IIT JEE Mathematics (Set Relations & Functions) Solved Question Papers for Examination 2013

Welcome to IIT JEE Mathematics Algebra (Set Relations & Functions) Solved Question Papers for Examination 2013 Section. You will find here IIT JEE Mathematics Algebra (Set Relations & Functions) Solved Question Papers for Examination 2013. These Question Papers will help you for better examination Preparation.

JEE Mathematics Algebra (Set Relations & Functions) Solved Question Papers:

Question 1:

Given are two sets A {1, 2, -2, 3} and B = {1, 2, 3, 5}. Is the function f(x) = 2x - 1 defined from A to B?

Answer :

Out of all the ordered pairs, the ordered pairs which are related by the function f(x) = 2x - 1 are {(1, 1), (2, 3), (3, 5) But for (-2) in A, we do not have any value in B. So, this function does not exist from
A->B.

Question 2:

A function f is defined as f: N -> N (where N is natural number set) and f(x) = x+2. Is this function ONTO?

Answer :

Since, N = {1, 2, 3, 4, .........} and A = B = N

For : A->B

When x = 1                f(x) = 3

When x = 2                f(x) = 4

So f(x) never assume values 1 and 2. So, B have two elements which do not have any pre-image in A. So, it is not an ONTO function.

Question 3 :

Find the range and domain of the function f(x) = (2x+1)/(x-1) and also find its inverse.

Answer :

This function is not defined for x = 1. So, domain of the function is
R -{1}.

Now, for finding the range

Let,(2x+3)/(x-1) = y

=> 2x + 3 = yx - y

=> yx - 2x = y + 3

=> (y - 2)x = y + 3

=> x =(y+3)/(t-2)

So, y cannot assume value 2

Range of f(x) is R - {2}.

Inverse is y =(x+3/x-2) .

Question 4:

Find domain and range of the function f(x) = (x2+2x+3)/(x2-3x+2)

Answer :

This function can be written as : f(x) =(x2+2x+3)/(x-1)(x-2) .

So, domain of f(x) is R - {1, 2}

For range, let (x2+2x+3)/(x2-3x+2) = y

=> (1 - y)x2 + (2x + 3y) x + 3 - 2y = 0

for x to be real, Discriminant of this equation must be > 0

D > 0

=> (2 + 3y)2 - 4(1 - y)(3 -2y) > 0

=> 4 + 9y2 + 12y - 4(3 + 2y2 - 5y) > 0

=> y2 + 32y - 8 > 0

=> (y + 16)2 - 264 > 0

=> y < - 16 - √264  or  y  > -  16 + √264.

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