Question 12 Marks
$f: \mathrm{A} \rightarrow \mathrm{B}, f(x)=1-4 x$ and $g: \mathrm{A} \rightarrow \mathrm{C}, g(x)=6 x+1$ and $\mathrm{A}=\{0,1,2\} .$ Are $f$ and $g$ equal functions ?
AnswerBoth the functions given here are defined on the same domain.
For $x=0, f(0)=1-4(0)=1$ and $g(0)=6(0)+1=1$
For $x=1, f(1)=1-4(1)=-3$ and $g(1)=6(1)+1=7$
Thus, $f(1) \neq g$ (1). Therefore, $f$ and $g$ are not equal functions.
View full question & answer→Question 22 Marks
$f: \mathrm{A} \rightarrow \mathrm{B}, \mathrm{A}=\{1,3\}, \mathrm{B}=\{1,4,9,16\}, f(x)=x^{2}$ and $g: \mathrm{A} \rightarrow \mathrm{B}, \mathrm{A}=\{1,3\}$, $\mathrm{B}=\{1,4,7,9,11\} g(x)=4 x-3 .$ Check the equality of the functions $f$ and $g .$
AnswerBoth the functions given here are defined on the same domain $A$.
$ \text { For } x=1, f(1) =1^{2}, g(1) =4(1)-3$
$ =1 =1$
$x=3, f(3) =3^{2}, g(3) =4(3)-3$
$ =9 =9 $
Thus, for each $x \in \mathrm{A}, f(x)=g(x)$.
Therefore, $f$ and $g$ are equal functions.
View full question & answer→Question 32 Marks
If $f=\mathbf{N} \rightarrow \mathbf{N}, f(x)=x^{2}$ and $g=\mathbf{Z}-\{0\} \rightarrow \mathbf{N}, g(x)=x^{2}$, can we say that $f$ and $g$ are equal functions ?
AnswerDomain of function $f$ is $\mathrm{D}_{f}=\mathrm{A}=\mathrm{N}$
Domain of function $g$ is $\mathrm{D}_{f}=\mathrm{A}=\mathrm{Z}-\{0\}$
Thus, both the functions are defined on different domains. So, they are not equal functions.
View full question & answer→Question 42 Marks
$(i) f: Z \rightarrow R$ and $f(x)=100$ then state the type of function $f$.
$(ii)$ State the type of the function between dates of a month and days of the week.
Answer$(1)$ Domain $\mathrm{A}=\mathrm{D}_{f}=\mathrm{Z}=\{\ldots . .,-2,-1,0,1,2, \ldots\}$ and co-domain $\mathrm{B}=\mathrm{R} .$ For every value of the domain, the image in $\mathrm{B}$ is same, which is " $100^{\prime \prime}$.
Hence, the given function is constant function.
$(2)$ In a month, there are four weeks. So, the four different days are repeated four times with different dates of a month.
Hence, it is a many—one function.
View full question & answer→Question 52 Marks
$f: \mathrm{Z} \rightarrow \mathbf{N} \cup\{0\}, f(x)=x^{2}$ then state the type of function $f$.
AnswerDomain $\mathrm{D}_{f}=\mathrm{Z}=\{\ldots-3,-2,-1,0,1,2,3, \ldots\}$
Co-domain $\mathrm{B}=\mathrm{N} \cup\{0\}=\{0,1,2,3, \ldots\}$
By substituting the values of $x=-2,-1,0,1,2$ the images are $4,1,0,1,4$
Thus, for two different values $\{-2,2\}$ and $\{-1,1\}$ of domain their images in co-domain are same. Hence,
the given function is many - one function.
View full question & answer→Question 62 Marks
State the type of function $f: \mathbf{N} \rightarrow \mathbf{N}, f(x)=x^{2}$
AnswerDomain $\mathrm{D}_{f}=\mathrm{N}=\{1,2,3, \ldots\}$
Co-domain $\mathrm{B}=\mathrm{N}=\{1,2,3, \ldots\}$
$f(x)=x^{2} .$ By substituting the values of $x=1,2,3, \ldots$ the images are $1,4,9, \ldots$
Thus, for two different values of domain their images in co-domain are different. Hence, the given
function is one - one function.
View full question & answer→Question 72 Marks
If $f(x)=\frac{x^2-4}{x-2}$ Where $x \in Z-\{2\}$ then find the value of $f(0)+f(1)-f(-2)$.
Answer$f(x)=\frac{x^2-4}{x-2}=\frac{(x+2)(x-2)}{x-2}=x+2$
$\therefore f(0)+f(1)+f(2)$
$=(0+2)+(1+2)-(-2+2)$
$=2+3-0$
$=5$
View full question & answer→Question 82 Marks
If domain of $f(x)=\frac{ x - 3 }{ x 4 }$ is $\{0,3,6\}$ then find its range.
