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Question

Let f(x)=Asin(πx2)+B, f(12)=2 and 10f(x)dx=2Aπ, then A and B are


Your Answer
A
π2,π2
Your Answer
B
2π,3π
Your Answer
C
0,4π
Correct Answer
D
4π,0

Solution

The correct option is D 4π,0
f(x)=Asin(πx2)+B
f(x)=Acos(πx2)π2
f(12)=Acos(π4)π2=2
A=4π
Now, 10f(x)dx=2Aπ
10[Asin(πx2)+B]dx=2Aπ
2Aπcos[πx2]10+B[x]10=2Aπ
B=0
Hence A=4π and B=0
 

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