Q.

Let y=f(x)=sin3π3cosπ324x3+5x2+132. Then, at x=1,

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a

2y+3π2y=0

b

y+3π2y=0

c

2y+3π2y=0

d

2y3π2y=0

answer is B.

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Detailed Solution

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f(x)=sin3π3cosπ324x3+5x2+132
f(1)=18
f1(x)=3sin2π3cosπ324x3+5x2+13/2π3cosπ3cosπ324x3+5x2+13/2                                                 ×sinπ324x3+5x2+13/2×π32324x3+5x2+11212x2+10x
f1(1)=314π33232×π32322(2)=3π216
2f1(1)+3π2f(1)=0

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