M2 DSE 2013

Q1
--

d

d
x

(
s
i
n
2
x
)

{\displaystyle \frac{d}{dx}(sin2x)}

=

lim

Δ
x
→
0

2
s
i
n

[
2
(
x
Δ
x
)
−
2
x
]

2

c
o
s

[
2
(
x
Δ
x
)
2
x
]

2

Δ
x

{\displaystyle =\lim\_{\Delta x\to 0} \frac{2sin\frac{[2(x+\Delta x)-2x]}{2}cos\frac{[2(x+\Delta x)+2x]}{2}}{\Delta x}}

=

lim

Δ
x
→
0

2

s
i
n
Δ
x

Δ
x

c
o
s
(
2
x
2
Δ
x
)

{\displaystyle =\lim\_{\Delta x\to 0} 2\frac{sin\Delta x}{\Delta x}cos(2x+2\Delta x)}

=
2
⋅
(
1
)
⋅
c
o
s
[
2
x
2
⋅
(
0
)
]

{\displaystyle =2\cdot(1)\cdot cos[2x+2\cdot(0)]}

=
2
c
o
s
2
x

{\displaystyle =2cos2x}

Q2
--

(
1
a
x

)
n

{\displaystyle (1+ax)^n}

=
1
n
a
x

n
(
n
−
1
)

2

a
2

x
2

.
.
.

{\displaystyle =1+nax+\frac{n(n-1)}{2}a^2x^2+...}

The coefficients of

x

{\textstyle x}

 and

x
2

{\textstyle x^2}

 are 20 and 180 respectively. So we have

{

n
# a
−
20

n
(
n
−
1
)

2

a
2

=
180

{\displaystyle \begin{cases} na=-20 & \\
\frac{n(n-1)}{2}a^2=180 \end{cases}}

On solving, we have

{

# a
−
2

# n
10

{\displaystyle \begin{cases} a=-2 \\
n=10 \end{cases}}

Q3
--

Let

P
(
n
)

{\textstyle P(n)}

 be the proposition

"

1

1

1
×
4

1

4
×
7

1

7
×
10

.
.
.

1

(
3
n
−
2
)
(
3
n
1
)

=

4
n
1

3
n
1

{\textstyle 1+\frac{1}{1\times 4}+\frac{1}{4\times 7}+\frac{1}{7\times 10}+...+\frac{1}{(3n-2)(3n+1)}=\frac{4n+1}{3n+1}}

for all positive integers

n

{\textstyle n}

."

For

# n
1

{\displaystyle n=1}

,

L.S.

=
1

1

1
×
4

=

5
4

{\displaystyle =1+\frac{1}{1\times4}=\frac{5}{4}}

R.S.

=

4
⋅
(
1
)
1

3
⋅
(
1
)
1

=

5
4

{\displaystyle =\frac{4\cdot(1)+1}{3\cdot(1)+1}=\frac{5}{4}}

L.S.=R.S.

⇒
P
(
1
)

{\displaystyle \Rightarrow P(1)}

 is true.
Next, suppose

P
(
n
)

{\displaystyle P(n)}

 is true for some positive integers

k

{\displaystyle k}

,

i.e.
"

1

1

1
×
4

1

4
×
7

1

7
×
10

.
.
.

1

(
3
k
−
2
)
(
3
k
1
)

=

4
k
1

3
k
1

{\textstyle 1+\frac{1}{1\times 4}+\frac{1}{4\times 7}+\frac{1}{7\times 10}+...+\frac{1}{(3k-2)(3k+1)}=\frac{4k+1}{3k+1}}

"

Then for

# n
k
1

{\displaystyle n=k+1}

,

L.S.

=
1

1

1
×
4

1

4
×
7

1

7
×
10

.
.
.

1

(
3
k
−
2
)
(
3
k
1
)

1

[
3
(
k
1
)
−
2
]
[
3
(
k
1
)
1
]

{\textstyle =1+\frac{1}{1\times 4}+\frac{1}{4\times 7}+\frac{1}{7\times 10}+...+\frac{1}{(3k-2)(3k+1)}+\frac{1}{[3(k+1)-2][3(k+1)+1]}}

=

4
k
1

3
k
1

1

(
3
k
1
)
(
3
k
4
)

{\displaystyle =\frac{4k+1}{3k+1}+\frac{1}{(3k+1)(3k+4)}}

=

4
k
1

3
k
1

1

(
3
k
1
)
(
3
k
4
)

{\displaystyle =\frac{4k+1}{3k+1}+\frac{1}{(3k+1)(3k+4)}}

=

(
4
k
1
)
(
3
k
4
)
1

(
3
k
1
)
(
3
k
4
)

{\displaystyle =\frac{(4k+1)(3k+4)+1}{(3k+1)(3k+4)}}

=

12

k
2

19
k
5

(
3
k
1
)
(
3
k
4
)

{\displaystyle =\frac{12k^2+19k+5}{(3k+1)(3k+4)}}

=

(
3
k
1
)
(
4
k
5
)

(
3
k
1
)
(
3
k
4
)

{\displaystyle =\frac{(3k+1)(4k+5)}{(3k+1)(3k+4)}}

=

4
k
5

3
k
4

{\displaystyle =\frac{4k+5}{3k+4}}

R.S.

=

4
(
k
1
)
1

3
(
k
1
)
1

{\textstyle =\frac{4(k+1)+1}{3(k+1)+1}}

=

4
k
5

3
k
4

{\displaystyle =\frac{4k+5}{3k+4}}

L.S.=R.S.

