Maths : Functions and Limits Problems

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