Difference between maximum and minimum values of (60sin⁡ α+pcos⁡ α) is 122 then p can be

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-61

-11

answer is B.

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Detailed Solution

E=60sin⁡ α+pcos⁡ αMaximum value of E=3600+p2Minimum value of E=−3600+p23600+p2+3600+p2=122∴ 23600+p2=122⇒ p2=121⇒ p=±11