Functions Test 38

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Functions Test 38

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Weekly Quiz Competition

Question 1
1 / -0

A function $$f$$ from the set of natural numbers to integers defined by $$f(n)=\begin{cases} \cfrac { n-1 }{ 2 } ,\quad \text{when n is odd} \\ -\cfrac { n }{ 2 } ,\quad \text{when n is even} \end{cases}$$ is

A
neither one-one nor onto

B
one-one but not onto

C
onto but not one-one

D
one-one and onto both

Solution

$$one-one$$ test of $$f:$$
Let $$x_1$$ and $$x_2$$ be any two elements in the domain $$(N).$$
$$Case\,I:$$ When both $$x_1$$ and $$x_2$$ are even.

Let $$f(x_1)=f(x_2)$$
$$\Rightarrow$$ $$\dfrac{-x_1}{2}=\dfrac{x_2}{2}$$
$$\Rightarrow$$ $$-x_1=-x_2$$
$$\Rightarrow$$ $$x_1=x_2$$

$$Case\,II:$$ When both $$x_1$$ and $$x_2$$ are odd.

Let $$f(x_1)=f(x_2)$$
$$\Rightarrow$$ $$\dfrac{x_1-1}{2}=\dfrac{x_2-1}{2}$$
$$\Rightarrow$$ $$x_1-1=x_2-1$$
$$\Rightarrow$$ $$x_1=x_2$$

$$Case\,III:$$ When $$x_1$$ be even and $$x_2$$ be odd.

Then, $$f(x_1)=\dfrac{-x_1}{2}$$ and $$f(x_2)=\dfrac{x_2-1}{2}$$
Then clearly,
$$\Rightarrow$$ $$x_1\ne x_2$$
$$\Rightarrow$$ $$f(x_1)\ne f(x_2)$$
From, all the cases, we can say that, $$f$$ is one-one.

$$onto$$ test of $$f:$$
Co-domain of $$f=Z=\{....,-3,-2,-1,0,1,2,3,...\}$$
Range of $$f=\left\{...,\dfrac{-2-1}{2},\dfrac{-(-2)}{2},\dfrac{-1-1}{2},\dfrac{0}{2},\dfrac{1-1}{2},\dfrac{-2}{2},\dfrac{3-1}{2},...\right\}$$
Range of $$f=\{...,-2,1,-1,0,0,-1,1,..\}$$
$$\Rightarrow$$ Co-domain of $$f=$$ Range of $$f$$
$$\therefore$$ $$f$$ is onto.
Question 2
1 / -0

Let $$f:Z\rightarrow Z$$ be given by $$f(x)=\begin{cases} \cfrac { x }{ 2 } ,\quad \text{if}\ x \ \text{is even} \\ 0,\quad \text{if }\ x \ \text{is odd} \end{cases}$$. Then, $$f$$ is

A
onto but not one-one

B
one-one but not onto

C
one-one and onto

D
neither one-one nor onto

Solution

$$Injectivity:$$
Let $$x_1$$ and $$x_2$$ be two elements in the domain $$(Z)$$, such that,
$$f(x_1)=f(x_2)$$
$$Case\,I:$$
Let both $$x_1$$ and $$x_2$$ be even.
Then, $$f(x_1)=f(x_2)$$
$$\Rightarrow$$ $$\dfrac{x_1}{2}=\dfrac{x_2}{2}$$
$$\Rightarrow$$ $$x_1=x_2$$
$$Case\,II:$$
Let both $$x_1$$ and $$x_2$$ be odd.
Then, $$f(x_1)=f(x_2)$$
$$\Rightarrow$$ $$0=0$$
Here, we cannot determine whether $$x_1=x_2.$$
So, $$f$$ is not one-one.

$$Surjectivity:$$
Let $$y$$ be an element in the co-domain $$(Z)$$, such that
Co-domain of $$f=Z=\{0,\pm 1,\pm 2,\pm 3,\pm 4,...\}$$
Range of $$f=\left\{0,0,\dfrac{\pm 2}{2},0,\dfrac{\pm 4}{2},...\right\}=\{0,\pm 1,\pm 2,....\}$$
$$\Rightarrow$$ Co-domain of $$f=$$ Range of $$f$$
$$\therefore$$ $$f$$ is onto.
.
Question 3
1 / -0

If $$g(x)={ x }^{ 2 }+x-2$$ and $$\cfrac { 1 }{ 2 } (g\circ f(x))=2{ x }^{ 2 }-5x+2$$, then $$f(x)$$ is equal to

A
$$2x-3$$

B
$$2x+3$$

C
$$2{x}^{2}+3x+1$$

D
$$2{x}^{2}-3x-1$$

Solution

We will solve this problem by the trial and error method.
Let us check option $$A$$ first.
If $$f(x)=2x-3$$
$$g(x)=x^2+x-2$$ [ Given ]
$$\Rightarrow$$ $$\dfrac{1}{2}(g\circ f)(x)=g[f(x)]$$
$$=\dfrac{1}{2}g(2x-3)$$

$$=\dfrac{1}{2}[(2x-3)^2+(2x-3)-2]$$

$$=\dfrac{1}{2}[4x^2+9-12x+2x-3-2]$$

$$=\dfrac{1}{2}[4x^2-10x+4]$$
$$=2x^2-5x+2$$
The given condition is satisfied by $$A.$$
Question 4
1 / -0

