Functions Test 26

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Functions Test 26

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Weekly Quiz Competition

Question 1
1 / -0

If $$f(x)=\dfrac{x+1}{x-1}$$ and $$g(x)=2x-1, f[g(x)]=$$

A
$$\dfrac{x-1}{x}$$

B
$$\dfrac{x}{x+1}$$

C
$$\dfrac{x+1}{x}$$

D
$$\dfrac{x}{x-1}$$

E
$$\dfrac{2x-1}{2x+1}$$

Solution

Given, $$f(x) = \dfrac { x+1 }{ x-1 } $$ and $$g(x) = 2x-1$$
$$\therefore f(g(x)) = f(2x-1)$$
$$ = \dfrac { 2x-1+1 }{ 2x-1-1 } $$
$$=\dfrac { 2x }{ 2x-2 } $$
$$=\dfrac { x }{ x-1 } $$
Question 2
1 / -0

If $$\left\{ \left( 7,11 \right) ,\left( 5,a \right) \right\} $$ represents a constant function, then the value of '$$a$$' is :

A
$$7$$

B
$$11$$

C
$$5$$

D
$$9$$

Solution

Since we know that in constant function image of every element is same.
Here, image of 7 is 11, so image of any element will be 11.
Therefore, image of 5 is too 11.
Hence, $$a=11$$
Question 3
1 / -0

The number of onto functions from the set $$\{1, 2, .........., 11\}$$ to set $$\{1, 2, ....., 10\}$$ is

A
$$5\times \underline{|11}$$

B
$$\underline{|10}$$

C
$$\dfrac{\underline{|11}}{2}$$

D
$$10\times \underline{|11}$$

Solution

Finding no. of such functions is same as putting 11 balls in 10 boxes with each box getting at least one ball. Then the only possible way is to put 2 balls in a box which can be done in $$^{11}C_2$$ ways.

$$^{11}C_{2}\times ^{9}C_{1} \times ^8C_1...................^1C_1$$

$$=\dfrac{11\times 10}{2}\times 9\times8\times7.................................$$

$$11!\times 5$$
Question 4
1 / -0

If $$f(x) = \log_{e}\left (\dfrac {1 + x}{1 - x}\right ), g(x) = \dfrac {3x + x^{3}}{1 + 3x^{2}}$$ and $$go f(t) = g(f(t))$$, then what is $$go f\left (\dfrac {e - 1}{e + 1}\right )$$ equal to?

A
$$2$$

B
$$1$$

C
$$0$$

D
$$\dfrac {1}{2}$$

Solution

$$f\left(\dfrac{e-1}{e+1}\right)=\log_e\left\lbrace\dfrac{1+\left(\dfrac{e-1}{e+1}\right)}{1-\left(\dfrac{e-1}{e+1}\right)}\right\rbrace=\log_e\left(\dfrac{2e}{2}\right)=\log_ee=1$$
$$\therefore gof\left(\dfrac{e-1}{e+1}\right)=g\left(f\left(\dfrac{e-1}{e+1}\right)\right)=g(1)=\dfrac{3+1}{1+3}=1$$
Question 5
1 / -0

The number of real linear functions $$f(x)$$ satisfying $$f(f(x))=x+f(x)$$ is

A
$$0$$

B
$$4$$

C
$$5$$

D
$$2$$

Solution

Since $$f(x)$$ is a linear function, so take $$f(x)=ax+b$$
Given $$f(f(x))=x+f(x)$$
$$\Rightarrow$$ $$a(ax+b)+b=x+(ax+b)$$
Comparing the coefficients we get,
$$ { a }^{ 2 }=a+1$$ and $$ab+b=b$$
$$a^{ 2 }-a-1=0$$ and $$ab=0\Rightarrow a\ne 0,b=0$$
$$\therefore a=\cfrac { 1\pm \sqrt { 5 } }{ 2 } $$ and $$b=0$$
Thus $$f(x)=\left( \cfrac { 1\pm \sqrt { 5 } }{ 2 } \right) x+0$$
So there are two possible such functions exists
Question 6
1 / -0

Consider the function $$f(x)=\displaystyle\frac{x-1}{x+1}$$. What is $$f(f(x))$$ equal to?

A
$$x$$

B
$$-x$$

C
$$-\displaystyle\frac{1}{x}$$

D
None of the above

Solution

$$f(x)=\dfrac{x-1}{x+1}$$

$$f(f(x))=\dfrac{\dfrac{x-1}{x+1} - 1}{\dfrac{x-1}{x+1}+1}$$

$$f(f(x)) = \dfrac{-1}{x}$$

hence $$Ans$$ is $$option C$$
Question 7
1 / -0

Let N denote the set of all non-negative integers and Z denote the set of all integers. The function $$ f : Z \rightarrow N$$ given by f(x) = |x| is :

A
One-one but not onto

B
Onto but not one-one

C
Both one-one and onto

D
Neither one-one nor onto

Solution

$$f: X\rightarrow Y$$
For a function to be one one or injective, every element in the domain is the image of at most one element of it's co-domain.
In simple words, no value of $$y$$ must be same for $$2$$ or more different values of $$x$$.
For $$f(x)=\left| x \right| $$, we see that $$f(a)=f(-a)$$, for $$a\in Z$$
Hence, the function is not one one

