Functions Test 22

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Functions Test 22

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Weekly Quiz Competition

Question 1
1 / -0

If f(x) = 2x+1 and g(x) = 3x-5 then find $\left ( fog \right )^{-1}\left ( 0 \right )$

A
5/3

B
3/2

C
2/3

D
3/5

Solution

$(fog)(x) = f(g(x)) = 2(3x-5) + 1 = 6x-10 + 1 = 6x-9$

Let $(fog)(x)= y$
Then, to find $(fog)^{-1} (x)$ , we find $x$ in terms of $y$

So, $6x-9 = y$
$=> x = \dfrac {y+9}{6}$

$=> (fog)^{-1} (x)= \dfrac {y+9}{6}$

$=> (fog)^{-1} (0)= \dfrac {0+9}{6} = \dfrac {3}{2}$
Question 2
1 / -0

If X = {2,3,5,7,11} and Y = {4,6,8,9,10} then find the number of one-one functions from X to Y

A
720

B
120

C
24

D
12

Solution

Both sets X and Y have the same number of elements.

When two sets have same number of $n$ elements, then the number of one to one functions from one set to the other is $n!$

So, number of one to one functions from X to Y $= 5 ! = 5 \times 4 \times 3 \times 2 \times 1 = 120$
Question 3
1 / -0

If $\displaystyle f\left ( x \right )=\left ( 1-x^{3} \right )^{\frac{1}{3}}$ , then find $fof(x)$

A
$\dfrac1x$

B
$x$

C
$x^2$

D
$x^3$

Solution

Given, $\displaystyle f\left ( x \right )=\left ( 1-x^{3} \right )^{\frac{1}{3}}$
Therefore, $\displaystyle fof\left ( x \right )=f\left [ f\left ( x \right ) \right ]$
$\displaystyle =f\left ( \left ( 1-x^{3} \right )^{\frac{1}{3}} \right )=\left [ 1-\left \{ \left ( 1-x^{3} \right )^{\frac{1}{3}} \right \}^{3} \right ]^{\frac{1}{3}}$
$\displaystyle =\left [ 1-\left ( 1-x \right )^{3} \right ]^{\frac{1}{3}}=\left ( x^{3} \right )^{\frac{1}{3}}=x$
Question 4
1 / -0

Find $\left( f\circ g \right) \left( 3 \right)$ when $f\left( x \right) =7x-6$ and $g\left( x \right) =5{ x }^{ 2 }-7x-6$ .

A
$-36$

B
$1014$

C
$-90$

D
$120$

Solution

We have, $f(x)=7x-6$ and $g(x)=5x^2-7x-6$
Then, $(f\circ g)(x)$
$=f(g(x))$
$=f(5x^2-7x-6)$
$=7(5x^2-7x-6)-6$
$=35x^2-49x-42-6$
$=35x^2-49x-48$
$\therefore (f\circ g)(3)$
$=35(3)^2-49(3)-48$
$=315-147-48$
$=120$
So, $\text{D}$ is the correct option.
Question 5
1 / -0

If f is a constant function and f(100)=100 then f(2007)=_____

A
2007

B
100

C
0

D
None of these

Solution

A constant function $f(x)$ will have the result as a constant for any value of $x$

So, if $f(100) = 100$
then $f(2007 )$ is also $100$
Question 6
1 / -0

If f(x) + f(1-x) = 10 then the value of $\displaystyle f\left ( \frac{1}{10} \right )+f\left ( \frac{2}{10} \right )+.........+f\left ( \frac{9}{10} \right )$

A
is 45

B
is 50

C
is 90

D
Cannot be determined

Solution

$f(\dfrac {1}{10}) + f(\dfrac {2}{10}) + f(\dfrac {3}{10}) + f(\dfrac {4}{10}) + f(\dfrac {5}{10}) + f(\dfrac {6}{10}) + f(\dfrac {7}{10}) + f(\dfrac {8}{10}) + f(\dfrac {9}{10})$

$= [f(\dfrac {1}{10}) + f(\dfrac {9}{10}) ] + [ f(\dfrac {2}{10}) + + f(\dfrac {8}{10}) ] + [ f(\dfrac {3}{10}) + + f(\dfrac {7}{10}) ] + [ f(\dfrac {4}{10}) + f(\dfrac {6}{10}) ] + f(\dfrac {5}{10}) ]$

