Functions Test 9

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Functions Test 9

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Weekly Quiz Competition

Question 1
1 / -0

The domain of ${f}({x})=\sqrt{2-\log_{3}(x-1)}$ is

A
$(2,12]$

B
$(-\infty,10]$

C
$(3,12]$

D
$(1,10]$

Solution

$f$ is defined only if term under root is positive.
$\Rightarrow 2-\log_{3}(x-1)\geq 0$
$\log_{3}(x-1)\leq 2$
$(x-1)\leq 9$
$x\leq 10$
$\log (x-1)$ is defined if $x-1>0 \Rightarrow x>1$
$\therefore$ Domain of $f$ is $(1,10]$ .
Question 2
1 / -0

The domain of the function defined by $\mathrm{f}({x})=^{(7-x)}P_{(x-3)}$ is

A
$\{3,7\}$

B
$\{3,4,5,6,7\}$

C
$\{3,4,5\}$

D
$\{1,2,3,4\}$

Solution

$f$ is defined only if $7-x\geq x-3$ and $7-x\geq 0$ .
$\Rightarrow x\leq 5$ and $x\geq 3$
$\therefore$ $f$ is defined at $x=\left \{ 3,4,5 \right \}$
Question 3
1 / -0

Let $S$ be set of all rational numbers. The functions $f:R\rightarrow R,\ g:R\rightarrow R$ are defined as
$$f(x)=\begin{cases}
0, & x \in S \\
1, & x \notin S
\end{cases}$$
$$g(x)=\begin{cases}
-1 & x\in S \\
0 & x\notin S
\end{cases}$$
then, $(fog) (\pi)+(gof)(e)=$

A
$-1$

B
$0$

C
$1$

D
$2$

Solution

$$f(x)=\begin{cases}
0, & x \text{ is rational} \\
1, & x \text{ is irrational}
\end{cases}$$
$$g(x)=\begin{cases}
-1 & x\text { is rational} \\
0& x\text { is irrational }
\end{cases}$$
$fog(\pi )=f(g(\pi ))=f(0)=0$
so $gof(e)=g(f(e))=g(1)=-1$
$\therefore fog(\pi )+gof(e)=-1$
Question 4
1 / -0

If $n\geq 1$ is any integer, $\mathrm{d}(n)$ denotes the number of positive factors of $n$ , then for any prime number $\mathrm{p},\ \mathrm{d}(\mathrm{d}(\mathrm{d}(\mathrm{p}^{7})))=$

A
$1$

B
$2$

C
$3$

D
$4$

Solution

For a number being $p^{n}$ the total number of positive factors excluding $1$ and the number itself will be $n-1$ .
Therefore total number of positive factors will be $n-1+2$
$=n+1$
Hence for $p^{7}$ we will have $8$ factors.
Now for $8=2^{3}$ .
Hence $8$ will have $4$ factors.
Now $4=2^{2}$ , hence the number of factors will be $3$ .
Question 5
1 / -0

The domain of $f(x) =\left ( {\sin }^{ -1 }x+{ \text{cosec} }^{ -1 }x\right )$ is

A
$[-1,1]$

B
$(-1, 1)$

C
$\{-1, 1\}$

D
$(-\infty, \infty)$
Question 6
1 / -0

lf $f:[-6,6]\rightarrow \mathbb{R}$ is defined by $f(x)=x^{2}-3$ for $x\in \mathbb{R}$ then
$(fofof)(-1)+(fofof)(0)+(fofof)(1)=$

A
$f(4\sqrt{2})$

B
$f(3\sqrt{2})$

C
$f(2\sqrt{2})$

D
$f(\sqrt{2})$

Solution

$fofof(-1)=fof(-2)=f(1)=-2$
$fofof(1)=fof(-2)=f(1)=-2$
$fofof(0)=fof(-3)=f(0)=33$
$\therefore fofof(-1)+fofof(1)+fofof(0)=29$
$=32-3$
$=(4\sqrt{2})^{2}-3$
$\Rightarrow f(4\sqrt{2})=(4\sqrt{2})^{2}-3$

Question 7

1 / -0

f

R\rightarrow R

is defined by

f(x)=\left\{\begin{array}{l}x+4 & x<-4\\3x+2 & -4\leq x<4\\x-4 & x\geq 4\end{array}\right.

then the correct matching of list I to List II is.

List - I	List - II
$\mathrm{A}) f(-5)+f(-4)=$	$\mathrm{i}) 14$
$\mathrm{B}) f(\|f(-8)\|)=$	ii $) 4$
$\mathrm{C}) f(f(-7)+f(3))=$	$\mathrm{i}\mathrm{i}\mathrm{i})-11$
$\mathrm{D}) f(f(f(f(0)))+1=$	$\mathrm{i}\mathrm{v})-1$
	v) $1$
	vi) $0$

A-iii , B-vi , C-ii , D- v

A-iii , B-iv , C-ii , D- vi

A-iv , B-iii , C-ii , D- i

A-ii , B-vi , C-v , D- ii

Solution

f(-5)=-5+4=-1  ;  f(-4)=3(-4)+2=-10

\therefore f(-5)+f(4)=-11

f(|f(-8)|)=f((-8+4))=f(4)=4-4=0

f(f(-7)) tf(3) = f(-7+4)+3(3)+2

=3(-3)+2+3(3)+2

=4

f(f(f(f(0))))+1=f(f(f(2)))+1=f(f(8))+1

=f(4)+1=0+1=1

Question 8
1 / -0

The domain of $\mathrm{f}(\mathrm{x})=\log_{\mathrm{x}}(9-\mathrm{x}^{2})$ is

A
$(-3,3)$

B
$(0,\infty)$

C
$(0,1) \cup (1,\infty)$

D
$(0,1) \cup (1,3)$

Solution

For $f$ to be defined $x>0$ and $x\neq 1$
$9-x^{2}\geq 0 \Rightarrow x\in (-3,3)$
$\therefore x\in (0,1)\cup (1,3)$
Question 9
1 / -0

