Let us examine through an example::
QUESTION: There were two cat aspirants Ajay and Vijay. Ajay keeps boasting about his quant percentiles( in 90’s).One day Vijay brought a question to Ajay, it said” There is a function f(x) = ax^^{2} +bx +c and f(4)= 100 . If the coefficients of a,b and c are distinct positive integers, then find the maximum value ( a+b+c) can attain ? “. Can you answer it faster than Ajay?
Options:
1) 82
2) 76
3)79
4)0
SOLUTION:
Ajay started scribbling in his rough paper, let us analyze his solution:
F(x) = ax^{2} +bx + c, what is the functional variable in this? x right !!
so if it is given that f(4) = 100 means that by putting x=4, the function will return a value of 100.
That is f(4)= 16a + 4b + c = 100 , we are good till now right!
But our requirement is finding the maximum value of (a+b+c), so the quantity with the largest coefficient has to be minimised,
16*1 +4*1 + 80 =100 (Keep in mind “positive integers” is mentioned a=1,b=1)
So the maximum value which Ajay got was 1+1+80= 82
The above result is certainly wrong and could fetch you a penalty of 1/3rd marks.
The term “DISTINCT POSITIVE INTEGERS ” is vital.
16 *1 + 4*2 + 76= 100
so maximum = (76+1+2) = 79
Hope the explanation is elaborative. Let’s move to our next segment.
There is a thumb rule for determining whether the given two functions are identical or not.
” The domain of f(x) should be equal to the domain of g(x) .”
Observe the following examples to get a clear scenario of the things.
Example 1: we will consider two fucntions f(x)= root(x^{2}) and g(x) = x
In f(x) for every real value of x , there is a value of f(x), which means x belongs to R
In G(x) for every real value of x, there is a value of g(x), which also means x belongs to R.
In both the cases, the domain is same. Hence, The Functions are “IDENTICAL”
Example2: Now again we have two fucntions f(x)= ln (x^{2} 5x +6) and g(x)= ln(x2) + ln(x3)
The ln values cannot be negative or zero . So we will analyse botgh cases.
x^{2} 5x +6x >0 ⇒ (x2)(x3)>0 ⇒ x belongs (infy,2)U(3,infy)
(x2)>0 and (x3)>0 ⇒ x>2 and x>3 ⇒ x belongs to (2, infy)
In both cases there are different domains ,so the fucntions are “NOT IDENTICAL”.
Questions from functions are always ghastly and require an extra effort over other topics especially in case of students without a mathematical background.
Questions related to greatest integer function have been seen very frequent in past few years in various exams and mocks.
So let us put an effort to unlock the mystery of Greatest Integer Function. Greatest Integer Function, [x] indicated the integral part of x which is nearest and smallest integer to x. It is also known as floor of x.
e.g. [9.638] = 9, [7] = 7, [0.286] = 0, [3.432] = 4, [0.432] = 1
In general, n = x < n+1 , where n is an integer, [x] = n
Now, we can say that for any real number x, x = I + f
where,
I = Integral Part denoted by [x]
f = Fractional Part denoted by {x}
Thus, {x} = x – [x]
Let us see some varied applications::
QUESTION: Find the value of [3/4] + [3/4 + 1/100] + [3/4+ 2/100] +……..+ [3/4 + 99/100].
SOLUTION:
[3/4] + [3/4+ 1/100] + [3/4+ 2/100] +… [3/4+25/100]+[3/4+26/100]+ ? + [3/4+ 99/100]
First of all the number of terms in the series is 100.
[3/4]=[0.75]=0
[3/4 + 1/100]= [0.75 + 0.01]=[0.76]=0
similarly the frist 25 terms will be equal to 0 (as 25th term is [3/4 + 25/100]=[3/4 +1/4]=1)
and from the 26th term each term will be equal to 1.
First 25 terms are 0 and the next 75 terms are 1.
So, sum of all the terms = (25 × 0) + (75 × 1) = 75
Interesting Formula 1 :
Difference between CI and SI for 2 and 3 years
respectively:
(CI)– (SI)= Pa^{2} (for two years)
(CI)– (SI)= Pa^{2} (a + 3) (for three years)
where, a = r/100
Interesting Formula 2:
A principal amount of X times in T years at S.I. It
will become Y times in:
Years = ((Y 1)/(X1)) * T
A principal amount of X times in T years at C.I. It
will become Y times in:
Years = T × n
where n is given by X^{n }= Y
How can you use the remainder theorem to evaluate polynomials?
Let’s first state the remainder theorem. The remainder theorem states the following:
If you divide a polynomial f(x) by (x – h), then the remainder is f(h).
The theorem states that our remainder equals f(h).
Therefore, we do not need to use long division, but just need to evaluate the polynomial when x = h to find the remainder.
A basic example to justify the above statement !
QUESTION: Given that, f(x) = x^{3} + 2x^{2} – 11x – 12
Find the remainder if f(x) is divided by (x – 2)
SOLUTION: The remainder theorem tells us that h = 2, so evaluate f(2). We replace x with 2 in the polynomial as follows:
f(2) = 2^{3} + 2(2)^{2} – 11(2) – 12
We then simplify: f(2) = 8 + 8 – 22 – 12 = 18
Now let us observe a moderate level question application ::
