A) 10
B) \[10\sqrt{2}\]
C) 5
D) \[5\sqrt{2}\]
Correct Answer: A
Solution :
[a] |
\[\tan \alpha \] and \[\tan \beta \] are roots of \[(k+1){{x}^{2}}-\sqrt{2}\lambda x\] |
\[-(1-k)=0\] |
\[\therefore \tan \alpha +\tan \beta =\frac{\sqrt{2}\lambda }{k+1}\] |
\[\tan \alpha +\tan \beta =\frac{k-1}{k+1}\] |
Now \[\tan (\alpha +\beta )=\frac{\frac{\sqrt{2}\lambda }{k+1}}{1-\left( \frac{k-1}{k+1} \right)}=\frac{\sqrt{2}\lambda }{2}=\frac{\lambda }{\sqrt{2}}\] |
\[\frac{{{\lambda }^{2}}}{2}=50\] |
\[\therefore \lambda =10\] |
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