Pure maths 2 and 3 questions??

prg SG

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??**

**how to do this?**

wee_ws SG

Hi,

Correct for both parts and well done!

Cheers,
Wen Shih

wee_ws SG

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 SG

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 SG

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 SG

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 SG

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