⇒
P
(
k
1
)

{\textstyle \Rightarrow P(k+1)}

 is true.

By the principle of mathematical induction,

P
(
n
)

{\textstyle P(n)}

 is true for all positive integers

n

{\displaystyle n}

.

Q4
--

a)

d
y

d
x

=

e
x

−
1

{\displaystyle \frac{dy}{dx}=e^x-1}

∫

d
y

d
x

d
# x
∫

e
x

−
1
d
x

{\displaystyle \int \frac{dy}{dx} dx=\int e^x-1 dx}

# y

e
x

−
x
C

{\displaystyle y=e^x-x+C}

At

(
1
,
e
)

{\textstyle (1,e)}

,

(
e
# )

e

(
1
)

−
(
1
)
C
⇒
# C
1

{\displaystyle (e)=e^{(1)}-(1)+C \Rightarrow C=1}

Therefore

# y

e
x

−
x
1

{\textstyle y=e^x-x+1}

 is the required equation.

b)
At

# x
0

{\textstyle x=0}

,

# y

e

(
0
)

−
(
0
)
# 1
2

{\displaystyle y=e^{(0)}-(0)+1=2}

d
y

d
x

# x
0

=

e

(
0
)

−
# 1
0

{\displaystyle \frac{dy}{dx}|\_{x=0}=e^{(0)}-1=0}

So the slope of the required tangent has a slope of 0 and an y-intercept of 2.
Therefore

# y
2

{\textstyle y=2}

 is the required tangent.

Q5
--

a)

Since the sign of

f
′

(
x
)

{\textstyle f'(x)}

 changes from

{\displaystyle +}

 to

−

{\displaystyle -}

 as

x

{\textstyle x}

 passes through 0 in the positive direction, a maximum point occurs at

# x
0

{\textstyle x=0}

.

Since the sign of

f
″

(
x
)

{\textstyle f''(x)}

 changes as

x

{\displaystyle x}

 passes through 1 and -1, inflexion points occur at

# x
1

{\textstyle x=1}

 and

# x
−
1

{\displaystyle x=-1}

.

At

# x
0

{\textstyle x=0}

,

f
(
0
# )

3
−
3
⋅
(
0

)
2

3
(
0

)
2

=
1

{\displaystyle f(0)=\frac{3-3\cdot(0)^2}{3+(0)^2}=1}

At

# x
1

{\textstyle x=1}

,

f
(
1
# )

3
−
3
⋅
(
1

)
2

3
(
1

)
2

=
0

{\displaystyle f(1)=\frac{3-3\cdot(1)^2}{3+(1)^2}=0}

At

# x
−
1

{\textstyle x=-1}

,

f
(
−
1
# )

3
−
3
⋅
(
−
1

)
2

3
(
−
1

)
2

=
0

{\displaystyle f(-1)=\frac{3-3\cdot(-1)^2}{3+(-1)^2}=0}

So a maximum point occurs at

(
0
,
1
)

{\textstyle (0,1)}

 and two inflexion points occur at

(
1
,
0
)

{\textstyle (1,0)}

 and

(
−
1
,
0
)

{\textstyle (-1,0)}

.

b)

Since

x
2

≥
0

{\textstyle x^2\geq 0}

 for any real number

x

{\textstyle x}

,

x
2

3
≥
3

{\textstyle x^2+3\geq 3}

 for any real number

x

{\textstyle x}

.

f
(
x
)

{\displaystyle f(x)}

 has no vertical asymptote.

Since

f
(
x
# )

3
−
3

x
2

3
3

x
2

=

−
3
(

x
2

3
)
12

3

x
2

=
−
3

12

3

x
2

{\textstyle f(x)=\frac{3-3x^2}{3+3x^2}=\frac{-3(x^2+3)+12}{3+x^2}=-3+\frac{12}{3+x^2}}

,

lim

x
→
±
∞

f
(
x
# )

lim

x
→
±
∞

−
3

12

3

x
2

=
−
3
(
0
# )
−
3

{\displaystyle \lim\_{x\to \pm\infty} f(x)=\lim\_{x\to \pm\infty} -3+\frac{12}{3+x^2}=-3+(0)=-3}

.
So

# y
−
3

{\textstyle y=-3}

 is a horizontal asymptote of

f
(
x
)

{\displaystyle f(x)}

.

No oblique asymptote.

c)

To be drawn

Q6
--

a)

Required area

=

∫
0
4

−

x
2

2

2
x
4
−
4
d
x

∫
4
5

4
−
(
−

x
2

2

2
x
4
−
4
)
d
x

{\displaystyle =\int\_0^4 -\frac{x^2}{2}+2x+4-4 dx+\int\_4^5 4-(-\frac{x^2}{2}+2x+4-4) dx}

=
−

x
3

6

0
4

x
2

0
4

x
3

6

4
5

−

x
2

4
5

=
−

64
6

16

125
6

−

64
6

−
25
16

{\displaystyle =-\frac{64}{6}+16+\frac{125}{6}-\frac{64}{6}-25+16}

=
−

13
2

{\displaystyle =-\frac{13}{2}}

 sq. units
b)

Required volume

=

π

∫
0
4

(
−

x
2

2

2
x
4
−
4

)
2

d
x
π

∫
4
5

[
4
−
(
−

x
2

2

2
x
4
)

