A) \[r\ne p\]
B) \[p=q=r\]
C) \[\underset{x\to 1}{\mathop{\lim }}\,\frac{\sqrt{1-\cos 2(x-1)}}{x-1}\]
D) \[\sqrt{2}\]
Correct Answer: B
Solution :
\[f(x)={{\sin }^{4}}x+{{\cos }^{4}}x\] \[={{({{\sin }^{2}}x+{{\cos }^{2}}x)}^{2}}-2{{\sin }^{2}}x{{\cos }^{2}}x\] \[=1-\frac{4{{\sin }^{2}}x{{\cos }^{2}}x}{2}\] \[=1-\frac{{{\sin }^{2}}2x}{2}\] \[=1-\frac{1}{4}(2{{\sin }^{2}}2x)\] \[=1\left( \frac{1-\cos 4x}{4} \right)\] \[=\frac{3}{4}+\frac{1}{4}\cos 4x\] hence function\[f(x)\]is increasing when \[f(x)>0f(x)=-\sin 4x>0\] \[\Rightarrow \] \[\sin 4x<0\] hence, \[\pi <4x<\frac{3\pi }{2}\] or \[\frac{\pi }{4}<x<\frac{3\pi }{8}\]
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