91 Q is clearly wrong
see a=c=p=r=1 , b=q=2
theres a contradiction
1. a, b, c, p, q, r are real nos. such that
ax2 + bx + c ≥ 0 and
px2 + qx + r ≥ 0 for all x ε R.
Prove that
apx2 + bqx + cr ≥ 0 for all x.
I know c, r, a and p > 0 and Discriminant ≤ 0 but im not able to prove it fr reqd. eqn.
2. Find all +ve integers n for which the quad eqn
has real roots for every choice of real numbers a1, a2, ... an+1.
I got the answer as n ≤ 4 , wanted to verify this.
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5 Answers
11Q1
Look as u have given 1st function as +ve 4 whole real line so
a>0,c>0,b≤0(as -b/2a will be +ve )
and similarly for 2nd function as there is
p>0,q>0,r≤0
then for the 3rd function we can see that here -bq/apcr≥0
so it has min value of 0 or greater than 0 and the function is concave up.
Hence
apx2 + bqx + cr ≥ 0 for all x.
9
1for, roots to be real we require:
(a_{1}^{2}+a_{2}^{2}+.....a_{n+1}^{2}) \geq a_{n+1}(a_{1}+a_{2}+......a_{k})
Now, by power mean :
\frac{(a_{1}^{2}+a_{2}^{2}+.....a_{n}^{2})}{n} \geq \left(\frac{a_{1}+....+a_{n}}{n}\right)^{2}
let (a_{1}+a_{2}+.....a_{n}) be x and let a_{n+1}
be y.
so, we have :
\frac{x^{2}}{n}+{y^{2}} \geq xy
applying AM GM we get the required condition as :n \leq 4
11@ cele this is the only case
actually 1st and the 2nd eq must be >0 only