Functions Test 27

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Functions Test 27

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Weekly Quiz Competition

Question 1
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 2
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 3
1 / -0

If $$fog = |\sin x|$$ and $$gof = \sin^{2}\sqrt {x}$$, then $$f(x)$$ and $$g(x)$$ are

A
$$f(x) = \sqrt {\sin x}, g(x) = x^{2}$$

B
$$f(x) = |x|, g(x) = \sin x$$

C
$$f(x) = \sqrt {x}, g(x) = \sin^{2}x$$

D
$$f(x) = \sin \sqrt {x}, g(x) = x^{2}$$

Solution

$$fog = f\left \{g(x)\right \} = |\sin x| = \sqrt {\sin^{2}x}$$

Also, $$gof = g\left \{f(x)\right \} = \sin^{2}\sqrt {x}$$

Obviously, $$\sqrt {\sin^{2}x} = \sqrt {g(x)}$$

and $$\sin^{2} \sqrt {x} = \sin^{2} \left \{f(x)\right \}$$
i.e., $$g (x) = \sin^{2}x$$

and $$f(x) = \sqrt {x}$$.
Question 4
1 / -0

If $$g(x)=\dfrac{1}{f(x)}$$ and $$f(x)=x, x\ne 0,$$ then which one of the following is correct?

A
$$f(f(f(g(g(f(x))))))=g(g(f(g(f(x)))))$$

B
$$f(g(f(g(g(f(g(x)))))))=g(g(f(g(f(x)))))$$

C
$$f(g(f(g(g(f(g(x)))))))=f(g(f(g(f(x)))))$$

D
$$f(f(f(g(g(f(x))))))=f(f(f(g(f(x)))))$$

Solution

From the given data, it is evident that $$gof(x)=\dfrac{1}{x}=fog(x)$$
$$f(f(f(g(g(f(x))))))=x$$ and $$g(g(f(g(f(x)))))=\dfrac{1}{x}$$
$$f(g(f(g(g(f(g(x)))))))=x$$ and $$g(g(f(g(f(x)))))=\dfrac{1}{x}$$
$$f(g(f(g(g(f(g(x)))))))=x$$ and $$f(g(f(g(f(x)))))=x$$
$$f(f(f(g(g(f(x))))))=x$$ and $$f(f(f(g(f(x)))))=\dfrac{1}{x}$$
Question 5
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 6
1 / -0

If $$g(x)=1+\sqrt{x}$$ and $$f\{g(x)\}=3+2\sqrt{x}+x$$, then $$f(x)$$ is equal to

A
$$1+2x^2$$

B
$$2+x^2$$

C
$$1+x$$

D
$$2+x$$

Solution

Given, $$g(x)=1+\sqrt{x}$$
and $$f\left \{ g(x) \right \}=3+2\sqrt{x}+x$$ ..... $$(i)$$
$$\Rightarrow f(1+\sqrt{x})=3+2\sqrt{x}+x$$
Put $$1+\sqrt{x}=y\Rightarrow x=(y-1)^2$$
$$\therefore f(y)=3+2(y-1)+(y-1)^2=3+2y-2+y^{2}-2y+1=2+y^2$$
$$\therefore f(x)=2+x^2$$
Question 7
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 8
1 / -0

The function $$f:A\rightarrow B$$ given by $$f(x) = x ,x\in A$$, is one to one but not onto. Then;

A
$$B\subset A$$

B
$$A=B$$

C
$$A'\subset B'\quad $$

D
$$A\subset B$$

E
$$A'\cap B'=\phi $$

Solution

$$f(x)=x$$ is identity function
It is given that $$f(x)$$ is one to one but not on to.
It means some element in $$B$$ has no pre image in $$A$$
$$A\subset B$$
Question 9
1 / -0

If $$f(x)=\sin ^{ 2 }{ x } +\sin ^{ 2 }{ \left( x+\cfrac { \pi }{ 3 } \right) } +\cos { x } \cos { \left( x+\cfrac { \pi }{ 3 } \right) } $$ and $$g\left( \cfrac { 5 }{ 4 } \right) =1$$, then $$g\circ f(x)$$ is equal to

