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QUESTION: 1

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

Solution:

**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**

QUESTION: 2

The gradient of the curve y = 2x^{3} –5x^{2} – 3x at x = 0 is

Solution:

QUESTION: 3

The derivative of y =

Solution:

QUESTION: 4

Solution:

QUESTION: 5

Solution:

QUESTION: 6

If y = x (x –1) (x – 2) then is

Solution:

QUESTION: 7

The gradient of the curve y – xy + 2px + 3qy = 0 at the point (3, 2 ) is and q are

Solution:

QUESTION: 8

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

Solution:

QUESTION: 9

The gradient of the curve y + px +qy = 0 at (1, 1) is 1/_{2}. The values of p and q are

Solution:

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

QUESTION: 10

If xy = 1 then y^{2} + dy/dx is equal to

Solution:

QUESTION: 11

The derivative of the function

Solution:

QUESTION: 12

Given e^{-}^{xy} –4xy = 0, can be proved to be

Solution:

QUESTION: 13

Solution:

QUESTION: 14

If log (x / y) = x + y, may be found to be

Solution:

QUESTION: 15

If f(x, y) = x^{3 }+ y^{3} – 3axy = 0, can be found out as

Solution:

QUESTION: 16

Given x = at^{2}, y = 2at; is calculated as

Solution:

QUESTION: 17

Given x = 2t + 5, y = t^{2} – 2; is calculated as

Solution:

QUESTION: 18

If y =

Solution:

QUESTION: 19

If x = 3t^{2} –1, y = t^{3} –t, then is equal to

Solution:

QUESTION: 20

The slope of the tangent to the curve y = at the point, where the ordinate and the abscissa are equal, is

Solution:

so the point is (√2,√2)

= -1

QUESTION: 21

The slope of the tangent to the curve y = x^{2} –x at the point, where the line y = 2 cuts the curve in the Ist quadrant, is

Solution:

QUESTION: 22

For the curve x^{2} + y^{2} + 2gx + 2hy = 0, the value of at (0, 0) is

Solution:

QUESTION: 23

Solution:

QUESTION: 24

If x^{y}.y^{x} = M, M is constant then is equal to

Solution:

QUESTION: 25

Given x = t + t^{–1} and y = t – t^{–1} the value of at t = 2 is

Solution:

QUESTION: 26

If x^{3} –2x^{2 }y^{2} + 5x +y –5 =0 then at x = 1, y = 1 is equal to

Solution:

QUESTION: 27

The derivative of x^{2} log x is

Solution:

QUESTION: 28

The derivative of

Solution:

QUESTION: 29

Solution:

QUESTION: 30

Solution:

QUESTION: 31

Solution:

QUESTION: 32

If f(x) = 3x^{2}, then F(x) =

Solution:

QUESTION: 33

Solution:

QUESTION: 34

Solution:

QUESTION: 35

Solution:

QUESTION: 36

If f(x) = x^{2} – 6x+8 then f ’(5) – f ’(8) is equal to

Solution:

QUESTION: 37

Solution:

QUESTION: 38

Solution:

QUESTION: 39

Solution:

QUESTION: 40

If f(x) = x^{k} and f’(1) = 10 the value of k is

Solution:

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