]
2

d
x

{\displaystyle =|\pi \int\_0^4 (-\frac{x^2}{2}+2x+4-4)^2 dx+\pi \int\_4^5 [4-(-\frac{x^2}{2}+2x+4)]^2 dx|}

=

π

∫
0
4

(
−

x
2

2

2
x

)
2

d
x
π

∫
4
5

(

x
2

2

−
2
x
)

]
2

d
x

{\displaystyle =|\pi \int\_0^4 (-\frac{x^2}{2}+2x)^2 dx+\pi \int\_4^5 (\frac{x^2}{2}-2x)]^2 dx|}

=

π

∫
0
5

(
−

x
2

2

2
x

)
2

d
x

{\displaystyle =|\pi \int\_0^5 (-\frac{x^2}{2}+2x)^2 dx|}

=

π

∫
0
5

x
4

4

−
2

x
3

4

x
2

d
x

{\displaystyle =|\pi \int\_0^5 \frac{x^4}{4}-2x^3+4x^2 dx|}

=

π
(

x
5

20

−

x
4

2

x
3

4

)

0
5

{\displaystyle =|\pi (\frac{x^5}{20}-\frac{x^4}{2}+\frac{x^3}{4})|\_0^5|}

=

123
π

12

{\displaystyle =\frac{123\pi}{12}}

 cu. units
Q7
--

a)

t
a
n
x

{\displaystyle tanx}

=

s
i
n
x

c
o
s
x

{\displaystyle =\frac{sinx}{cosx}}

=

s
i
n
x

c
o
s
x

×

2
c
o
s
x

2
c
o
s
x

{\displaystyle =\frac{sinx}{cosx}\times \frac{2cosx}{2cosx}}

=

2
s
i
n
x
c
o
s
x

2
c
o

s
2

x

{\displaystyle =\frac{2sinxcosx}{2cos^2x}}

=

s
i
n
2
x

1
c
o
s
2
x

{\displaystyle =\frac{sin2x}{1+cos2x}}

b)

t
a
n
y

{\displaystyle tany}

=

s
i
n
2
y

1
c
o
s
2
y

{\displaystyle =\frac{sin2y}{1+cos2y}}

 (By (a))

=

s
i
n
2
y

1
c
o
s
2
y

×

s
e
c
2
y

s
e
c
2
y

{\displaystyle =\frac{sin2y}{1+cos2y}\times \frac{sec2y}{sec2y}}

=

t
a
n
2
y

s
e
c
2
y
1

{\displaystyle =\frac{tan2y}{sec2y+1}}

=

s
i
n
4
y

1
c
o
s
4
y

s
e
c
2
y
1

{\displaystyle =\frac{\frac{sin4y}{1+cos4y}}{sec2y+1}}

 (By (a))

=

s
i
n
4
y

(
1
c
o
s
4
y
)
(
s
e
c
2
y
1
)

×

s
e
c
4
y

s
e
c
4
y

{\displaystyle =\frac{sin4y}{(1+cos4y)(sec2y+1)}\times \frac{sec4y}{sec4y}}

=

t
a
n
4
y

(
s
e
c
4
y
1
)
(
s
e
c
2
y
1
)

{\displaystyle =\frac{tan4y}{(sec4y+1)(sec2y+1)}}

=

s
i
n
8
y

1
c
o
s
8
y

(
s
e
c
4
y
1
)
(
s
e
c
2
y
1
)

{\displaystyle =\frac{\frac{sin8y}{1+cos8y}}{(sec4y+1)(sec2y+1)}}

 (By (a))

=

s
i
n
8
y

(
1
c
o
s
8
y
)
(
s
e
c
4
y
1
)
(
s
e
c
2
y
1
)

×

c
o
s
2
y
c
o
s
4
y

c
o
s
2
y
c
o
s
4
y

{\displaystyle =\frac{sin8y}{(1+cos8y)(sec4y+1)(sec2y+1)}\times \frac{cos2ycos4y}{cos2ycos4y}}

=

s
i
n
8
y
c
o
s
2
y
c
o
s
4
y

(
1
c
o
s
8
y
)
(
1
c
o
s
4
y
)
(
1
c
o
s
2
y
)

{\displaystyle =\frac{sin8ycos2ycos4y}{(1+cos8y)(1+cos4y)(1+cos2y)}}

Q8
--

# M

[

1

k

0

0

1

1

k

0

0

]

,
k
≠
0

{\displaystyle M=\begin{bmatrix} 1 & k & 0 \\ 0 & 1 & 1 \\ k & 0 & 0 \end{bmatrix}, k \ne 0 }

a)

d
e
t
# M

{\displaystyle detM=}

1

1

1

0

0

−
k

0

1

k

0

0

0

1

k

0

{\displaystyle 1 \begin{vmatrix} 1 & 1 \\ 0 & 0 \end{vmatrix} - k \begin{vmatrix} 0 & 1 \\ k & 0 \end{vmatrix} + 0 \begin{vmatrix} 0 & 1 \\ k & 0 \end{vmatrix}}

=

k
2

{\displaystyle =k^2}

M

−
1

{\displaystyle M^{-1}}

=

1

d
e
t
M

[

0

k

−
k

0

0

k
2

k

−
1

1

]

T

{\displaystyle =\frac{1}{detM}{\begin{bmatrix} 0 & k & -k \\ 0 & 0 & k^2 \\ k & -1 & 1 \end{bmatrix}}^T}

=

1

(

k
2

)

[

0

0

k

k

0

−
1

−
k

k
2

1

]