A
$$0$$

B
$$1$$

C
$$\sin { { 1 }^{ o } } $$

D
None of these

Solution

Given, $$f(x)=\sin ^{ 2 }{ x } +\sin ^{ 2 }{ \left( x+\cfrac { \pi }{ 3 } \right) } +\cos { x } \cos { \left( x+\cfrac { \pi }{ 3 } \right) } $$
$$=\sin ^{ 2 }{ x } +\left( \sin { x } \cos { \cfrac { \pi }{ 3 } } +\cos { x } \sin { \cfrac { \pi }{ 3 } } \right) +\cos { x } \left( \cos { \cfrac { \pi }{ 3 } } -\sin { x } \sin { \cfrac { \pi }{ 3 } } \right) $$
$$=\sin ^{ 2 }{ x } +\cfrac { \sin ^{ 2 }{ x } }{ 4 } +\cfrac { 3\cos ^{ 2 }{ x } }{ 4 } +\cfrac { 2\sqrt { 3 } }{ 2.2 } \sin { x } \cos { x } +\cfrac { \cos ^{ 2 }{ x } }{ 2 } -\cos { x } \sin { x } \cfrac { \sqrt { 3 } }{ 2 } $$
$$=\cfrac { 5 }{ 4 } \left( \sin ^{ 2 }{ x } +\cos ^{ 2 }{ x } \right) =\cfrac { 5 }{ 4 } \quad $$
Therefore, $$ g\circ f(x)=g(f(x))=g\left( \cfrac { 5 }{ 4 } \right) =1\quad \quad $$
Question 10
1 / -0

If $$f(x)=ax+b $$ and $$g(x)=cx+d$$, then $$f\left( g(x) \right) =g\left( f(x) \right) \Leftrightarrow$$

A
$$f(a)=g(c)$$

B
$$f(b)=g(b)$$

C
$$f(d)=g(b)$$

D
$$f(x)=g(a)$$

Solution

We have $$f(x)=ax+b,g(x)=cx+d$$
Therefore, $$ f\left\{ g(x) \right\} =g\left\{ f(x) \right\} \Leftrightarrow f(cx+d)=g(ax+b)$$
$$\Leftrightarrow a(cx+d)+b=c(ax+b)+d$$
$$\Leftrightarrow ad+b=cb+d\Leftrightarrow f(d)=g(b)$$

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Functions Test 27

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Functions Test 27

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Question 1

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 2

$$8^3x$$

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

$$8x^3$$

$$8x$$

Solution

Question 3

$$f(x) = \sqrt {\sin x}, g(x) = x^{2}$$

$$f(x) = |x|, g(x) = \sin x$$

$$f(x) = \sqrt {x}, g(x) = \sin^{2}x$$

$$f(x) = \sin \sqrt {x}, g(x) = x^{2}$$

Solution

Question 4

$$f(f(f(g(g(f(x))))))=g(g(f(g(f(x)))))$$

$$f(g(f(g(g(f(g(x)))))))=g(g(f(g(f(x)))))$$

$$f(g(f(g(g(f(g(x)))))))=f(g(f(g(f(x)))))$$

$$f(f(f(g(g(f(x))))))=f(f(f(g(f(x)))))$$

Solution

Question 5

Only (I)

Only (I) and (II)

Only (III) and (IV)

Only (I) and (IV)

Solution

Question 6

$$1+2x^2$$

$$2+x^2$$

$$1+x$$

$$2+x$$

Solution

Question 7

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

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

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

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

Solution

Question 8

$$B\subset A$$

$$A=B$$

$$A'\subset B'\quad $$

$$A\subset B$$

$$A'\cap B'=\phi $$

Solution

Question 9

$$0$$

$$1$$

$$\sin { { 1 }^{ o } } $$

None of these

Solution

Question 10

$$f(a)=g(c)$$

$$f(b)=g(b)$$

$$f(d)=g(b)$$

$$f(x)=g(a)$$

Solution

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