1)if a,b,c  are real no's such that   then ab+bc+ca lies in the interval

a)[1/2 , 2]       b)[-1, 2]     c)[-1/2 , 1]      d)[-1 , 1/2]

2)if  then d values of are:

a)3  b)2  c)1  d)not

3)if a,b are the roots of   and c,d are the roots of  then d value of

(a-c)(b-c)(a-d)(b-d)is:

a) b) c)  d)

2.take 2 to the other side and thn do cubing both sides



ull get



 

 hence 2 wud be the answer

1.

a+b+c minimum value can be 0

put  

ab+bc+ac=-1/2

whn a=b=c

ab+bc+ac=1

hence the answer will be[-1/2,1]

in such ques

always give priority to solve it by putting values 

\even if  u know d correct way but y go all d way round

i m now solving each of these by d method which takes least time

i)(a+b+c)^2  =  1  + 2(ab+bc+ac)

ab+bc+ac = [(a+b+c)^2 - 1]/2

(a+b+c)^2 >=0

(a+b+c)^2 -1 >=-1

ab+bc+ac>=-1/2

by looking at d options

left limit  is satisfied only by  c) so its  correct

next

ii)  ADITI already DID it

iii)a+b=-p ,ab=1

    c+d=-q  ,cd=1

(a-c)(b-c)  =ab -(a+b)c +c^2

                  =1+pc+c^2

!!ly  (a-d)(b-d) = 1+pc +d^2

ur ans= 1+pc +dasde....................and solve it u ll get it

easier way

let p=2,q=-2

so

a=b=-1

c=d=1

ur  ans =-2*-2*2*2

=16

which is [2-(-2)]^2

=(p-q)^2

so dats it

Source of the problems?

for 3Q) more simple solution

a+b=-p,ab=1a=1/b      c+d=-q,cd=1c=1/d

1/b+b=-p,1/d+d=-q

now subtract both the upper eqn. we get

(bd-1)(b-d)/bd=q-p

Now reduce (a-c)(b-c)(a-d)(b-d) in terms of b,d

u will get the answer as (p-q)^2

If u find the answer right/satisfactory plzzzz vote me