4 Marks Each

Question 14 Marks

Obtain domain, co-domain and range for the following functions :
$(1)$ $f: \mathrm{A} \rightarrow \mathrm{B}, \mathrm{A}=\{-1,0,1\}, \mathrm{B}=\{1,2,3,4,5,6,7\}, f(x)=2 x+5, x \in \mathrm{A}$
$(2)$ $g: \mathrm{A} \rightarrow \mathrm{N}, \mathrm{A}=\{-1,2,3,4\}, g(x)=3 x+5, x \in \mathrm{A}$
$(3)$ $h: \mathrm{P} \rightarrow \mathrm{S}, \mathrm{P}=\{-2,-1,0,1\}, \mathrm{S}=\{-4,-3,-2,-1\}, h(x)=x-2, x \in \mathrm{P}$
$(4)$ $k: \mathrm{A} \rightarrow \mathrm{Z}, \mathrm{A}=\left\{-\frac{1}{2}, 0, \frac{1}{2}\right\}, k(\mathrm{x})=4 x^{2}+3, x \in \mathrm{A}$

Answer

$(1)$ Here, Domain $\mathrm{A}=\mathrm{D}_{f}=\{-1,0,1\}$
Co-domain $\mathrm{B}=\{1,2,3,4,5,6,7\}$
Now, for every $ x \in \mathrm{A}, \text { we find } f(x)=2 \mathrm{x}+5$
For $x=-1, f(-1)=2(-1)+5=-2+5=3$
For $x=0, f(0)=2(0)+5=0+5=5$
For $x=1, f(1)=2(1)+5=2+5=7$
$\therefore$ Range of the function $ \mathrm{R}_{f}=\{f(-1), f(0), f(1)\}=\{3,5,7\}$
$(2)$ Here, Domain $\mathrm{A}=\mathrm{D}_{f}=\{-1,2,3,4\}$
Co-domain $\mathrm{B}=\mathrm{N}$
Now, for every $ x \in \mathrm{A},$ we find $g(x)=3 x+5 .$
For $x=-1, g(-1)=3(-1)+5=2$
For $x=2, g(2)=3(2)+5=11$
For $x=3, g(3)=3(3)+5=14$
For $x=4, g(4)=3(4)+5=17$
$\therefore$ Range of the function $\mathrm{R}_{f}=\{2,11,14,17\}$ which is subset of co-domain.
$(3)$ Here, Domain $\mathrm{A}=\mathrm{P}=\mathrm{D}_{f}=\{-2,-1,0,1\}$
Co-domain $\mathrm{B}=\mathrm{S}=\{-4,-3,-2,-1\}$
Now, for every $x \in \mathrm{P},$ we find $h(x)=x-2 \text {. }$
For $x=-2, h(-2)=-2-2=-4$
For $x=-1, h(-1)=-1-2=-3$
For $x=0, h(0)=0-2=-2$
For $x=1, h(1)=1-2=-1$
$\therefore$ Range of the function$\mathrm{R}_{f}=\{-4,-3,-2,-1\}$
Here, the range and co-domain are same.
$(4)$ Here, Domain $\mathrm{A}=\mathrm{D}_{f}=\left\{-\frac{1}{2}, 0, \frac{1}{2}\right\}$
Co-domain $\mathrm{B}=\mathrm{Z}$
Now, for every $x \in \mathrm{A}$, we find $k(x)=4 x^{2}+3$.
For $x=-\frac{1}{2}, k\left(-\frac{1}{2}\right)=4\left(-\frac{1}{2}\right)^{2}+3=1+3=4$
For $x=0, k(0)=4(0)^{2}+3=0+3=3$
For $x=\frac{1}{2}, k\left(\frac{1}{2}\right)=4\left(\frac{1}{2}\right)^{2}+3=1+3=4$
$\therefore \quad$ Range of the function $R_{f}=\{3,4\} .$
Here, the range is a subset of co-domain.

