ALGEBRA - 6

SNAPSHOT

 Functions – Domain and Range

 Piecewise defined functions
 Composite functions
 Even and Odd function
 Periodic function
 Inverse of a function
 Function as a series
I would like to ask what would you call an expression such 2x 3  3x2 + 5x + 1? Your answer
would most probably be ‘an algebraic expression’ and you would be right. What is x here? A
variable whose value YOU will have to input. So for x = 1, the value of your algebraic
expression would be 5. For x = 2, the value would be 15. If you plot the values against the
input value of the variable x, you would get a graph such as the one shown below:

This is a simple graph on the x-y axis which intersects the y axis at y = 1 for x = 0. There
are some basic observations that we can make about the graph:
 What values can we put in the expression? Clearly, all the real values, i.e. values
which lie on the number line, can be inserted into the expression. To draw the graph
in the x-y plane we need to put values which we CAN find on the x-axis. This graph
can have all the values from  to .
 For every unique value of x, the expression will give a unique value only, i.e. if we
plot the graph completely, at each value of x, we will have only a unique value of y.
There cannot be two values for a single value of x.
 What are the values of the expression we can obtain by putting the values of x?
Clearly, the values we can obtain will also be from  to .

Sketching the graph of an expression

Let’s have a small look at curve sketching here. Let’s try and sketch the graph of the
expression x3 – 4x2 + 1.
There are two ways we can go about it-

Input Values: Probably the best method for curve sketching for a beginner is to put
different values for the variable and check the output values of the expression. The table
below shows some of the values we can keep.

What do we see? That the value of the expression is increasing as x is increasing from
negative to zero, decreases again as x increases from zero to positive, and then increases
again. Although, putting values gives us a fairly good idea about the shape of the graph, it
is a tedious method, especially so when the increase or decrease of the graph occurs at
longer intervals and we do not know how many values to put. To understand the
characteristics of the graph accurately we combine this method with our second method.
Finding the derivatives: You would need a small knowledge of calculus here, that of
dy d2y
finding the first ( ) and second ( ) derivative of the expression. Let us denote first
dx dx 2
and the second derivative as f’(x) and f”(x), respectively. Observe the following rules:

 If the first derivative f' is positive, then the function f is increasing.

 If the first derivative f' is negative, then the function f is decreasing.

Let f be a continuous near a.

 If f'(a) = 0 and f''(a) > 0, then f has a local minimum at a.
 f'(a) = 0 and f''(a) < 0, then f has a local maximum at a.

When a graph has a local minimum, the curve is concave upward (and thus lies above the
tangent lines) at the minimum. Similarly, the curve is concave downward at a local
maximum.

Now let’s see the derivatives of the expression x3 – 4x2 + 1:

d(x3  4x2  1) 2 3 2
 3x2  8x and d (x  4x  1)  6x  8 .
dx dx2
8
Now f’(x) = 3x2  8x = 0 at x = 0 and x = .
3
f”(x) = 8 (<0) at x = 0 therefore the graph has a local
maximum at x = 0
8
and f”(x) = 8 (>0) at x = therefore the graph has a local
3

8
minimum at x = .
3

Now that we know the graph, we can draw it also. The graph is shown in the adjacent
figure.

Combining these two methods, we can sketch the graph of expressions like 2sinx + 5, 2x +
1, 6x2 x + 1 and so on.

Coming back to my original question, what would you call expression such as 2sinx + 5 or
6x2 x + 1? You would answer ‘sinusoidal’ expression and ‘polynomial,’ respectively. You
would be right of course. But what would you call an expression such as 2sinx + 6x 2 + 1?
The expression shares the same characteristics as the ones we have discussed-
 The input values in the expression would be ‘real’ values, values which lie on the
number lines.
 The output values will also be real values.
 For a particular value of x, we get only one value of the expression.
 We can define the range of the real values for the input and output.