{\displaystyle =\frac{1}{(k^2)}\begin{bmatrix} 0 & 0 & k \\ k & 0 & -1 \\ -k & k^2 & 1 \end{bmatrix}}

b)

M

[

x

1

z

]

=

[

2

2

1

]

{\displaystyle M\begin{bmatrix} x \\ 1 \\ z \end{bmatrix}=\begin{bmatrix} 2 \\ 2 \\ 1 \end{bmatrix}}

[

1

k

0

0

1

1

k

0

0

]

[

x

1

z

]

=

[

2

2

1

]

{\displaystyle \begin{bmatrix} 1 & k & 0 \\ 0 & 1 & 1 \\ k & 0 & 0 \end{bmatrix} \begin{bmatrix} x \\ 1 \\ z \end{bmatrix}=\begin{bmatrix} 2 \\ 2 \\ 1 \end{bmatrix}}

{

x
# k
2
.
.
.
.
.
.
(
1
)

1
# z
2
.
.
.
.
.
.
(
2
)

k
# x
1
.
.
.
.
.
.
(
3
)

{\displaystyle \begin{cases} x+k=2 ......(1) & \\
1+z=2 ......(2) & \\
kx=1 ......(3) \end{cases}}

On solving (1) and (3), we have

# k
1

{\textstyle k=1}

.

Q9
--

a)

[

1

−
a

1

2

2

1
−
2
a

2
−
b

a
4

3

1
−
3
a

3
−
a
b

4

]

∼

[

1

−
a

1

2

0

1

−
b

a

0

1

−
a
b

−
2

]

{\displaystyle \left[\begin{array}{rrr|r}
1 & -a & 1 & 2 \\
2 & 1-2a & 2-b & a+4 \\
3 & 1-3a & 3-ab & 4
\end{array}\right]\sim \left[\begin{array}{rrr|r}
1 & -a & 1 & 2 \\
0 & 1 & -b & a \\
0 & 1 & -ab & -2
\end{array}\right]}

We require the system to have infinitely solutions, i.e.

{

# a
−
2
.
.
.
.
.
.
(
1
)

−
# b
−
a
b
.
.
.
.
.
.
(
2
)

{\displaystyle \begin{cases} a=-2 ......(1) & \\
-b=-ab ......(2) & \end{cases}}

From (2),

# a
1

{\displaystyle a = 1}

 or

# b
0

{\displaystyle b=0}

, but since we have already required

# a
−
2

{\displaystyle a=-2}

, we reject the solution

# a
1

{\displaystyle a = 1}

.

Therefore, for infinitely many solutions we are left with

{

# a
−
2

# b
0

{\displaystyle \begin{cases} a=-2 & \\
b=0 & \end{cases}}

b)
Since

(
E
)

{\displaystyle (E)}

 has infinitely many solutions,

# a
−
2

{\displaystyle a=-2}

 and

# b
0

{\displaystyle b=0}

.

[

1

−
a

1

2

2

1
−
2
a

2
−
b

a
4

3

1
−
3
a

3
−
a
b

4

]

∼

[

1

−
a

1

2

0

1

−
b

a

0

1

−
a
b

−
2

]

{\displaystyle \left[\begin{array}{rrr|r}
1 & -a & 1 & 2 \\
2 & 1-2a & 2-b & a+4 \\
3 & 1-3a & 3-ab & 4
\end{array}\right]\sim \left[\begin{array}{rrr|r}
1 & -a & 1 & 2 \\
0 & 1 & -b & a \\
0 & 1 & -ab & -2
\end{array}\right]}

(From (a))

∼

[

1

−
(
−
2
)

1

2

0

1

−
(
0
)

(
−
2
)

0

1

−
(
−
2
)
(
0
)

−
2

]

∼

[

1

2

1

2

0

1

0

−
2

0

1

0

−
2

]

{\displaystyle \sim \left[\begin{array}{rrr|r}
1 & -(-2) & 1 & 2 \\
0 & 1 & -(0) & (-2) \\
0 & 1 & -(-2)(0) & -2
\end{array}\right]\sim \left[\begin{array}{rrr|r}
1 & 2 & 1 & 2 \\
0 & 1 & 0 & -2 \\
0 & 1 & 0 & -2
\end{array}\right]}

∼

[

1

2

1

2

0

1

0

−
2

0

0

0

0

]

⇒

{

# y
−
2
−
t

# x
2
−
t
−
2
(
−
2
−
t
# )
6
t

{\displaystyle \sim \left[\begin{array}{rrr|r}
1 & 2 & 1 & 2 \\
0 & 1 & 0 & -2 \\
0 & 0 & 0 & 0
\end{array}\right]\Rightarrow \begin{cases} z=t & \\
y=-2-t & \\
x=2-t-2(-2-t)=6+t & \end{cases}}

 for any real number t.
## Q10

a)

O
N

→

{\displaystyle \overrightarrow{ON}}

=

k

k
1

O
A

→

1

k
1

O
B

→

{\displaystyle =\frac{k}{k+1}\overrightarrow{OA}+\frac{1}{k+1}\overrightarrow{OB}}

=

k

k
1

(
2

ı
^

)

1

k
1

(

ı
^

2

ȷ
^

)

{\displaystyle =\frac{k}{k+1}(2\hat{\imath})+\frac{1}{k+1}(\hat{\imath}+2\hat{\jmath})}

