Question 9:
--- 9(a) ---
9(a) | 1 1 1 1 1
− −
2x2 +13x 2 =3x 2 +12 all  x2 x−6x2 +5=0 | *M1 | OE
Equating the two expressions in x and then multiplying each
1 1
term by x2 or by their substitution for x2.
Coefficients need to be retained but condone +/– sign errors.
1
Allow x2 replaced by x.
 1  1  6 36−415
x2 −1x2 −5 [= 0] or [x=]
  
2
   | DM1 | OE
Solving their three-term quadratic.
Alternative method for first 2 marks of Question 9(a)
1 1 1 1 1
− −
2x2 +13x 2 =3x 2 +12 all  x2 leading to 2x+10=12x2 | *M1 | Equating the two expressions in x and isolating their term in
1
x2.
(2x+10)2 =144x leading to [4] ( x2 −26x+25 )=0 
26 676−4125
leading to [4](x−25)(x−1) [= 0] or [x=]
2 | DM1 | OE
Squaring both sides, rearranging and solving a three-term
quadratic.
3
x=1 and 25 , y=15 and 1 2
5 | A1, A1 | A1 for both x-values and A1 for both y values.
If M1DM0 scored then SCB1B1 is available for final
answers.
4 | Answers without working score 0/4
Question | Answer | Marks | Guidance
--- 9(b) ---
9(b) |  1   1 1  1 1
− − −
Area = 3x 2 +12−2x2 +13x 2 dx =−2x2 +12−10x 2
   
      | M1 | Attempt to integrate, defined by at least one correct fractional
power, and subtract – condone the wrong way round.
 3   1 
 2x2   10x2 
=− +12x− 
3 1
   
 2   2  | B1 B1 | B1 for either { }.
B1 for completely correct integration of their expression
following through +/– sign errors from the subtraction.
 4 3 1
 − (their 25) 2 +12(their 25)−20(their 25) 2 −
 3 
 4 3 1
 − (their1 ) 2 +12(their1 )−20(their1 ) 2
 3  | M1 | OE
Substitution of their positive limits from part (a) in their
integrated expression, defined by at least one correct
fractional power, and subtraction.
Question | Answer | Marks | Guidance
9(b) | Alternative method for first 4 marks of Question 9(b)
 1   1 1
− −
Area = 3x 2 +12  dx −2x2 +13x 2  dx
   
    | M1 | Attempt to integrate, defined by at least one correct fractional
power, and subtract – condone the wrong way round.
 1   3 1 
3x2  2x2 13x2
=  +12x −  + 
1 3 1
   
 2   2 2  | B1 B1 | OE
One mark for each correct expression.
 1   1 
6(their 25) 2 +12(their 25)  − 6(their1 ) 2 +12(their1 )  −
   
4 3 1 4 3 1
 (their 25) 2 +26(their 25) 2 −  (their1 ) 2 +26(their1 ) 2
3  3  | M1 | OE
Substitution of their positive limits from part (a) in both of
their integrated expressions, defined by at least one correct
fractional power, and subtraction.
128 2
[Area =] ,42 , 42.7
3 3 | A1 | AWRT
If M1B1B1M0 then SC B1 available for correct final answer.
Condone negative answer if corrected.
5 | Condone the presence of π for the first 4 marks but use of
 y2 scores 0/5
Question | Answer | Marks | Guidance