The domain of $\displaystyle f(x)=\sqrt{\log\left(\frac{1}{|\sin x|}\right)}$ is

A
$(-\infty,\infty)$

B
$R-\displaystyle \{\frac{n\pi}{2}:n\in Z\}$

C
$\mathrm{R}-\{\mathrm{n}\pi :\mathrm{n}\in \mathrm{Z}\}$

D
$R-\displaystyle \{n\pi+(-1)2\frac{\pi}{2}:n\in Z\}$

Solution

$\displaystyle \frac{1}{|\sin x|}\geq 1$
$\therefore \log$ is always defined except at which denominator is zero,i.e, when $\sin x=0 \Rightarrow x=n\pi$
$\therefore$ $f$ is defined an $R-n\pi \ \ \forall n$ .
Question 10
1 / -0

lf $g(f(x)) =|\sin \mathrm{x}|,f(g(x)) =(\sin\sqrt{\mathrm{x}})^{2}$ , then

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

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

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

D
${f}, {g}$ cannot be determined

Solution

$g(f(x))=|\sin x|=\sqrt{\sin^{2}x}$ .......(i)
$f(g(x))=\sin^{2}\sqrt{x}$ .......(ii)

Comparing $i$ and $ii$

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

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

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

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

Therefore
$\Rightarrow$ $g(x)=\sqrt{x}$

$\Rightarrow$ $f(x)=\sin^{2}x$

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Functions Test 9

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Question 1

(2,12](2,12](2,12]

(−∞,10](-\infty,10](−∞,10]

(3,12](3,12](3,12]

(1,10](1,10](1,10]

Solution

Question 2

{3,7}\{3,7\}{3,7}

{3,4,5,6,7}\{3,4,5,6,7\}{3,4,5,6,7}

{3,4,5}\{3,4,5\}{3,4,5}

{1,2,3,4}\{1,2,3,4\}{1,2,3,4}

Solution

Question 3

−1-1−1

000

111

222

Solution

Question 4

111

222

333

444

Solution

Question 5

[−1,1][-1,1][−1,1]

(−1,1)(-1, 1)(−1,1)

{−1,1}\{-1, 1\}{−1,1}

(−∞,∞)(-\infty, \infty)(−∞,∞)

Question 6

f(42)f(4\sqrt{2})f(42​)

f(32)f(3\sqrt{2})f(32​)

f(22)f(2\sqrt{2})f(22​)

f(2)f(\sqrt{2})f(2​)

Solution

Question 7

A-iii , B-vi , C-ii , D- v

A-iii , B-iv , C-ii , D- vi

A-iv , B-iii , C-ii , D- i

A-ii , B-vi , C-v , D- ii

Solution

Question 8

(−3,3)(-3,3)(−3,3)

(0,∞)(0,\infty)(0,∞)

(0,1)∪(1,∞)(0,1) \cup (1,\infty)(0,1)∪(1,∞)

(0,1)∪(1,3)(0,1) \cup (1,3)(0,1)∪(1,3)

Solution

Question 9

(−∞,∞)(-\infty,\infty)(−∞,∞)

R−{nπ2:n∈Z}R-\displaystyle \{\frac{n\pi}{2}:n\in Z\}R−{2nπ​:n∈Z}

R−{nπ:n∈Z}\mathrm{R}-\{\mathrm{n}\pi :\mathrm{n}\in \mathrm{Z}\}R−{nπ:n∈Z}

R−{nπ+(−1)2π2:n∈Z} R-\displaystyle \{n\pi+(-1)2\frac{\pi}{2}:n\in Z\}R−{nπ+(−1)22π​:n∈Z}

Solution

Question 10

f(x)=sin⁡2x,g(x)=x{f}({x})=\sin^{2} {x},{g}({x})=\sqrt{{x}}f(x)=sin2x,g(x)=x​

f(x)=sin⁡x,g(x)=∣x∣{f}({x})=\sin x,g({x})=|{x}|f(x)=sinx,g(x)=∣x∣

f(x)=x2,g(x)=sin⁡x{f}({x})={x}^{2},{g}({x})=\sin\sqrt{{x}}f(x)=x2,g(x)=sinx​

f,g{f}, {g}f,g cannot be determined

Solution

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$(2,12]$

$(-\infty,10]$

$(3,12]$

$(1,10]$

$\{3,7\}$

$\{3,4,5,6,7\}$

$\{3,4,5\}$

$\{1,2,3,4\}$

$-1$

$0$

$1$

$2$

$1$

$2$

$3$

$4$

$[-1,1]$

$(-1, 1)$

$\{-1, 1\}$

$(-\infty, \infty)$

$f(4\sqrt{2})$

$f(3\sqrt{2})$

$f(2\sqrt{2})$

$f(\sqrt{2})$

$(-3,3)$

$(0,\infty)$

$(0,1) \cup (1,\infty)$

$(0,1) \cup (1,3)$

$(-\infty,\infty)$

$R-\displaystyle \{\frac{n\pi}{2}:n\in Z\}$

$\mathrm{R}-\{\mathrm{n}\pi :\mathrm{n}\in \mathrm{Z}\}$

$R-\displaystyle \{n\pi+(-1)2\frac{\pi}{2}:n\in Z\}$

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

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

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

${f}, {g}$ cannot be determined