Functions Test 25

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Functions Test 25

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Weekly Quiz Competition

Question 1
1 / -0

If $$f(x) = 4x - 3$$ and $$g(x) = x - 4$$, determine which of the following composite function has a value of $$-11$$.

A
$$f(g(2))$$

B
$$g(f(2))$$

C
$$g(f(3))$$

D
$$f(g(3))$$

E
$$f(g(4))$$

Solution

Given, $$f(x)= 4x-3, g(x)=x-4$$
$$\Rightarrow f(g(2)) = f(-2) = -11$$
$$\Rightarrow g(f(2))=g(5) = 1$$
$$\Rightarrow g(f(3)) = g(9) = 5$$
$$\Rightarrow f(g(3)) = f(-1) = -7$$
$$\Rightarrow f(g(4)) = f(0) = -3$$
Question 2
1 / -0

If $$f(x) = x^{2} - 10$$ and $$g(x) = 4x + 3$$, calculate the value of $$f(g(2))$$.

A
$$-24$$

B
$$-21$$

C
$$12$$

D
$$27$$

E
$$111$$

Solution

Given, $$f(x)=x^2-10$$ and $$g(x)=4x+3$$
$$\therefore g(2) = 4 \times 2 + 3 = 8+3 = 11$$
Then, $$f(g(2)) = f(11) = { 11 }^{ 2 }-10=121-10=111$$
Question 3
1 / -0

Let f : $$N \rightarrow N$$ defined by $$f(n)=\left\{\begin{matrix}
\dfrac{n+1}{2} & \text{if }\, n \, \text{is odd} \\
\dfrac{n}{2} & \text{if}\, n \, \text{is even}
\end{matrix}\right.$$
then $$f$$ is.

A
Many-one and onto

B
One-one and not onto

C
Onto but not one-one

D
Neither one-one nor onto

Solution

By substituting values of $$n$$, the elements in the range of function are,
$$\{1,2,3,4,5,6,...............\}$$, i.e. set of all natural number
So the range and co-domain are same, ergo function is onto.
Also $$f(1)=f(2)$$, so the function is many one.

Hence the given function is many-one and onto
Question 4
1 / -0

If $$f: R\rightarrow R$$ is defined by $$f(x) = \dfrac {x}{x^{2} + 1}$$, find $$f(f(2))$$

A
$$\dfrac {1}{29}$$

B
$$\dfrac {10}{29}$$

C
$$\dfrac {29}{10}$$

D
$$29$$

Solution

We have, $$f(x)=\dfrac{x}{x^2+1}$$
So, $$f(2)=\dfrac{2}{2^2+1}=\dfrac{2}{5}$$
Hence $$f(f(2))=f\left(\dfrac{2}{5}\right)=\dfrac{\frac{2}{5}}{\left(\frac{2}{5}\right)^2+1}=\dfrac{2\cdot 5}{5^2+2^2}=\dfrac{10}{29}$$
Question 5
1 / -0

If $$f(x) = 3x - 5$$ and $$g(x) = x^2 + 1, f [g(x)] =$$

A
$$3x^2-5$$

B
$$3x^2+6$$

C
$$x^2-5$$

D
$$3x^2-2$$

E
$$3x^2+5x-2$$

Solution

Given $$f(x) = 3x-5$$ and $$g(x) = {x}^{2}+1$$
$$\therefore f(g(x)) = f({x}^{2}+1) $$
$$= 3({x}^{2}+1)-5 $$
$$= 3{x}^{2}-2$$
Question 6
1 / -0

If the function $$f : R \rightarrow R$$ is defined by $$f(x) = (x^2+1)^{35} \forall \in R$$, then $$f$$ is

A
One-one but not onto

B
Onto but not one-one

C
Neither one-one nor onto

D
Both one-one and onto

Solution

$$f(x)=(x^2+1)^{35}$$

$$f'(x)=35(x^2+1)^{34}(2x)=70(x^2+1)^{34}x$$

So for $$x<0,$$ $$f'(x)<0$$ and for $$x>0, f'(x)>0$$

That means the function is increasing for $$x>0$$ and decreasing for $$x>0$$
So the function is many one,

