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The Curve possessing the property that t...

The Curve possessing the property that the intercept made by the tangent at any point of the curve on they-axis is equal to square of the abscissa of the point of tangency, is given by

A

square of the abscissa of the point of tangency

B

square root of the abscissa of the point of tangency

C

cube of the abscissa of the point of tangency

D

cube root of the abscissa of the point of tangency

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The correct Answer is:
C
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ARIHANT MATHS-DY / DX AS A RATE MEASURER AND TANGENTS, NORMALS -Exercise (Single Option Correct Type Questions)
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  2. The graphs y=2x^(3)-4x+2and y=x^(3)+2x-1 intersect in exactly 3 distin...

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  3. In which of the following functions is Rolles theorem applicable? (a)f...

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  4. The figure shows a right triangle with its hypotenuse OB along the y-a...

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  5. Number of positive integral value(s) of a for which the curve y=a^(x) ...

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  6. Given f(x)=4-(1/2-x)^(2/3),g(x)={("tan"[x])/x ,x!=0 1,x=0 h(x)={x},k(...

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  7. If the function f(x)=x^(4)+bx^(2)+8x+1 has a horizontal tangent and a...

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  8. Coffee is coming out from a conical filter, with height and diameter b...

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  9. A horse runs along a circle with a speed of 20k m//h . A lantern is at...

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  10. Water runs into an inverted conical tent at the rate of 20 ft^3/ min a...

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  11. Let f(x)=x^3-3x^2+2x Find f'(x)

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  12. The Curve possessing the property that the intercept made by the tange...

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  13. If f(x) is continuous and differentible over [-2, 5] and -4lef'(x)le3 ...

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  14. A curve is represented parametrically by the equations x=t+e^(at) and ...

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  15. At any two points of the curve represented parametrically by x=a (2 co...

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  16. Let F(x)=int(sinx)^(cosx)e^((1+sin^(-1)(t))dt on [0,(pi)/(2)], then

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  17. Given f'(1)=1and f(2x)=f(x)AAxgt0.If f'(x) is differentiable, then th...

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  18. Let f(x)a n dg(x) be two functions which are defined and differentiabl...

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  19. Let S be a square with sides of length x. If we approximate the change...

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  20. Consider f(x)=int1^x(t+1/t)dt and g(x)=f'(x) If P is a point on the cu...

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