A function $$f$$ from the set of natural numbers to the set of integers defined by
$$f(n)=\begin{cases} \cfrac { n-1 }{ 2 } ,\quad \text{when n is odd} \\ -\cfrac { n }{ 2 } ,\quad \text{when n is even} \end{cases}$$

A
neither one-one nor onto

B
one-one but not onto

C
onto but not one-one

D
one-one and onto both

Solution

$$one-one$$ test of $$f:$$
Let $$x_1$$ and $$x_2$$ be any two elements in the domain $$(N).$$
$$Case\,I:$$ When both $$x_1$$ and $$x_2$$ are even.

Let $$f(x_1)=f(x_2)$$
$$\Rightarrow$$ $$\dfrac{-x_1}{2}=\dfrac{x_2}{2}$$
$$\Rightarrow$$ $$-x_1=-x_2$$
$$\Rightarrow$$ $$x_1=x_2$$

$$Case\,II:$$ When both $$x_1$$ and $$x_2$$ are odd.

Let $$f(x_1)=f(x_2)$$
$$\Rightarrow$$ $$\dfrac{x_1-1}{2}=\dfrac{x_2-1}{2}$$
$$\Rightarrow$$ $$x_1-1=x_2-1$$
$$\Rightarrow$$ $$x_1=x_2$$

$$Case\,III:$$ When $$x_1$$ be even and $$x_2$$ be odd.

Then, $$f(x_1)=\dfrac{-x_1}{2}$$ and $$f(x_2)=\dfrac{x_2-1}{2}$$
Then clearly,
$$\Rightarrow$$ $$x_1\ne x_2$$
$$\Rightarrow$$ $$f(x_1)\ne f(x_2)$$
From, all the cases, we can say that, $$f$$ is one-one.

$$onto$$ test of $$f:$$
Co-domain of $$f=Z=\{....,-3,-2,-1,0,1,2,3,...\}$$
Range of $$f=\left\{...,\dfrac{-2-1}{2},\dfrac{-(-2)}{2},\dfrac{-1-1}{2},\dfrac{0}{2},\dfrac{1-1}{2},\dfrac{-2}{2},\dfrac{3-1}{2},...\right\}$$
Range of $$f=\{...,-2,1,-1,0,0,-1,1,..\}$$
$$\Rightarrow$$ Co-domain of $$f=$$ Range of $$f$$
$$\therefore$$ $$f$$ is onto.
Question 5
1 / -0

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

$$M=\left\{A=\begin{bmatrix} a & b\\ c& d\end{bmatrix}: a,b,c,d\in R\right\}$$
$$f:M\rightarrow R$$ is given by $$f(A)=|A|$$
$$Injectivity:$$
$$f\left(\begin{bmatrix} 0 & 0\\0& 0\end{bmatrix} \right )=\begin{vmatrix} 0&0\\0&0\end{vmatrix}=0$$

and $$f\left(\begin{bmatrix} 1 & 0\\0& 0\end{bmatrix} \right )=\begin{vmatrix} 1&0\\0&0\end{vmatrix}=0$$

$$\Rightarrow$$ $$f\left(\begin{bmatrix} 0&0\\0&0\end{bmatrix}\right)=f\left(\begin{bmatrix}1&0\\0&0\end{bmatrix}\right)=0$$
So, $$f$$ is not one-one.
$$Surjectivity:$$
Let
$$f(A)=y,\,A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$$
$$\Rightarrow$$ $$\begin{vmatrix} a& b\\c&d\end{vmatrix}=y$$

$$\Rightarrow$$ $$ad-bc=y$$
$$\Rightarrow$$ $$a,b,c,d\in R$$ implies $$y\in R$$
$$\Rightarrow$$ CoDomain=Range.
$$\therefore$$ $$f$$ is onto.
Question 6
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The function $$f:R\rightarrow R$$ defined by $$f(x)=(x-1)(x-2)(x-3)$$ is

A
one-one but not onto

B
onto but not one-one

C
both one and onto

D
neither one-one nor onto

Solution

$$f(x)=(x-1)(x-2)(x-3)$$

$$one-one$$ test:
$$\Rightarrow$$ $$f(1)=(1-1)(1-2)(1-3)=0$$
$$\Rightarrow$$ $$f(2)=(2-1)(2-2)(2-3)=0$$
$$\Rightarrow$$ $$f(3)=(3-1)(3-1)(3-3)=0$$
$$\Rightarrow$$ $$f(1)=f(2)=f(3)=0$$
We can see, $$1,2,3$$ has same image $$0.$$
$$\therefore$$ $$f$$ is not one-one.

