2 Marks Questions

Question 12 Marks

Draw the graph of constant function $f: R \rightarrow R ; f(x)$ $=2 \forall x \in R$. Also, find its domain and range.

Answer

Given, $f: R \rightarrow R ; f(x)=2 \forall x \in R$

Domain $=R$ and Range $=\{2\}$.

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Question 22 Marks

Find the range of the function $f(x)=1-|x-2|$

Answer

$\begin{array}{l}\text { Given, } \quad f(x)=1-|x-2| \\ \text { We know that, }|x-2| \geq 0 \Rightarrow-|x-2| \leq 0 \\ \Rightarrow \quad 1-|x-2| \leq 1 \\ \Rightarrow \quad r(x) \leq 1 \\ \therefore \text { Range of } f(x)=(-\infty, 1] .\end{array}$

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Question 32 Marks

Find the domain of the function $f(x)=\frac{1}{\log (4-x)}$.

Answer

Given, $f(x)=\frac{1}{\log (4-x)}$
To determine the domain of $f(x)$, it should be real. Therefore $\log (4-x)$ should be defined and should not be equal to zero.
$
\begin{array}{lrl}
\text { i.e., } & 4-x & >0 \Rightarrow x<4\quad \ldots(i) \\
\text { Also, } & \log (4-x) & \neq 0 \Rightarrow x \neq 3\quad \ldots(ii)
\end{array}
$
From eqs. (i) and (ii), domain is $(-\infty, 3) \cup(3,4)$.

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Question 42 Marks

Find the range of the function $f(x)=\sqrt{x^2+4}$.

Answer

$\begin{array}{l}\text { Range of } f \text { : Let } y=f(x) \\ \qquad y=\sqrt{x^2+4} \\ \qquad x=\sqrt{y^2-4} \\ x \text { is defined, } y^2-4 \geq 0 \Rightarrow(y+2)(y-2) \geq 0 \\ \therefore \text { Range of } f=(-\infty,-2] \cup[2, \infty) .\end{array}$

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Question 52 Marks

Let $f$ be the subset of $Z \times Z$ defined by $f=\{(a b, a+b): a, b \in Z\}$. Is $f$ a function from $Z$ to Z ? Justify your answer.

Answer

Given, $f=\{(a b, a+b): a, b \in Z\}$
Taking $a=b=1$, we have $(a b, a+b)=(1,2) \in f$
Taking $a=b=-1$, we have $(a b, a+b)=(1,-2) \in f$
$\Rightarrow f$-image of 1 is not unique
Hence, $f$ is not a function.

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Question 62 Marks

Find the range of the function $f(x)=1-|x-2|$.

Answer

$
\text { Given, } \quad f(x)=1-|x-2|
$
We know that, $|x-2| \geq 0 \Rightarrow-|x-2| \leq 0$
$
\begin{array}{lr}
\Rightarrow 1-|x-2| \leq 1 \\
\Rightarrow f(x) \leq 1 \\
\therefore \text { Range of } f(x)=(-\infty, 1] .
\end{array}
$

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Question 72 Marks

Find the domain of function $f(x)=[x]+x$.

Answer

Given, $f(x)=[x]+x$ i.e., $f(x)=h(x)+g(x)$
where, $h(x)=[x]$ and $g(x)=x$
The domain of $h=R$ and The domain of $g=R$
$\therefore$ Domain of $f=R$.

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Question 82 Marks

Find the domain and range of real function \[f(x)=-|x|\]

Answer

$\begin{array}{l}\text { Given, } f(x)=-|x| \\ \text { Clearly, } f(x)=-|x| \leq 0, \forall x \in R \\ \therefore \text { Domain }=R \\ \text { and Range }=\{x \in R : x \leq 0\}=R^{-} \cup\{0\} .\end{array}$

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Question 92 Marks

Find the domain for which the functions defined by $f(x)=3 x^2-1$ and $g(x)=3+x$ are equal.

Answer

Given, $f(x)=3 x^2-1$ and $g(x)=3+x$
$
\begin{array}{l}
\text { Since, } \quad f(x)=g(x) \\
\Rightarrow \quad 3 x^2-1=3+x \\
\Rightarrow \quad 3 x^2-x-4=0 \\
\Rightarrow \quad(x+1)(3 x-4)=0 \\
\Rightarrow \quad x=-1 \text { and } x=\frac{4}{3} \\
\therefore \quad \text { Domain }=\left\{-1, \frac{4}{3}\right\} .
\end{array}
$

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Question 102 Marks

Find the range of the function $f(x)=x^2+2$.

Answer

Given, $f(x)=x^2+2$
Let $y=f(x)$
$y=x^2+2$
$x^2=y-2 \Rightarrow x=\sqrt{y-2}$
Clearly, $x$ will take real values, if
$y-2 \geq 0$
$y \geq 2$
$\therefore$ Range of $y=[2, \infty)$.

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Question 112 Marks

Find the domain of the function $f$ is given by
$f(x)=\frac{x^2+2 x+1}{x^2-8 x+12}$

Answer

$\therefore$ Domain of $f=R-\left\{x: x^2-8 x+12=0\right\}$
i.e., for function to be defined
$x^2-8 x+12 \neq 0$
$\begin{array}{l}\Rightarrow \quad x^2-6 x-2 x+12 \neq 0 \\ \Rightarrow \quad x(x-6)-2(x-6) \neq 0\end{array}$
$\begin{array}{lrl}\Rightarrow \quad (x-2)(x-6) \neq 0 \\ \Rightarrow \quad x \neq 2 \text { and } x \neq 6\end{array}$
$\therefore$ Domain $=R-\{2,6\}$

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Question 122 Marks

Find the range of function $f(x)=1+3 \cos 2 x$.

Answer

$\begin{array}{lrl}\text { Given, } f(x) =1+3 \cos 2 x \\ \text { We know that, } -1 \leq \cos 2 x \leq 1 \\ \Rightarrow -3 \leq 3 \cos 2 x \leq 3 \\ \Rightarrow -3 \leq 1+3 \cos 2 x \leq 1+3 \\ \Rightarrow -2 \leq 1+3 \cos 2 x \leq 4 \\ \Rightarrow \text { Range of } f(x) =[-2,4]\end{array}$

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Question 132 Marks

Justify that the relation $t=\{(x, 3) \mid x$ is a real number $\}$ is a function.

Answer

Given, $t=\{(x, 3) \mid x$ is a real number $\}$
The given relation is a function because every element in the domain has the image 3 .

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Applied Maths 2 Marks Questions

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