For want of a better word, we call these expressions sharing the same characteristics a
function. So simply stated, a function is a device or an operator which gives a real number
output for a real number input, with one rule- it gives a single output for a single input. The
input, denoted by x, is called the independent variable and the output, denoted by y, is
called the dependent variable. A function is commonly denoted as f(x). Therefore, in our
example above
f(x) = 2sinx + x + 3. The graph of the function is shown in the
adjacent figure. The following can be observed about this
function-
 You can input all values of x in the range  to . The
range of values of the independent variable (x) that we can
take is known as the domain of the function. Here, the
domain of the function is  to .
 The range of values of the dependent variable (y) that we
can take is known as the range of the function. Here, the
range of the function is  to .
 It does not happen that for a single value of x we get two
different values of y. Therefore, if we draw a vertical line for any value of x in
the domain it will intersect the curve only once.

Find the domain and range of the function f(x) = 4  x2 .

2
Answer- we can see that the expression 4  x will become negative if x > 2 or x < 2.

Root of a negative number is not a real number, therefore the domain of this function is 2
 x  2. We will take only positive values of f(x) for each value of x. The value of f(x) would
vary from 0 to 2. Therefore the range of this function is [0, 2].

1
Find the domain and range of the function f(x) = .
1 x
Answer- As the roots of a negative number cannot be real, the domain of x would be for x
 0. The maximum value of f(x) would be 1 for x = 0. Therefore, the range of the function
f(x) = (0, 1].

Piecewise- Defined Functions

Till we have only seen functions which are continuous, i.e. functions you can draw without
having to raise your pencil from the paper. That is because the functions are defined by a
single rule. But how would you define the function given in the following figure?

Let us first see how the function is behaving. We can see that the value of the function is 0
for x < 0 and x > 5. Therefore, one rule is f(x) = 0 for x < 0 and x > 5. From x = 0 to x =
3
2, the value of f(x) changes from 0 to 3. Therefore, one more rule is f(x) = x for 0  x 
2

2. When x is greater than 2 but less than or equal to 5, the value of f(x) varies from 0 to 2.
2x  4
Therefore, f(x) = for 2 < x  5. Now we can put these rules all in one-
3
 3
2 x when 0  x  2

fx  2x  4 x when 2  x  5

 3
0 otherwise


There! We have created a rule for a piecewise defined function. Would you like to see some
more piecewise-defined functions?
x for x  0
 Modulus function- The modulus function x can be defined as x   .
x for x  0
 Greatest Integer Function- The greatest integer function, for any real number x,
gives the greatest integer less than or equal to x.
 Least integer function- The least integer function, for any real number x, gives the
smallest integer greater than or equal to x.

x 0 x 1
Find the maximum value of the function f(x)   .
2  x 1  x  2
Answer- We can see that f(x) follows the graph of the line y = x for x = 0 to 1 and the
graph of y = 2  x for x = 1 to 2. We can see that the maximum value occurs at the
intersection of these two graphs at x = 1. The maximum value of f(x) = f(1) = 1

Let f(x) be a function with domain 0  x  1 defined as

Find the area between the curve y = f(f(f(x))) and the coordinate axes.
Answer- The graph of the function is drawn as shown below

1
Area is the sum of the area of 8 triangles with base 1/8 and height 1. Total area = 8 × ×
2
1 1
×1=
8 2
Let f(x) be a function such that

For how many values of x is f(f(f(x))) = 19?

Answer- f(f(f(x))) = 19  f(f(x)) = 38  f(x) = 37 or 76  x = 74 (for f(x) = 37), 75 or
152 (for f(x) = 76). Therefore, three values of x are possible.

