A) 1
B) \[\frac{1}{\sqrt{2}}\]
C) \[\frac{1}{3}\]
D) \[\frac{1}{2}\]
Correct Answer: D
Solution :
| [d] Let \[{{\cos }^{-1}}\sqrt{p}+{{\cos }^{-1}}\sqrt{1-p}\] |
| \[+{{\cos }^{-1}}\sqrt{1-q}=\frac{3\pi }{4}\] .. (i) |
| Let \[a={{\cos }^{-1}}\sqrt{p}b={{\cos }^{-1}}\sqrt{1-p}\] and \[c={{\cos }^{-1}}\sqrt{1-q}\] |
| \[\Rightarrow \cos a=\sqrt{p},\cos b=\sqrt{1-p},\cos c=\sqrt{1-q}\] |
| \[\Rightarrow {{\cos }^{2}}a=p,{{\cos }^{2}}b=1-p,{{\cos }^{2}}c=1-q\] |
| Now, \[{{\sin }^{2}}a=1-{{\cos }^{2}}a=1-p\] |
| \[\Rightarrow \sin a=\sqrt{1-p},\] |
| \[{{\sin }^{2}}b=1-{{\cos }^{2}}b=1-1+p\Rightarrow \sin b=\sqrt{p}\] |
| \[{{\sin }^{2}}c=1-{{\cos }^{2}}c=1-1+q=q\Rightarrow \sin c=\sqrt{q}\] |
| \[\therefore \] equation (i) can be written as |
| \[a+b+c=\frac{3\pi }{4}\Rightarrow a+b=\frac{3\pi }{4}-c\] |
| Take cos on each side, we get |
| \[\cos (a+b)=cos\left( \frac{3\pi }{4}-c \right)\] |
| \[\Rightarrow \cos a\,\cos b-\sin a\,\sin b\] |
| \[=\cos \left\{ \pi -\left( \frac{\pi }{4}+c \right) \right\}=-\cos \left( \frac{\pi }{4}+c \right)\] |
| Put values of \[\cos a,cosb\] and \[\sin a,\sin b,\] we get |
| \[\sqrt{p}.\sqrt{1-p}-\sqrt{1-p}\sqrt{p}\] |
| \[=-\left( \frac{1}{\sqrt{2}}\sqrt{1-q}-\frac{1}{\sqrt{2}}\sqrt{q} \right)\] |
| \[\Rightarrow 0=\sqrt{1-q}-\sqrt{q}\Rightarrow \sqrt{1-q}=\sqrt{q}\] |
| Squaring on both side;\[\Rightarrow 1-q=q\] |
| \[\Rightarrow 1=2q\Rightarrow q=\frac{1}{2}\] |
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