Question

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

​

Answer 1

Recall the expansion of a cubic polynomial having roots p, q and r.

$\longrightarrow(x-p)(x-q)(x-r)=x^3-(p+q+r)x^2+(pq+qr+rp)x-pqr$

Taking $x=1,$

$\longrightarrow(1-p)(1-q)(1-r)=1-(p+q+r)+(pq+qr+rp)-pqr$

We're given,

• $p+q+r=2\quad\quad\dots(1)$

• $p^2+q^2+r^2=30\quad\quad\dots(2)$

• $pqr=10\quad\quad\dots(3)$

Squaring (1),

$\longrightarrow (p+q+r)^2=2^2$

$\longrightarrow p^2+q^2+r^2+2(pq+qr+rp)=4$

From (2),

$\longrightarrow 30+2(pq+qr+rp)=4$

$\longrightarrow pq+qr+rp=-13$

Now,

$\longrightarrow(1-p)(1-q)(1-r)=1-(p+q+r)+(pq+qr+rp)-pqr$

$\longrightarrow(1-p)(1-q)(1-r)=1-2-13-10$

$\longrightarrow\underline{\underline{(1-p)(1-q)(1-r)=-24}}$

Hence (B) is the answer.

Answer 2

$\: {\boxed{\underline{\tt{ \orange{Required \: \: answer \: \: is \: \: as \: \: follows:-}}}}}$

Option (B) --- ( -24 )

★GIVEN:-

• $\sf{p+q+r = 2}$

• $\sf{ {p}^{2} + {q}^{2} + {r}^{2} = 30 }$

• $\sf{pqr = 10}$

TO FIND:-

• $\sf{(1 - p) (1 - q)(1 - r)}$

★SOLUTION:-

$: \longrightarrow \displaystyle \sf{(1 - p) (1 - q)(1 - r)}$

$: \longrightarrow \displaystyle \sf{(1 - r)(1 - q - p + pq)}$

$:\longrightarrow\displaystyle\sf{1 - q - p + pq - r + rq + rp - rpq} \\ \\ :\longrightarrow\displaystyle\sf{1 - p - q - r + pq + rq + rp - rpq} \\ \\ :\longrightarrow\displaystyle\sf{1 - (p + q + r) + pq + rq + rp - rpq━━(1)} \\ \\ :\longrightarrow\displaystyle\sf{ {(p + q + r)}^{2} = {p}^{2} + {q}^{2} + {r}^{2} + 2(pq + rq + rp) } \\ \\ :\longrightarrow\displaystyle\sf{4 = 30 + 2(pq + rq + rp)} \\ \\ :\longrightarrow\displaystyle\sf{ - 26 = 2(pq + rq + rp)} \\ \\:\longrightarrow\displaystyle\sf{ (pq + rq + rp) = - 13━━(2)} \\ \\ \displaystyle\sf{ put \: value \: of \: (2) \: in \: (1)} \\ \\ : \longrightarrow \displaystyle \sf{1 - 2 - 13 - 10} \\ \\ { \boxed{ \underline{ \pink{ \huge{ - 24}}}}}$