# 3 Practice Free Response Problems...

## 3 Practice Free Response Problems...

When you are posting a solution to any one of these problems, please give the year and problem number.

Given the function

(a) Describe the symmetry of the graph of

(b) Find the domain of

(c) Find all values of

(d) Write a formula for f

A particle moves along the x-axis in such a way that its position at time t for t≥0 is given by

x(t) = 1/3t

(a) Show that at time t = 0, the particle is moving to the right.

(b) Find all values of t for which the particle is moving to the left.

(c) What is the position of the particle at t = 3?

(d) When t = 3, what is the total distance the particle has traveled?

The line x =

(a) If a line tangent to the cubic at point

(b) Write the equations of the two tangent lines described in (a).

**1975 AB 1**:Given the function

*f*defined by f(x) = ln(x^{2}- 9).(a) Describe the symmetry of the graph of

*f*.(b) Find the domain of

*f*.(c) Find all values of

*x*such that f(x) = 0.(d) Write a formula for f

^{-1}(x), the inverse function of*f*, for x > 3.**1975 AB 2**:A particle moves along the x-axis in such a way that its position at time t for t≥0 is given by

x(t) = 1/3t

^{3}- 3t^{2}+ 8t.(a) Show that at time t = 0, the particle is moving to the right.

(b) Find all values of t for which the particle is moving to the left.

(c) What is the position of the particle at t = 3?

(d) When t = 3, what is the total distance the particle has traveled?

**1975 AB 5**:The line x =

*c*where*c*> 0 intersects the cubic y = 2x^{3}+3x^{2}- 9 at point*P*and the parabola y = 4x^{2}+ 4x + 5 at point*Q*.(a) If a line tangent to the cubic at point

*P*is parallel to the line tangent to the parabola at point*Q*, find the value of*c*where*c*> 0.(b) Write the equations of the two tangent lines described in (a).

## Re: 3 Practice Free Response Problems...

**1975 AB 1:**

a) The function f(x) = ln(x

^{2}-9) is reflected across the y-axis.

b) The domain of ƒ is x < -3 or x > 3.

ln(x) exists when x > 0. So x

^{2}-9 = 0 ⇒ x = ± 3

c)0 = ln(x

^{2}-9)

e

^{0}= e

^{ln(x² -9)}

1 = x

^{2}-9

x

^{2}= 10

x = ±√(10)

d)x = ln(y

^{2}-9)

e

^{x}= e

^{ln(y^2-9)}

e

^{x}= (y

^{2}-9)

e

^{x}+9 = y

^{2}

√(e

^{x}+9) = y = f

^{-1}

(Ok the corrected stuff is in orange. Wow....I can't believe I did that for part d xP I shouldn't do math late at night.)

**1975 AB 2:**

a) d/dt (1/3t

^{3}- 3t

^{2}+ 8t) = t

^{2}- 6t +8 = v(t)

(0)

^{2}- 6(0) +8 = 8;

∴ v(t) = 8 > 0

If v(t) > 0, then the particle moves to the right.

b) If v(t) < 0, then the particle moves to the left.

t

^{2}- 6t +8 < 0

(t - 4)(t - 2) < 0

2 < t < 4

∴ the particle is moving to the left when 2 < t < 4.

c) x(t) = 1/3t

^{3}- 3t

^{2}+ 8t

1/3(3)

^{3}- 3(3)

^{2}+ 8(3) = 6

If t = 3, then x = 6.

d)When t = 3, then the particle has traveled 6 units. (....Is this a trick question or something?)

Last edited by ebilmonky on Sun Jun 13, 2010 4:59 am; edited 2 times in total

**ebilmonky**- Posts : 8

Join date : 2010-06-02

## Re: 3 Practice Free Response Problems...

Parts (a) and (b) JUSTIFY your ANSWERS. Show the work that leads up to your conclusion.

(d) Look at your second to third lines...what's wrong?

(c) Correct!

(d) Look at your second to third lines...what's wrong?

(c) Correct!

## Re: 3 Practice Free Response Problems...

1975 AB1

d)

x=ln(y

e

e

√(e

d)

x=ln(y

^{2}-9)e

^{x}=y^{2}-9e

^{x}+9=y^{2}√(e

^{x}+9)=y=f^{-1}(x)**jodo_059**- Posts : 11

Join date : 2010-06-02

## Re: 3 Practice Free Response Problems...

1975 AB1

a.) f(x)=ln(x^2-9) let x=-x

f(-x)=ln(-x^2-9)

f(-x)=ln(x^2-9) =f(x)=ln(x^2-9)

because f(-x)=f(x) it is an even graph so it reflects about the y axis

b.) x^2-9=0

x^2=3

x=±√9

x=± 3

so x≥3 and x≤-3

c.) 0=ln(x^2-9)

e^0=e^ln(x^2-9) --> e^0=1

1=x^2-9

10=x^2

x=±√10

d.) f(x)=ln(x^2-9); f(x)=y ; switch x and y

x=ln(y^2-9)

