Relations and Functions Test

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Relations and Functions Test - 46

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Weekly Quiz Competition

Question 1
1 / -0

Let $$\displaystyle f:R \rightarrow R $$ be defined as $$\displaystyle f(x)= x^{2}+5x+9$$ then $$\displaystyle f^{-1}(8) $$ equals to

A
$$\displaystyle \left \{ \frac{-5+\sqrt{20}}{2},\frac{-5-\sqrt{21}}{2} \right \}$$

B
$$\displaystyle \left \{ \frac{-5+\sqrt{21}}{2} ,\frac{-5-\sqrt{21}}{2}\right \}$$

C
$$\displaystyle \left \{ \frac{5-\sqrt{21}}{2} ,\frac{21-\sqrt{5}}{2}\right \}$$

D
Does not exist.

Solution

Let $$\displaystyle f^{-1}(8)= x, $$ then $$\displaystyle f(x)= 8$$

$$\implies x^{2}+5x+9= 8 $$

$$\implies x^{2}+5x+1= 0 $$

$$\displaystyle

\Rightarrow x= \frac{-5\pm \sqrt{25-4}}{2} $$

$$\displaystyle x=

\frac{-5+\sqrt{21}}{2},\frac{-5-\sqrt{21}}{2}\in R $$

Hence

$$\displaystyle f^{-1}(8)= \left \{ \frac{-5+\sqrt{21}}{2}

,\frac{-5-\sqrt{21}}{2}\right \}$$
Question 2
1 / -0

If $$\displaystyle f(x)= (x-1)+(x+1)$$ and
$$\displaystyle g(x)= f\left \{ f(x) \right \}$$ then $$\displaystyle {g}'(3)$$

A
equals $$1$$

B
equals $$0$$

C
equals $$3$$

D
equals $$4$$

Solution

Simplifying,$$f(x)$$ we get

$$f(x)=2x$$

Hence

$$f(f(x))=2(f(x))$$

$$=2(2x)$$

$$=4x$$

Hence

$$g(x)=f(f(x))$$

$$=4x$$

Thus

$$g'(x)=4$$

Hence $$g'(3)=4$$ .
Question 3
1 / -0

If $$\displaystyle f\left ( x \right )=x+\tan x$$ and $$\displaystyle g^{-1}=f$$ then $$\displaystyle g{}'\left ( x \right )$$ equals

A
$$\displaystyle \frac{1}{2+\left [ g\left ( x \right )+x \right ]^{2}}$$

B
$$\displaystyle \frac{1}{1+\left [ g\left ( x \right )-x \right ]^{2}}$$

C
$$\displaystyle \frac{1}{2+\left [ g\left ( x \right )-x \right ]^{2}}$$

D
$$\displaystyle \frac{1}{2-\left [ g\left ( x \right )-x \right ]^{2}}$$

Solution

$$f\left( x \right) =x+\tan { x } \dashrightarrow \left( i \right) \\ { g }^{ -1 }(x)=f(x)\dashrightarrow \left( ii \right) \\ f\left( g(x) \right) =g(x)+\tan { g(x) } \quad \left( from\quad equation\quad \left( i \right) \quad \right) \\ { g }^{ -1 }\left( g(x) \right) =g(x)+\tan { g(x) } \left( from\quad equation\quad \left( ii \right) \quad \right) \\ x=g(x)+\tan { g(x) } \quad \dashrightarrow \left( iii \right) \quad \left( { \because \quad g }^{ -1 }\left( g(x) \right) =x \right) \\ $$
Differentiating the above equation w.r.t x,
$$\\ 1=g^{ ' }\left( x \right) +(\sec ^{ 2 }{ g(x) } )g^{ ' }\left( x \right) \\ g^{ ' }\left( x \right) =\dfrac { 1 }{ 1+\sec ^{ 2 }{ g(x) } } =\dfrac { 1 }{ 2+\tan ^{ 2 }{ g\left( x \right) } } \quad \left( \because \sec ^{ 2 }{ g(x) } =1+\tan ^{ 2 }{ g\left( x \right) } \quad \right) \\ from\quad eqn\quad \left( iii \right) ,\quad \tan { g(x) } =x-g\left( x \right) \\ \therefore g^{ ' }\left( x \right) =\dfrac { 1 }{ 2+{ (x-g\left( x \right) ) }^{ 2 } } $$
Question 4
1 / -0

