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1

WB JEE 2010

MCQ (Single Correct Answer)

The function $$f(x) = \sec \left[ {\log \left( {x + \sqrt {1 + {x^2}} } \right)} \right]$$ is

A
odd
B
even
C
neither odd nor even
D
constant

Explanation

$$\because$$ $$f(x) = \sec \left[ {\log \left( {x + \sqrt {1 + {x^2}} } \right)} \right]$$

$$\therefore$$ $$f(x) = \sec [\log \{ - x + \sqrt {1 + {{( - x)}^2}} \} ]$$

$$ = \sec \left[ {\log {{(\sqrt {1 + {x^2}} - x)(\sqrt {1 + {x^2}} + x)} \over {(\sqrt {1 + {x^2}} + x)}}} \right]$$

$$ = \sec \left[ {\log \left( {{{1 + {x^2} - {x^2}} \over {x + \sqrt {1 + {x^2}} }}} \right)} \right] = \sec \log {(x + \sqrt {1 + {x^2}} )^{ - 1}}$$

$$ = \sec \{ - \log (x + \sqrt {1 + {x^2}} )\} = \sec [\log (x + \sqrt {1 + {x^2}} )\} = f(x)$$

$$\therefore$$ f(x) is an even function.

2

WB JEE 2010

MCQ (Single Correct Answer)

If A = {1, 2, 3, 4}, B = {1, 2, 3, 4, 5, 6} are two sets, and function f : A $$\to$$ B is defined by f(x) = x + 2 $$\forall$$ x$$\in$$ A, then the function f is

A
bijective
B
onto
C
one-one
D
many-one

Explanation

$$\therefore$$ f(x) = x + 2

Here, f is one-one only. f is not onto. [$$\because$$ 1, 2 have no pre-image]

3

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

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

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

4

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

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