Functions Test 36

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Functions Test 36

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Weekly Quiz Competition

Question 1
1 / -0

Let $$E=\left\{1,2,3,4\right\}$$ and $$F=\left\{1,2\right\}$$. Then the number of onto functions from $$E$$ to $$F$$ is

A
$$14$$

B
$$16$$

C
$$12$$

D
$$8$$

Solution

The total number of ways of distributing $$(1,2)$$ considering them as 2 objects among $$(1,2,3,4)$$ considering them as 4 people is equal to total no. of Onto functions from E to F.

There can be two ways of distributing $$(1,2),$$

Either we give both to one people

Or give $$1$$ to one people and $$2$$ to another.

So, Total no. of ways $$=4×1+^4C_2×2=16$$
Question 2
1 / -0

Number of one-one functions from A to B where $$n(A)=4, n(B)=5$$.

A
$$4$$

B
$$5$$

C
$$120$$

D
$$90$$

Solution

$$n(A)=4$$ and $$n(B)=5$$

For one-one mapping
4 elements can be selected out of 5 elements of set B in $${}^5C_4$$ ways
and then those 4 selected elements can be mapped with 4 elements of set A in $$4!$$ ways.

Number of one-one mapping from $$A$$ to $$B$$ $$={}^5C_4\times4!={}^5P_4=\dfrac{5!}{(5-4)!}=5!=120$$
Question 3
1 / -0

If $$f\left( x \right) =\sqrt { { x }^{ 2 }-4 } $$ and $$g\left( x \right) =\dfrac { x-1 }{ x-3 } $$ then number of integer elements, which are not in the domain of the function $$(f.g)(x)$$ equals

A
3

B
4

C
5

D
None of these

Solution
Question 4
1 / -0

Let $$f(x)=x^ {135}+x^ {125}-x^ {115}+x^ {5}+1$$. If $$f(x)$$ divided by $$x^ {3}-x$$, then the remainder is some function of $$x$$ say $$g(x)$$. Then $$g(x)$$ is an:-

A
one-one function

B
many one function

C
into function

D
onto function

Solution
Question 5
1 / -0

If f : $$R\rightarrow S$$, defined by f(x) =sin x -$$\sqrt{3}$$ cos x +1, is onto, then the interval of S is

A
[0, 3]

B
[-1, 1]

C
[0, 1]

D
[-1, 3]

Solution
Question 6
1 / -0

If $$ f(x) = \sqrt{2-x} $$ and $$ g(x) = \sqrt{1-2x} , $$ then the domain of $$ log(x) $$ is :

A
$$ ( - \infty , 1/2] $$

B
$$ (1/2 , \infty ] $$

C
$$ ( - \infty , 3/2] $$

D
None of these

Solution

$$f(x)=\sqrt{2-x} \quad g(x)=\sqrt{1-2 x}$$
Domain of $$\operatorname{fog}(x)$$ is equal to the domain of $$g(x)$$
$$\begin{array}{l}1-2 x \geq 0 \\ x \leq \frac{1}{2}\end{array}$$
$$x \in\left(-\infty, \frac{1}{2}\right]$$
Question 7
1 / -0

If $$ f : R \rightarrow R $$ be given by $$ f(x)=\left(3-x^{3}\right)^{\dfrac{1}{3}}, $$ then $$fof(x)$$ is

A
$$
x^{\dfrac{1}{3}}
$$

B
$$
1^{3}
$$

C
x

D
$$
\left(3-x^{3}\right)
$$

Solution

Given, $$ f(x)=\left(3-x^{3}\right)^{\dfrac{1}{3}} $$.
Now,
$$fof(x)$$
$$=f[f(x)]$$

$$ =\left(3-[f(x)]^{3}\right)^{\dfrac{1}{3}}, $$

$$ =\left(3-[(3-x^3)^{\dfrac{1}{3}}]^{3}\right)^{\dfrac{1}{3}}, $$

$$ =\left(3-[(3-x^3)]\right)^{\dfrac{1}{3}}, $$

$$=[x^3]^{\dfrac{1}{3}}$$

$$=x$$.
Question 8
1 / -0

Let : $$R\rightarrow R$$ defined as $$f\left( x \right) =\dfrac { x\left( x+1 \right) \left( { x }^{ 4 }+1 \right) +{ 2x }^{ 4 }+{ x }^{ 2 }+2 }{ { x }^{ 2 }+x+1 } $$