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Question 24 Marks

Verify whether relations between the elements of the sets given below are functions or not.
(1) $\mathrm{A}=\{1,2,3,4\}, \mathrm{B}=\{3,5,7,9\}$, and relation is $f(x)=2 x+1, x \in \mathrm{A}$
(2) $\mathrm{P}=\left\{-\frac{1}{2}, 0,1\right\}, \mathrm{S}=\{10\}$, and rule is $k(x)=10, x \in \mathrm{P}$
(3) $\mathrm{A}=\{2,5,6\}, \mathrm{B}=\left\{1, \frac{3}{2}, \frac{9}{5}, \frac{11}{7}, \frac{13}{6}\right\}$, and rule is $y=\frac{2 x-1}{x+1}, x \in \mathrm{A}$
(4) $\mathrm{B}=\{-1,0,1,3\}, \mathrm{C}=\{-5,-3,-1,1,3\}$, and rule is $h(x)=2 x+3, x \in \mathrm{B}$

Answer

(1) We shall determine the image for all $x \in \mathbf{A}$.
The relation is $f(x)=2 x+1$
For $x=1, f(1)=2(1)+1=3$
For $x=2, f(2)=2(2)+1=5$
For $x=3, f(3)=2(3)+1=7$
For $x=4, f(4)=2(4)+1=9$
Thus, for each element of set $\mathrm{A}$, there exists one and only one element in set $\mathrm{B}$ i.e. each element of set $\mathrm{A}$ is related with one and only one element of set $\mathrm{B}$ by the relation $f(x)=2 x+1$. Hence, $f(x)=2 x+1$ is a function.
(2) We shall determine the image for all $x \in P$ for $k(x)=10$
For $x=-\frac{1}{2}, k\left(-\frac{1}{2}\right)=10$
For $x=0, k(0)=10$
For $x=1, k(1)=10$
So, for every $x \in P$, there exists an element " 10 " in set $\mathrm{S}$.
Hence the relation $k(x)=10$ is a function.
(3) We shall determine the image for all $x \in \mathrm{A}$, for $y=\frac{2 x-1}{x+1}$
For $x=2, y=\frac{2(2)-1}{2+1}=\frac{4-1}{3}=\frac{3}{3}=1$
For $x=5, y=\frac{2(5)-1}{5+1}=\frac{10-1}{6}=\frac{9}{6}=\frac{3}{2}$
For $x=6, y=\frac{2(6)-1}{6+1}=\frac{12-1}{7}=\frac{11}{7}$
So for every $x \in \mathrm{A}$, there exists a unique coresponding element in set $\mathrm{B}$.
Hence, the relation $y=\frac{2 x-1}{x+1}$ is a function.
(4) We shall determine the image for all $x \in$ B. The relation is $h(x)=2 x+3$
For $x=-1, h(-1)=2(-1)+3=-2+3=1$
For $x=0, h(2)=2(0)+3=0+3=3$
For $x=1, h(1)=2(1)+3=2+3=5$, which not an element of set C
So, there is no corresponding element for $x=1$ in set $\mathrm{C}$.
Hence, the relation $h(x)=2 x+3$ is not a function.

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Question 34 Marks

If $g(x)=\sqrt{36-x^{2}},-6 \leq x \leq 6 ;$ find $g(0), g(-3), g$ $(5), g(-6)$ and $g(2) .$

Answer

$g(0)=6, g(-3)=3 \sqrt{3}, g(5)=\sqrt{11}, g(-6)=$ $0, g(2)=4 \sqrt{2}$

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Question 44 Marks

If $\mathrm{f}: R \rightarrow R$ and $\mathrm{f}(\mathrm{x})=5 x^{2}-7 x+9$, find $\mathrm{f}(0)$, $\mathrm{f}(1), \mathrm{f}(-2)$ and $f\left(\frac{1}{2}\right) .$

Answer

$\mathrm{f}(0)=9, \mathrm{f}(1)=7, \mathrm{f}(-2)=43, f\left(\frac{1}{2}\right)=\frac{27}{4}$

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