Relations and Functions Test

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Relations and Functions Test - 59

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Weekly Quiz Competition

Question 1
1 / -0

Let $$A=\left\{ x\in R|x\ge 0 \right\} $$. A function $$f:A\rightarrow A$$ is defined by $$f(x)={ x }^{ 2 }$$. Which one of the following is correct?

A
The function does not have inverse

B
$$f$$ is its own inverse

C
The function has an inverse but $$f$$ is not its own inverse

D
None of the above

Solution

$$f\left( x \right) ={ x }^{ 2 }\\ \Longrightarrow f^{ -1 }\left( x \right) =\sqrt { x } $$
Clearly, $$f^{ -1 }\left( x \right) $$ exists in the domain of definition of $$f$$.
Also, $$f\left( x \right)$$ is not its own inverse.
Option C is the answer.
Question 2
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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 3
1 / -0

If $$f : (3, 4) \rightarrow (0, 1)$$ is defined by $$f(x)=x-\left[x\right]$$, where $$\left[x\right]$$ denotes the greatest integer function, then $${f}^{-1}(x)$$ is

A
$$\dfrac{1}{x-\left[x\right]}$$

B
$$\left[x\right]-x$$

C
$$x-3$$

D
$$x+3$$

Solution

$$f\left( x \right) =x-\left[ x \right] \\ f:(3,4)\rightarrow (0,1)$$
For $$x\in (3,4)$$
$$\left[ x \right] =3\\ \Rightarrow f\left( x \right) =x-3\\ y=x-3\\ x=y+3\\ f^{ -1 }\left( x \right) =x+3$$
So option $$D$$ is correct.
Question 4
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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
Question 5
1 / -0

Consider the following statements :
Statement 1 : The function $$f:R \rightarrow R$$ such that $$f(x)=x^3$$ for all $$x\in R$$ is one-one.
Statement 2 : $$f(a) = f(b) \Rightarrow a=b$$ for all $$a, b \in R$$ if the function $$f$$ is one-one.
Which one of the following is correct in respect of the above statements?

A
Both the statements are true and Statement 2 is the correct explanation of Statement 1.

B
Both the statements are true and Statement 2 is not the correct explanation of Statement 1.

C
Statement 1 is true but Statement 2 is false.

D
Statement 1 is true but Statement 2 is true.

Solution

Solution:
Statement $$1$$ is correct, since $$f(x)=x^3$$
For any $$x,y\in R$$
$$f(x)=f(y)$$
$$\Rightarrow x^{3}=y^3$$
$$\Rightarrow x=y$$
$$\therefore\ f$$ is one-one $$\forall x\in R.$$
$$\because f(a)=f(b)\Longrightarrow a^3=b^3\Longrightarrow a=b$$
Statement $$2$$ is also correct and is the correct explanation of statement $$1.$$
Hence, A is the correct option.
Question 6
1 / -0

On the set $$Z$$, of all integers $$\ast$$ is defined by $$a\ast b = a + b - 5$$. If $$2\ast (x\ast 3) = 5$$ then $$x =$$

A
$$0$$

B
$$3$$

C
$$5$$

D
$$10$$

Solution

Given $$a*b=a+b-5$$ where $$*$$ is a unique operator
So $$2*(x*3)=2*(x+3-5)=2*(x-2)=2+x-2-5=x-5$$
Its value is given as $$5$$. so $$x-5=5 \implies x=10$$
Question 7
1 / -0

If $$f(x) = 8x^3, g(x) = x^{1/3}$$, then fog (x) is

A
$$8^3x$$

B
$$(8x)^{1/3}$$

C
$$8x^3$$

D
$$8x$$

Solution

$$f\left( x \right) =8{ x }^{ 3 }$$

$$g\left( x \right) ={ x }^{ { 1 }/{ 3 } }$$

$$fog\left( x \right) =f\left[ g\left( x \right) \right] $$
$$=f\left[ { x }^{ { 1 }/{ 3 } } \right] $$

$$=8{ \left[ { x }^{ { 1 }/{ 3 } } \right] }^{ 3 }$$

$$=8x$$
Question 8
1 / -0

Let $$f (x) = \sqrt {2 - x - x^2}$$ and g(x) = cos x. Which of the following statements are true?
(I) Domain of $$f((g(x))^2) = $$ Domain of f(g(x))
(II) Domain of f(g(x)) + g(f(x)) = Domain of g(f(x))
(III) Domain of f(g(x)) = Domain of g(f(x))
(IV) Domain of $$g((f(x))^3) = $$ Domain of f(g(x))

A
Only (I)

B
Only (I) and (II)

C
Only (III) and (IV)

D
Only (I) and (IV)

