Question 1:
--- 1(a) ---
1(a) | 1−tanh2 x≡sech 2x
ex −e−x  2
1−tanh2 x=1− 
ex +e−x
 | Replaces the tanh x on the lhs with a
correct expression in terms of
exponentials. | B1
(ex +e−x)2 −(ex −e−x)2 (e2x +2+e−2x)−(e2x −2+e−2x) 2e2x×2e−2x
= = ore.g.
(ex +e−x)2 (ex +e−x)2 (ex +e−x)2
Attempts to find common denominator and expand numerator | M1
 4 
=   =sech2 x*
(ex +e−x)2
 | Obtains the rhs with no errors. | A1cso
(3)
ALT 1 | 1−tanh2 x=(1−tanhx)(1+tanhx)
 ex −e−x  ex −e−x 
=1− 1+ 
ex +e−x ex +e−x
   | Uses the difference of 2 squares on the lhs
and replaces the tanh x with a correct
expression in terms of exponentials. | B1
 2e−x  2ex 
=  
ex +e−x ex +e−x
 | Attempt to find common denominators and
simplify numerators. | M1
 4 
=   =sech2 x*
(ex +e−x)2
 | Obtains the rhs with no errors. | A1cso
ALT 2 | 4
sech2 x=
(ex +e−x)2 | Replaces the sech x on the rhs with a
correct expression in terms of
exponentials. | B1
(e2x +2+e−2x)−(e2x −2+e−2x) (ex +e−x)2 −(ex −e−x)2
= =
(ex +e−x)2 (ex +e−x)2
Attempts to express the “4” in terms of the denominator. | M1
ex −e−x  2
=1−  =1−tanh2 x*
ex +e−x
 | Obtains the lhs with no errors. | A1cso
(b) | 2sech2 x+3tanhx =3⇒ 2(1−tanh2 x)+3tanhx =3
⇒2tanh2 x−3tanhx+1=0
Uses sech2 x=1−tanh2 xand forms a 3 term quadratic in tanh x | M1
(2tanhx−1)(tanhx−1)=0⇒tanhx=... | Solves 3TQ by any valid method
including calculator. | M1
1
tanhx= → x=ln 3
2 | 1 1 1
ln 3. Accept ln3, − ln
2 2 3
And no other answers. | A1
(3)
ALT |  4  ex −e−x 
2sech2 x+3tanhx=3⇒2 +3 =3
(ex +e−x)2 ex +e−x
 
⇒8+3(e2x −e−2x)=3(e2x +2+e−2x)⇒...
Substitutes the correct exponential forms, attempts to eliminate fractions and collect terms | M1
1
6e−2x =2⇒e−2x =
3 | Rearranges to reach e−2x =... | M1
x = ln 3 | 1 1 1
ln 3. Accept ln3, − ln
2 2 3
And no other answers. | A1
Total 6
2. | y = 9−x2, 0≤ x≤3
(a) | dy x
=−
dx 9−x2 | Correct derivative in any form. | B1
Note that the derivative may be obtained implicitly after squaring e.g.
dy dy x
y = 9−x2 ⇒ y2 =9−x2 ⇒2y =−2x⇒ =−
dx dx 9−x2
⌠ x2
Length of C=  1+ dx
9−x2
⌡ | ⌠ 2
dy
dy
Uses  1+   dx with their
⌡ dx dx | M1
Note that the above may be obtained via the implicit route as e.g.
⌠ dy 2 ⌠ x2 ⌠ x2
 1+   dx=  1+ dx=  1+ dx
⌡ dx ⌡ y2 ⌡ 9−x2
In which case the B1 is implied.
⌠ 9 x  x 
= dx=3arcsin (+c ) or −3arccos (+c ) 
⌡ 9−x2 3  3 
3
⌠ 9
 dx =3arcsin(1)−3arcsin(0) ( or −3arccos(1)+3arccos(0))
⌡ 9−x2
0
Finds common denominator, integrates to obtain arcsin… or arccos…
and applies the limits 0 and 3. | M1
3π
= *
2 | Obtains the printed answer with no errors.
This mark should be withheld if there is no
evidence at all of the limits being applied. | A1
Special case:
x
+ dy
If is obtained for score B0M1M1A1 if otherwise correct but allow full
9−x2 dx
recovery in (b)
(4)
(b) | Surface Area
⌠  9 
= 2π 9−x2   dx
 9−x2 
⌡   | ⌠ 2
dy
dy
Uses 2πy 1+   dx with their
⌡ dx dx | M1
=∫ 3 6πdx=6π[ x ]3 =...
0 0 | Integrates to obtain kx and applies the limits
0 and 3. Condone omission of the lower
limit. | M1
=18π | 18πcao | A1
(3)
Total 7
3. |  3 1 p
 
