Download SSCE 1693 Part A (55%) Answer All Questions. 1. Prove the identity

SSCE 1693
Part A
(55%)
Answer All Questions.
π
.
2
Use this identity to find the value of x that satisfies the equation
1. Prove the identity
sin−1 x + cos−1 x =
2 sin−1 x − 3 cos−1 x =
π
.
6
[6 marks]
2. Evaluate
Z
x3 − sinh−1 x
√
dx.
1 + x2
[6 marks]
3. Differentiate
2x
tanh
1+x
with respect to x and give your answer in terms of x.
−1
4. Test the convergence of the series given by
∞
X
r=2
[5 marks]
1
.
r(ln r)2
[5 marks]
5. Determine the convergence or divergence of the integral
Z ∞
t e−2t dt.
0
[6 marks]
6. Find the equation of the plane that contains the points A(1, 2, 3),
B(−1, 0, 7), and C(7, 9, 4).
[5 marks]
2
SSCE 1693
2
. Determine if
1 − sin θ
the point (−2, 0◦ ) is on the graph and find the equivalent Cartesian
equation. Hence sketch the graph.
7. A graph is given by the polar equation
r=
[6 marks]


3 −3
6


8. Determine the eigenvalues of the matrix A =  0
2 −8 
0
0 −2
 
3
and show that
 
1
is an eigenvector of A.
0
[5 marks]
9. Find the intersection points between the cardioid r = 2(1 + cos θ) and
π
the line θ = .
3
[5 marks]
1
1
1
=
+ .
z
z1 z2
Give your answer in the form of a + bi. Hence, find the modulus and
argument of z, such that −π ≤ arg(z) ≤ π.
10. Given that z1 = 1 + i and z2 = −2 + 3i, find z such that
[6 marks]
3
SSCE 1693
Part B
(45%)
Answer Any Three Questions.
11. Given the lines l1 and l2 ;
l1 : x = 1 + 2t, y = −1 + t, z = 2 + 4t
l2 :
x+2
−y
=
= z + 1.
4
3
(a) Show that l1 and l2 are skewed by showing that they do not
intersect and not parallel.
[5 marks]
(b) Find the equation of the plane containing the line l1 and parallel
to the plane containing the line l2 . Hence obtain the shortest
distance between the lines l1 and l2 .
[6 marks]
(c) Find the acute angle between the line l1 and the plane
3x + 5y + 4z = 6.
12. (a) Given w = 1 −
[4 marks]
√
3 i,
(i) express w in polar form.
[2 marks]
√
5
(ii) find all the roots of z = 1 − 3 i in the form of a + i b. Show
all the roots on an Argand diagram.
[6 marks]
(b) Use de Moivre’s theorem to show that
sin 3 θ = 3 sin θ − 4 sin3 θ.
Hence solve 24x − 32x3 = 1.
[7 marks]
4
SSCE 1693
13. Given that r2 = 16 cos 2 θ.
(a) Test the symmetries of the above equation.
[3 marks]
(b) Construct a table for (r, θ) with appropriate values and sketch the
graph of r2 = 16 cos 2 θ. (Use the polar grid provided)
[4 marks]
(c) Sketch the graph r = 2 cos θ on the same diagram.
[2 marks]
(d) Find the intersection points between the curves r2 = 16 cos 2 θ
and r = 2 cos θ.
[6 marks]
14. (a) Use Gauss Jordan elimination method solve the system of linear
equations given by AX = B where




x
1 −1
1
 


A= 2
1 −3 , X =  y , and
z
−1
4 −6


4


B =  2 .
−10
Comment on the solutions of the above system of linear equations.
Is the solution a line or a plane?.
[8 marks]
(b) Use the adjoint method to obtain the inverse matrix of P and
hence solve the system of linear equations given by P X = C
where




