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A normal is drawn at a point P(x , y) of...

A normal is drawn at a point `P(x , y)` of a curve. It meets the x-axis and the y-axis in point `A` AND `B ,` respectively, such that `1/(O A)+1/(O B)` =1, where `O` is the origin. Find the equation of such a curve passing through `(5. 4)`

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ARIHANT MATHS ENGLISH-DIFFERENTIAL EQUATION -Exercise (Questions Asked In Previous 13 Years Exam)
  1. A normal is drawn at a point P(x , y) of a curve. It meets the x-axis ...

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  2. If f:R-{-1}toR and f is differentiable function satisfies: f((x)+f(y...

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  3. A solution curve of the differential equation (x^2+xy+4x+2y+4)((dy)/(d...

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  4. "Let "f:(0.oo)rarrR" be a differentiable function such that "f'(x)=2-...

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  5. Let y(x) be a solution of the differential equation (1+e^(x))y^(')+ye^...

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  6. Consider the family of all circles whose centers lie on the straight l...

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  7. The function y=f(x) is the solution of the differential equation (d...

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  8. Let f:[1/2,1]->R (the set of all real numbers) be a positive, non-cons...

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  9. A curve passes through the point (1,(pi)/(6)). Let the slope of the cu...

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  10. Let f:[0,1]rarrR (the set of all real numbers) be a function. Suppose ...

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  11. Let f:[0,1]rarrR be a function. Suppose the function f is twice differ...

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  12. Let f:[0,1]rarrR (the set of all real numbers) be a function. Suppose ...

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  13. Let f:[0,1]toR (the set of all real numbers ) be a function. Suppose t...

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  14. If y(x) satisfies the differential equation y^(prime)-ytanx=2xs e c...

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  15. Let y^(prime)(x)+y(x)g^(prime)(x)=g(x)g^(prime)(x),y(0),x in R , wher...

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  16. Let f: R to R be a continuous function which satisfies f(x)= int0^xf(...

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  17. Let a solution y=y(x) of the differential equation xsqrt(x^(2)-1) dy-...

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  18. If a curve y=f(x) passes through the point (1,-1) and satisfies the di...

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  19. Let y(x) be the solution of the differential equation (xlogx)(dy)/(dx)...

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  20. Let the population of rabbits surviving at a time t be governed by t...

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  21. At present, a firm is manufacturing 2000 items. It is estimated tha...

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