Let $$\displaystyle f:N\rightarrow Y$$ be a function defined as $$f(x)=4x+3$$ where $$\displaystyle Y=\left \{ y \in N:y=4x+3 \right \}$$ for some $$\displaystyle x\in N$$ such that $$f$$ is invertible then its inverse is

A
$$\displaystyle g\left ( y \right )=4+\frac{y+3}{4}$$

B
$$\displaystyle g\left ( y \right )=\frac{2y-3}{4}$$

C
$$\displaystyle g\left ( y \right )=\frac{3y+4}{3}$$

D
$$\displaystyle g\left ( y \right )=\frac{y-3}{4}$$

Solution

Let $$\displaystyle y=f(x)\Rightarrow x= f^{-1}(y)$$ ........(A)
$$\displaystyle \therefore y=4x+3$$
$$\displaystyle \Rightarrow 4x=y-3 \Rightarrow x=\frac{y-3}{4}$$
$$\displaystyle \Rightarrow f^{-1}(y)=\frac{y-3}{4}$$ .......(using A)
$$\displaystyle \Rightarrow g(y)=\frac{y-3}{4}$$
Question 5
1 / -0

Let f(x)=tan x, x$$\displaystyle \epsilon \left [ -\frac{\pi }{2},\frac{\pi }{2} \right ]$$ and $$\displaystyle g\left (x \right )=\sqrt{1-x^{2}}$$ Determine $$g o f(1)$$.

A
1

B
0

C
-1

D
not defined

Solution
Question 6
1 / -0

If $$\displaystyle f\left ( x \right )=\frac{ax+b}{cx+d}$$ and $$\displaystyle \left ( fof \right )x=x,$$ then d=?

A
$$a$$

B
$$-a$$

C
$$b$$

D
$$-b$$

Solution

$$\displaystyle \left ( fof \right )x=x\Rightarrow f\left [ f\left ( x \right ) \right ]=x$$ or $$\displaystyle f\left [ \frac{ax+b}{cx+d} \right ]=x$$ or $$\displaystyle\frac{a\left [ \frac{ax+b}{cx+d}\right ]+b}{c\left [\frac{ax+b}{cx+d} \right ]+d}=x$$ $$\displaystyle \therefore \frac{x\left ( a^{2}+bc \right )+b\left ( a+d \right )}{cx\left ( a+d \right )+\left ( bc+d^{2} \right )}=x$$ Clearly if $$\displaystyle a+d=0$$ or $$\displaystyle d=-a$$ then in that case $$\displaystyle L.H.S.=\frac{x\left ( a^{2}+bc \right )+0}{0+\left ( bc+a^{2} \right )}=x$$ $$\displaystyle \because d=-a$$
Question 7
1 / -0

The total number of injective mappings from a set with $$m$$ elements to a set with $$n$$ elements,$$\displaystyle m\leq n,$$ is

A
$$\displaystyle m^{n}$$

B
$$\displaystyle n^{m}$$

C
$$\displaystyle \frac{n!}{\left ( n-m \right )!}$$

D
$$\displaystyle n!$$

Solution

Let $$A=\{a_1,a_2, a_3.....a_m\}$$
and $$B=\{b_1,b_2, b_3.....b_n\}$$ where $$m \le n$$
Given $$f:A\rightarrow B$$ be an injective mapping.
So, for $$a_1 \in A$$, there are $$n$$ possible choices for $$f(a_1)\in B$$.
For $$a_2 \in A$$, there are $$(n-1)$$ possible choices for $$f(a_2)\in B$$.
Similarly for $$a_m \in A$$, there are $$(n-m-1)$$ choices for $$f(a_m)\in B$$
So, there are $$n(n-1)(n-2).....(n-m-1)=\dfrac{n!}{(n-m)!}$$ injective mapping from $$A$$ to $$B.$$
Question 8
1 / -0

Given $$\displaystyle f\left ( x \right )=\log \left ( \frac{1+x}{1-x} \right )$$ and $$\displaystyle g\left ( x \right )=\frac{3x+x^{3}}{1+3x^{2}}, fog (x)$$ equals

