Functions Test 10

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Functions Test 10

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Weekly Quiz Competition

Question 1
1 / -0

The domain of $$f(x)=\displaystyle \frac{1}{[x]-x}$$ is

A
$$R$$

B
$$Z$$

C
$$R-Z$$

D
$$Q-{0}$$

Solution

$$[\mathrm{x}]-\mathrm{x}\neq 0$$
Which means $$x$$ must not be an integer.
The required domain is $$R - Z$$
Hence, option 'C' is correct.
Question 2
1 / -0

lf $${f}\left({x}\right)=\sin^{2}{x}+\sin^{2}\left({x}+\displaystyle \dfrac{\pi}{3}\right)+ \cos x \cos \left({x}+\displaystyle \dfrac{\pi}{3}\right)$$ and $${g}\left(\displaystyle\dfrac{5}{4}\right)=1$$, $$g\left(1\right) = 0 $$ then $$\left({g}{o}{f}\right)\left({x}\right)=$$

A
$$1$$

B
$$0$$

C
$$\sin x$$

D
Data is insufficient

Solution

$$\displaystyle {f}\left({x}\right)=\sin^{2}{x}+\left(\sin{x}\cos\frac{\pi}{3}+ \cos{x} \displaystyle \sin\frac{\pi}{3}\right)^{2}+ {c}{o}{s} {x}\left(\displaystyle \cos {x}\cos\frac{\pi}{3}-\sin {x}\sin\frac{\pi}{3}\right)$$
$$\displaystyle =\sin ^{ 2 }{ x } +{ \left[ \frac { \sin{ x } }{ 2 } +\frac { \sqrt { 3 } \cos{ x } }{ 2 } \right] }^{ 2 }+\frac { \cos^{ 2 }{ x } }{ 2 } -\frac { \sqrt { 3 } }{ 2 } \cos x \sin x$$
$$\displaystyle =\sin ^{ 2 }{ x } +\frac { \sin ^{ 2 }{ x } }{ 4 } +\frac { 3 }{ 4 } \cos ^{ 2 }{ x } +\frac { \sqrt { 3 } }{ 2 } \sin x\cos{ x }+\frac { \cos ^{ 2 }{ x } }{ 2 } -\frac { \sqrt { 3 } }{ 2 } \sin { x\cos { x } } $$
$$=\displaystyle \frac{5}{4}\left(\sin^{2}{x}+\cos^{2}{x}\right)=\frac{5}{4}$$
$$\displaystyle \therefore $$ $$[{g}{o}{f}]\left({x}\right)={g}[{f}\left({x}\right)]={g}\left(\displaystyle \frac{5}{4}\right)=1$$
Question 3
1 / -0

If $$ f : R \rightarrow R$$ is defined by $$f(x)=2x-2,$$ then $$(f\circ f) (x) + 2 =$$

A
$$f(x)$$

B
$$2f(x)$$

C
$$3f(x)$$

D
$$-f(x)$$

Solution

Let $$f : R \rightarrow R$$ is defined by $$f(x)=2x-2$$. Then
$$(f\circ f)(x)+2= f[f(x)]+2$$
$$=f(2x-2)+2$$
$$=2(2x-2)-2+2$$
$$=2f(x)$$
Question 4
1 / -0

If $$f(x) =\displaystyle \frac{x}{\sqrt{1-x^2}}, g(x)=\frac{x}{\sqrt{1+x^2}}$$ then $$(f\circ g)(x) =$$

A
$$\displaystyle \frac{x}{\sqrt{1-x^2}}$$

B
$$\displaystyle \frac{x}{\sqrt{1+x^2}}$$

C
$$\displaystyle \frac{1-x^2}{\sqrt{1-x^2}}$$

D
$$x$$

Solution

Let $$x= \tan \theta$$
$$(f\circ g)(x)=f[g(x)] = f[g(\tan \theta)]$$.
Since $$\displaystyle g(x) = \frac{x}{\sqrt{1+x^2}}$$, we have
$$f[g(\tan \theta)]=\displaystyle f \left [ \frac{\tan \theta}{\sqrt{1+\tan^2 \theta}} \right ] = f \left [ \frac{\tan \theta}{\sec \theta} \right ]=f (\sin \theta)$$
Also since $$f(x)=\cfrac{x}{\sqrt{1-x^2}}$$ , we have
$$\displaystyle f (\sin \theta) = \frac{\sin \theta}{\sqrt{1-\sin^2 \theta}}$$
$$=\tan \theta =x$$
Question 5
1 / -0

If $$f(x)=\log x, g(x) = x^3$$ then $$f[g(a)]+f[g(b)]= $$

A
$$f[g(a)+g(b)]$$

B
$$f[g(ab)]$$

C
$$g[f(ab)]$$

D
$$g[f(a)+f(b)]$$

Solution

Given that $$g(x)=x^3$$. Therefore,
$$f[g(a)]+f[g(b)]=f(a^3)+f(b^3)$$.
Since $$f(x)=\log x$$, we have
$$f(a^3)+fb^3=\log a^3+\log b^3$$
$$=\log (a^3b^3)=\log ((ab)^{3})$$
$$=f[g(ab)]$$
Question 6
1 / -0

