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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Question 1

{−5+202,−5−212}\displaystyle \left \{ \frac{-5+\sqrt{20}}{2},\frac{-5-\sqrt{21}}{2} \right \}{2−5+20​​,2−5−21​​}

{−5+212,−5−212}\displaystyle \left \{ \frac{-5+\sqrt{21}}{2} ,\frac{-5-\sqrt{21}}{2}\right \}{2−5+21​​,2−5−21​​}

{5−212,21−52}\displaystyle \left \{ \frac{5-\sqrt{21}}{2} ,\frac{21-\sqrt{5}}{2}\right \}{25−21​​,221−5​​}

Does not exist.

Solution

Question 2

equals 111

equals 000

equals 333

equals 444

Solution

Question 3

12+[g(x)+x]2\displaystyle \frac{1}{2+\left [ g\left ( x \right )+x \right ]^{2}}2+[g(x)+x]21​

11+[g(x)−x]2\displaystyle \frac{1}{1+\left [ g\left ( x \right )-x \right ]^{2}}1+[g(x)−x]21​

12+[g(x)−x]2\displaystyle \frac{1}{2+\left [ g\left ( x \right )-x \right ]^{2}}2+[g(x)−x]21​

12−[g(x)−x]2\displaystyle \frac{1}{2-\left [ g\left ( x \right )-x \right ]^{2}}2−[g(x)−x]21​

Solution

Question 4

g(y)=4+y+34\displaystyle g\left ( y \right )=4+\frac{y+3}{4}g(y)=4+4y+3​

g(y)=2y−34\displaystyle g\left ( y \right )=\frac{2y-3}{4}g(y)=42y−3​

g(y)=3y+43\displaystyle g\left ( y \right )=\frac{3y+4}{3}g(y)=33y+4​

g(y)=y−34\displaystyle g\left ( y \right )=\frac{y-3}{4}g(y)=4y−3​

Solution

Question 5

1

0

-1

not defined

Solution

Question 6

aaa

−a-a−a

bbb

−b-b−b

Solution

Question 7

mn\displaystyle m^{n}mn

nm\displaystyle n^{m}nm

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

n!\displaystyle n!n!

Solution

Question 8

−f(x)-f(x)−f(x)

3f(x)3f(x)3f(x)

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

none of these

Solution

Question 9

5−(1−x5)1/35-(1-x^{5})^{1/3}5−(1−x5)1/3

5+(1−x5)1/35+(1-x^{5})^{1/3}5+(1−x5)1/3

5+(1+x5)1/35+(1+x^{5})^{1/3}5+(1+x5)1/3

5−(1+x5)1/35-(1+x^{5})^{1/3}5−(1+x5)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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$\displaystyle \left \{ \frac{-5+\sqrt{20}}{2},\frac{-5-\sqrt{21}}{2} \right \}$

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

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

equals $1$

equals $0$

equals $3$

equals $4$

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

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

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

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

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

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

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

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

$a$

$-a$

$b$

$-b$

$\displaystyle m^{n}$

$\displaystyle n^{m}$

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

$\displaystyle n!$

$-f(x)$

$3f(x)$

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

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

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

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

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