substituting the values we obtain a condition : 4a^2-7a+3 >=0. Solving that we obtain that a should lie outside [3/4,1].
Please tell me whether the answer and the solution is correct or wrong.