Solution

Domain of $$f(g(x))$$ is R
$$\because f(g(x)) = 2- \cos x - \cos x^2 \geq 0$$
$$\Rightarrow(\cos x + 2)(\cos x - 1) \leq 0$$
$$\Rightarrow$$ $$-2$$ $$\leq$$ $$\cos x$$ $$\leq$$ $$1$$
$$\therefore$$ x $$\in$$ R
Domain of $$g(f(x))$$ is [$$-2$$, $$1$$]
$$\because$$ In $$\cos \sqrt{2 - x - x^2}$$
$$2 - x - x^2$$ $$\geq$$ $$0$$
Domain of $$f((g(x))^2)$$ is R
since $$2$$ - $$\cos^2 x$$ - $$\cos^4 x$$ $$\geq$$ $$0$$
($$\cos^2 x$$ + $$2$$)($$\cos^2 x$$ -$$1$$) $$\leq$$ $$0$$
$$-1$$ $$\leq$$ $$\cos x$$ $$\leq$$ $$1$$
$$\Rightarrow$$ x$$\in$$ R
Domain of $$g((f(x))^3)$$ is same as Domain of $$g(f(x))$$
i.e. [$$-2$$,$$1$$]
From above discussion statement $$\textrm{I}$$ is true and statements $$\textrm{II, III}$$ and $$\textrm{IV}$$ are false.
Question 9
1 / -0

If $$f:R\rightarrow R, g:R \rightarrow R$$ be two functions given by $$f(x)=2x-3$$ and $$g(x)=x^3+5$$, then $$(fog)^{-1}(x)$$ is equal to

A
$$\begin{pmatrix}\dfrac{x+7}{2}\end{pmatrix} ^{\dfrac{1}{3}}$$

B
$$\begin{pmatrix}\dfrac{x-7}{2}\end{pmatrix} ^{\dfrac{1}{3}}$$

C
$$\begin{pmatrix}x-\dfrac{7}{2}\end{pmatrix} ^{\dfrac{1}{3}}$$

D
$$\begin{pmatrix}x+\dfrac{7}{2}\end{pmatrix} ^{\dfrac{1}{3}}$$

Solution

Given,
$$f(x)=2x-3$$ and $$g(x)=x^3+5$$
Now, $$(fog)(x)=f(g(x))=2(x^3+5)-3$$
$$=2x^3+10-3=2x^3+7$$
Let, $$(fog)(x)=y,$$ then,
$$y=2x^3+7$$
$$\Longrightarrow x=\left(\cfrac{y-7}{2}\right)^{\cfrac 13}$$
$$\therefore fog^{-1}(x)=\left(\cfrac{x-7}{2}\right)^{\cfrac 13}$$
Hence, B is the correct option.
Question 10
1 / -0

If $$f : R \rightarrow R$$ is defined by $$f(x) = x^{3}$$ then $$f^{-1}(8) =$$

A
$$\left \{2\right \}$$

B
$$\left \{2, 2\omega, 2\omega^{2}\right \}$$

C
$$\left \{2, -2\right \}$$

D
$$\left \{2, 2\omega\right \}$$

Solution

The inverse function of the given function is given by interchanging the $$x$$ and $$y$$ and finding the function (the inverse function which is now $$y$$) in terms of $$x$$.
For $$f(x)=x^3$$
Let $$f(x)=y\Rightarrow x = f^{-1}(y)$$
$$y=x^3$$
$$x=y^{\tfrac { 1 }{ 3 } }$$
$$\therefore f^{-1}(y)=$$ $$y^{\tfrac { 1 }{ 3 } }$$
$$f^{-1}(8)=8^{\tfrac { 1 }{ 3 } }=2\times (1)^{\tfrac13}$$
We know that the cube roots of unity $$(1)$$ are $$1, \omega$$ and $$\omega^2$$
$$\therefore f^{-1}(8) = \{2,2\omega, 2\omega^2\}$$
Hence, option $$B$$ is correct.

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Relations and Functions Test - 59

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Relations and Functions Test - 59

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SHARING IS CARING

Question 1

The function does not have inverse

$$f$$ is its own inverse

The function has an inverse but $$f$$ is not its own inverse

None of the above

Solution

Question 2

$$125$$

$$375$$

$$250$$

$$147$$

Solution

Question 3

$$\dfrac{1}{x-\left[x\right]}$$

$$\left[x\right]-x$$

$$x-3$$

$$x+3$$

Solution

Question 4

One-one and onto

Neither one-one nor onto

One-one but not onto

Onto but not one-one

Solution

Question 5

Both the statements are true and Statement 2 is the correct explanation of Statement 1.

Both the statements are true and Statement 2 is not the correct explanation of Statement 1.

Statement 1 is true but Statement 2 is false.

Statement 1 is true but Statement 2 is true.

Solution

Question 6

$$0$$

$$3$$

$$5$$

$$10$$

Solution

Question 7

$$8^3x$$

$$(8x)^{1/3}$$

$$8x^3$$

$$8x$$

Solution

Question 8

Only (I)

Only (I) and (II)

Only (III) and (IV)

Only (I) and (IV)

Solution

Question 9

$$\begin{pmatrix}\dfrac{x+7}{2}\end{pmatrix} ^{\dfrac{1}{3}}$$

$$\begin{pmatrix}\dfrac{x-7}{2}\end{pmatrix} ^{\dfrac{1}{3}}$$

$$\begin{pmatrix}x-\dfrac{7}{2}\end{pmatrix} ^{\dfrac{1}{3}}$$

$$\begin{pmatrix}x+\dfrac{7}{2}\end{pmatrix} ^{\dfrac{1}{3}}$$

Solution

Question 10

$$\left \{2\right \}$$

$$\left \{2, 2\omega, 2\omega^{2}\right \}$$

$$\left \{2, -2\right \}$$

$$\left \{2, 2\omega\right \}$$

Solution

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