1i cant think of any other method :( ..
Q1. Find the values of a for which the inequation x^2+ax+a^2+6a<0 is satisfied for all x ε (1,2)
ans: (-7-√45)/2 < a < -4+2√3
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11 Answers
1as coefficient of x2>0 the graph of the function will be concave upwards ..
the graph will intersect the x axis at atmost two points
it has to for the above inequation to be satisfied ..
so first D>0 ...
then the second condition is that (1,0) and (2,0) must be contained in the region (α,0) and (β,0)
where α and β are the two roots
so we must have 1>α and 2<β...
then ou will get range of a
106But in that method we dont get the range of a as given in the answer. btw it is from AD Gupta so i dont kno whether it is correct or i mite be making a mistake in the calculations :(
1as its an upward concave parabola, if u can visualise, for all values of x in the interval to have a negetive we just need dat 1, and 2 sud have a negetive F(x)...
or F(1)= a2+7a +1<0
F(2)=a2+8a+4<0
both simultanously, we get the interval of a required as
\left(\frac{-7-\sqrt{45}}{2},\frac{-8+\sqrt{48}}{2} \right)
mite be some silly mistakes, but am sure about the concept..
cheers!!!
106Q2. If a2+b2+c2=1 then ab+bc+ca lies in the interval:
(a) [1/2,2]
(b) [-1,2]
(c) [-1/2,1]
(d) [-1,1/2]
1we have a2+b2+c2-ab-bc-ca=1/2((a-b)2+(b-c)2+(c-a)2) >=0
thus ab+bc+ca <=1
further we have (a+b+c)2>=0
thus
2(ab+bc+ca)>=-1
thus >=-1/2
1(a+b+c)2 ≥ 0
gives 1 + 2 (ab+bc+ca) ≥ 0
so ab+bc+ca ≥ -1/2
A.M ≥ G.M gives
a2 + b2 + c2 ≥ ab+bc+ca
so ab+bc+ca ≤ 1
EDIT : yeah im a slow typer