For a function $$f: X\rightarrow Y$$, to be surjective,
every element $$y$$ in the co-domain Y must be linked with at least one element $$x$$ in the domain.
Every element in the co-domain of $$f(x)=\left| x \right| $$ is linked to at-least one element in domain.
Thus, $$f(x)=\left| x \right| $$ is onto but not one one.
Hence, b is correct.
Question 8
1 / -0

If $$f:[0, \infty)\to [0,\infty)$$ and $$f(x) = \dfrac{x}{1+x}$$, then $$f$$ is

A
One-one and onto

B
One-one but not onto

C
Onto but not one-one

D
Neither one-one nor onto

Solution

Here, $$f:[0, \infty) \rightarrow [0, \infty)$$ i.e, domain is $$[0, \infty)$$ and codomain is $$[0, \infty)$$
For one-one
$$f(x) = \dfrac{x}{1+x}$$
$$f'(x) = \dfrac{1}{(1+x)^2} > 0, \forall x \in [0, \infty)$$
$$\therefore f(x)$$ is increasing in this domain. Thus $$f(x)$$ is one-one in its domain
For onto (we find range)
$$f(x) = \dfrac{x}{1+x}$$ i.e. $$y =\dfrac{x}{1+x}$$
$$\Rightarrow y+yx=x$$
$$\Rightarrow x=\dfrac{y}{1-y}$$
$$\Rightarrow x=\dfrac{y}{1-y} \ge 0$$ as $$x \ge 0$$
$$\therefore 0 \le y \neq 1$$ and $$y<1$$
$$\therefore$$ Range $$\neq$$ Codomain
$$\therefore f(x)$$ is one-one but not onto
Question 9
1 / -0

Let $$f(x)=\dfrac{x+1}{x-1}$$ for all $$x \neq 1$$.
Let
$$f^1(x)=f(x), f^2(x)=f(f(x))$$ and generally
$$f^n(x)=f(f^{n-1}(x)) $$ for $$n > 1$$
Let $$P= f^1(2)f^2(3)f^3(4)f^4(5)$$
Which of the following is a multiple of P ?

A
$$125$$

B
$$375$$

C
$$250$$

D
$$147$$

Solution

$$f^{ 1 }(x)=\dfrac { x+1 }{ x-1 } $$

$$f^{ 2 }(x)=\dfrac { \dfrac { x+1 }{ x-1 } +1 }{ \dfrac { x+1 }{ x-1 } -1 }$$

$$=\dfrac { 2x }{ 2 } =x$$
Now, $$f^{3}(x)=f(f^{2}(x))=f(x)=f^{1}(x)$$
This will repeat so,
$$f^{ 3 }(x)=f^{ 1 }(x)$$ $$ f^{ 4 }(x)=f^{ 2 }(x)$$
We have,
$$f^{ 1 }(2)=3,$$ $$ f^{ 2 }(3)=3,$$ $$ f^{ 3 }(4)=\dfrac { 5 }{ 3 }, $$ $$f^4(5)=5$$
So, $$P= 75$$ and $$375$$ is the multiple of $$75$$.
Hence, the answer is $$375$$.
Question 10
1 / -0

Let $$M$$ be the set of all $$2 \times 2$$ matrices with entries from the set of real numbers $$R$$. Then the function $$ f : M \rightarrow R$$ defined by $$f\left( A \right) =\left| A \right|$$ for every $$ A\in M$$, is

A
One-one and onto

B
Neither one-one nor onto

C
One-one but not onto

D
Onto but not one-one

Solution

$$ f : M \rightarrow R$$ defined by $$f\left( A \right) =\left| A \right|$$ for every $$ A\in M$$
The function is the determinant of the matrix
We know that, two different matrices can have a same determinant
For example, $$A=\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$$ and $$B=\begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}$$
Then $$|A|=1=|B|$$, but $$A\neq B$$
So, the function will not be one-one.
Now, determinants can have any real values, so range of the function will be $$R$$.
Thus, the function will be onto.
Hence, option D is correct

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Re-Attempt Weekly Quiz Competition

Functions Test 26

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Functions Test 26

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Question 1

$$\dfrac{x-1}{x}$$

$$\dfrac{x}{x+1}$$

$$\dfrac{x+1}{x}$$

$$\dfrac{x}{x-1}$$

$$\dfrac{2x-1}{2x+1}$$

Solution

Question 2

$$7$$

$$11$$

$$5$$

$$9$$

Solution

Question 3

$$5\times \underline{|11}$$

$$\underline{|10}$$

$$\dfrac{\underline{|11}}{2}$$

$$10\times \underline{|11}$$

Solution

Question 4

$$2$$

$$1$$

$$0$$

$$\dfrac {1}{2}$$

Solution

Question 5

$$0$$

$$4$$

$$5$$

$$2$$

Solution

Question 6

$$x$$

$$-x$$

$$-\displaystyle\frac{1}{x}$$

None of the above

Solution

Question 7

One-one but not onto

Onto but not one-one

Both one-one and onto

Neither one-one nor onto

Solution

Question 8

One-one and onto

One-one but not onto

Onto but not one-one

Neither one-one nor onto

Solution

Question 9

$$125$$

$$375$$

$$250$$

$$147$$

Solution

Question 10

One-one and onto

Neither one-one nor onto

One-one but not onto

Onto but not one-one

Solution

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