If the sum of coefficients in the expansion of (1−xsinθ+x2)n is P, then
A
minimum value of P=1
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B
minimum value of P=−1
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C
maximum value of P=3n
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D
maximum value of P=2n
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Solution
The correct options are A minimum value of P=1 C maximum value of P=3n Sum of coefficients in (1−xsinθ+x2)n is (1−sinθ+1)n (putting x=1) This sum is greatest when sinθ=−1, Therefore maximum value of sum of coefficient will be 3n And sum will be minimum when sinθ=1 Therefore minimum value will be 1