Pure maths 2 and 3 questions??

• prg

  12 Dec 08, 23:12

  QUESTION:

  curve y=e^-2x -3x  crosses x-axis at A (a,0) and y-axis at B(0,1) 

  Curve:

  :  

   

  (a)Write down an equation satisfied by a.

  if  y=e^-2x -3x   then x=a and y =0

   **is this correct**

  (b) show that tangent at A meets y-axis at point whose y-coordinates is 

  for eq of tangent at A  if curve: y=e^-2x -3x

  y - y1 =m (x - x1)

  x1= a     y1=0

  

   

  

  if x= 0 (y-axis)

  y-coord.:    **is this the correct way??**

  (c) Show that d^2 y / dx^2) >0  and using results from parts (a) and (b), deduce that 6a^2 + 3a  < 1

   

  **how to do this?**

• wee_ws

  13 Dec 08, 05:26

  Hi,

    Correct for both parts and well done!

  Cheers,
  Wen Shih

• wee_ws

  13 Dec 08, 07:06

  Hi,

    To show that 2nd derivative > 0, find 2nd derivative and then use the fact that the exponential function is always positive.

    To show the inequality, compare the gradient of tangent with the gradient of the line AB.

    Try it!

  Cheers,
  Wen Shih

• prg

  13 Dec 08, 14:42

  ok so how to do: using results from parts (a) and (b), deduce that 6a^2 + 3a  < 1

   

  compare which gradient of tangent with which gradient of line AB??

  Basically, the result for (a) is y=e^(-2x) - 3x    (when x= a and y=0)    0=e^(-2a) - 3a

   and result for (b) is curve's gradient: dy /dx = -2e^(-2a) - 3

                                     eq of tangent at A: y = (-2e^(-2a) -3) (x-a)

  when x=0 tangent at A meets y-axis at point whose y-coordinates is 2ae^-2a + 3a

  HOW TO COMPARE THESE TWO RESULTS??

• wee_ws

  13 Dec 08, 18:47

  Hi,

    Sorry, we should compare

    the gradient of the line joining point A and the y-coordinate shown in (b)

    with

    the gradient of the line joining point A and B.

    Thus,

    {2ae^(-2a) + 3a}/a < 1/a

    To simplify to the desired inequality, we need to use the fact that e^(-2a) = 3a which comes from (i).

    Hope it is clearer now.

  Cheers,
  Wen Shih

• prg

  14 Dec 08, 00:08

  ok great it is clearer now, but why is it {2ae^(-2a) + 3a}/a < 1/a

  where did u get: /a , < and  1/a ?? Sorry if i sound stupid!!!

   

• wee_ws

  14 Dec 08, 07:23

  Hi,

    Both sides of the inequality are ratios because we are looking at gradients, i.e. y-value divided by x-value. Both gradients have the same x-value, i.e. a.

    If you indicate the y-intercept on the given diagram, you will observe that it is lower than point B. Thus gradient on LHS is less than gradient on RHS. Try to draw both gradients and convince yourself of the inequality. I believe the diagram is given to help the student visualise the problem better :)

    Thanks!

  Cheers,
  Wen Shih