A
odd and one-one

B
even and one-one

C
many to one and even

D
many to one and neither even nor odd
Question 9
1 / -0

If domain of $$f$$ is $$D _ { 1 }$$ and domain of $$g$$ is $$D _ { 2 }$$ then domain of $$f + g$$ is :

A
$$ \left( D _ { 1 } / D _ { 2 } \right)$$

B
$$D _ { 1 } - \left( D _ { 1 } / D _ { 2 } \right)$$

C
$$D _ { 2 } / \left( D _ { 2 } / D _ { 1 } \right)$$

D
$$D _ { 1 } \cap D _ { 2 }$$

Solution

$${\textbf{Step - 1 : Use addition rule of two functions}}{\text{.}}$$

$${\text{Given two functions f and g such that domain of f is }}{{\text{D}}_1}{\text{ and domain of g is }}{{\text{D}}_2}.$$

$${\text{Now, }}\left( {{\text{f + g}}} \right)\left( {\text{x}} \right) = {\text{f}}\left( {\text{x}} \right) + {\text{g}}\left( {\text{x}} \right).$$

$${\textbf{Step - 2 : Deduce the domain of f + g}}{\text{.}}$$

$${\text{The domain of f + g must satisfy the individual domain of both f and g}}{\text{.}}$$

$${\text{which is only possible for the domain }}{{\text{D}}_1} \cap {{\text{D}}_2}.$$

$${\text{i}}{\text{.e the domain of f + g is }}{{\text{D}}_1} \cap {{\text{D}}_2}.$$

$${\textbf{Final answer: the domain of f + g is }}{{\textbf{D}}_1} \cap {{\textbf{D}}_2}.$$
Question 10
1 / -0

Let f : $$R\rightarrow R$$ be a function defined by f(x) = $${ x }^{ 3 }+{ x }^{ 2 }+3x+sin\times .$$ Then f is.

A
one-one & onto

B
one-one & into

C
many one & onto

D
many one & into

Solution

$$f(x)=x^{3}+x^{2}+3 x+\sin x$$
$$Suppose$$
$$f(x_{1})=f(x_{2})$$
$$x_{1}^{3}+x_{1}^{2}+3 x_{1}+\sin x_{1}=x_{2}+x_{2}+\sin x-\sin x_{2}$$
$$Equating\;above\;equations\;to \;0$$
$$x_{1}^{3}+x_{1}^{2}+3 x_{1}+\sin x_{1}=x_{2}+x_{2}+\sin x-\sin x_{2}=0$$
$$Solving\;them$$
$$It\; will\; not\; prove\; that$$
$$x_{1}=x_{2}$$
$$So,\;f(x)\;is\;one-one$$
$$Range\;f(x)=Real$$
$$Co-domain\;f(x)=Real$$
$$Range=Co-domain$$
$$Hence,\; f(x) \;is \;onto$$
$$So,\;option\; C \;is\;correct.$$

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Re-Attempt Weekly Quiz Competition

Functions Test 36

Result

Functions Test 36

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Rank

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SHARING IS CARING

Question 1

$$14$$

$$16$$

$$12$$

$$8$$

Solution

Question 2

$$4$$

$$5$$

$$120$$

$$90$$

Solution

Question 3

3

4

5

None of these

Solution

Question 4

one-one function

many one function

into function

onto function

Solution

Question 5

[0, 3]

[-1, 1]

[0, 1]

[-1, 3]

Solution

Question 6

$$ ( - \infty , 1/2] $$

$$ (1/2 , \infty ] $$

$$ ( - \infty , 3/2] $$

None of these

Solution

Question 7

$$ x^{\dfrac{1}{3}} $$

$$ 1^{3} $$

x

$$ \left(3-x^{3}\right) $$

Solution

Question 8

odd and one-one

even and one-one

many to one and even

many to one and neither even nor odd

Question 9

$$ \left( D _ { 1 } / D _ { 2 } \right)$$

$$D _ { 1 } - \left( D _ { 1 } / D _ { 2 } \right)$$

$$D _ { 2 } / \left( D _ { 2 } / D _ { 1 } \right)$$

$$D _ { 1 } \cap D _ { 2 }$$

Solution

Question 10

one-one & onto

one-one & into

many one & onto

many one & into

Solution

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$$
x^{\dfrac{1}{3}}
$$

$$
1^{3}
$$

$$
\left(3-x^{3}\right)
$$