1
Find the domain of the function f(x)  , where [x] denotes the greatest
  [x]
integer less than or equal to x.

Answer- We know that the square root of negative quantities is imaginary. Therefore, for
the output (range) to be real, the quantity under the square root should be positive.
Therefore, [x] should be less than  = 3.14.. Therefore, as long as x is less than [x] would
be 3 or less. Therefore, the domain of the function consists of all values of x less than 4.
Therefore, Domain = ( , 4)

1
Find the domain of the function f(x)  , where [x] denotes the greatest
[x  2]
integer less than or equal to x.

Answer- The above function will give real values for all values of x except those values
where the denominator becomes zero. We cannot divide any number by zero. Therefore
when 2  x < 3, 0  x  2 < 1 and the greatest integer will give zero as the value.
Therefore, the domain of the function will contain all values of x except when 2  x < 3.
Therefore, Domain = R  [2, 3), where R stands for the set of all real numbers.

Find the domain of the function .

Answer- As we only need to avoid having negative numbers inside the square root, we
should ask ourselves the question “are there any values of x where the quantity under the
square root will become negative?” Now for positive values of x, we know that x 2  [x]2. We
only need to consider the negative values. If the negative values are integer, we know that
the quantity under the square root will be zero. Let’s see non-integer negative values of x.
For example, let’s take x =  3.6.
. Therefore, whenever
we take negative non-integer values of x, the quantity under the root will negative.
Therefore, the domain of x will be all non-negative values of x plus negative integers.
Therefore, Domain = I u [0, ), where I denotes the set of negative integers.

Methods for Finding Range of a Function

 Sketch the graph of the function: This will more often than not give you an idea
about the range. Examine the critical points in the domain and see how the curve
behaves at those points. Keep some elementary values and check the behaviour.
 Solve for independent variable: I would like to thank Deepanshu, a regular TGite,
for pointing out this method.
1. Put f(x) = y
2. Calculate x in terms of y means solve to obtain x = P(y)
3. Calculate domain of P(y)
4. This domain is nothing but range of F(x).
 x
For example, let f(x) = . Find the range of f(x).
x 1

x y
Answer- Let f(x) = y =  x  . Now for x to be real, the denominator
x 1 1 y

cannot be zero. Therefore, y cannot be equal to 1. Therefore, y can take all real
values except 1. Therefore, the range of f(x) would equal to R  1, where R stands for
the set of all real numbers. You can see the graph of the function in the adjacent figure.
Drawing a graph gives a good idea about how a function is going to behave.

x
Find the range of the function f(x) 
1  x2
x 1  4y2
Answer- Let f(x) = y = .  x2y x + y = 0  x  1  . Now x would be real
1  x2 2y

for y  0 and
1 1
1  4y2  0  y  [  , 0) u (0, ]
2 2

Now that we know the basics of functions, let us see the basic rule for sketching the curve
of functions at hand

How to Draw Graph of F(x)

 Find the domain, range and any symmetry f(x) has.
dy d2y
 Calculate. and
dx dx2
dy
 Determine the critical points of f(x) (  0 ) and use to find if f(x) has local
dx
d2y d2y
maximum or minimum at these points ( > 0 for minimum and < 0 for

minimum).

 Find the inflection points, if any ( = 0) where the curve changes concavity.
dx2
 Determine if the function is increasing or decreasing.
 Identify any asymptotes, i.e. lines which are tangent to the curve at infinity.
 Plot important points such as intercepts on axes, critical points etc.

Draw the graph of the function f(x) = x3(x + 2).

Answer- We can see that
 the domain of the function consists of all real values of x.
 the function intersects the x-axis at x = 0 and x = 2, and it is
negative between these two points.
df(x)
 2x2(2x  3) 3
 which is equal to zero at x = 0 and x =  .
dx 2
3
The second derivative is negative at x =  which means that the
2
curve has a local minimum at this point.
 d2f(x)
  12x(x  1) which is equal to zero at x = 0 and x = 1. That means that the
dx2

curve changes concavity at these two points. The graph of the curve is shown in the
adjacent figure. See how well our deductions fit with the actual graph of the curve.