3 −3
6
x


 
P = 3
2 −8 , X =  y , and
1
2 −2
z
5


−11


C =  11 .
11
[7 marks]
FORMULA
SSCE 1693
Hyperbolic
Trigonometric
2
2
cos x + sin x = 1
1 + tan2 x = sec2 x
cot2 x + 1 = cosec2 x
sin(x ± y) = sin x cos y ± cos x sin y
ex − e−x
2
ex + e−x
cosh x =
2
2
2
cosh x − sinh x = 1
1 − tanh2 x = sech2 x
coth2 x − 1 = cosech2 x
sinh(x ± y) = sinh x cosh y ± cosh x sinh y
cosh(x ± y) = cosh x cosh y ± sinh x sinh y
tanh x ± tanh y
tanh(x ± y) =
1 ± tanh x tanh y
sinh 2x = 2 sinh x cosh x
cosh 2x = cosh2 x + sinh2 x
= 2 cosh2 x − 1
= 1 + 2 sinh2 x
2 tanh x
tanh 2x =
1 + tanh2 x
sinh x =
cos(x ± y) = cos x cos y ∓ sin x sin y
tan x ± tan y
tan(x ± y) =
1 ∓ tan x tan y
sin 2x = 2 sin x cos x
cos 2x = cos2 x − sin2 x
= 2 cos2 x − 1
= 1 − 2 sin2 x
2 tan x
tan 2x =
1 − tan2 x
2 sin x cos y = sin(x + y) + sin(x − y)
2 sin x sin y = − cos(x + y) + cos(x − y)
2 cos x cos y = cos(x + y) + cos(x − y)
Inverse Hyperbolic
√
sinh x = ln(x + x2√+ 1), −∞ < x < ∞
2
cosh−1 x = ln(x
+ x− 1), x ≥ 1
1+x
1
, −1 < x < 1
tanh−1 x = ln
2
1−x
Logarithm
−1
ax = ex ln a
logb x
loga x =
logb a
6
FORMULA
SSCE 1693
Differentiations
Differentiations
d
[k] = 0,
dx
k
stant.
Integrations
Z
k dx = kx + C,
con-
k constant.
Z
d n
[x ] = nxn−1
dx
d x
[e ] = ex .
dx
Z
d
1
[ln |x|] = .
dx
x
Z
d
[cos x]
dx
− sin x.
Z
xn+1
+ C,
xn dx =
n+1
n 6= −1.
x
x
e dx = e + C.
=
dx
= ln |x| + C.
x
d
[sec x]
dx
sec x tan x.
d
[sinh x]
dx
cosh x.
d
[tanh x]
dx
sin x dx = − cos x + sech2 x.
C.
d
[coth x]
dx
=
Z
2
d
−cosech
x.
[sin x] = cos x.
cos x dx = sin x + C.
dx
d
[sech x]
sec x dx = tan x + C.
dx
=
Z
−sech x tanh x.
= cosec2 x dx
=
d
[cosech x]
− cot x + C.
dx
=
−cosech x coth x.
d
[tan x] = sec2 x.
dx
d
[cot x]
dx
−cosec2 x.
Z
Z
=
sec x tan x dx = sec x +
C.
Z
d
[cosec x]
dx
=
−cosec x cot x.
d
[cosh x]
dx
sinh x.
Integrations
cosec x cot x dx
= −cosec x + C.
Z
=
sinh x dx = cosh x + C.
Z
cosh x dx = sinh x + C.
=
Z
sech2 x dx = tanh x + C.
Z
cosech2 x dx
=
− coth x + C.
Z
2
7
=
sech x tanh x dx
= −sech x + C.
Z
cosech x coth x dx
= −cosech x +
C.
FORMULA
SSCE 1693
Differentiations of Inverse Functions
1
d
[sin−1 x] = √
, |x| < 1.
dx
1 − x2
d
−1
[cos−1 x] = √
, |x| < 1.
dx
1 − x2
d
1
.
[tan−1 x] =
dx
1 + x2
Integrations Resulting in Inverse Functions
Z
d
−1
[cot−1 x] =
.
dx
1 + x2
d
1
√
[sec−1 x] =
, |x| > 1.
dx
|x| x2 − 1
d
−1
√
, |x| > 1.
[cosec−1 x] =
dx
|x| x2 − 1
d
1
.
[sinh−1 x] = √
2
dx
x +1
d
1
[cosh−1 x] = √
, |x| > 1.
dx
x2 − 1
d
1
[tanh−1 x] =
, |x| < 1.
dx
1 − x2
dx
= tan−1 (x) + C.
1 + x2
Z
dx
√
= sec−1 (x) + C.
|x| x2 − 1
Z
Z
Z
d
−1
[sech−1 x] = √
, 0 < x < 1.
dx
x 1 − x2
d
−1
√
[cosech−1 x] =
, x 6= 0.
dx
|x| 1 + x2
8
dx
= sin−1 (x) + C.
1 − x2
Z
Z
d
1
[coth−1 x] =
, |x| > 1.
dx
1 − x2
√
√
dx
= sinh−1 (x) + C.
x2 + 1
√
dx
= cosh−1 (x) + C, x > 0.
x2 − 1
1
dx
= ln
1 − x2
2
1
dx
= ln
2
x −1
2
1+x
1−x
x−1
x+1
+ C, |x| < 1.
+ C, |x| > 1.
Z
dx
√
= −sech−1 (x) + C, |x| < 1.
x 1 − x2
Z
dx
√
= −cosech−1 x + C, x 6= 0.
2
|x| 1 + x