A) 1
B) \[\frac{1}{2}\]
C) \[\frac{1}{4}\]
D) Put \[{{\cos }^{-1}}x=y\] and \[x\to 1\,\Rightarrow \,\,y\to 0.\] \[\underset{x\to 1}{\mathop{\lim }}\,\,\frac{1-\sqrt{x}}{{{({{\cos }^{-1}}x)}^{2}}}=\underset{y\to 0}{\mathop{\lim }}\,\,\frac{1-\sqrt{\cos y}}{{{y}^{2}}}\] Now rationalizing it, we get \[\underset{y\to 0}{\mathop{\lim }}\,\,\frac{(1-\cos y)}{{{y}^{2}}(1+\sqrt{\cos y})}\] \[=\underset{y\to 0}{\mathop{\lim }}\,\,\frac{1-\cos y}{{{y}^{2}}}\,.\,\underset{y\to 0}{\mathop{\lim }}\,\,\frac{1}{1+\sqrt{\cos y}}=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}.\]
Correct Answer: D
Solution :
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