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Match the items of List I with items of List II and choose the correct option.
Column-I Column-II
(I) The number of real solution(s) of the equation $x^{2} - | x | - 2 = 0$ is/are (P) $0$
(II) If one of the roots is the square of the other for the equation $3 x^{2} + p x + 3 = 0, p > 0$ then $| p | =$ (Q) $1$
(III) The number of real value(s) of $k$ for which the system of equations $(k + 1) x + 8 y = 4 k$ and $k x + (k + 3) y = 3 k - 1$ have infinitely many solutions is/are (R) $2$
(IV) Number of real root(s) of $x^{2} - \frac{2}{x - 2} = 4 - \frac{2}{x - 2}$ is/are (S) $3$
(T) $6$

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$(I) - (S); (I I) - (R); (I I I) - (Q); (I V) - (R)$

$(I) - (S); (I I) - (P); (I I I) - (Q); (I V) - (Q)$

$(I) - (R); (I I) - (S); (I I I) - (Q); (I V) - (Q)$

$(I) - (R); (I I) - (S, T); (I I I) - (Q); (I V) - (R)$

answer is C.

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Detailed Solution

(I) ${| x |}^{2} - | x | - 2 = 0 \Rightarrow (| x | + 1) (| x | + 2) = 0 | x | = 2 \Rightarrow x = \pm 2$
(II) Let $α, α^{2}$ be roots
product of root $α . α^{2} = \frac{3}{3}$
$\Rightarrow α = 1, ω, ω^{2}$
If $α = 1$ then p = –6 not acceptable as p > 0
if $α = ω, α^{2} = ω^{2}$ then p = 3
(III) $\frac{K + 1}{K} = \frac{8}{K + 3} = \frac{4 K}{3 K - 1} \Rightarrow K = 1$
(IV) $x^{2} - \frac{2}{x - 2} = 4 - \frac{2}{x - 2} \Rightarrow x = \pm 2$ but at $x = 2, \frac{2}{x - 2}$ is not defined.

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