M = 1 1 2
 
 
 −1 p 2
(a) | 3 1 p
detM = 1 1 2
−1 p 2
=3(2−2p)−1(2+2)+ p(p+1) | Attempts determinant. Requires at least 2
correct “terms”. May use other
rows/columns or rule of Sarrus. | M1
= p2 −5p+2 | Correct simplified determinant. | A1
p2 −5p+2 = 0⇒ p =... | Solves 3TQ | M1
5± 17
2 | Correct values. | A1
(4)
(b) |  2−2p 4 p+1 
 
Minors (2− p2) 6+ p (3p+1)
 
 
 2− p (6− p) 2  | Attempts the matrix of minors. If there is
any doubt look for at least 6 correct
elements. May be implied by their matrix
of cofactors. | M1
(B1 on
EPEN)
 2−2p −4 p+1 
 
Cofactors  −(2− p2) 6+ p −(3p+1) 
 
 2− p −(6− p) 2  | Attempts cofactors. | M1
Correct matrix | A1
2−2p p2 −2 2− p
1  
M−1 =  −4 6+ p p−6
p2 −5p+2
 p+1 −3p−1 2 
  | Transposes matrix of cofactors and
divides by determinant. | M1
Follow though their det M from part (a)
but the adjoint matrix must be correct. | A1ft
(5)
Total 9
f(x)= xarccosx, −1≤ x≤1,
x
f′( )=arccosx−
x
1−x2
M1: Differentiates using the product rule to obtain an expression of the form:
x
arccosx±
1−x2
A1: Correct derivative | M1A1
0.5 π− 3
f′(0.5)=arccos0.5− =
1−0.52 3 | π− 3 π 1
−
oe e.g.
3 3 3 | A1
(3)
(ii) | g(x) =arctan(e2x)
2e2x
g'(x)=
e4x +1
M1: Differentiates using the chain rule to obtain an expression of the form:
ke2x
( )2
e2x +1
A1: Correct derivative in any form | M1A1
2
g'(x)= =sech(2x)
e2x +e−2x | Introduces sech(2x). Depends on previous
M. | dM1
g''(x)=−2sech(2x)tanh(2x) | Differentiates sech(u)→±sechutanhu
Depends on both previous M’s. | dM1
Correct expression. | A1
(5)
(ii)
ALT 1 | 2e2x
g'(x)=
e4x +1
M1: Differentiates using the chain rule to obtain an expression of the form:
ke2x
( )2
e2x +1
A1: Correct derivative in any form | M1A1
4e2x( 1+e4x) −4e4x×2e2x
g''(x)=
(e4x +1)2 | Differentiates using quotient or product
rule. Depends on first M. | dM1
4e2x −4e6x −4(e2x −e−2x)
= =
(e4x +1)2 (e2x +e−2x)2 | Multiply through by e−4x. Depends on
both previous M’s. | dM1
2 e2x −e−2x
=−2
e2x +e−2x e2x +e−2x
=−2sech2xtanh2x | Correct expression. | A1
Note that the first derivative may be obtained implicitly in either method e.g.
( ) dy dy 2e2x
y =arctan e2x ⇒ tan y =e2x ⇒sec2 y =2e2x ⇒ =
dx dx 1+ ( e2x )2
Total 8
5. | I =∫secn xdx, n≥0
n