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Hi, I've worked some maths questions and I found problems in the following parts. Thank you for helping.

1)Given that f:x --> 2x² + 8x -10 for the domain x>=k,

i)find the keast value of k for which f is one-one.

2)The equation of a curve is y = x² - 3x + 4

i)Show that the whole of the curve lies above the x-axis.

ii)Find the set of values of x for which x² - 3x + 4 is a decreasing function of x.

The equation of a line is y + 2x = k, where k is a constant.

iii)Find the value of k for which the line is a tangent to the curve.

3)The diagram(which I am unable to show here) shows the curve y = x³ - 3x² -9x + k, where k is a constant. The curve has a minimum point on the x-axis.

i)Find the value of k.

So I do it by dy/dx=3x²-6x-9, then dy/dx=0

Solving 3x²-6x-9=0 gives x=3 or -1. And I choose the value 3(because the minimum point shown on the graph is on the positive x side) to substitute to the equation of the curve and get k=27, is this right?

ii)Find the coordinates of the maximum point of the curve.

iii)State the set of values of x for which x³ - 3x² - 9x + k is a decreasing function of x.

(How to do these two?)

4)A curve is such that dy/dx = 4[6-2x]^(-0.5), and P(1, 8) is a point on the curve.

i)The normal to the curve at the point P meets the coordinate axes at Q and at R. Find the coordinates of the mid-point of QR.

I know how to find the equation of the normal to the curve, but then I am unable to proceed from there, how do we find the coordinates of Q and R? Then I know how to find the coordinates of the mid-point of QR.

Thank you.

1)Given that f:x --> 2x² + 8x -10 for the domain x>=k,

i)find the keast value of k for which f is one-one.

2)The equation of a curve is y = x² - 3x + 4

i)Show that the whole of the curve lies above the x-axis.

ii)Find the set of values of x for which x² - 3x + 4 is a decreasing function of x.

The equation of a line is y + 2x = k, where k is a constant.

iii)Find the value of k for which the line is a tangent to the curve.

3)The diagram(which I am unable to show here) shows the curve y = x³ - 3x² -9x + k, where k is a constant. The curve has a minimum point on the x-axis.

i)Find the value of k.

So I do it by dy/dx=3x²-6x-9, then dy/dx=0

Solving 3x²-6x-9=0 gives x=3 or -1. And I choose the value 3(because the minimum point shown on the graph is on the positive x side) to substitute to the equation of the curve and get k=27, is this right?

ii)Find the coordinates of the maximum point of the curve.

iii)State the set of values of x for which x³ - 3x² - 9x + k is a decreasing function of x.

(How to do these two?)

4)A curve is such that dy/dx = 4[6-2x]^(-0.5), and P(1, 8) is a point on the curve.

i)The normal to the curve at the point P meets the coordinate axes at Q and at R. Find the coordinates of the mid-point of QR.

I know how to find the equation of the normal to the curve, but then I am unable to proceed from there, how do we find the coordinates of Q and R? Then I know how to find the coordinates of the mid-point of QR.

Thank you.

aha..ive done some of the questions before...CIE maths 9709 ryte....

1)(i) try completing the square and hence u can find the minimum value of k (i.e x)...i got my ans k=-2

2)(i)complete the square

(ii)use the value from (i)...the ans will be x<3/2

(iii)find dy/dx of the curve....the gradient of the line and the curve is the same

3(i)ya i think u are ryte...

(ii)use the other value of x (i.e x=-1) and substitute k=27 inthe eqn.

(iii)i'm not sure how to to this...maybe if i see the graph i can figure it out..

4(i) i)The normal to the curve at the point P meets the coordinate axes at Q and at R. Find the coordinates of the mid-point of QR.

use the eqn. of normal...substitute y=0 into the eqn. and find x.....next substitute x=0 into the eqn. and find y....then use the 2 coordinates to find the mid-point

1)(i) try completing the square and hence u can find the minimum value of k (i.e x)...i got my ans k=-2

2)(i)complete the square

(ii)use the value from (i)...the ans will be x<3/2

(iii)find dy/dx of the curve....the gradient of the line and the curve is the same

3(i)ya i think u are ryte...

(ii)use the other value of x (i.e x=-1) and substitute k=27 inthe eqn.

(iii)i'm not sure how to to this...maybe if i see the graph i can figure it out..

4(i) i)The normal to the curve at the point P meets the coordinate axes at Q and at R. Find the coordinates of the mid-point of QR.

use the eqn. of normal...substitute y=0 into the eqn. and find x.....next substitute x=0 into the eqn. and find y....then use the 2 coordinates to find the mid-point

1) Find the minimum value of f(x), which I find to be at x= -2 so take x>=-2 then f(x) is strictly increasing and so is injective (1-1)

2)

(i) Just show that it has complex roots

(ii)Differentiate, then it is the values of x so that 2x-3<0

3)(i) Diffrentiate at get 3x^2 -6x -9 = 0

=> x = -1 and x = 3 and sub it in and I get k=27, since

x^3 - 3x^2 - 9x + k = 0 at x=3

(iii) Differentiate, and we want

3x^2 - 6x - 9 =< 0

=> (3x + 3)(x - 3) =< 0

=>3x+3 =< 0 and x-3 >=0

=> x=< -1 and x>=3 which is not possible, so the other possibility is

x-3 =< 0 and 3x+3>= 0

=> x=< 3 and x >= -1

=> -1=<x=<3 and the function should be decreasing in this range

2)

(i) Just show that it has complex roots

(ii)Differentiate, then it is the values of x so that 2x-3<0

3)(i) Diffrentiate at get 3x^2 -6x -9 = 0

=> x = -1 and x = 3 and sub it in and I get k=27, since

x^3 - 3x^2 - 9x + k = 0 at x=3

(iii) Differentiate, and we want

3x^2 - 6x - 9 =< 0

=> (3x + 3)(x - 3) =< 0

=>3x+3 =< 0 and x-3 >=0

=> x=< -1 and x>=3 which is not possible, so the other possibility is

x-3 =< 0 and 3x+3>= 0

=> x=< 3 and x >= -1

=> -1=<x=<3 and the function should be decreasing in this range

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