Composite Functions
As we said earlier, a function can said to be an operator. For example, the plus sign ‘’ is an
addition operator. In this regard, a function can operate on itself or on other functions. Let
us say that there are two functions, f(x) and g(x), such that f(x) = 2x + 1 and g(x) = x 2 
3.

 f(f(x)) = f(2x + 1) = 2(2x + 1) + 1 = 4x + 3.

 f(g(x)) = f(x2  3) = 2(x2  3) + 1 = 2x2  5.

As we can deduce, we can operate the function as many times as we want.

The operation of function has the same rule as that of real numbers-
(f + g)(x) = f(x) + g(x)
(f  g)(x) = f(x)  g(x)
(fg)(x) = f(x) g(x)

1
If f(x)  , x  0, and f(g(x)) = g(x), then what is the value of f(g(x))  g(x)?
x
1 1
Answer- As f(x)  , f(g(x)) =  f(g(x))  g(x) = 1

x g(x)

If the functions f(x) and g(x) are defined as follows

 3
2x  3 for x 
 2
f(x)  
3
0 for x 

 2
3x  2 for 0  x  2
g(x)   2
x  1 for 2  x  5
find the value of f(g(2))  g(f(3))
Answer- f(g(2))  g(f(3)) = f(22 + 1)  g(2  3  3) = f(5)  g(3) = 7  10 = 3.

If f (x + y) = f (x) + f (y) + f (x)  f (y) and f (1) = 3 then find f (3).

1. 63 2. 48 3. 53 4. 36
Answer- Keeping x = y = 1 in the equation, we get
f(2) = f(1) + f(1) + f(1)  f(1)  f(2) = 15. Now keeping x = 2, y = 1 in the equation
f(3) = f(2) + f(1) + f(2)  f(1)  f(3) = 63.

Let f(x) be a function satisfying f(x) f(y) = f(xy) for all real x, y. If f(2) = 4, then
1
what is the value of f( )?(CAT 2008)
2
1
Answer- Keeping x = 1 and y = 2, we get f(1)  f(2) = f(2)  f(1) = 1. Now keeping x =
2
1 1 1
and y = 2 in the equation we get f( )  f(2) = f(1)  f( )=
2 2 4
Even and Odd Functions

Both even and odd functions are function which have an axis of symmetry- y-axis for even
functions and the line y = x for odd functions. Even and odd can be mathematically defined
as follows:

A function f(x) is an
Even function if f(x) = f(x), for example f(x) = x2 + 1 is an even function.
Odd function if f(x) = f(x), for example f(x) = x3 + x is an odd function.

In the figure above are shown graphs of two functions, one odd- x3 + x- and one even- x2 +
2. We can see that the line y = x is the axis of symmetry for the first and the line x = 0 is
the axis of symmetry for the second.

 If f(x) is even and g(x) is even  f(g(x)) is an even function

 If f(x) is odd and g(x) is odd  f(g(x)) is an odd function
 If f(x) is even and g(x) is odd  f(g(x)) is an even function
 If f(x) is odd and g(x) is even  f(g(x)) is an even function

The function f(x) = log(x  x2  1) is

(a) an even function
(b) an odd function
(c) neither odd nor even
(d) cannot be determined

Answer- Now f(x) = log(x  x2  1)  f(x) + f(x) = log(x  x2  1)+

= log(x2 + 1  x2)
= log1 = 0. As f(x) + f(x) = 0, the function is an odd function.

Periodic Function
A function f(x) is called a periodic function if there is a positive number p such that f(x + p)
= f(x) for every x. The smallest such value of p is known as the period of f(x). For example,
the period of trigonometric functions- sinx, cosx, tanx etc. - is 2.
Let g(x) be a function such that g(x + 1) + g(x − 1) = g(x) for every real x. Then
for what value of p is the relation g(x + p) = g(x) necessarily true for every real x?
(CAT 2005)
1. 5 2. 3 3. 2 4. 6
Answer- Let g(x  1) = a and g(x) = b  g(x + 1) = b  a. Now, keeping x + 1 in place of x
we get
 g(x + 2) + g(x) = g(x + 1)  g(x + 2) + b = b  a  g(x + 2) = a.
Again, keeping x + 2 in place of x we get
 g(x + 3) + g(x + 1) = g(x + 2)  g(x + 3) + b  a =  a  g(x + 3) = b.
Now we can see that g(x) = b and g(x + 3) = b, therefore, g(x + 3 + 3) = (b) = b  g(x
+ 6) = b
Therefore, p = 6.

