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1

WB JEE 2010

MCQ (Single Correct Answer)

Let R be the set of real numbers and the mapping f : R $$\to$$ R and g : R $$\to$$ R be defined by f(x) = 5 $$-$$ x2 and g(x) = 3x $$-$$ 4, then the value of (fog)($$-$$1) is

A
$$-$$ 44
B
$$-$$ 54
C
$$-$$ 32
D
$$-$$ 64

Explanation

$$fog( - 1) = f\{ g( - 1)\} = f\{ 3( - 1) - 4\} $$

$$ = f( - 7) = 5 - {( - 7)^2} = - 44$$

2

WB JEE 2009

MCQ (Single Correct Answer)

The domain of definition of the function $$f(x) = \sqrt {1 + {{\log }_e}(1 - x)} $$ is

A
$$ - \infty < x \le 0$$
B
$$ - \infty < x \le {{e - 1} \over e}$$
C
$$ - \infty < x \le 1$$
D
$$x \ge 1 - e$$

Explanation

$$f(x) = \sqrt {1 + {{\log }_e}(1 - x)} $$

Value of f(x) is real when

$$1 + {\log _e}(1 - x) \ge 0$$ and $$1 - x > 0$$

$$ \Rightarrow {\log _e}(1 - x) \ge - 1$$ and $$x < 1$$

$$ \Rightarrow {\log _e}(1 - x) \ge {\log _e}{e^{ - 1}}$$ and $$x < 1$$

$$ \Rightarrow 1 - x \ge {1 \over e}$$ and $$x < 1 \Rightarrow x \le {{e - 1} \over e}$$ and $$x < 1$$

or, $$x \le {{e - 1} \over e}$$

3

WB JEE 2009

MCQ (Single Correct Answer)

A mapping from IN to IN is defined as follows:

$$f:IN \to IN$$

$$f(n) = {(n + 5)^2},\,n \in IN$$

(IN is the set of natural numbers). Then

A
f is not one-to-one
B
f is onto
C
f is both one-to-one and onto
D
f is one-to-one but not onto

Explanation

$$f(n) = {(n + 5)^2}$$

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

$$ \Rightarrow {n_1} + 5 \ne {n_2} + 5 \Rightarrow {({n_1} + 5)^2} \ne {({n_2} + 5)^2} \Rightarrow f({n_1}) \ne f({n_2})$$

$$\therefore$$ f is one-one

$$\because$$ n $$\in$$ N i.e., n = 1, 2, 3, .....

$$\therefore$$ f(1) = 36, f(2) = 49, .....

$$\therefore$$ Range = {f(1), f(2), f(3), .....}

= {36, 49, .....} $$\ne$$ N (codomain)

$$\therefore$$ f is one to one but not onto.

4

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

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