$= 10 + 10 + 10 + 10 + 5 = 45$
Question 7
1 / -0

If $f (x) = 2x - 1$ and $g (x) = 3x + 2$ , then find $(fog) (x)$ :

A
$2 (3x + 1)$

B
$2 ( 3x + 2)$

C
$3 (2x + 1 )$

D
$3 ( 3x + 1 )$

Solution

$fog(x)=2(3x+2)-1=6x+4-1=6x+3=3(2x+1)$
Question 8
1 / -0

The domain of the function f(x)= log |x - 1| is ........

A
$R - {1, -1}$

B
$R - {1}$

C
$R - {-1}$

D
$R-{ 0 }$

Solution

Log is defined for all positive numbers but undefined for $0$ . Hence $x$ can attain any value other than those which cause expression $(x-1)$ to be null, that is, $x$ can not be $1$ .
Question 9
1 / -0

If f = {(1,3) (2,1) (3,4) (4,2)} and g = {(1,2) (2,3) (3,4) (4,1)} then find n(fog)

A
12

B
16

C
4

D
5

Solution

we can find $n(fog)$ by finding the range of the domain of $f$
In $(1,3)$ as the range $3$ is mapped to $4$ in $g$ , the ordered pair here for $fog$ is $(1,4)$
In $(2,1)$ as the range $1$ is mapped to $2$ in $g$ , the ordered pair here for $fog$ is $(2,2)$
In $(3,4)$ as the range $4$ is mapped to $1$ in $g$ , the ordered pair here for $fog$ is $(3,1)$
In $(4,2)$ as the range $2$ is mapped to $3$ in $g$ , the ordered pair here for $fog$ is $(4,3)$

As we have $4$ ordered pairs for $fog$ , we have $n(fog) = 4$
Question 10
1 / -0

If $f(x)\, =\, (p\, -\, x^n)^{1/n},\, p\, >\, 0$ and $n$ is a positive integer, then $f(f(x)) =$

A
$x$

B
$x^n$

C
$p^{1/n}$

D
$p\, -\, x^n$

Solution

Given, $f(x)=(p-x^{n})^{\dfrac{1}{n}}$

$\therefore f(f(x))=(p-((p-x^{n})^{\dfrac{1}{n}})^{n})^{\dfrac{1}{n}}$

$=(p-(p-x^{n}))^{\dfrac{1}{n}}$

$=(x^{n})^{\dfrac{1}{n}}$

$=x$

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Re-Attempt Weekly Quiz Competition

Functions Test 22

Result

Functions Test 22

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Question 1

5/3

3/2

2/3

3/5

Solution

Question 2

720

120

24

12

Solution

Question 3

1x\dfrac1xx1​

xxx

x2x^2x2

x3x^3x3

Solution

Question 4

−36-36−36

101410141014

−90-90−90

120120120

Solution

Question 5

2007

100

0

None of these

Solution

Question 6

is 45

is 50

is 90

Cannot be determined

Solution

Question 7

2(3x+1)2 (3x + 1)2(3x+1)

2(3x+2)2 ( 3x + 2)2(3x+2)

3(2x+1)3 (2x + 1 )3(2x+1)

3(3x+1)3 ( 3x + 1 )3(3x+1)

Solution

Question 8

R−1,−1R - {1, -1}R−1,−1

R−1R - {1}R−1

R−−1R - {-1}R−−1

R−0R-{ 0 }R−0

Solution

Question 9

12

16

4

5

Solution

Question 10

xxx

xnx^nxn

p1/np^{1/n}p1/n

p − xnp\, -\, x^np−xn

Solution

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$\dfrac1x$

$x$

$x^2$

$x^3$

$-36$

$1014$

$-90$

$120$

$2 (3x + 1)$

$2 ( 3x + 2)$

$3 (2x + 1 )$

$3 ( 3x + 1 )$

$R - {1, -1}$

$R - {1}$

$R - {-1}$

$R-{ 0 }$

$x$

$x^n$

$p^{1/n}$

$p\, -\, x^n$