Inverse of a Function
Simply put, an inverse of a function y = f(x) is a function in the reverse direction x = f 1y,
i.e. with range as the input and domain as the output.

How to find the inverse of a function? For example, how do I find the inverse of the function
2x  3
f(x) = y = ?
x 5
Here are the steps-
2x  3
 Write the original function y =
x 5
5y  3
 Solve for x  x =
y 2

5x  3
 Switch x and y  f(x) = y =
x 2

Remember that a function and its inverse are mirror images of each other along the line y
=x

If the inverse of the function f (x) = ax + b is f -1 (x) = bx + a, then the values of a

and b are
1. 1, 1 2. –1, 1 3. 1, -1 4. –1, -1
y b  f-1 (x) = x  b = bx + a (given). Comparing
Answer- Let f(x) = y = ax + b  x = 
a a a a
1 b
b  a
coefficients both sides, we get and . Solving, we get a = 1 and b = 1.
a
a

Function as a Series

2f(n)  1
Let f(n) be a function such that f(n  1)  for n = 1, 2, 3, … and f(1) = 2.
2
Then f(101) is equal to

1 1
Answer- Arranging the equation we get f(n + 1)  f(n) =  f(2)  f(1) =
, f(3)  f(2) =
2 2
1 1
, … f(101)  f(100) = . Adding all the equations we get f(101)  f(1) = 50  f(101) =
2 2
52.
A function f(x) satisfies f(1) = 3600, and f(1) + f(2) + ... + f(n) = n2f(n), for all
positive integers n > 1. What is the value of f(9)? (CAT 2007)
Answer- Writing the equation for n + 1 instead of n we get f(1) + f(2) + ... + f(n + 1) = (n
+ 1)2f(n + 1). Subtracting the original equation f(1) + f(2) + ... + f(n) = n2f(n) from the
above equation we get
n
f(n + 1) = (n + 1)2f(n + 1)  n2f(n)  f(n 1)  f(n). Therefore,
n 2
8 8 7 8 7 6 5 4 3 2 1
f(9)  f(8)   f(7)  ...     f(1) = 80
   
 

10 10 9 10 9 8 7 6 5 4 3

9x 1  2   881 
If f(x)  , then the value of f  f  ...  f will be
    
9x  3  882   882   882 
9
9x 1x x
Answer- f(x)   f(1  x)  9  9 
9

3

9x  3 91x  3 9
3 9  3  9x 3  9x
9x

9x + 3 = 1
Therefore, f(x) + f(1 - x) =
9x  3 3  9x

1 2  880 
Therefore, f   881   f  ...  1 . There are 440 such pairs and the
f f
       
 882   882   882   882
middle
 441 term
  would
1  1 be 1   881  1
f f  . Therefore, f  2  ...  f = 440 + =440.5
f
         
 882   2  2  882   882   882  2

For a function f(x), f(x) + f(x  1) = x2 and f(19) = 94. Find f(94)
Answer- f(x) = x2  f(x  1)  f(94) = 942  f(93) = 942  (932  f(92)) = 942  932 + (922
 f(91) = …
= 942  932 + 922  912 + 902  892 + … + 202  f(19)

