Maths Assignment: Domain, Function, Convergence, and Triangle Analysis

This Maths assignment solution addresses several key concepts. It begins by defining the domain of a function and provides examples. It analyzes the properties of functions, including one-to-one functions, and provides examples to illustrate the concepts. The assignment also covers convergence and divergence of sequences, determining whether given sequences converge or diverge. Additionally, it explores the concept of triangle similarity, proving the similarity of triangles and deriving lengths of line segments within the triangles. The solution provides step-by-step explanations and justifications for each problem, making it a valuable resource for students studying these topics. The assignment solution is contributed by a student to be published on the website Desklib, a platform which provides all the necessary AI based study tools for students.

MATHS
[DATE]

Question 1
(a) Domain
Domain f ( x ) = √ x2−5 x
x2−9
Domain of a given function is the argument value at which the function is well defined and
real.
Non-negative value for radicals
x ←3or 0 ≤ x ≤ 3∨x ≥5
√ f ( x ) =f ( x ) ≥ 0
x2 −5 x
x2 −9 ≥ 0
x ( x−5)
( x−3 ) (x +3) ≥ 0
x ←3∨x=0∨0< x <3∨x=5∨x >5
Merge overlapping intervals
x ←3∨0 ≤ x< 3∨x ≥5
Function is undefined points
x=3 , x=−3
Solution
Domain = x ←3 ,∨0 ≤ x<3∨x ≥ 5
Interval Notation = (−∞,−3 ) ∪ [ 0 , 3 ) ∪¿
(b) If g(x) is an odd function which means it would be surely symmetric about the origin
Also,
f (−x ) =−f ( x )
1

g ( 0 ) would be zero .
Example:
y=x3
y=x
y=sin x
(c) Graph of function
|x|+| y|=1+ x
| y|=1+ x−|x|
It can be seen that for every value of x in domain, there would be two values of y and hence,
it can be said that it is not a function of x.
(d) True
If
an> 0
2

lim
n → ∞
an =l
Let, l<0∧take e=−l
|an−l|<e
|an−l|<−l
an−l<−l
Now,
an< 0 Which comes contradiction and hence, l>0is TRUE.
(e) False
Let an= { 1 if n is odd
−1if nis even
This sequence is bounded but has two limit and thus, it is divergent. Hence, it is FALSE
(f) False
Let
an=−1
n
It is increasing sequence & lim
n → ∞
an =
lim
n → ∞
−1
n =0
Therefore, it is convergent and hence, it is FALSE.
Question 2
(a) Need to prove AED∧BEC are similar tringles
Radius of circle is a which means
OC=OB=OF=OD=a
3

OE=c
CE=b
¿ ACF=θ
In ∆ AED∧∆ BEC
Now,
¿ AED=¿ BEC
¿ ADE=¿ BCE
¿
¿ DAE=¿ EBC
Through, A-A-A similarity,
∆ AED ∆ BECHere, AED∧BEC are similar tringles (Proved).
(b) Need to prove that length of AE and BE are 2 a cos θ−b and a−c
c2=a2+ b2−2 ab cos θ
In ∆ FAC =90 °
cos θ= AC
FC
cos θ= AE+ EC
2 a
4

2 a cos θ=AE+ EC
2 a cos θ=AE+b
AE=2 a cos θ−b (Proved)
BE=OB−OE
BE=a−c(Proved)
In ∆ AED ∆ BEC
AE
BE = DE
CE
2 a cosθ−b
a−c = a+c
b
2 ab cosθ−b2= ( a2−c2 )
c2=a2+ b2−2 ab cos θ (Proved)
Question 3
f ( x )=x3 + x
(a) Graph
It is apparent from the graph that for different values of x, the function would have different
image and hence, function f(x) = one to one
5

(b) f ' (x)
f ( x )=x3 + x
f ' ( x )=3 x2+ 1
Here,
f ' ( x ) >0 at x ∈ R
Here, f (x) is increasing function so f(x) is termed as one to one function.
(c) Let
f ( x 1 )=f ( x 2 )
x 13 +x 1=x 23 + x 2
x 13 +x 1−x 23 −x 2=0
( x 1−x 2 ) + ( x 13−x 23 ) =0
( x 1−x 2 ) ( x 12 + x 1 x 2+x 22 ) + ( x 1−x 2 )=0
( x 1−x 2 ) {( ( x 12 + x 1 x 2+ x 22 )+ 1 ) }=0
6

x 1 , x 2 ∈ R
x 12 +x 1 x 2+ x 22+ 1≠ 0
Here,
( x 1−x 2 )=0
x 1=x 2
f ( x 1 )=f ( x 2 )
x 1=x 2
Hence, it can be concluded that function f(x) is one to one.
(d) Let x 1 ≠ x 2
x 13 ≠ x 23
x 1 , x 2 ∈ R
x 13 + x 1≠ x 23 + x 2
f ( x 1 ) ≠ f ( x 2 )
Now,
x 1 ≠ x 2
f ( x 1 ) ≠ f ( x 2)
For different values of x, the function must possess different values and hence, f(x) would be
one to one.
(e) Graph
f ' ( x )=3 x2+ 1
7

Domain: The function has no undefined points nor domain constraints. Hence, the domain
would be −∞ < x <∞ .
Interval notation: (−∞, ∞)
Range: The set of values of the dependent variable for which a function is defined.
Vertex of 3 x2+1is (0,1)
f ( x ) ≥1
Interval notation: ¿
Question 4
(a) Converge or diverge
an= ( 1+2n )
1+3n
8

∑
n=0
∞ (1+2n )
1+3n
Apply series ratio test
Converge
lim
n →∞
( 1+ 2n )
1+3n =
lim
n → ∞ ( 1
3n + ( 2
3 )
n
)
lim
n → ∞ ( 1
3n +1 )
lim
n → ∞
( 1+2n )
1+3n =0
1 =0
(b) Converge or diverge
an= n!
nn
∑
n=0
∞ n !
nn
Apply series ratio test
Converge
lim
n → ∞
n!
nn =0
(c) Converge or diverge
an=n sin ( 1
n )
Apply series ratio test
∑
n=0
∞
n sin ( 1
n )
9

Diverge
lim
n → ∞
n sin ( 1
n ) =lim
n → ∞
sin ( 1
n )
1
n
¿ lim
n → ∞
−cos ( 1
n )
n2
−1
n2
¿ lim
n → ∞
cos (1
n )
lim
n → ∞
n sin ( 1
n )=1
10

(g)
(h)
11