Also the range of $$f(x)$$ is $$[1,\infty)$$ which is not same as codomain
Hence $$f$$ is into function

Therefore option $$C$$ is correct choice
Question 7
1 / -0

Let $$f(x) = 2^{100}x+1$$
$$g(x) = 3^{100}x+1$$
Then the set of real numbers x such that $$f(g(x)) = x$$ is

A
Empty

B
A singleton

C
A finite se with more than one element

D
Infinite

Solution

$$f(g(x)) =2^{100}(3^{100}x+1)+1=x$$

$$\Rightarrow x(6^{100}-1)=-(1+2^{100})$$

Clearly there is just one of value of x satisfying this equation. Option B is correct.
Question 8
1 / -0

If k is a positive constant different from 1, which of the following could be the graph of $$\displaystyle y-x=k(x+y)$$ in the xy-plane?

A

B

C

D

Solution

to predict graph of
$$y-x=k(x+y)$$
$$\Rightarrow y-x=k(x+y)$$
$$\Rightarrow y-x =kx+ky$$
$$\Rightarrow y=\left(\dfrac{1+k}{\bot-l}\right)x$$
we get the equation of form $$y=m$$
where $$m=\dfrac{\bot+k}{\bot -k}$$
this graph is a line that passes through the origin.
(option B ) is the only graph that fulfil the criteria.
Question 9
1 / -0

The Set $$A$$ has $$4$$ elements and the Set $$B$$ has $$5$$ elements then the number of injective mappings that can be defined from $$A$$ to $$B$$ is

A
$$144$$

B
$$72$$

C
$$60$$

D
$$120$$

Solution

$$\textbf{Step 1: Number of injective function from set A to B.}$$

$$\text{Given,}$$

$$\text{Number of elements in A is 4}$$

$$\& \text{ Number of elements in B is 5 }$$

$$\because \text{For injective functions distinct elements has distinct image}$$

$$\text{First element of A has 5 choices in B}$$

  $$\text{Second element of A has 4 choices in B}$$

  $$\text{Third element of A has 3 choices in B}$$

  $$\text{Last element of A has 2 choices in B}$$

$$\text{Total Injective functions = }5\times 4\times 3\times 2$$
$$=120$$

$$\textbf{Hence, Option D is correct.}$$
Question 10
1 / -0

Let $$f : R\rightarrow R$$ be defined by $$f(x) = \dfrac {1}{x} \ \ \forall \ x \ \in \ R$$, then $$f$$ is _____

A
One-one

B
Onto

C
Bijective

D
$$f$$ is not defined

Solution

The function $$f(x)=\dfrac{1}{x}$$ is not defined for $$x=0$$

But in the question, domain is given $$R$$

Therefore given function is not defined

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Re-Attempt Weekly Quiz Competition

Functions Test 25

Result

Functions Test 25

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SHARING IS CARING

Question 1

$$f(g(2))$$

$$g(f(2))$$

$$g(f(3))$$

$$f(g(3))$$

$$f(g(4))$$

Solution

Question 2

$$-24$$

$$-21$$

$$12$$

$$27$$

$$111$$

Solution

Question 3

Many-one and onto

One-one and not onto

Onto but not one-one

Neither one-one nor onto

Solution

Question 4

$$\dfrac {1}{29}$$

$$\dfrac {10}{29}$$

$$\dfrac {29}{10}$$

$$29$$

Solution

Question 5

$$3x^2-5$$

$$3x^2+6$$

$$x^2-5$$

$$3x^2-2$$

$$3x^2+5x-2$$

Solution

Question 6

One-one but not onto

Onto but not one-one

Neither one-one nor onto

Both one-one and onto

Solution

Question 7

Empty

A singleton

A finite se with more than one element

Infinite

Solution

Question 8

Solution

Question 9

$$144$$

$$72$$

$$60$$

$$120$$

Solution

Question 10

One-one

Onto

Bijective

$$f$$ is not defined

Solution

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