=

2
k
1

k
1

ı
^

2

k
1

ȷ
^

{\displaystyle =\frac{2k+1}{k+1}\hat{\imath}+\frac{2}{k+1}\hat{\jmath}}

b)

M
B

→

{\displaystyle \overrightarrow{MB}}

=

O
B

→

−

O
M

→

{\displaystyle =\overrightarrow{OB}-\overrightarrow{OM}}

=

O
B

→

−

1
2

O
A

→

{\displaystyle =\overrightarrow{OB}-\frac{1}{2}\overrightarrow{OA}}

=
(

ı
^

2

ȷ
^

)
−
[

1
2

(
2

ı
^

)
]

{\displaystyle =(\hat{\imath}+2\hat{\jmath})-[\frac{1}{2}(2\hat{\imath})]}

=
2

ȷ
^

{\displaystyle =2\hat{\jmath}}

O
A

→

⋅

M
B

→

=
0

{\displaystyle \overrightarrow{OA}\cdot\overrightarrow{MB}=0}

 since

ı
^

⊥

ȷ
^

{\displaystyle \hat{\imath}\perp\hat{\jmath}}

⇒
∠
P
M
# A

90

∘

{\displaystyle \Rightarrow \angle PMA=90^{\circ}}

∠
A
N
O

{\displaystyle \angle ANO}

=

180

∘

−
∠
P
M
A

{\displaystyle =180^{\circ}-\angle PMA}

 (opp.

∠

{\displaystyle \angle}

, cyclic quad.)

=

180

∘

−
(

90

∘

# )

90

∘

{\displaystyle =180^{\circ}-(90^{\circ})=90^{\circ}}

⇒

O
N

→

⊥

A
B

→

⇒

O
N

→

⋅

A
B

→

=
0

{\displaystyle \Rightarrow \overrightarrow{ON}\perp\overrightarrow{AB}\Rightarrow \overrightarrow{ON}\cdot\overrightarrow{AB}=0}

A
B

→

{\displaystyle \overrightarrow{A B}}

=

O
B

→

−

O
A

→

{\displaystyle =\overrightarrow{OB}-\overrightarrow{OA}}

=
(

ı
^

2

ȷ
^

)
−
(
2

ı
^

)

{\displaystyle =(\hat{\imath}+2\hat{\jmath})-(2\hat{\imath})}

=
−

ı
^

2

ȷ
^

{\displaystyle =-\hat{\imath}+2\hat{\jmath}}

O
N

→

⋅

A
B

→

=
0

{\displaystyle \overrightarrow{ON}\cdot\overrightarrow{AB}=0}

(

2
k
1

k
1

ı
^

2

k
1

ȷ
^

)
⋅
(
−

ı
^

2

ȷ
^

# )
0

{\displaystyle (\frac{2k+1}{k+1}\hat{\imath}+\frac{2}{k+1}\hat{\jmath})\cdot(-\hat{\imath}+2\hat{\jmath})=0}

−

2
k
1

k
1

4

k
1

=
0

{\displaystyle -\frac{2k+1}{k+1}+\frac{4}{k+1}=0}

3
−
2
# k
0

{\displaystyle 3-2k=0}

# k

3
2

{\displaystyle k=\frac{3}{2}}

## Q11

a)

d

d
θ

[
ln
⁡
(
s
e
c
θ
t
a
n
θ
)
]

{\displaystyle \frac{d}{d\theta}[\ln(sec\theta+tan\theta)]}

=

1

s
e
c
θ
t
a
n
θ

s
e
c
θ
t
a
n
θ
s
e

c
2

θ

{\displaystyle =\frac{1}{sec\theta+tan\theta}sec\theta tan\theta+sec^2\theta}

=

s
e
c
θ
t
a
n
θ

s
e
c
θ
t
a
n
θ

s
e
c
θ

{\displaystyle =\frac{sec\theta+tan\theta}{sec\theta+tan\theta}sec\theta}

=
s
e
c
θ

{\displaystyle =sec\theta}

Therefore

∫
s
e
c
θ
d
# θ
∫

d

d
θ

ln
⁡
(
s
e
c
θ
t
a
n
θ
)
d
θ
# C
ln
⁡
(
s
e
c
θ
t
a
n
θ
)
C

{\displaystyle \int sec\theta d\theta=\int\frac{d}{d\theta}\ln(sec\theta+tan\theta)d\theta+C=\ln(sec\theta+tan\theta)+C}

b)
i)

∫

d
u

u
2

−
1

{\displaystyle \int \frac{du}{\sqrt{u^2-1}}}

Let

# u
s
e
c
θ

{\displaystyle u=sec\theta}

 for

0
≤
θ
≤

π
2

{\displaystyle 0\le\theta\le\frac{\pi}{2}}

, so that

d
# u
s
e
c
θ
t
a
n
θ
d
θ

{\displaystyle du=sec\theta tan\theta d\theta}

.