You all can calculate now 

 ASSIGNMENT

1. If P(x,y) = smallest prime number greater than sum of x and y,

If Q(x,y) = quotient when x is divided by y,
L(x,y) = LCM of x and y
A(x,y) = smallest natural number greater than the arithmetic mean of x and y. Which of
the following statement is true?
a. P[L{Q(19,4),A(2,3)},A(5,20)] =28
b. Q[A{P(4,16),L(7,14)},P(1,2)] =4
c. A[P{Q(22,7),A(9,12)},L(4,7)] = 23
d. A[P{Q(22,7),A(9,12)},L(4,7)] = 19
e. None of these

 x  x   x   x  5x
2. Find number of solutions of x satisfying,     , where x is between 0
10    20  30  
       40  24

and 1000, Where [x] represents greatest integer less than or equal to x.
a. 6 b. 8 c. 0 d. 16

3. If 2x + 9y  90, x  0 and y  0, x and y are integers, find the total number of

solutions.
a. 257 b. 256 c. 125 d. 128

4. Find the number of solutions of the equation x 4 - 5x = 0.

a. 0 b. 1 c. 4 d. 6

5. Find the number of solutions of the equation x 4 - 4x = 0.

a. 2 b. 3 c. 1 d. Infinite
2  3
6. Find the number of solutions of the equation sin x – x  0
 
 2 
 a. 2 b. 3 c. 1 d. None

  1 2

7. Find the number of solutions of the equation x   1  2 0 x =0, where x i s positive.
 
x 
 a. 2 b. 3 c. 1 d. None

y(x  y)
8. If f(x, y) is a natural number and defined as f(x, y)  , then which of the
(x  y)2  (x  y)2
following is possible?
a. x and y are both odd or both even.
b. x and y are relatively prime.
c. There is only one unique value of (x,y).
d. There are two unique values of x and y.
e. None of the above.
9. Find the area of the region confined between the graphs |x| =|y| and |x|=a.
a. 2a2 b. 4a2 c. a2 d. 8a2
12. Let s be the set of all pairs (i, j) where 1  i  j  n and n  4 . Any two distinct numbers of
s are called as friends if they have one constituent of the pair in common and enemies
otherwise. For example if n=4, then s={(1,2),(1,3),(1,4),(2,3),(2,4),(3,4)}, here (1,2)
and (1,3) are friends and (1,3) and (2,4) are enemies. Now for a general value of n,
how many enemies will each member of s have?
n2  5n  6 n2  7n  14 n2  3n  2
a. 2n-7 b. c. d. n - 3 e.
2 2 2

13. In the above question for a general value of n, consider any two members of s that are
friends. How many other members of s will be common friends of both these members?
a. n(n  3) n2  7n  16 n2  5n  8
2 b. c. n – 2 d. e. 2n – 6
2 2

14. If a1 =p and b1=q, where p and q are positive quantities and

an=pbn-1, bn =qbn-1, where n >1 and even
an=pan-1, bn =qan-1, where n >1 and odd
Find the value of an + bn if n is even
n n n n n n
1 1
a. q(pq)2 (p  q) b. q2 (p  q)2 c. q(pq)2 (p  q)2 d. q2 (p  q) e. None

15. In the above question if p=1/3 and q =2/3, then what is the smallest odd n such that,
an +bn < 0.01
a. 11 b. 9 c. 15 d. 7 e. 13

16. In a tournament there are n teams T1, T2, T3, ….Tn, where n>5. Each team consists of k
players, where k>3. The following pairs of the team have one player in common: T1& T2,
T2&T3, … ….Tn-1&Tn and Tn&T1. No other pair of teams has any player in common. How
many players are participating in the tournament considering all the n teams together?
a. n(k-1) b. n(k-2) c. (n-1)(k-1) d. k(n-1) e. k(n-2)

17. Consider the set S ={ 2,3,4,5…..(2n+1)}, where n is a positive integer more than 2007.
if x is defined as the average of the odd integers in S and y is defined as the average of
the even integers in S. Find (x –y).
n
a. b. n  1 c. 1 d. 2008 e. 0
2 2n

