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in Complex number and Quadratic equations by (35 points)
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14. Number of real roots of the equation \(\sqrt x+\sqrt{x-\sqrt{1-x}}=1\) is

(A) 0

(B) 1

(C) 2

(D) 3

15. Number of real solutions of the equation  x2 + \(\frac{x^2}{(x-1)^2}\) = 8 

(A) 1

(B) 2

(C) 3

(D) 4

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1 Answer

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(14) Correct option is (A) 0

\(\sqrt x+\sqrt{x-\sqrt{1-x}}=1\)

For domain, we have

\(\geq\) 0 & 1 - x \(\geq\) 0 (\(\because\) Domain of root function \(\sqrt x\) is x \(\geq\) 0)

⇒  x \(\geq\) 0 & x \(\leq\) 1-----(1)

Also,  x - \(\sqrt{1-x}\) \(\geq\)0

⇒ x \(\leq\) \(\sqrt{1-x}\)

⇒ x2 \(\leq\) 1 - x (By squaring both sides)

⇒ x2 + x - 1 \(\leq\) 0

⇒ \(\frac{-(1+\sqrt5)}2\leq x\leq\frac{\sqrt5-1}2\)----(2)

\(\therefore \) From  (1) & (2), we can conclude that domain of given function is

\(\leq\) x \(\leq\) \(\frac{\sqrt5-1}2\) or [0, \(\frac{\sqrt5-1}2\)],

Now, \(\sqrt x + \sqrt{x-\sqrt{1-x}}=1 \) 

⇒ \(\sqrt{x-\sqrt{1-x}}=1-\sqrt x\) 

⇒ x - \(\sqrt{1-x}\) = 1 + x - 2\(\sqrt x\) (By squaring on both sides)

⇒ -\(\sqrt{1-x}=1-2\sqrt x\) 

⇒ 1 - x = (1 - 2\(\sqrt x\))2 = 1 + 4x - 4\(\sqrt x\)

⇒ 4\(\sqrt x\) = 5x ⇒ \(\sqrt x\) = 5/4

⇒ x = \(\frac{25}{16}\not\in\) [0, \(\frac{\sqrt5-1}2\)]

\(\therefore\) Number of real roots of given equation is 0.

(15) Correct option is (C) 3

 x2 + \(\frac{x^2}{(x-1)^2}\) = 8

⇒ x2 (x - 1)2(x2 + 2x - 2) = 0

⇒ x = 2, \(\frac{-2\pm\sqrt{4-4\times1\times-2}}2\) 

 = 2, \(\frac{-2\pm\sqrt{12}}2\) 

 = 2, \(\frac{-2\pm\sqrt{3}}2\) 

 = 2, -1\(\pm\sqrt3\) 

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