Question

8. For the equation x-1=√(a-x^2) which of the
following statements is false?
(A) the solution is real if a >1/2
(B) there is no solution if a < 1
(C) x={1+√(2a-1)}/2 is a solution if a >= 1
(D) x={1-√(2a-1)}/2 is a solution if a>1/2
​

Answer 1

Answer:

b statement is false.

Step-by-step explanation:

$x-1=$$\sqrt{ (a- x^{2} )}$

$(x-1)^{2} =a-x^{2}$

$x^{2} -2x+1=a-x^{2} \\2x^{2} -2x+(1-a)=0\\using x=\frac{-b+\sqrt{b^{2} -4ac} }{2a} and \frac{-b-\sqrt{b^{2} -4ac} }{2a}$

solution exists if$\sqrt{b^{2} -4ac} } \geq 0$

therefore

${b^{2} -4ac} \geq 0\\2^{2} -4*2*(1-a)\geq 0\\4 - 8*(1-a)\geq 0\\4-8+8a\geq 0\\8a\geq 4\\a\geq \frac{1}{2}$

therefore solution exists for $\frac{1}{2}\leq a\leq 1$