QUESTION: The polynomial p(x) satisfies
p(−x) = −p(x). When p(x) is divided by x − 3 the remainder is 6. Find the remainder when p(x) is divided by x^{2} − 9.
SOLUTION: Now first inference that can be drawn is p(x)=p(x) which implies that it is an odd function.
According to Remainder structure, N = dq+r (N is dividend, d is dividor, q is quotient, r is remainder)
so ,p(x)=(x3)*k+6 ((x3) is dividend, k is divisor, 6 is remainder)
when p(x) mod x^{2}9
p(x)=m*(x^{2}9)+ax+b (as if the degree of divisor is x^{2} ,then remainder will be of lesser degree i.e ax+b)
now at x=3
p(x)=3a+b=6
and x=3 remainder will be 6
p(3)=3a+b=6
2b=0
so b=0 and a=2
remainder= ax+b= 2x
QUESTION: If f(x) = x^{4} + x^{3} + x^{2} + x + 1, where x is a positive integer greater than 1.
What will be the remainder if f(x^5) is divided by f(x) ?
SOLUTION:
f(x) = x^{4} + x^{3} + x^{2} + x + 1.
By replacing x with x5 we get,
f(x^{5}) = x^{20} + x^{15} + x^{10} + x^{5} + 1.
Let’s take x = 2.
f (x) = f(2) = 31.
f (x^{5}) = f(32) = 2^{20} + 2^{15} + 2^{10} + 2^{5}+1.
f(32) mod f(2) = 2^{20} + 2^{15} + 2^{10} + 2^{5}+1 mod 31
= (2^{5})4 + (2^{5})3+(2^{5})2 +(2^{5}) + 1 mod 31
= 1+1+1+1+1
= 5.
NOTE: You can try the same with x=3 and cross verify.But if assume x=1 , we will be lost.
Let us see the application of functions in algebra/Quadratic
QUESTION: If 5 is root of a quadratic function of f(x)=0 and 6*f(2)=f(4)
find another root of f(x)=0
SOLUTION: Firstly, basics on Quadratic eqn , It has only two roots.
One of the root is given to us (ie 5) which means on substituting the value of 5
in the given f(x) it will result a 0.
Now, f(x) = ax^{2} +bx+c = 0
f(2) = 4a +2b +c =0
f(4)= 16a +4b+c=0
using the relation 6 * f(2)= f(4)
24a+12b +6c = 16a+4b+c
8a+8b+5c=0
and 25a+5b+c=0;
we will get a relationship between a and b, a/b = ( 17/117)
Let the other root be y
sum of roots = 5+y = ( b/a)….eqn 2
solving both the eqns y = 32 /17
Hope the explanation is !!
If we take sets and analyze the number of functions or relations that can be obtained, we do stump on some standard results.
Set A has m elements.
Set B has n elements.
(1) If m>n (Number of elements in Set A is greater than the number of elements in Set B)
Number of One to One function =0 ;
If n>m ((Number of elements in Set B is greater than the number of elements in Set A)
Number of One to One function = P(n,m) ;
(2) Number of functions = n^{m} ;
Number of relations = 2^{(m*n)} ;
(3) No of Many to One functions = Number of functions – Number of One to One
=n^{m} P(n,m)
(4)If n>m Number of Onto functions =0 ;
If m>n Number of Onto functions= summation (r = 1 to n) (1)^{(nr)} *C(n,r)*r^{m};.
Number of Functions
A = (a,e,i,o,u), B = ( 1,2,3)
A = (a,e,i), B = (1,2,3,4,5)
QUESTION: If f(x − 1) + f(x + 1) = f(x) and f(2) = 6, f(0) = 1, then what is the value of f(50) ?
1) −7
2) 6
3) 1
4) 7
SOLUTION: Since we know both f(0)=1 and f(2)=6, we can find f(1).
By plugging the value of x=1
f(1) = f(0) + f(2) = 7
f(2) = f(1) + f(3)
f(3) = f(2) – f(1)
= 6 – 7
= −1
Also, f(3) = f(2) + f(4)
f(4) = f(3) – f(2) = −7
Continuing in a similar way, we can find out
f(0) = 1
f(1) = 7
f(2) = 6
f(3) = −1
f(4) = −7
f(5) = −6
f(6) = 1
f(7) = 7 and so on
After every 6 integral values of x, f(x) repeats itself.
f(6) = f(12) = f(18 ) = f(24) = f(30) = f(36) = f(42) = f(48 ) = 1
f(49) = 7
f(50) = 6
QUESTION: If f(x) = 4^{x} / (4^{x} + 2)
Then find the value of f(1/100)+f(2/100)+ …+ f(99/100).
SOLUTION :
In these type of questions , an important observation has to be made regarding the pattern involved in the “required part”
(1/100 +99/100) = 1 and 99/100 = (1 1/100) so f(x) and f(1x) has a certain relation.
f (x) = 4^{x}/4^{x} + 2………(1)
f (1 – x) = 4^{(1 – x)} / 4^{(1 – x)} + 2. ……..(2)
= 4/4 + 2.4^{x}
= 2/2 + 4^{x}.
On adding both the equations (1)and (2), we get
f(x) + f(1 – x) = [4^{x}/4^{x} + 2] + [2/4^{x} + 2 ]
= 1.
Hence a generalisation is obtained,
f(1 / 100) + f(1 – 1/100) = f(1) + f(99/100) = 1.
f (2/100) + f( 1 – 2/100) = f(2) + f(98/100) = 1.
Similarly f(49/100) + f(50/100) = 1.
Hence f(1/100) + f(2/100) + f(3/100)………………………..f(99/100)
= 49.
194 videos156 docs192 tests

Practice Questions for Sets Doc  10 pages 
Important Formulae: Set Theory Doc  9 pages 
Introduction Set Theory Video  11:44 min 
1. What is a function in algebra? 
2. How do you determine if a relation is a function? 
3. What is the difference between a dependent variable and an independent variable in a function? 
4. How do you find the domain of a function? 
5. Can a function have more than one dependent variable? 
194 videos156 docs192 tests

Practice Questions for Sets Doc  10 pages 
Important Formulae: Set Theory Doc  9 pages 
Introduction Set Theory Video  11:44 min 

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