18. If Tn  (1)n1(Tn1  1) & T0  1 , find the value of S199 , where Sn  T1  T2  ...Tn

a. 100 b. -100 c. -199 d. -99 e. -101

x
19. If y  , find the maximum value of y. Given that y > 0.
4x2  1 5x  9
 1 1 1 1
a. b. c. d. e. None of these
30 29 27 28

20. If A, B, C and D are four positive integers having sum of 4. Find the minimum possible
value of the sum of their reciprocals.
1 6 1
a. 4 b. 2 c. 3 d. 4 e. None of these
4 7 3
21. If ap qbqrcr p  1 , where a, b, c, p, q & r are rational numbers, which of the following is
true?
a. a=b=c
b. a=b=c is not necessary
c. p=q=r
d. Either (a) or (c)
e. None of these

22. If f(x,y,z) = xyz, where x, y and z are positive numbers. Find the greatest value of the
f(x,y,z), given that xy + yz + zx =12.
a. 8 b. 64 c. 4 d. 32 e. None of these

23. If 2f(xy)  f(x)y  f(y)x , where x and y are real numbers. If f(1)=a where a  1 , find
n

the value of (a  1) f(i)

i 0

a. an1  1 b. an1  1 c. an1  1 d. an1  1 e. None of these

24. If f(x)f(y) = f(x) + f(y) +f(xy)-2 and f(2) =5, find f(5).
a. 24 b. 25 c. 2 d. 26 e. None of these

25. If x1, x2,….x100 are positive integers such that xi + xi+1=k for all i, where k is a
constant. If x10 =1, find x1
a. k b. k+1 c. k-1 d. 0 e. Indeterminable

26. A function y=f(N) is defined for all natural numbers, as the sum of the digits of ‘N’. If k
is a natural number such that f5(k) = 1 and k> f(k)> f2(k)> f3(k)> f4(k)> f5(k). where
f2(k) = f(f(k)), f3(k)=f(f(f(k))) etc. What is the minimum number of digits that k can
have?
(1)11111….111(22 times) +1
(2)22222….222(22 times) +1
(3)23
(4)222…….222(22 times)
(5)None of these.

27. A function F(N) takes a natural number N as its input and gives output as the sum of N
and the number formed by reversing the digits of N. Another function G(N) takes a
natural number N as its input and gives output as the difference of N and the number
formed by reversing the digits of N. If k is a three digit natural number with its digits in
either ascending or descending order, how many values can F(G(k) take?
a. 1 b. 2 c. 3 d. 4 e. None of these.

28. Find the number of integral solutions of the equation |x-3| +|y-5|=11
a. 36 b. 44 c. 55 d. 60 e. 72
29. Which of the following statements is/ are true?
(a) Area formed by the graphs | x  3 |  4 and| y  4 | 6 is 96 square units.
(b) If x & y are integers, the number of solutions of |x-3|+|y-5|=11, is 44.
(c) There is no solution to the equation 10x + 10-x <2, where x is a real number.

a. only a b. only b c.only c d. a and b e. a,b,c

30. If f(x) =  , then find the value of f(3) + f(7) + f(13) + f(21).
a. 12 b. 13 c. 14 d. 15 e. None of these

31. Which of the following is true?

a. If the graph f(x) =|x| is reflected in the line x=1, to obtain the graph of new function
g(x), then g(x+2) =|x|.
b. If g(x) =|x|-2, then g(x) can be obtained if graph of |x| is reflected in y=2 and then
this image is reflected in y=1
c. If |x| is reflected in y=-1 and then this image is reflected in y=-2, we obtain graph of
new function g(x) =|x|-2.
a. only a b. only b c. only c d. a, b, c e. none
 1   2   32 
162 x
32. If the function f(x) = , then find the value of  f   f    ...  f  .
162 x 16 32  32 32 
      
