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If cosy=xcos(a+y). And dx/dy =k/1+(x)2-2xcosa...... Then find value of k Now my main question is..For above que. We should diff. w.r.t "y" and we get k=sina.. ..And i see same que in diff book like this mainor.. .... If cosy=xcos(a+y). And dy/dx =k/1+(x)2-2xcosa...... Then find value of k ,and for this we should diff w.r.t "x"..And we get same ans. K=sina then for this que. We prove dy/dx=dx/dy... Thanks

4 years ago


Answers : (1)


may be the question is wrong or may be the value ofdy/dx ...or dx/dy are equal to 1 at k = sina

3 years ago

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Lim x->0 (xcosx-log(1+x))/x^2.Limx->0 (xcosx - x )/x^2 (by elimination )Lim x->0 (cosx - x)/xWhich is finally equal to zero but while using expansion and L hospital`s rule i get the answer...
Lim x->0 (xcosx-log(1+x))/x^2 from this to the below written statement Limx->0 (xcosx - x )/x^2 How you get this ??
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limit x tends to infinty :(2^x-1)/(2^x-1) ..please find it s answer..
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Avani Patidar 10 months ago
Integrate log(sinx+cosx)dx from -pi/4 to pi/4. someone plz help
This is a tricky question First put x = -x and add Now put 2x = t which implies, dx = dt/2 Now refer to the following steps to solve for the simplified integral
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A circle passes through the point (3,(7/2)^0.5) and touches the line pair x^2 – y^2 -2x+1=0. the co ordinates of centre of circle are? (4,0) (6,0)
OK so first we solve for the lines whihc gives the two lines as and now plotting them on the coordinate axes, it’s easy to see by symmetry that the center of circle would lie on x-...
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