If x−1=a−x2 has real solution for x then

x−1=a−x2Right hand side is non negative ∀x∈R such that a−x2≥0So left hand side x−1≥0⇒x≥1(x−1)=a−x2(x−1)2=a−x2x=1±2a−12the root x=1−2a−12 does not satisfy x≥1So it is not a root and x=1+2a−12 satisfies the equation for a≥1

If x−1=a−x2 has real solution for x then

x−1=a−x2Right hand side is non negative ∀x∈R such that a−x2≥0So left hand side x−1≥0⇒x≥1(x−1)=a−x2(x−1)2=a−x2x=1±2a−12the root x=1−2a−12 does not satisfy x≥1So it is not a root and x=1+2a−12 satisfies the equation for a≥1

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If $x - 1 = \sqrt{a - x^{2}}$ has real solution for $x$ then

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$a \in R$

$a \geq 1$

$a > \frac{1}{2}$

$a < \frac{1}{2}$

answer is B.

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Detailed Solution

$x - 1 = \sqrt{a - x^{2}}$
Right hand side is non negative $\forall x \in R$ such that $a - x^{2} \geq 0$
So left hand side $x - 1 \geq 0$
$\Rightarrow x \geq 1$
$(x - 1) = \sqrt{a - x^{2}}$
${(x - 1)}^{2} = a - x^{2}$
$x = \frac{1 \pm \sqrt{2 a - 1}}{2}$
the root $x = \frac{1 - \sqrt{2 a - 1}}{2}$ does not satisfy $x \geq 1$
So it is not a root and $x = \frac{1 + \sqrt{2 a - 1}}{2}$ satisfies the equation for $a \geq 1$