TG EAMCET 2 May 2025 Shift 2 Paper

Given,α,β,γare the roots of the equationsx3−Px2+Qx−R=0
α+β+γ‌=P\∴‌αβ+βγ+γα=Q
Andαβγ=R
Also,(α−2)2,(β−2)2,(γ−2)2are the roots of equations,x3−5x2+4x=0
Now,x2−5x2+4x=0
⇒x(x2−5x+4)=0
⇒x(x−1)(x−4)=0
⇒x=0,1,4
∴(α−2)2=0⇒α=2\(β−2)2=1⇒β−2=±1⇒β=1,3𝜕
And(γ−2)2=4⇒γ−2=±2⇒γ=0,4
Now,
P+Q+R=(α+β+γ)+(αβ+βγ+γα)+αβγ
Case I whenα=2,β=1,γ=0
then,
P+Q+R‌=(2+1+0)+(2+0+0)
‌=5
Case II whenα=2,β=1,γ=4
then,
P+Q+R=(2+1+4)+(2+4+8)+(2×1×4)=29
Case III whenα=2,β=3,γ=0then
P+Q+R‌=(2+3+0)+(6+0+0)+0
‌=11
Case IV whenα=2,β=3,γ=4
then
P+Q+R‌=(2+3+4)+(6+12+8)+24
‌=59
Clearly, least value ofP+Q+Ris 5 .