∫

d
u

u
2

−
1

{\displaystyle \int \frac{du}{\sqrt{u^2-1}}}

=
∫

s
e
c
θ
t
a
n
θ
d
θ

s
e

c
2

θ
−
1

{\displaystyle =\int \frac{sec\theta tan\theta d\theta}{\sqrt{sec^2\theta-1}}}

=
∫

s
e
c
θ
t
a
n
θ
d
θ

t
a

n
2

θ

{\displaystyle =\int \frac{sec\theta tan\theta d\theta}{\sqrt{tan^2\theta}}}

=
∫
s
e
c
θ

{\displaystyle =\int sec\theta}

=
ln
⁡
(
s
e
c
θ
t
a
n
θ
)
C

{\displaystyle =\ln(sec\theta+tan\theta)+C}

 (By (a))

=
ln
⁡
(
u

u
2

−
1

)
C
,
u
≥
1

{\displaystyle =\ln(u+\sqrt{u^2-1})+C, u\ge 1}

Diagram to be inserted here

ii)

∫
0
1

2
x

4

x
2

x
4

3

d
x

{\displaystyle \int\_0^1 \frac{2x}{\sqrt{4x^2+x^4+3}} dx}

=

∫
0
1

2
x

x
4

4

x
2

4
−
1

d
x

{\displaystyle =\int\_0^1 \frac{2x}{\sqrt{x^4+4x^2+4-1}} dx}

=

∫
0
1

d
(

x
2

2
)

(

x
2

2

)
2

−
1

{\displaystyle =\int\_0^1 \frac{d(x^2+2)}{\sqrt{(x^2+2)^2-1}}}

=
ln
⁡
[
(

x
2

2
)

(

x
2

2

)
2

−
1

]

0
1

{\displaystyle =\ln[(x^2+2)+\sqrt{(x^2+2)^2-1}]|\_0^1}

 (By (b))

=
ln
⁡
[

(
(
1

)
2

2
)

(
(
1

)
2

2

)
2

−
1

(
(
0

)
2

2
)

(
(
0

)
2

2

)
2

−
1

]

{\displaystyle =\ln[\frac{((1)^2+2)+\sqrt{((1)^2+2)^2-1}}{((0)^2+2)+\sqrt{((0)^2+2)^2-1}}]}

=
ln
⁡
(

3
2

2

2

3

)

{\displaystyle =\ln(\frac{3+2\sqrt{2}}{2+\sqrt{3}})}

=
ln
⁡
(

3
2

2

2

3

2
−
s
q
r
t

3

2
−
s
q
r
t

3

)

{\displaystyle =\ln(\frac{3+2\sqrt{2}}{2+\sqrt{3}}\frac{2-sqrt{3}}{2-sqrt{3}})}

=
ln
⁡
[

(
3
2

2

)
(
2
−
s
q
r
t

3

)

4
−
3

]

{\displaystyle =\ln[\frac{(3+2\sqrt{2})(2-sqrt{3})}{4-3}]}

=
ln
⁡
(
4

2

−
3

3

−
2

6

6
)

{\displaystyle =\ln(4\sqrt{2}-3\sqrt{3}-2\sqrt{6}+6)}

c)

# t
t
a
n
ϕ

{\displaystyle t=tan\phi}

# 1
s
e

c
2

ϕ

d
ϕ

d
t

{\displaystyle 1=sec^2\phi\frac{d\phi}{dt}}

d
ϕ

d
t

=
c
o

s
2

ϕ

{\displaystyle \frac{d\phi}{dt}=cos^2\phi}

d
ϕ

d
t

=
(

1

1

t
2

)
2

=

1

1

t
2

{\displaystyle \frac{d\phi}{dt}=(\frac{1}{\sqrt{1+t^2}})^2=\frac{1}{1+t^2}}

So

d
# ϕ

d
t

1

t
2

{\displaystyle d\phi=\frac{dt}{1+t^2}}

∫
0

π
4

t
a
n
ϕ

1
2
c
o

s
2

ϕ

d
ϕ

{\displaystyle \int\_0^{\frac{\pi}{4}} \frac{tan\phi}{\sqrt{1+2cos^2\phi}} d\phi}

=

∫
0
1

t

1

2

1

t
2

(

d
t

1

t
2

)

{\displaystyle =\int\_0^1 \frac{t}{\sqrt{1+\frac{2}{1+t^2}}} (\frac{dt}{1+t^2})}

=

∫
0
1

t
d
t

(
1

t
2

)
2

2
(
1

t
2

)

{\displaystyle =\int\_0^1 \frac{t dt}{\sqrt{(1+t^2)^2+2(1+t^2)}}}

=

∫
0
1

t
d
t

(
1

t
2

)
2

2
(
1

t
2

)
1
−
1

{\displaystyle =\int\_0^1 \frac{t dt}{\sqrt{(1+t^2)^2+2(1+t^2)+1-1}}}

=

∫
0
1

t
d
t

[
(
1

t
2

)
1

]
2

−
1

{\displaystyle =\int\_0^1 \frac{t dt}{\sqrt{[(1+t^2)+1]^2-1}}}

=

1
2

∫
0
1

d
(

t
2

2
)

(

t
2

2

)
2

−
1

{\displaystyle =\frac{1}{2}\int\_0^1 \frac{d(t^2+2)}{\sqrt{(t^2+2)^2-1}}}

=

1
2

ln
⁡
(
4

2

−
3

3

−
2

6

6
)

{\displaystyle =\frac{1}{2}\ln(4\sqrt{2}-3\sqrt{3}-2\sqrt{6}+6)}

 (By (b)(ii))
## Q12

a)

i)

T

{\displaystyle T}

=

x
7

(
40
−
x

)
2

30
2

1.4

{\displaystyle =\frac{x}{7}+\frac{\sqrt{(40-x)^2+30^2}}{1.4}}

=

x
5

(
40
−
x

)
2

900

7

{\displaystyle =\frac{x+5\sqrt{(40-x)^2+900}}{7}}

ii)

d
T

d
x

=

1
5
⋅

1
2

[
(
40
−
x

)
2

900

]

