(14) Correct option is (A) 0
\(\sqrt x+\sqrt{x-\sqrt{1-x}}=1\)
For domain, we have
x \(\geq\) 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
0 \(\leq\) 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\(\pm\sqrt3\)