557 559 559
a. b. c. d. 0 e. none of these
117 117 119

33. If (3 + a + a² + a³)5 = C  C a  C a  .....C a , where C0, C1, C2, …, C15 are integers,
2 15

0 1 2 15
15
then find the value of C .
i1
i

a. 243 b. 7776 c. 7533 d. 7775 e. none of these

34. Which of the following is true?

  2013 
  2011 
III. 2012  
a. II only b. III only c. I and II d. I and III e. none of these

35. If Tn  Tn1 1 1 , where n is a positive integer. If T1  3, then the value of

2

T1  T2  T3 T9  T10
a. 1024 b. 2048 c. 22048 d. 22048 +1 e. 21024 -1

36. How many of the functions are odd

 ), sin3x, x2 + |x|, a x  a x
Log ( x +
a x  a x

a. 1 b. 2 c. 3 d. 0 e. 4
 37. Which of the following is true?
(a) a, b ( two positive real numbers) are expressed as the sum of m positive and n
positive real numbers respectively as follows:
a  S1  S2  S3  ....Sm and b  t1  t2  t3  ....tn also
[a] [S1] [S2 ] [S3 ] [Sm ]  4 and [b] [t 1 ] [t 2 ] [t 3 ] ....[t m ]  3 , where [x] denotes

the greatest integer less than or equal to x, then minimum possible value of m+n
is 9
6
(b) if a, b, c >0 a+b+c =6 and f(x)  1 , the minimum value of f(a).f(b).f(c) is 20.
x

(c) The area enclosed by the graph |x-1| +|y-1|=2 is 8 square units.

a. only a b. only b & c c. only c d. a , b, c e. None

38. If f(t) = f(t-1) + f(t+1), find the minimum value of k such that f(t + k) + f(t) =0
a. 2 b. 3 c. 4 d. 5 e. 6

39. If f(t) = f(t-1) + f(t+1), find the minimum value of k such that f(t + k) =f(t) .
a. 3 b. 2 c. 6 d. 5 e. None of these

40. If an = an-1 + an+1, where an represents nth term of a sequence. If a0 =1, a1 =1.23, find
the value of a0 + a2 + a3 +… .... a1002.
a. 0 b. 1 c. 1.23 d. 2.23 e. None of these

41. Find the number of points with integral coordinates inside the region x + 2y  50, x  0 ,
y 0
a. 625 b. 650 c. 676 d. 1275 e. 702

42. If [x] represents greatest integer function of x, find the simplified value of
5  5  1  5  2  5 19 9
       .... 


    
     

2  2 10 0 2 10 0 2 10 0 
       
a. 598 b. 599 c. 602 d. 600
e. 601
 
43. The function f (x) = logx  x x ... x x x  where the root is taken x – 1 times and x 
  
 

2. So f2 = log2( 2) f3= log3  3 3  . Then f 7 is equal to:

127 b. 63 c. 31 255
a. 64 32 d.
128 256

44. If f(x) = x2/(x2 – 1) then 31f(30) f(29) f(3)f(2) is equal to

a. 93 b. 60 c. 61 d. 100
45. Let f (x) = log (x + (x2 + 1)). Then which of the following statements are true?
I. The domain of f (x) is all real numbers.
II. The graph of f (x) contains the origin (0, 0)
III. f (-x) = - f (x)
a. II and III b. I and II c. I and III d. I, II and III

46. The function f (x) = ex/5. The inverse function f-1 (x) of the function f (x) is:
a. lnx + ln5 b. 5lnx c. lnx/5 d. ln(x + 5)

47. The domain of the function f (x) = (log (x – 2)) is:

a. [1, ) b. [3, ) c. (2,) d. R

 x 
48. Given f  3x. Find f (1).
 
2x  3 
a. –5 b. –9 c. 3 d. –3

49. The number of different functions that can be defined from the domain
D = {a, b, c, d, e} onto a range R = {1, 2, 3, 4} is:
a. 1024 b. 625 c. 240 d. 600

50. If f (x) = ax + b and f (f ( f (x))) = 27x + 52, then a + b is equal to:

a. 5 b. 7 c. 9 d. 6