−

1
2

[
2
⋅
(
40
−
x
)
⋅
(
−
1
)
]

7

{\displaystyle \frac{dT}{dx}=\frac{1+5\cdot\frac{1}{2}[(40-x)^2+900]^{-\frac{1}{2}}[2\cdot(40-x)\cdot(-1)]}{7}}

When

T

{\displaystyle T}

 is minimum,

d
T

d
x

=
0

{\displaystyle \frac{dT}{dx}=0}

. So,

(
0
# )
1
5
⋅

1
2

[
(
40
−
x

)
2

900

]

−

1
2

[
2
⋅
(
40
−
x
)
⋅
(
−
1
)
]

{\displaystyle (0)=1+5\cdot\frac{1}{2}[(40-x)^2+900]^{-\frac{1}{2}}[2\cdot(40-x)\cdot(-1)]}

# 0
1
−
5
[
(
40
−
x

)
2

900

]

−

1
2

(
40
−
x
)
]

{\displaystyle 0=1-5[(40-x)^2+900]^{-\frac{1}{2}}(40-x)]}

# 1
5
[
(
40
−
x

)
2

900

]

−

1
2

(
40
−
x
)
]

{\displaystyle 1=5[(40-x)^2+900]^{-\frac{1}{2}}(40-x)]}

(
40
−
x

)
2

# 900
[
5
(
40
−
x
)

]
2

{\displaystyle (40-x)^2+900=[5(40-x)]^2}

[
(
40
−
x

)
2

900
# ]
25
(
40
−
x

)
2

{\displaystyle [(40-x)^2+900]=25(40-x)^2}

24
(
40
−
x

)
2

−
# 900
0

{\displaystyle 24(40-x)^2-900=0}

(
40
−
x

)
2

−

900
24

=
0

{\displaystyle (40-x)^2-\frac{900}{24}=0}

2

x
2

−
160
x
# 3125
0

{\displaystyle 2x^2-160x+3125=0}

# x

160

2
⋅
2

±

(
−
160

)
2

−
4
⋅
(
2
)
⋅
(
3125
)

2
⋅
2

{\displaystyle x=\frac{160}{2\cdot2}\pm\frac{\sqrt{(-160)^2-4\cdot(2)\cdot(3125)}}{2\cdot2}}

# x
40
±

75
2

=
40
±
5

3
2

{\displaystyle x=40\pm\frac{75}{2}=40\pm5\frac{3}{2}}

First derivative test hand-drawn table here

So,

# x
40
5

3
2

{\displaystyle x=40+5\sqrt{\frac{3}{2}}}

 for minimum

T

{\displaystyle T}

.

Q
B

{\displaystyle QB}

=

[
40
−
(
40
5

3
2

)

]
2

30
2

{\displaystyle =\sqrt{[40-(40+5\sqrt{\frac{3}{2}})]^2+30^2}}

 (Pyth. thm.)

=

1875
2

{\displaystyle =\sqrt{\frac{1875}{2}}}

=
25

3
2

2

2

{\displaystyle =25\sqrt{\frac{3}{2}}\frac{\sqrt{2}}{\sqrt{2}}}

=

25

6

2

{\displaystyle =\frac{25\sqrt{6}}{2}}

 m
b)
i)

# M

[

a

b

c

d

]

,
t
r
(
M
)
≡
a
d
.

{\displaystyle M=\begin{bmatrix} a & b \\ c & d \end{bmatrix}, tr(M)\equiv a+d. }

B
A

B

−
1

=

[

1

0

0

3

]

{\displaystyle BAB^{-1}=\begin{bmatrix} 1 & 0 \\ 0 & 3 \end{bmatrix}}

i)

M
# N

[

a

b

c

d

]

[

e

f

g

h

]

=

[

a
e
b
g

a
f
b
h

c
e
d
g

c
f
d
h

]

{\displaystyle MN=\begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} e & f \\ g & h \end{bmatrix}=\begin{bmatrix} ae+bg & af+bh \\ ce+dg & cf+dh \end{bmatrix}}

N
# M

[

e

f

g

h

]

[

a

b

c

d

]

=

[

a
e
f
c

e
b
f
d

g
a
c
h

g
b
h
d

]

{\displaystyle NM=\begin{bmatrix} e & f \\ g & h \end{bmatrix}\begin{bmatrix} a & b \\ c & d \end{bmatrix}=\begin{bmatrix} ae+fc & eb+fd \\ ga+ch & gb+hd \end{bmatrix}}

t
r
(
M
N
# )
a
e
b
g
c
f
d
h

{\displaystyle tr(MN)=ae+bg+cf+dh}

t
r
(
N
M
# )
a
e
f
c
g
b
h
# d
t
r
(
M
N
)

{\displaystyle tr(NM)=ae+fc+gb+hd=tr(MN)}

ii)

t
r
(
B
A

B

−
1

)

{\displaystyle tr(BAB^{-1})}

=
t
r
(
A

B

−
1

B
)

{\displaystyle =tr(AB^{-1}B)}

 (By (a)(i))

=
t
r
(
A
I
)

{\displaystyle =tr(AI)}

=
t
r
(
A
)

{\displaystyle =tr(A)}

Since

t
r
(
B
A

B

−
1

# )
1
# 3
4

{\displaystyle tr(BAB^{-1})=1+3=4}

,

t
r
(
A
# )

{\displaystyle tr(A)=4.}

iii)

d
e
t
(
B
A

B

−
1

)

