IIT JEE , AIEEE Forum - Mathematics , Algebra

kislay

3ax^2+2bx+c+dcosx doesnt changes its sign . any where in its domain

find the relation..obtained at such.
ans.
b²<3a(c-|d|)

nishantsingh89

-d<=dcosx<=d

3ax^2 + 2bx +c + dcosx doesn't change its sign

so if a>0 { the parabola concave upwards again we don't want it to cut x axis because then sign will change so it should lie wholly above x axis}

min value of (3ax^2 + 2bx +c) >d           { if cos x =-1 then the expression should be                                                             +ve}

min value of 3ax^2 +2bx+c is -D/4A = 12ac-4b^2/12a

12ac -4b^2 >12ad
4b^2<12a(c-d)

b^2 < 3a(c-d) ...................(2)

now when a<0 { parabola will be concave downwards also we don't want any real roots so the parabola lies wholly below x axis}

so max value of 3ax^2 + 2bx + c + d<0
max value of 3ax^2 + 2bx +c <-d             { if cosx = 1 then too the expression must be negative}

max value of 3ax^2 + 2bx + c = -D/4A = 12ac-4b^2/12a

12ac - 4b^2 / 12a <-d

12ac - 4b^2>-12ad         { a is negative here so multiplying by 12a on both sides changes inequality sign!!}

4b^2<12a(c+d)

b^2 < 3a(c+d) ....................(1)

now here i assumed to be that d>0 , but when in case where d<0
we will get the same results
b^2<3a(c+d)
b^2<3a(c-d)

now when d>0 we get b^2<3a(c-d) = b^2<3a(c-|d|)
when d<0 we get b^2<3a(c+d) = b^2<3a(c-(-d)) = b^2<3a(c-|d|)
answer

kislay

HEY NISHANT HERE IS UR SOLN., I HAVE FEW DOUBTS ABT . IT.
now when a<0 { parabola will be concave downwards also we don't want any real roots so the parabola lies wholly below x axis}

so max value of 3ax^2 + 2bx + c + d<0
max value of 3ax^2 + 2bx +c <-d             { if cosx = 1 then too the expression must be negative}

max value of 3ax^2 + 2bx + c = -D/4A = 12ac-4b^2/12a

12ac - 4b^2 / 12a <-d

12ac - 4b^2>-12ad         { a is negative here so multiplying by 12a on both sides changes inequality sign!!}

4b^2<12a(c+d)

b^2 < 3a(c+d) ....................(1)

now here i assumed to be that d>0 , but when in case where d<0
we will get the same results but in reverse order as obtained above
b^2<3a(c+d)
b^2<3a(c-d)

IN EQN. 1. d has been assumed to be positive, in next highlighted eqn. it has been assumed to be negative.there fore the combinatin of all the four equations sud give
b²<3a(c+|d|)

nishantsingh89

actually you will not get results in reverse order rather in same order , i had corrected that earlier, sorry got myself a bit confuse

when d<0 , instead of putting cosx = -1(as in case d>0) , you put cosx=1 you get the same result, and same for the second case

hope i cleared the muddle now :)

kislay

see we hav to get the max. value in rhs, from which b^2 sud be less than. so we hav to put
cosx=1 when d>0
and cosx =-1 when d<0.
isn't it????

nishantsingh89

no, see when a>0, d>0
the parabola aove x axis, concave upwards,
(3ax^2 + 2bx +c)+dcosx doen't have to change sign

now if both (3ax^2+2bx + c) & dcosx are positive the expression on the whole wil never be negative

3ax^2 + 2bx + c is always positive,
let value of 3ax^2 + 2bx + c =y (y is positive)
y+dcosx >0

when dcosx>0 the expression is always true

now minimum value of dcosx is -d
if y-d>0
then exp is always positive
dcosx = -d
cosx = -1

similarly for case when d<0

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