$$onto$$ test:
Let $$y$$ be an element in the co-domain $$R,$$ such that
$$y=f(x)$$
$$\Rightarrow$$ $$y=(x-1)(x-2)(x-3)$$
Since, $$y\in R$$ and $$x\in R$$.
$$\therefore$$ $$f$$ is onto.
Question 7
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Let $$S$$ be the set of all real roots of the equation, $${ 3 }^{ x }\left( { 3 }^{ x }-1 \right) +2=\left| { 3 }^{ x }-1 \right| +\left| { 3 }^{ x }-2 \right| $$. Then $$S$$:

A
contains at least four elements

B
is a singleton

C
is an empty set

D
contains exactly two elements

Solution

$$3^x (3^x - 1) + 2 = |3^x - 1| + |3^x - 2|$$

Let $$3^x = t$$
So, $$t (t - 1) + 2 = | t - 1| + |t - 2|$$
$$\Rightarrow t^2 - t + 2 = |t - 1|+ |t - 2|$$

$$f(t)=t^2-t+2 \quad\quad\text{[green curve]}$$
$$g(t)=|t-1|+|t+2|\quad\quad\text{[blue curve]}$$

we plot $$f(t)=t^2-t+2$$ and $$g(t)=|t-1|+|t+2|$$
$$\text{these two curves intersect at two points but only one root is positive.}$$
$$\text{As} \,3^x \,\text{is always positive, therefore only positive value of} \,t\, \text{will be the solution.}$$

$$\text{Therefore, we have only one solution.}$$
Question 8
1 / -0

If $$f(x) = \dfrac{x+1}{x-1}$$, then the valueof $$f(f(x))$$ is equal to

A
$$x$$

B
$$0$$

C
$$-x$$

D
$$1$$

Solution

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

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

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

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

$$=\dfrac{2x}{2}$$

$$=x$$
Question 9
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Let $$f : x \rightarrow y $$ be such that $$f(1) = 2$$ and $$f(x + y) = f(x) f(y)$$ for all natural numbers x and y. If $$\displaystyle \sum_{k= 1}^n f(a + k) = 16 (2^n - 1)$$ , then a is equal to

A
$$3$$

B
$$4$$

C
$$5$$

D
$$6$$

E
$$7$$

Solution

We have,
$$f(1)=2$$ and $$f(x+y)=f(x).f(y)$$
Now, $$f(2)=f(1+1)=f(1).f(1)=2.2=2^2$$
$$f(3)=f(2+1)+f(2).f(1)=2^2.2=2^3$$
and so on
$$\therefore f(x)=2^n$$......(i)
Now, we have
$$\displaystyle \sum^n_{k=1}f(a+k)=16(2^n-1)$$
$$\Rightarrow f(a+1)+f(a+2)+----f(a+n)=16(2^n-1)$$
$$\Rightarrow f(a).f(1)+f(a).f(2)+-----f(a).f(n)=16(2^n-1)$$
$$\Rightarrow f(a)=[f(1)+f(2)+---f(n)]=16(2^n-1)$$
$$\Rightarrow f(a)[2+2^2+----+2^n]=16(2^n-1)$$

$$\Rightarrow f(a).[2\dfrac{(2^n-1)}{2-1}]=16(2^n-1)$$
$$\Rightarrow 2f(a).(2^n-1)=16.(2^n-1)\Rightarrow f(a)=8$$
$$\Rightarrow 2^a=8[\because f(x)=2^n\Rightarrow f(a)=2^a]$$
$$\Rightarrow 2^a=2^3=a=3$$
Question 10
1 / -0

If $$f(x)=\dfrac{(4x+3)}{(6x-4)}, x\neq \dfrac{2}{3}$$ then $$(f o f)(x)=?$$

A
$$x$$

B
$$(2x-3)$$

C
$$\dfrac{4x-6}{3x+4}$$

D
None of these

Solution

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Re-Attempt Weekly Quiz Competition

Functions Test 38

Result

Functions Test 38

Score

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Rank

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SHARING IS CARING

Question 1

neither one-one nor onto

one-one but not onto

onto but not one-one

one-one and onto both

Solution

Question 2

onto but not one-one

one-one but not onto

one-one and onto

neither one-one nor onto

Solution

Question 3

$$2x-3$$

$$2x+3$$

$$2{x}^{2}+3x+1$$

$$2{x}^{2}-3x-1$$

Solution

Question 4

neither one-one nor onto

one-one but not onto

onto but not one-one

one-one and onto both

Solution

Question 5

one-one and onto

neither one-one nor onto

one-one but not onto

onto but not one-one

Solution

Question 6

one-one but not onto

onto but not one-one

both one and onto

neither one-one nor onto

Solution

Question 7

contains at least four elements

is a singleton

is an empty set

contains exactly two elements

Solution

Question 8

$$x$$

$$0$$

$$-x$$

$$1$$

Solution

Question 9

$$3$$

$$4$$

$$5$$

$$6$$

$$7$$

Solution

Question 10

$$x$$

$$(2x-3)$$

$$\dfrac{4x-6}{3x+4}$$

None of these

Solution

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