If f(x)=Asin⁡πx2+B,f′(1/2)=2 and∫01 f(x)dx=2Aπthen the constants A and B are respectively

f′(x)=Aπ2cos⁡π2x⇒2=f′12=Aπ22so A=4π Also 2Aπ=∫01 f(x)dx=−A2πcos⁡π2x+Bx01=B+2Aπ ⇒ B=0

If f(x)=Asin⁡πx2+B,f′(1/2)=2 and∫01 f(x)dx=2Aπthen the constants A and B are respectively

f′(x)=Aπ2cos⁡π2x⇒2=f′12=Aπ22so A=4π Also 2Aπ=∫01 f(x)dx=−A2πcos⁡π2x+Bx01=B+2Aπ ⇒ B=0

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If $f (x) = A \sin \frac{π x}{2} + B, f^{'} (1 / 2) = \sqrt{2}$ and
$\int_{0}^{1} f (x) d x = \frac{2 A}{π}$ then the constants $A$ and $B$ are respectively

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$π / 2, π / 2$

$2 / π, 3 / π$

$0, - 4 / π$

$4 / π, 0$

answer is D.

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Detailed Solution

$f^{'} (x) = A \frac{π}{2} \cos \frac{π}{2} x \Rightarrow \sqrt{2} = f^{'} (\frac{1}{2}) = \frac{A π}{2 \sqrt{2}}$

so $A = \frac{4}{π}$ Also

$\frac{2 A}{π} = \int_{0}^{1} f (x) d x = - A \frac{2}{π} \cos \frac{π}{2} x + {B x|}_{0}^{1}$

$= B + \frac{2 A}{π} \Rightarrow B = 0$

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If f(x)=Asin⁡πx2+B,f′(1/2)=2 and∫01 f(x)dx=2Aπthen the constants A and B are respectively

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If $f (x) = A \sin \frac{π x}{2} + B, f^{'} (1 / 2) = \sqrt{2}$ and
$\int_{0}^{1} f (x) d x = \frac{2 A}{π}$ then the constants $A$ and $B$ are respectively