A) \[{{x}^{2}}-hx+({{h}^{2}}-1)=0\]
B) \[2{{x}^{2}}-2hx+({{h}^{2}}-1)=0\]
C) \[{{x}^{2}}-hx+2({{h}^{2}}-1)=0\]
D) \[{{x}^{2}}-2hx+({{h}^{2}}-1)=0\]
Correct Answer: B
Solution :
Given, \[\sin \theta +cos\theta =h\] ??(i) On squaring both sides, we get \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta +2\sin \theta \,\cos \theta ={{h}^{2}}\] \[\Rightarrow \] \[2\,\sin \theta \,\cos \theta ={{h}^{2}}-1\] \[\Rightarrow \] \[\sin \theta \,\cos \theta =\frac{{{h}^{2}}-1}{2}\] ?.(ii) The quadratic equation having the roots \[\sin \theta \] and \[\cos \theta \] is \[{{x}^{2}}-(\sin \theta +\cos \theta )x+\sin \theta \,\cos \theta =0\] \[\Rightarrow \] \[{{x}^{2}}-hx+\left( \frac{{{h}^{2}}-1}{2} \right)=0\] [from Eqs. (i) and (ii)] \[\Rightarrow \] \[2{{x}^{2}}-2hx+({{h}^{2}}-1)=0\]
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