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Difference between maximum and minimum values of `(60sin alpha+p cos alpha)` is 122 then p can beA. 61B. 11C. `-61`D. `-11` |
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Answer» Correct Answer - B::D `E= 60 sin alpha + pcosalpha ` Maximum value of `E = sqrt(3600 + p^(2))` Minimum value of E `=-sqrt(3600 + p^(2))` `sqrt(3600 + p^(2)) + sqrt(3600 + p^(2)) = 122" "` (given) `therefore 2 sqrt ( 3600 + p^(2)) = 122` `rArr p^(2) = 121` `rArr p = pm 11` |
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