If $$f:R \rightarrow R$$ and $$g : R \rightarrow R$$ are defined by $$f(x)=2x+3$$ and $$g(x)=x^2+7$$, then the values of $$x$$ such that $$g(f(x)) =8$$ are:

A
$$1, 2$$

B
$$-1, 2$$

C
$$-1, -2$$

D
$$1, -2$$

Solution

Since $$f(x)=2x+3$$,
$$g[f(x)]=8 \Rightarrow g(2x+3)=8$$.
Also since $$g(x)=x^2+7$$,
$$g(2+3x)=8\Rightarrow (2x+3)^2+7=8$$
$$\Rightarrow 2x+3 = \pm 1$$
$$\Rightarrow x=-1$$ or $$-2$$.
Question 7
1 / -0

If $$f : R \rightarrow R$$ and $$g :R \rightarrow R$$ are defined by $$f(x) = x -[x]$$ and $$g(x) = [x]$$ for $$x \in R$$, where $$[x]$$ is the greatest integer not exceeding $$x$$, then for every $$x \in R, f(g(x)) =$$

A
$$x$$

B
$$0$$

C
$$f(x)$$

D
$$g(x)$$

Solution

Since $$f(x)=x-[x]$$, we have
$$f[g(x)] =g(x)-[g(x)]$$
Also since $$g(x)=[x]$$,we have
$$g(x)-[g(x)]=[x] - [[x]]$$.
Here, $$[x]$$ is the greatest integer not exceeding $$x$$. Therefore,
$$[[x]]=[x]$$ and hence,
$$[x]-[[x]]=[x]-[x]=0$$
Question 8
1 / -0

If $$y=f(x) = \dfrac{2x-1}{x-2}$$, then $$f(y)=$$

A
$$x$$

B
$$y$$

C
$$2y-1$$

D
$$y-2$$

Solution

$$\displaystyle f(y) = \frac{2y-1}{y-2} = \frac{2 \left ( \dfrac{2x-1}{x-2} \right ) -1}{\left ( \dfrac{2x-1}{x-2} \right ) -2}$$
$$=\displaystyle \frac{4x-2-x+2}{2x-1-2x+4} = \frac{3x}{3} = x$$
Question 9
1 / -0

Find the domain of $$ e^x$$.

A
N

B
R

C
Z

D
All of the above

Solution
Question 10
1 / -0

A mapping function $$f:X\rightarrow Y$$ is one-one, if

A
$$f({ x }_{ 1 })\neq f({ x }_{ 2 })\ $$for all $$ { x }_{ 1 },{ x }_{ 2 }\in X$$

B
$$f({ x }_{ 1 })=f({ x }_{ 2 })\Rightarrow { x }_{ 1 }={ x }_{ 2 }$$ for all $${ x }_{ 1 },{ x }_{ 2 }\in X$$

C
$${ x }_{ 1 }={ x }_{ 2 }\Rightarrow f({ x }_{ 1 })=f({ x }_{ 2 })$$ for all $${ x }_{ 1 },{ x }_{ 2 }\in X$$

D
none of these

Solution

For a function to be one one
$$f(x_{1})=f(x_{2})$$
Implies that
$$x_{1}=x_{2}$$
Hence the answer is option B

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Re-Attempt Weekly Quiz Competition

Functions Test 10

Result

Functions Test 10

Score

-

Rank

-

SHARING IS CARING

Question 1

$$R$$

$$Z$$

$$R-Z$$

$$Q-{0}$$

Solution

Question 2

$$1$$

$$0$$

$$\sin x$$

Data is insufficient

Solution

Question 3

$$f(x)$$

$$2f(x)$$

$$3f(x)$$

$$-f(x)$$

Solution

Question 4

$$\displaystyle \frac{x}{\sqrt{1-x^2}}$$

$$\displaystyle \frac{x}{\sqrt{1+x^2}}$$

$$\displaystyle \frac{1-x^2}{\sqrt{1-x^2}}$$

$$x$$

Solution

Question 5

$$f[g(a)+g(b)]$$

$$f[g(ab)]$$

$$g[f(ab)]$$

$$g[f(a)+f(b)]$$

Solution

Question 6

$$1, 2$$

$$-1, 2$$

$$-1, -2$$

$$1, -2$$

Solution

Question 7

$$x$$

$$0$$

$$f(x)$$

$$g(x)$$

Solution

Question 8

$$x$$

$$y$$

$$2y-1$$

$$y-2$$

Solution

Question 9

N

R

Z

All of the above

Solution

Question 10

$$f({ x }_{ 1 })\neq f({ x }_{ 2 })\ $$for all $$ { x }_{ 1 },{ x }_{ 2 }\in X$$

$$f({ x }_{ 1 })=f({ x }_{ 2 })\Rightarrow { x }_{ 1 }={ x }_{ 2 }$$ for all $${ x }_{ 1 },{ x }_{ 2 }\in X$$

$${ x }_{ 1 }={ x }_{ 2 }\Rightarrow f({ x }_{ 1 })=f({ x }_{ 2 })$$ for all $${ x }_{ 1 },{ x }_{ 2 }\in X$$

none of these

Solution

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