1.

Difference between maximum and minimum values of `(60sin alpha+p cos alpha)` is 122 then p can beA. 61B. 11C. `-61`D. `-11`

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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