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Difference between maximum and minimum values of `(60sin alpha+p cos alpha)` is 122 then p can be

A

61

B

11

C

`-61`

D

`-11`

Text Solution

Verified by Experts

The correct Answer is:
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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