e^x=e^ln(y^2-9)

e^x=y^2-9

e^x+9=y^2

y=±√e^x+9=f-1(x)

f -1(x)=±√e^x+9

a.) f(x)=ln(x^2-9) let x=-x

f(-x)=ln(-x^2-9)

f(-x)=ln(x^2-9) =f(x)=ln(x^2-9)

because f(-x)=f(x) it is an even graph so it reflects about the y axis

b.) x^2-9=0

x^2=3

x=±√9

x=± 3

so x≥3 and x≤-3

c.) 0=ln(x^2-9)

e^0=e^ln(x^2-9) --> e^0=1

1=x^2-9

10=x^2

x=±√10

d.) f(x)=ln(x^2-9); f(x)=y ; switch x and y

x=ln(y^2-9)

e^x=e^ln(y^2-9)

e^x=y^2-9

e^x+9=y^2

y=±√e^x+9=f-1(x)

f -1(x)=±√e^x+9

**chiko**- Posts : 2

Join date : 2010-06-04

## Re: 3 Practice Free Response Problems...

1975 AB 5:

(a) Tangents parallel when derivatives are equal

d/dx 2x^3 + 3x^2 - 9 = d/dx 4x^2 + 4x + 5

6x^2 + 6x = 8x + 4

6x^2 - 2x - 4 = 0

3x^2 - x - 2 = 0

(3x + 2)(x - 1) = 0

x = 1 or x = -2/3

x = c; c > 0 so c = 1

(b) 6x^2 + 6x = 8x + 4

6(1)^2 + 6(1) = 8(1) + 4

12 = 12

Slope = 12

y = 2x^3 + 3x^2 - 9____________y = 4x^2 + 4x + 5

y = 2(1)^3 + 3(1)^2 - 9_________y = 4(1)^2 + 4(1) + 5

y = -4________________________y = 13

y + 4 = 12(x - 1)_______________y - 13 = 12(x - 1)

Hopefully it's right ... how do make exponents and spaces?

(a) Tangents parallel when derivatives are equal

d/dx 2x^3 + 3x^2 - 9 = d/dx 4x^2 + 4x + 5

6x^2 + 6x = 8x + 4

6x^2 - 2x - 4 = 0

3x^2 - x - 2 = 0

(3x + 2)(x - 1) = 0

x = 1 or x = -2/3

x = c; c > 0 so c = 1

(b) 6x^2 + 6x = 8x + 4

6(1)^2 + 6(1) = 8(1) + 4

12 = 12

Slope = 12

y = 2x^3 + 3x^2 - 9____________y = 4x^2 + 4x + 5

y = 2(1)^3 + 3(1)^2 - 9_________y = 4(1)^2 + 4(1) + 5

y = -4________________________y = 13

y + 4 = 12(x - 1)_______________y - 13 = 12(x - 1)

Hopefully it's right ... how do make exponents and spaces?

**G-Unit**- Posts : 3

Join date : 2010-06-11

## NEW PROBLEMS...

**AGAIN...NO CALCULATORS!!!**Note that 1975 AB 2 is still open:

**1975 AB 6/BC 2**:

Let

*R*be the region in the first quadrant bounded by the graphs of x

^{2}/9 + y

^{2}/81 = 1 and 3x + y = 9.

(a) Set up but

__do not evaluate__an integral representing the area of

*R*. Express the integrand a s a function of a single variable.

(b) Set up but

__do not evaluate__an integral representing the volume of the solid generated when

*R*is rotated about the

__. Express the integrand as a function of a single variable.__

*x - axis*(c) Set up but

__do not evaluate__an integral representing the volume of the solid generated when

*R*is rotated about the

__. Express the integrand as a function of a single variable.__

*y - axis***1975 BC 7**:

(a) For what value of

*m*is the line y =

*m*x tangent to the graph of y = ln

*x*.

(b) Prove that the graph of y = ln

*x*lies entirely below the graph of the line found in part (a).

(c) Use the results of (b) to show that e

^{x}≥x

^{e}for x > 0.

**1976 AB 3/BC 2**:

Let

*R*be the region bounded by the curves

*f(x)*= 4/x and

*g(x)*= (x - 3)

^{2}.

(a) Find the area of

*R*.

(b) Find the volume of the solid generated by revolving

*R*about the

__.__

*x - axis****CHALLENGE PROBLEM #1**

*This is REALLY HARD*

A snowplow can remove snow at a constant rate (in ft

^{3}/min). One day, there was no snow on the ground at sunrise, but sometime in the morning it began snowing at a steady rate. At noon, the plow began to remove snow. It had cleared 2 miles of snow between noon and 1 PM, and 1 more mile of snow between 1 PM and 2 PM. At what time did it start snowing?

HINTS:

(1) The velocity that the snowplow travels is inversely proportional to the amount of time since it started snowing.