A
$$-f(x)$$

B
$$3f(x)$$

C
$$\displaystyle \left [ f\left ( x \right ) \right ]^{3}$$

D
none of these

Solution

Given $$f(x)\, =\, \log\, \left( \displaystyle \dfrac{1\, +\, x}{1\, -\,

x} \right)$$ and $$g(x)\, =\, \displaystyle \dfrac{3x\, +\, x^{3}}{1\,

+\, 3x^{2}}$$

$$\therefore fog (x)\, =\, \log \left( \displaystyle \dfrac{1\, +\, \dfrac{3x\, +\,

x^{3}}{1\, +\, 3x^{2}}}{1\, -\,\displaystyle \dfrac{3x\, +\, x^{3}}{1\,

+\, 3x^{2}}} \right)$$

$$=\, \log\, \left(\displaystyle \dfrac{(1\, +\,

x)^{3}}{(1\, -\, x)^{3}} \right)\, =\, 3\, \log\, \left(\displaystyle

\dfrac{1\, +\, x}{1\, -\, x} \right)\, =\, 3\, f(x)$$
Question 9
1 / -0

The inverse of the function$$\displaystyle f(x)=(1-(x-5)^{3})^{1/5} $$is

A
$$5-(1-x^{5})^{1/3}$$

B
$$5+(1-x^{5})^{1/3}$$

C
$$5+(1+x^{5})^{1/3}$$

D
$$5-(1+x^{5})^{1/3}$$

Solution

Let $$y=(1-(x-5)^{3})^{\frac{1}{5}}$$

$$y^{5}=1-(x-5)^{3}$$

$$(x-5)^{3}=1-y^{5}$$

$$x-5=(1-y^{5})^{\frac{1}{3}}$$

$$x=5+(1-y^{5})^{\frac{1}{3}}$$

Replacing $$x$$ with $$y$$ gives us

$$f^{-1}(x)=5+(1-x^{5})^{\frac{1}{3}}$$
Question 10
1 / -0

Are the following sets of ordered pairs functions? If so, examine whether the mapping is surjective or injective :
{(x, y): x is a person, y is the mother of x}

A
injective (one- one ) and surjective (into)

B
injective (one- one ) and not surjective (into)

C
not injective (one- one ) and surjective (into)

D
not injective (one- one ) and not surjective (into)

Solution

We have {(x, y) : x is a person, y is the mother of x}. Clearly each person 'x' has only one biological mother. So above set of ordered pair is a function. Now more than one person may have same mother. So function is many-one and surjective.

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Relations and Functions Test - 46

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Relations and Functions Test - 46

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Question 1

$$\displaystyle \left \{ \frac{-5+\sqrt{20}}{2},\frac{-5-\sqrt{21}}{2} \right \}$$

$$\displaystyle \left \{ \frac{-5+\sqrt{21}}{2} ,\frac{-5-\sqrt{21}}{2}\right \}$$

$$\displaystyle \left \{ \frac{5-\sqrt{21}}{2} ,\frac{21-\sqrt{5}}{2}\right \}$$

Does not exist.

Solution

Question 2

equals $$1$$

equals $$0$$

equals $$3$$

equals $$4$$

Solution

Question 3

$$\displaystyle \frac{1}{2+\left [ g\left ( x \right )+x \right ]^{2}}$$

$$\displaystyle \frac{1}{1+\left [ g\left ( x \right )-x \right ]^{2}}$$

$$\displaystyle \frac{1}{2+\left [ g\left ( x \right )-x \right ]^{2}}$$

$$\displaystyle \frac{1}{2-\left [ g\left ( x \right )-x \right ]^{2}}$$

Solution

Question 4

$$\displaystyle g\left ( y \right )=4+\frac{y+3}{4}$$

$$\displaystyle g\left ( y \right )=\frac{2y-3}{4}$$

$$\displaystyle g\left ( y \right )=\frac{3y+4}{3}$$

$$\displaystyle g\left ( y \right )=\frac{y-3}{4}$$

Solution

Question 5

1

0

-1

not defined

Solution

Question 6

$$a$$

$$-a$$

$$b$$

$$-b$$

Solution

Question 7

$$\displaystyle m^{n}$$

$$\displaystyle n^{m}$$

$$\displaystyle \frac{n!}{\left ( n-m \right )!}$$

$$\displaystyle n!$$

Solution

Question 8

$$-f(x)$$

$$3f(x)$$

$$\displaystyle \left [ f\left ( x \right ) \right ]^{3}$$

none of these

Solution

Question 9

$$5-(1-x^{5})^{1/3}$$

$$5+(1-x^{5})^{1/3}$$

$$5+(1+x^{5})^{1/3}$$

$$5-(1+x^{5})^{1/3}$$

Solution

Question 10

injective (one- one ) and surjective (into)

injective (one- one ) and not surjective (into)

not injective (one- one ) and surjective (into)

not injective (one- one ) and not surjective (into)

Solution

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