11.a --8a2/3
[/img]\int_{0}^{a}{\sqrt{4ax} dx}=[4/3\sqrt{a}x^{3/2}]_{0}^a=8/3a^2
2. √1-x2 + √1-y2 = a(x-y) then dy/dx is
a) (1-x2)/(1-y2) b) (1-y2)/ (1-x2) c) (√1-x2)/(1-y2) d) √1-y2)/ (1-x2)
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3 Answers
29Ans 4...- √a2 + b2 < asinx + bcosx < √a2 + b2..
so answer is √2.
Ans 5...d
Ans 3..d...use -dx/dy for finding the normal..and point is in first quadrant..
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