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If `cos^4theta+alpha`are the roots of the equation `x^2+2b x+b=0a n dcos^2theta+beta,sin^2theta+betaa r e`the roots of the equation `x^2+4x+2=0,`then values of `b`are`2`b. `-1`c. `-2`d. `2`A. 2B. -1C. -2D. 1 |
Answer» Correct Answer - 1,2 We have ` cos ^(2) theta - sin^(2() theta = cos 2 theta` `rArr cos ^(4) theta theta - sin ^(4) theta = cos 2 theta ` `rArr (-2b)^(2) - 4b = - (-4)^(2) - 4xx2` (since L.H.S. is differnece of roots of first equation and R.H.S. is diffenece of roots of second equation). or ` 4b^(2) - 4b = 16 - 8= 8 ` or ` 4b^(2) - 4b - 8 = 0` or ` b^(2) - b - 2 = 0` or ` (b + 1)(b - 2) = 0` or ` b = 2, - 1` |
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