Question

If p & q are distinct reals, then $2\left[\left(x-p\right)\left(x-q\right)+\left(p-x\right)\left(p-q\right)+\left(q-x\right)(q-p\right)]=(p-q{)}^2+(x-p{)}^2+(x-q{)}^2$ is satisfied for

• A. no value of x
• B. exactly one value of x
• C. exactly two values of x
• D. infinite values of x

Answer 1

The correct option is D infinite values of x

LHS = $2[(x-p)(x-q)+(p-x)(p-q)+(q-x)(q-p)]$ = $2(x^2-xq-px+pq+p^2-pq-px+qx+q^2-pq-qx+px)$ = $2(x^2+p^2+q^2-px-qx-pq)$ = $({2x}^2+{2p}^2+{2q}^2-2px-2qx-2pq)$
RHS = ${(p-q)}^2+{(x-p)}^2+{(x-q)}^2$ = $p^2+q^2-2pq+x^2{+p}^2-2px+x^2+q^2-2qx$ = $({2x}^2+{2p}^2+{2q}^2-2px-2qx-2pq)$ = LHS
Since LHS=RHS,therefore all values of $x$ satisfy the equation $\Rightarrow$ $x\in R$
Hence,(D) is the correct choice.