Functions Test 49

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Functions Test 49

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Weekly Quiz Competition

Question 1
1 / -0

Let $$f:\{x, y , z\} \rightarrow \{1, 2, 3\}$$ be a one-one mapping such that only one of the following three statements and remaining two are false : $$f(x) \neq 2, f(y) =2, f(z) \neq 1$$, then

A
$$f(x) > f(y) > f(z)$$

B
$$f(x) < f(y) < f(z)$$

C
$$f(y) < f(y) < f(z)$$

D
$$f(y) < f(z) < f(x)$$

Solution

$$f:\left\{ x,y,z \right\} \rightarrow \left\{ 1,2,3 \right\} $$
$$I-f\left( x \right) \neq 2$$
$$II-f\left( y \right) =2$$
$$III-f\left( z \right) \neq 1$$
1. Let statement I is true, $$\therefore $$ other two are false
$$\therefore f\left( x \right) \neq 2,f\left( y \right) \neq 2$$
$$\therefore f\left( z \right) =1$$
$$f\left( x \right) \& f\left( y \right) $$ can be $$2$$ or $$3$$
We don't know exact value of $$f\left( x \right) \& f\left( y \right) $$
2. Let statement II is true
$$f\left( y \right) =2\quad \therefore f\left( z \right) =1\quad $$ [$$\because f\left( z \right) \neq 1$$ is false]
$$f\left( x \right) =3$$
$$f\left( x \right) >f\left( y \right) >f\left( z \right) $$
Question 2
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Domain of definition of the function $$f\left( x \right) =\sqrt { 2\sin ^{ -1 }{ \left( 2x \right) +\dfrac { \pi }{ 3 } } }$$, for real value x, is

A
$$\left[ -\dfrac { 1 }{ 4 } ,\dfrac { 1 }{ 2 } \right]$$

B
$$\left[ -\dfrac { 1 }{ 2 } ,\dfrac { 1 }{ 2 } \right]$$

C
$$\left( -\dfrac { 1 }{ 2 } ,\dfrac { 1 }{ 9 } \right)$$

D
$$\left[ -\dfrac { 1 }{ 4 } ,\dfrac { 1 }{ 4 } \right]$$

Solution
Question 3
1 / -0

The domain of definition of the function $$f(x) = \left \{x\right \}^{\left \{x\right \}} + [x]^{[x]}$$ is (where $$\left \{\cdot \right \}$$ represents fractional part and $$[\cdot ]$$ represents greatest integral function).

A
$$R - I$$

B
$$R - [0, 1)$$

C
$$R - \left \{I\cup (0, 1)\right \}$$

D
$$I\cup (0, 1)$$

Solution
Question 4
1 / -0

Let $$f$$ be real valued function defined by $$f(x)=\sin ^{ -1 }{ \left( \cfrac { 1-\left| x \right| }{ 3 } \right) } +\cos ^{ -1 }{ \left( \cfrac { \left| x \right| -3 }{ 5 } \right) } $$. Then domain of $$f(x)$$ is given by

A
$$\left[ -4,4 \right] $$

B
$$\left[ 0,4 \right] $$

C
$$\left[ -3,3 \right] $$

D
$$\left[ -5,5 \right] $$

Solution
Question 5
1 / -0

$$f:\left( 0,\infty \right) \rightarrow \left( 0,\infty \right) $$ is defined by $$f(x)=\begin{cases} { 2 }^{ x },\quad x\in \left( 0,1 \right) \\ { 5 }^{ x },\quad x\in [1,\infty ) \end{cases}$$ is

A
one-one but not onto

B
onto but not one-one

C
neither one-one nor onto

D
bijective

Solution
Question 6
1 / -0

Let $$A=\left\{ { a }_{ 1 },{ a }_{ 2 },{ a }_{ 3 },{ a }_{ 4 }{ a }_{ 5 },{ a }_{ 6 } \right\} $$ and $$B=\left\{ { b }_{ 1 },{ b }_{ 2 },{ b }_{ 3 } \right\} $$. The number of functions of $$f:A\rightarrow B$$ such that it is onto and there are exactly three elements in $$A$$ such that $$f(A)=b$$, is

A
$$75$$

B
$$90$$

C
$$100$$

D
$$120$$

Solution

Since there should be exactly 3 elements in $$ A $$ such that
$$ f(A) = b $$
the function $$ f$$ is a bijection from a subset $$ A_1 $$ of $$ A $$ to $$ B $$., each having 3 elements.

