AP EAMCET Engineering 22 Apr 2019 Shift 1 Paper

The given function is,
ϕ (x) =

x

(x2+1)(x+1)

ϕ(a)ϕ(b)ϕ(c) =

abc

(1+a)(1+b)(1+c)(1+a2) (1+b2)(1+c2)

Since, a.b.c are the roots of given cubic equation so,
ab+bc+ca=−3 ....(I)
a+b+c=0 .....(II)
abc=−λ .....(II)
From equation (II),
(a+b+c)2=0
a2+b2+c2+2(ab+bc+ca)=0
a2+b2+c2=6
Similarly,
a2b2+b2c2+c2a2=9
This gives us,
ϕ(a)ϕ(b)ϕ(c) =

−λ

(1+0−3−λ)(1+6+9+λ2)

=

−λ

(−2−λ)(λ2+16)

= λ(2+λ)(λ2+16 )