Question 9:
--- 9 (a) ---
9 (a) | ( )
( 2 x + 1 )  ( 1 2 x + 4 ) − 2 6 x 2 + 4 x − 2
f ( x ) =
( 2 x + 1 ) 2 | M1 A1
12x2 +12x+8
= o.e
( 2x+1 )2 | A1
(3)
(b) | 2
1 2  2 + 1 2  22 + 8  1 6 
At x = 2  f ( x ) = =
( 2  2 + 1 ) 5 | M1
5
Full method of normal y − 6 = − ( x − 2 )
1 6 | dM1
1 6 y − 9 6 = − 5 x + 1 0  1 6 y + 5 x = 1 0 6 * | A1*
(3)
(c) | 1 5
Any correct value A = 3 , B = o r D = −
2 2 | B1
6 x 2 + 4 x − 2 = ( A x + B ) ( 2 x + 1 ) + D  Values of A, B and D | M1
− 5
1
2
3 x + + o.e
2 2 x + 1 | A1
(3)
(d) | −5
 1 2 3 2 1 5
3x+ + dx= x + x− ln(2x+1)
2 2x+1 2 2 4 | M1, A1
ft
Area under curve =
 3 x 2 + 1 x − 5 ln ( 2 x + 1 )  2 =  3  2 2 + 1  2 − 5 ln (5 )  −  3   1  2 + 1   1  − 5 ln  5  
2 2 4 13 2 2 4 2 3 2 3 4 3 | dM1
2
Area of R = 13 3 x + 1 + − 5 21 d x + 1  6   1 0 6 − 2 
2 2 x + 2 5 | M1
9 6 4 5
Area of R = − l n 3
1 5 4 | A1
(5)
Total 14
A1: Correct (unsimplified) expression for f ( x ) . Look for
• f ( x ) =
( 2 x + 1 )  ( 1 2 x
(
+
2
4
x
)
+
−
1
2
2 )
(
6 x 2 + 4 x − 2
)
via the quotient rule
( )
• f(x) =( 2x+1 )−1( 12x+4 )+ 6x2 +4x−2 −2 ( 2x+1 )−2 via the product rule
• 3+5(2x+1)−2following division. Condone an incorrect value for A.
A1: Simplified answers such as
4
(
3 x
( 2
2
x
+
+
3
1
x
)
+
2
2
)
,
1 2
4
x
x
2
2
+
+
1
4
2
x
x
+
+
1
8
or 3 + 5 ( 2 x + 1 ) − 2 following correct A.
ISW following a correct answer
(b)
M1: Substitutes x = 2 into their attempt at f ( x ) .
If f ( x ) is incorrect then look for correctly embedded 2’s or a correct value.
dM1: Full method for finding the equation of the normal at ( 2 , 6 ) . Look for y − 6 = −
f
1
( 2 )
( x − 2 )
A1*: y − 6 = −
1
5
6
( x − 2 ) o.e with at least one correct intermediate line leading to 16y+5x=106
If y = m x + c is used look for 6 = −
1
5
6
 2 + c  c =
1
1
0 6
6
You can allow y = −
1
5
6
x +
1
1
0 6
6
going straight to
the given answer. ISW following sight of correct answer. It must follow a correct derivative in (a)
(c)
B1: Any correct value for A = 3 , B =
1
2
o r D = −
5
2
.
They may be embedded within an expression e.g. 3x in the quotient of a division sum
M1: Correct method for finding all three constants.
Via comparison/inspection look for 6 x 2 + 4 x − 2 = ( A x + B ) ( 2 x + 1 ) + D  Values of A, B and D
Via division look for a linear quotient and a constant remainder
A1: 3 x + 1
2
+
−
2 x
5
+
2
1
o.e such as 3 x + 1
2
−
2 ( 2
5
x + 1 )
which may be seen within an integral in (d).
It is not just for values of A, B and D.
ISW following a correct answer (e.g. they may go on to  2
PMT
but then the A1 ft in (d) would be unavailable)
(d)
M1: Attempts to integrate their A x + B +
2
D
x + 1
. Look for two terms in the correct form, one of which must
be the ln term. Award for ... x 2 + .... + k l n ( 2 x + 1 ) or . . . + . . . x + k l n ( 2 x + 1 )
Also note that kln(4x+2)and any multiple of (2x+1)within the ln is also correct
A1ft: 3
2
x 2 + 1
2
x − 5
4
l n ( 2 x + 1 ) o.e. but follow through on their values of A, B and D
dM1: A correct method of finding the area under the curve using limits of
1
3
and 2 either way around.
Dependent upon the previous M mark. Award if the intention is clear.
M1: For a correct method of finding the area of R. It is scored for adding the area under the curve to the
correct calculation for the area of the triangle. The integration of f(x) need not be correct but the limits
must be applied the correct way around. The area of the triangle may be done by integration. Look for
2
 1 106  1 106  1 1 106 
f(x)dx+ 6 −2=I(x) 2
1
+ 6 −2= I(2)−I + 6 −2=...
2  5  2  5  3 2  5 
3
1
3
where I(x) is their attempt at

f ( x ) d x
A1:
9
1
6
5
4
−
5
4
l n 3
PMT
PMT
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