{\displaystyle det(BAB^{-1})}

=
d
e
t
B
d
e
t
A
d
e
t
(

B

−
1

)

{\displaystyle =detBdetAdet(B^{-1})}

=
d
e
t
B
d
e
t
(

B

−
1

)
d
e
t
A

{\displaystyle =detBdet(B^{-1})detA}

=
d
e
t
(
B

B

−
1

)
d
e
t
A

{\displaystyle =det(BB^{-1})detA}

=
d
e
t
(
I
)
d
e
t
A

{\displaystyle =det(I)detA}

=
(
1
)
⋅
d
e
t
A

{\displaystyle =(1)\cdot detA}

=
d
e
t
A

{\displaystyle =detA}

Since

d
e
t
(
B
A

B

−
1

# )
1
×
3
−
0
×
# 0
3

{\displaystyle det(BAB^{-1})=1\times3-0\times0=3}

,

d
e
t
# A

{\displaystyle detA=3.}

b)

# C

[

p

q

r

s

]

{\displaystyle C=\begin{bmatrix} p & q \\ r & s \end{bmatrix}}

C

[

x

y

]

=

λ
1

[

x

y

]

{\displaystyle C\begin{bmatrix} x \\ y \end{bmatrix}=\lambda\_1\begin{bmatrix} x \\ y \end{bmatrix}}

 and

C

[

x

y

]

=

λ
2

[

x

y

]

{\displaystyle C\begin{bmatrix} x \\ y \end{bmatrix}=\lambda\_2\begin{bmatrix} x \\ y \end{bmatrix}}

 for

[

x

y

]

≠
0

{\displaystyle \begin{bmatrix} x \\ y \end{bmatrix}\ne0}

,

λ
1

≠

λ
2

{\displaystyle \lambda\_1\ne\lambda\_2}

.
i)

C

[

x

y

]

=

λ
i

[

x

y

]

{\displaystyle C\begin{bmatrix} x \\ y \end{bmatrix}=\lambda\_i\begin{bmatrix} x \\ y \end{bmatrix}}

 for

# i
1
,
2

{\displaystyle i = 1,2}

(
C
−

λ
i

)

[

x

y

]

=

[

0

0

]

{\displaystyle (C-\lambda\_i)\begin{bmatrix} x \\ y \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \end{bmatrix}}

[

p
−
l
a
m
b
d

a
i

q

r

s
−
l
a
m
b
d

a
i

]

[

x

y

]

=

[

0

0

]

{\displaystyle \begin{bmatrix} p-lambda\_i & q \\ r & s-lambda\_i \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \end{bmatrix}}

As we have required

[

x

y

]

{\displaystyle \begin{bmatrix} x \\ y \end{bmatrix}}

 to be non-trivial, i.e.

[

x

y

]

≠
0

{\displaystyle \begin{bmatrix} x \\ y \end{bmatrix}\ne0}

, we must have

p
−

λ
1

q

r

s
−

λ
1

=

p
−

λ
2

q

r

s
−

λ
2

=
0

{\displaystyle \begin{vmatrix} p-\lambda\_1 & q \\ r & s-\lambda\_1 \end{vmatrix}=\begin{vmatrix} p-\lambda\_2 & q \\ r & s-\lambda\_2 \end{vmatrix}=0}

.
ii)

p
−

λ
i

q

r

s
−

λ
i

=
0

{\displaystyle \begin{vmatrix} p-\lambda\_i & q \\ r & s-\lambda\_i \end{vmatrix}=0}

 for

# i
1
,
2

{\displaystyle i = 1,2}

(
p
−

λ
i

)
(
s
−

λ
i

)
−
q
# r
0

{\displaystyle (p-\lambda\_i)(s-\lambda\_i)-qr=0}

p
s
−
p

λ
i

−
s

λ
i

λ
i
2

−
q
# r
0

{\displaystyle ps-p\lambda\_i-s\lambda\_i+\lambda\_i^2-qr=0}

λ
i
2

−
(
p
s
)

λ
i

(
p
s
−
q
r
# )
0

{\displaystyle \lambda\_i^2-(p+s)\lambda\_i+(ps-qr)=0}

Notice that

d
e
t
# C
p
s
−
q
r

{\displaystyle detC=ps-qr}

 and

t
r
# C
p
s

{\displaystyle trC=p+s}

.

So,

λ
i
2

−
t
r
C

λ
i

d
e
t
# C
0

{\displaystyle \lambda\_i^2-trC\lambda\_i+detC=0}

 for

# i
1
,
2

{\displaystyle i = 1,2}

.

λ
1

{\displaystyle \lambda\_ 1}

 and

λ
2

{\displaystyle \lambda\_2}

 are therefore the roots of the equation

λ
2

−
t
r
C
λ
d
e
t
# C
0

{\displaystyle \lambda^2-trC\lambda+detC=0}

.

c)

λ
2

−
t
r
C
λ
d
e
t
# C
0

{\displaystyle \lambda^2-trC\lambda+detC=0}

 (By (b)(ii))

λ
2

−
(
4
)
λ
(
3
# )
0

{\displaystyle \lambda^2-(4)\lambda+(3)=0}

 (By (a)(ii) and (iii))

(
λ
−
1
)
(
λ
−
3
# )
0

{\displaystyle (\lambda-1)(\lambda-3)=0}

# λ
1

{\displaystyle \lambda=1}

 or

3

{\displaystyle 3}