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Question
If `int_0^alpha(3x^2+2x+1)dx=14` then `alpha=`
(A) 1
(B) 2
(C) –1
(D) –2
Solution
(B) 2
`int_0^alpha (3x^2+2x+1)dx=14`
`[x^3+x^2+x]_0^alpha=14`
`alpha^3+alpha^2+alpha-14=0`
`(alpha-2)(alpha^2+3alpha+7)=0`
But `alpha^2+3alpha+7=0` does not have real roots
`alpha=2`
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