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The equation $x^{2} - a = \sqrt{x + a}$ has real or imaginary roots depending on values of a. List-I represents the nature of root and List-II represents the corresponding exhaustive values of a. match the following Column-I with Column-II
Column-I Column-II
A No real root P $(0, 1) \cup {- \frac{1}{4}}$
B One real root Q $(- \infty, - \frac{1}{4})$
C Exactly two real roots R $(- \frac{1}{4}, 0] \cup [1, \infty)$
D Atleast two real roots S $(0, \frac{3}{4}) \cup {- \frac{1}{4}}$
T $(- \frac{1}{4}, 0] \cup [\frac{3}{4}, \infty)$
The Correct option is:

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$A \to Q; B \to P; C \to T; D \to S$

$A \to Q; B \to P; C \to R; D \to R$

$A \to Q; B \to S; C \to T; D \to T$

$A \to P; B \to S; C \to T; D \to T$

answer is A.

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Detailed Solution

SOL-
A) $(P) x^{2} - a = x$
$x^{2} - x - a = 0 D < 0 1 + 4 a < 0 a < - \frac{1}{4}$
Question Image
$(Q) - \sqrt{a} < - a \sqrt{a} > a a > a^{2} a^{2} - a < 0 a (a - 1) < 0 0 < a < 1$

$(R) x^{2} - x - a = 0 1 + 4 a > 0, - a > 0 a > - \frac{1}{4}, a < 0 - \frac{1}{4} < a \leq 0 - a < - \sqrt{a} a > \sqrt{a}, a^{2} > a a (a - 1) > a \geq 1$

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