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Choose the most appropriate option (a) (b) (c) or (d)
The gradient of the curve y = 2x^{3} –3x^{2} – 12x +8 at x = 0 is
ANSWER : a
Solution : 2x^3  3x^2 12x + 8 = 0
dy/dx = 6x^2  6x 12 = 0
(At x=0) = 6(0)^2  6(0)  12
= 12
The gradient of the curve y = 2x^{3} –5x^{2} – 3x at x = 0 is
The derivative of y =
If y = x (x –1) (x – 2) then is
The gradient of the curve y – xy + 2px + 3qy = 0 at the point (3, 2 ) is and q are
The curve y^{2} = ux^{3} + v passes through the point P(2, 3) and = 4 at P. The values of u and v are
The gradient of the curve y + px +qy = 0 at (1, 1) is 1/_{2}. The values of p and q are
y+px+qy=0
y+qy=px
(1+q)y=px
y=[p/(1+q)]x
This is a linear function, so has constant gradient at all points on the curve. Hence
p/(1+q)=1/2
2p=(1+q)
But there is an issue: you have stated that the curve's gradient is 1/2 at the point (1,1) but the curve does not cross through this point! Regardless of our choices for p and q satisfying the expressions above this paragraph, the equation of the curve will always simplify to y=0.5x, which crosses through the origin (0,0), as well as (1,0.5) and (2,1)  but not (1,1).
For your curve to pass through (1,1), we would need to add a constant term, like so:
y+px+qy=1/2
If xy = 1 then y^{2} + dy/dx is equal to
The derivative of the function
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so the point is (√2,√2)
= 1
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If x^{3} –2x^{2 }y^{2} + 5x +y –5 =0 then at x = 1, y = 1 is equal to
The derivative of x^{2} log x is
If f(x) = 3x^{2}, then F(x) =
If f(x) = x^{2} – 6x+8 then f ’(5) – f ’(8) is equal to
If f(x) = x^{k} and f’(1) = 10 the value of k is
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