(2) Note that ds/dt = k/t for some constant k (where s is the distance traveled by the plow).

*All Challenge Problems are about 100,000 times harder than the actual AP TEST.

## 1975 AB 2

a) x(t) =1/3t

v(t) = t

v(0) = 8.

8>0, therefore particle is moving to the right.

^{3}- 3t^{2}+8tv(t) = t

^{2}- 6t +8v(0) = 8.

8>0, therefore particle is moving to the right.

**shakalohana**- Posts : 17

Join date : 2010-06-04

## 1975 AB 6/BC 2

1975 AB 6/BC 2

The two equations are of a vertical ellipse and a line which create a sort of diagonal half oval in the first quadrant with its tips on points (0,9) and (3,0)

a) I used a veritcal rectangular strip so its Δx so therefore everything must be changed into only x so solve for y and do top(ellispse) minus bottom(line).

Therefore solve for y for both equations and you get:

y=√(81-9x

y=-3x+9

Then you plug it in and get:

∫ √(81-9x

b) Again i used a vertical rectangular strip and got a washer where R is the ellipse and r is the line. Plugging it in you get:

∫ π(√(81-9x

c)Again i used a vertical rectangular strip and got a cylinder where r=x and h=ellispse-line. Plugging it in you get:

∫ 2πx[√(81-9x

BTW THOSE THINGS THAT LOOK LIKE n'S R PI.

The two equations are of a vertical ellipse and a line which create a sort of diagonal half oval in the first quadrant with its tips on points (0,9) and (3,0)

a) I used a veritcal rectangular strip so its Δx so therefore everything must be changed into only x so solve for y and do top(ellispse) minus bottom(line).

Therefore solve for y for both equations and you get:

y=√(81-9x

^{2})y=-3x+9

Then you plug it in and get:

∫ √(81-9x

^{2})-(-3x+9) dx where a=0, b=3b) Again i used a vertical rectangular strip and got a washer where R is the ellipse and r is the line. Plugging it in you get:

∫ π(√(81-9x

^{2}))^{2}-(-3x+9)^{2}dx where a=0, b=3c)Again i used a vertical rectangular strip and got a cylinder where r=x and h=ellispse-line. Plugging it in you get:

∫ 2πx[√(81-9x

^{2})-(-3x+9)]dx where a=0 and b=3BTW THOSE THINGS THAT LOOK LIKE n'S R PI.

**seikochan**- Posts : 1

Join date : 2010-06-02

## Re: 3 Practice Free Response Problems...

1975 AB 2

(a) Correct - 2/2

Finish the rest now...

1975 AB 6/BC 2

Absolutely CORRECT! GREAT JOB! Score - 9/9

(a) Correct - 2/2

Finish the rest now...

1975 AB 6/BC 2

Absolutely CORRECT! GREAT JOB! Score - 9/9

## Re: 3 Practice Free Response Problems...

the rest!

b)

v(t) = t

(t-2)(t-4) = 0

t = 2, 4

so t moves to the left when greater than or equal to 2 and less than 4 because when it was at 0 it was moving to the right and so at 2 and 4 is when the particle stops and turns around.

c) so the position function is x(t) = 1/3t

and plugging in 3 for t, i got 15.

d) from when t=0 to t=3, i know that the particle only turns around once, which is at t=2.

so i have to do |x(2) - x(0)| + |x(3)- x(2)|. and i got 25/3 + 20/3 = 55/3

b)

v(t) = t

^{2}- 6t + 8 = 0(t-2)(t-4) = 0

t = 2, 4

so t moves to the left when greater than or equal to 2 and less than 4 because when it was at 0 it was moving to the right and so at 2 and 4 is when the particle stops and turns around.

c) so the position function is x(t) = 1/3t

^{3}- 3t^{2}+ 8tand plugging in 3 for t, i got 15.

d) from when t=0 to t=3, i know that the particle only turns around once, which is at t=2.

so i have to do |x(2) - x(0)| + |x(3)- x(2)|. and i got 25/3 + 20/3 = 55/3

**shakalohana**- Posts : 17

Join date : 2010-06-04

## Re: 3 Practice Free Response Problems...

Part (b) only asks for the time that the particle turns to the left, so v(t) < 0 ⇒ 2 < t < 4.

Part (c) OMG...TRY AGAIN!

Part (d) Correct!

Score: 8/9 Fix part (c)

Part (c) OMG...TRY AGAIN!

Part (d) Correct!

Score: 8/9 Fix part (c)

## Re: 3 Practice Free Response Problems...

lolol did i calculate wrong!?! o.o

haha ok let me do this slowly...

1/3(27) - 3(9) + 8(3) = 9 - 27 + 24 = 6

hahahaha ok it's SIX.

BETTER!

haha ok let me do this slowly...

1/3(27) - 3(9) + 8(3) = 9 - 27 + 24 = 6

hahahaha ok it's SIX.

BETTER!

**shakalohana**- Posts : 17

Join date : 2010-06-04

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