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1

### WB JEE 2009

MCQ (Single Correct Answer)

For any two sets A and B, A $$-$$ (A $$-$$ B) equals

A
B
B
A $$-$$ B
C
A $$\cap$$ B
D
Ac $$\cap$$ Bc

## Explanation

A $$-$$ (A $$-$$ B)

= A $$\cap$$ (A $$\cap$$ Bc)c = A $$\cap$$ (Ac $$\cup$$ B)

= $$\phi$$ $$\cup$$ (A $$\cap$$ B) = A $$\cap$$ B

2

### WB JEE 2008

MCQ (Single Correct Answer)

Let A = {1, 2, 3} and B = {2, 3, 4}, then which of the following relations is a function from A to B?

A
{(1, 2), (2, 3), (3, 4), (2, 2)}
B
{(1, 2), (2, 3), (1, 3)}
C
{(1, 2), (2, 3), (3, 3)}
D
{(1, 2), (2, 3), (3, 4)}

## Explanation

If a relation is function from set A to set B, then first element of ordered pairs in relation should not be repeated and second element of ordered pairs should be in set B. So Relation {(1, 3), (2, 3), (3, 3)} is a function from A = {1, 2, 3} to B = {2, 3, 4}.

3

### WB JEE 2008

MCQ (Single Correct Answer)

A function f : A $$\to$$ B, where A = {x/$$-$$1 $$\le$$ x $$\le$$ 1} and B = {y/1 $$\le$$ y $$\le$$ 2} is defined by the rule y = f(x) = 1 + x2. Which of the following statement is then true?

A
f is injective but not surjective
B
f is surjective but not injective
C
f is both injective and surjective
D
f is neither injective nor surjective

## Explanation

$$\because$$ $$-$$ 1 $$\le$$ x $$\le$$ 1 $$\Rightarrow$$ 0 $$\le$$ x2 $$\le$$ 1

$$\Rightarrow$$ 1 $$\le$$ x2 + 1 $$\le$$ 2 (adding 1)

$$\Rightarrow$$ 1 $$\le$$ f(x) $$\le$$ 2 $$\Rightarrow$$ Rf = [1, 2] = Co-domain B

$$\therefore$$ f is surjective.

Let $$x = - {1 \over 2},{1 \over 2} \in A$$ $$\because$$ $$- {1 \over 2} \ne {1 \over 2} \Rightarrow {\left( { - {1 \over 2}} \right)^2} = {\left( {{1 \over 2}} \right)^2}$$

$$\Rightarrow {1 \over 4} = {1 \over 4} \Rightarrow 1 + {1 \over 4} = 1 + {1 \over 4} \Rightarrow f\left( { - {1 \over 2}} \right) = f\left( {{1 \over 2}} \right)$$

$$\therefore$$ f is not one-one

($$\because$$ $${x_1} \ne {x_2} \Rightarrow f({x_1}) = f({x_2})$$ then f is not one-one)

$$\therefore$$ f is surjective but not injective.

4

### WB JEE 2008

MCQ (Single Correct Answer)

The mapping f : N $$\to$$ N given by f(n) = 1 + n2, n $$\in$$ N where N is the set of natural numbers, is

A
One to one and onto
B
Onto but not one-to-one
C
One-to-one but not onto
D
Neither one-to-one nor onto

## Explanation

Range of mapping = {f(1), f(2), f(3), ....}

= {2, 5, 10, ....} $$\ne$$ N (co-domain)

$$\therefore$$ mapping is not onto

If n = n1, n2 $$\in$$ N such that n1 $$\ne$$ n2

$$\Rightarrow n_1^2 \ne n_2^2$$ ($$\because$$ n1 and n2 are +ve)

$$\Rightarrow 1 + n_1^2 \ne 1 + n_2^2$$

$$\Rightarrow f({n_1}) \ne f({n_2})$$

$$\therefore$$ Mapping is one-one

$$\therefore$$ Mapping is one-one and not onto.

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