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IGCSE MATHS Doubts

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princess12:
thanxs
what about q 7

Dark Prince:
Can some1 tell me how to do part (e) of this Qs....(attached below)....tnkx

Ghost Of Highbury:

--- Quote from: Adzel on May 16, 2010, 06:12:32 pm ---o/n 2007 p4 q3 (explaination please)

Thanks in advance...

--- End quote ---

Ans 3ai) Point C --> Its x-co-ordinate is 0 and it lies in the curve y = x^2 + 1.

thus, just substitute the value x=0 in the equation --> y = 0^2 + 1 = 1

C --> (0,1)

3aii) with y-axis --> y = 4 - x.

any point in y-axis has the co-ordinate x=0.

apply x=0 in y = 4-x --> y = 4-0

thus, with y-axis --> (0,4)

with x-axis --> y = 4-x

any point in x-axis has y-co-ordinate = 0

apply y=0 in y = 4-x

0 = 4-x

x = 4

point --> (4,0)

b) y = mx + c , here m = gradient and c is y intercept.

write y = 4-x in this form

y = -x + 4    or     y = -1(x) + 4

here m = -1

gradient = -1

c) this means find the x-values for which the point on the curve has a negative gradient. in this, a parabola, all points in the part of the curve in the second quadrant has a negative gradient. Thus, when x<0, gradient = negative.

d) the two graphs are y = x^2 + 1     and   y = 4-x

meaning , x^2 + 1 = 4-x

x^2 + 1 -4 +x = 0

x^2 +x - 3 = 0

------------------

e) apply x = +/-

u shud get 1.3 and -2.3

 The answers i.e 1.3 and -2.3 are the x-cordinates of the point B and A respectively.

y = 4-x

y = 4-1.3 = 2.7

y = 4-x

y = 4+2.3 = 6.3

A - (-2.3, 6.3)         B - (1.3, 2.7)

(x1+x2)/2 = (-2.3+1.3)/2 = -0.5

(y1+y2)/2 = (6.3+2.7)/2 = 4.5

midpoint = (-0.5,4.5)


the_grim_reaper:

--- Quote ---for these kind of ques u hav to use some common sense .. even im kinda bad at sequences =\ unfortunately there are chances of this topic comin up .. because it didnt come in p2

i got this one for my mock test and i solved it by trial n error .. one thing is good tht they already mention the ans. so i try everything to get the ans

   n (n+1)
= -------   + (n + 1)
      2
=n2 + 3n + 2
--------------
      2
ans is not (n +1 )2 so i tried adding again

=n2 + 3n + 2                      n (n+1)   
-----------------        +    ---------
         2                                 2

=(n +1 )2

got it .. sorry i couldnt give u a proper explanation =\
--- End quote ---


Let me take a shot at this.
All right, in the previous part it is stated that the formula for the nth term is n(n+1)/k where we've found k to be 2.

So,  if n is n LOL then its n(n+1)/2 and if n is (n+1) so we have (n+1)((n+1)+1)/2 (replace n by n+1 since that is the nth term here)..

Next just add them
   n(n+1)     (n+1)(n+2)
= ------- + -----------
       2           2
= (n^2+n) + (n^2+2n+n+2)
 --------------------------
                  2
= n^2+n^2+n+2n+n+2
----------------------
              2
= 2n^2+4n+2
-------------
          2
Taking 2 as common, you have =2(n^2+2n +1)
                                             ------------
                                                     2
After simplification, you have n^2+ 2n +1 which is simplified becomes (n+1)(n+1) which is equal to (n+1)^2     


As far sequences are concerned, we all suck at em' mate. It's a universal issue ;D.

Adzel:

--- Quote from: A@di on May 16, 2010, 06:44:41 pm ---Ans 3ai) Point C --> Its x-co-ordinate is 0 and it lies in the curve y = x^2 + 1.

thus, just substitute the value x=0 in the equation --> y = 0^2 + 1 = 1

C --> (0,1)

3aii) with y-axis --> y = 4 - x.

any point in y-axis has the co-ordinate x=0.

apply x=0 in y = 4-x --> y = 4-0

thus, with y-axis --> (0,4)

with x-axis --> y = 4-x

any point in x-axis has y-co-ordinate = 0

apply y=0 in y = 4-x

0 = 4-x

x = 4

point --> (4,0)

b) y = mx + c , here m = gradient and c is y intercept.

write y = 4-x in this form

y = -x + 4    or     y = -1(x) + 4

here m = -1

gradient = -1

c) this means find the x-values for which the point on the curve has a negative gradient. in this, a parabola, all points in the part of the curve in the second quadrant has a negative gradient. Thus, when x<0, gradient = negative.

d) the two graphs are y = x^2 + 1     and   y = 4-x

meaning , x^2 + 1 = 4-x

x^2 + 1 -4 +x = 0

x^2 +x - 3 = 0

------------------

e) apply x = +/-

u shud get 1.3 and -2.3

 The answers i.e 1.3 and -2.3 are the x-cordinates of the point B and A respectively.

y = 4-x

y = 4-1.3 = 2.7

y = 4-x

y = 4+2.3 = 6.3

A - (-2.3, 6.3)         B - (1.3, 2.7)

(x1+x2)/2 = (-2.3+1.3)/2 = -0.5

(y1+y2)/2 = (6.3+2.7)/2 = 4.5

midpoint = (-0.5,4.5)




--- End quote ---

Thanx

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