Number of ways to select 3 elements from $$ A $$ $$ =\; ^6C_3 $$

Number of bijection from $$ A_1 $$ to $$ B $$ $$ = 3 ! $$

Hence, the number of functions $$ =\; ^6C_3 \; 3!$$ $$ = 120 $$
Question 7
1 / -0

Domain of the function $$f(x) = \dfrac {x - 3}{(x - 1)\sqrt {x^{2} - 4}}$$ is

A
$$(1, 2)$$

B
$$(-\infty, -2) \cup (2, \infty)$$

C
$$(-\infty, -2)\cup (1, \infty)$$

D
$$(-\infty, \infty) - \left \{1, \pm 2\right \}$$
Question 8
1 / -0

Let $$f(x) = \dfrac {x}{1 - x}$$ and let $$\alpha$$ be a real number. If $$x_{0} = \alpha, x_{1} = f(x_{0}), x_{2} = f(x_{1}), ....$$ and $$x_{2011} = - \dfrac {1}{2012}$$ then the value of $$\alpha$$ is

A
$$\dfrac {2011}{2012}$$

B
$$1$$

C
$$2011$$

D
$$-1$$

Solution

$$f(x)=\dfrac {x}{1-x}$$ also $$x_0=\alpha$$
$$x_1 =f(x_0)=f(\alpha)=\dfrac {\alpha}{1-\alpha}$$
$$x_2 =f(x_1)=f\left (\dfrac {\alpha}{1-\alpha}\right)=\dfrac {\dfrac {\alpha}{1-\alpha}}{1-\dfrac {\alpha}{1-\alpha}}=\dfrac {\alpha}{1-2\alpha}$$
$$x_3 =f(x_2)=f\left (\dfrac {\alpha}{1-2\alpha}\right)=\dfrac {\dfrac {\alpha}{1-2\alpha}}{1-\dfrac {\alpha}{1-2\alpha}}=\dfrac {\alpha}{1-3\alpha}$$
Analysis the values $$x_1, x_2, x_3$$, we can say that $$x_{2011}=\dfrac {\alpha}{1-20011\alpha}=-\dfrac {1}{2012}$$
$$\Rightarrow \ 2012\alpha =-1+2011\alpha$$
$$\Rightarrow \ 2012\alpha -2011\alpha =-1$$
$$\Rightarrow \ \boxed {\alpha =-1}$$
Question 9
1 / -0

Domain of $$f\left( x \right) =\sqrt { 2{ \left\{ x \right\} }^{ 2 }-3\left\{ x \right\} +1 }$$ where $$\left\{ . \right\}$$ denotes the fractional part, in $$\left\{ . \right\}$$ is

A
$$\left[ -1,1 \right] \sim \left( \dfrac { 1 }{ 2 } ,1 \right)$$

B
$$\left[ -1,-\dfrac { 1 }{ 2 } \right] \bigcup \left[ 0,\dfrac { 1 }{ 2 } \right] \bigcup \left\{ 1 \right\}$$

C
$$\left[ -1,\dfrac { 1 }{ 2 } \right]$$

D
$$\left[ -\dfrac { 1 }{ 2 } ,1 \right]$$

Solution
Question 10
1 / -0

Consider set $$A={1,2,3,4}$$ and set $$B={0,2,4,6,8}$$, then the number of one-one function set $$A$$ to set $$B$$ in which $$f(i)\neq i$$ is,

A
$$84$$

B
$$78$$

C
$$42$$

D
$$24$$