Functions Test 22

Result

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

Score

-

Rank

-

SHARING IS CARING

Question 1

5/3

3/2

2/3

3/5

Solution

Question 2

720

120

24

12

Solution

Question 3

$$\dfrac1x$$

$$x$$

$$x^2$$

$$x^3$$

Solution

Question 4

$$-36$$

$$1014$$

$$-90$$

$$120$$

Solution

Question 5

2007

100

0

None of these

Solution

Question 6

is 45

is 50

is 90

Cannot be determined

Solution

Question 7

$$2 (3x + 1)$$

$$2 ( 3x + 2)$$

$$3 (2x + 1 )$$

$$3 ( 3x + 1 )$$

Solution

Question 8

$$R - {1, -1}$$

$$R - {1}$$

$$R - {-1}$$

$$R-{ 0 }$$

Solution

Question 9

12

16

4

5

Solution

Question 10

$$x$$

$$x^n$$

$$p^{1/n}$$

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

Solution

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