If p + q +r = 2, p²+ q² + r² = 30 and pqr = 10, then the value of (1 – p) (1 – q)(1 – r) will be ??? be [NSEJS 2017-18] Options: A. -18 B.-24 C. -27 D. -35 About the author Jade
[tex] \: {\boxed{\underline{\tt{ \orange{Required \: \: answer \: \: is \: \: as \: \: follows:-}}}}}[/tex] Option (B) — ( –24 ) ★GIVEN:- [tex] \sf{p+q+r = 2}[/tex] [tex] \sf{ {p}^{2} + {q}^{2} + {r}^{2} = 30 }[/tex] [tex] \sf{pqr = 10}[/tex] TO FIND:- [tex] \sf{(1 – p) (1 – q)(1 – r)}[/tex] ★SOLUTION:- [tex] : \longrightarrow \displaystyle \sf{(1 – p) (1 – q)(1 – r)}[/tex] [tex] : \longrightarrow \displaystyle \sf{(1 – r)(1 – q – p + pq)}[/tex] [tex] :\longrightarrow\displaystyle\sf{1 – q – p + pq – r + rq + rp – rpq} \\ \\ :\longrightarrow\displaystyle\sf{1 – p – q – r + pq + rq + rp – rpq} \\ \\ :\longrightarrow\displaystyle\sf{1 – (p + q + r) + pq + rq + rp – rpq━━(1)} \\ \\ :\longrightarrow\displaystyle\sf{ {(p + q + r)}^{2} = {p}^{2} + {q}^{2} + {r}^{2} + 2(pq + rq + rp) } \\ \\ :\longrightarrow\displaystyle\sf{4 = 30 + 2(pq + rq + rp)} \\ \\ :\longrightarrow\displaystyle\sf{ – 26 = 2(pq + rq + rp)} \\ \\:\longrightarrow\displaystyle\sf{ (pq + rq + rp) = – 13━━(2)} \\ \\ \displaystyle\sf{ put \: value \: of \: (2) \: in \: (1)} \\ \\ : \longrightarrow \displaystyle \sf{1 – 2 – 13 – 10} \\ \\ { \boxed{ \underline{ \pink{ \huge{ – 24}}}}}[/tex] Reply
Recall the expansion of a cubic polynomial having roots p, q and r. [tex]\longrightarrow(x-p)(x-q)(x-r)=x^3-(p+q+r)x^2+(pq+qr+rp)x-pqr[/tex] Taking [tex]x=1,[/tex] [tex]\longrightarrow(1-p)(1-q)(1-r)=1-(p+q+r)+(pq+qr+rp)-pqr[/tex] We’re given, [tex]p+q+r=2\quad\quad\dots(1)[/tex] [tex]p^2+q^2+r^2=30\quad\quad\dots(2)[/tex] [tex]pqr=10\quad\quad\dots(3)[/tex] Squaring (1), [tex]\longrightarrow (p+q+r)^2=2^2[/tex] [tex]\longrightarrow p^2+q^2+r^2+2(pq+qr+rp)=4[/tex] From (2), [tex]\longrightarrow 30+2(pq+qr+rp)=4[/tex] [tex]\longrightarrow pq+qr+rp=-13[/tex] Now, [tex]\longrightarrow(1-p)(1-q)(1-r)=1-(p+q+r)+(pq+qr+rp)-pqr[/tex] [tex]\longrightarrow(1-p)(1-q)(1-r)=1-2-13-10[/tex] [tex]\longrightarrow\underline{\underline{(1-p)(1-q)(1-r)=-24}}[/tex] Hence (B) is the answer. Reply
[tex] \: {\boxed{\underline{\tt{ \orange{Required \: \: answer \: \: is \: \: as \: \: follows:-}}}}}[/tex]
Option (B) — ( –24 )
★GIVEN:-
TO FIND:-
★SOLUTION:-
[tex] : \longrightarrow \displaystyle \sf{(1 – p) (1 – q)(1 – r)}[/tex]
[tex] : \longrightarrow \displaystyle \sf{(1 – r)(1 – q – p + pq)}[/tex]
[tex] :\longrightarrow\displaystyle\sf{1 – q – p + pq – r + rq + rp – rpq} \\ \\ :\longrightarrow\displaystyle\sf{1 – p – q – r + pq + rq + rp – rpq} \\ \\ :\longrightarrow\displaystyle\sf{1 – (p + q + r) + pq + rq + rp – rpq━━(1)} \\ \\ :\longrightarrow\displaystyle\sf{ {(p + q + r)}^{2} = {p}^{2} + {q}^{2} + {r}^{2} + 2(pq + rq + rp) } \\ \\ :\longrightarrow\displaystyle\sf{4 = 30 + 2(pq + rq + rp)} \\ \\ :\longrightarrow\displaystyle\sf{ – 26 = 2(pq + rq + rp)} \\ \\:\longrightarrow\displaystyle\sf{ (pq + rq + rp) = – 13━━(2)} \\ \\ \displaystyle\sf{ put \: value \: of \: (2) \: in \: (1)} \\ \\ : \longrightarrow \displaystyle \sf{1 – 2 – 13 – 10} \\ \\ { \boxed{ \underline{ \pink{ \huge{ – 24}}}}}[/tex]
Recall the expansion of a cubic polynomial having roots p, q and r.
[tex]\longrightarrow(x-p)(x-q)(x-r)=x^3-(p+q+r)x^2+(pq+qr+rp)x-pqr[/tex]
Taking [tex]x=1,[/tex]
[tex]\longrightarrow(1-p)(1-q)(1-r)=1-(p+q+r)+(pq+qr+rp)-pqr[/tex]
We’re given,
Squaring (1),
[tex]\longrightarrow (p+q+r)^2=2^2[/tex]
[tex]\longrightarrow p^2+q^2+r^2+2(pq+qr+rp)=4[/tex]
From (2),
[tex]\longrightarrow 30+2(pq+qr+rp)=4[/tex]
[tex]\longrightarrow pq+qr+rp=-13[/tex]
Now,
[tex]\longrightarrow(1-p)(1-q)(1-r)=1-(p+q+r)+(pq+qr+rp)-pqr[/tex]
[tex]\longrightarrow(1-p)(1-q)(1-r)=1-2-13-10[/tex]
[tex]\longrightarrow\underline{\underline{(1-p)(1-q)(1-r)=-24}}[/tex]
Hence (B) is the answer.