Answer$f(x)=\frac{x-3}{x+4^{\prime}} \text { Domain }=\{0,3,6\}$
$\therefore f(0)=\frac{0-3}{0+4}=-\frac{3}{4}$
$f(3)=\frac{3-3}{3+4}=\frac{0}{7}=0$
$f(6)=\frac{6-3}{6+4}=\frac{3}{10}$
Range $R_f=\{f(x) \mid x \in$ domain $\}=\{f(0), f(3), f(6)\}$
$\therefore R_f=\left\{-\frac{3}{4^{\prime}}, 0, \frac{3}{10}\right\}$
View full question & answer→Question 92 Marks
If $P: A \rightarrow B, P(x)=2 x-3$ and $R_f=\{-2,-1,0\}$, then find domain of the function.
Answer$P(x)=2 x-3$
$R_f=\{-2,-1,0\}-1=2 x-3 \quad 0=2 x-3$
$-2=2 x-3 \quad-1+3=2 x \quad 3=2 x$
$-2+3=2 x \quad 2=2 x \quad x=$
$1=2 x \quad x=1$
$x=$
Therefore, domain of the function $=\{,1,\}$
View full question & answer→Question 102 Marks
$h: A \rightarrow B, A=\{1,2,3\} \text {, }$
$B=(3,4,5,6,7,8\} h(x)=x+5$. State the type of function $h$.
Answer$h: A \rightarrow B$
$A=\{1,2,3\} \text { and } h(x)=x+5$
$\therefore h(1)=1+5=6, h(2)=2+5=7, h(3)=3+5=8$
$B=\{3,4,5,6,7,8\}$
For the elements $1,2,3$ of domain $h(1) \neq h(2) \neq h(3)$.
Therefore, function $h$ is one-one function.
View full question & answer→Question 112 Marks
$K: X \rightarrow Y, X=\{t \in Z,-3 \leq t \leq 3\} ; Y=\{a \mid a \in N, 1 \leq a \leq 20\} ; k(t)=t^2+2$. State the type of function $K$.
Answer$K : X \rightarrow Y$
Domain $X =\{t \mid t \in Z,-3 \leq t \leq 3\}$
$=\{-3,-2,-1,0,1,2,3\}$
$K ( t )=t^2+2$
$\therefore K (-3)=(-3)^2+2=9+2=11 .$
$K (3)=(3)^2+2=11,$
$K(-2)=(-2)^2+2=4+2=6 .$
$K (2)=(2)^2+2=6 .$
$K (-1)=(-1)^2+2=1+2=3 .$
$K (1)=(1)^2+2=3 .$
$K (0)=0+2=2$
Co - domain $Y=\{a \mid a \in N \Rightarrow 1, a \leq 20\}$
$=\{2,3,4,5,6,7,8,9,10,11,12 .............. 18,19,20\}$
Here, for two different elements $(-3,3)$ of domain, their images in co-domain are $k(-3)=(3)=11$. Therefore, function $k$ is manv one function.
View full question & answer→Question 122 Marks
$g: A \rightarrow N, A=\{x \mid x \in N, 1 < x \leq 4\} ; g(x)=x+1$. Find range of function $g$.
Answer$A=\{x \mid x \in N, 1 < x \leq 4\}$
$\therefore A=\{2,3,4\} \Rightarrow \text { Domain }$
$g(x)=x+1$
$\therefore g(2)=2+1=3, g(3)=3+1=4, g(4)=4+1=5$
Range of function $R_f=\{g(x) \mid x \in A\}$
$=\{g(2), g(3), g(4)\}=\{3,4,5\}$
View full question & answer→Question 132 Marks
Define range of a function.
AnswerSuppose $f : A B.$ then a set of images or functional values $f$ all the elements of set $A$ is called range of function $f$. It is denoted by $R_f$.
$R_f=\{f(x) \mid x \quad A\}$
The range of a function may be co-domain itself or a subset of co-domain.
View full question & answer→Question 142 Marks
Define domain and co-domain of a function.
AnswerDomain of a function: Suppose, $f: A \rightarrow B$, the set $A$ Is called domain of the function $f$. It is denoted by $D_f$. Co domain of a function: Suppose, $f : A \rightarrow B$, th e set $B$ is called the co-domain of the function $f$.
View full question & answer→Question 152 Marks
If $f(x)=x^3-2 x+\frac{1}{x}$ is a real function then find the value of $f(3)+f(-3)$.
Answer$f(x)=x^3-2 x+\frac{1}{x}$
$\therefore f(-3)=(-3)^3-2(-3)+\frac{1}{(-3)}=-27+6-\frac{1}{3}=-21-\frac{1}{3}=\frac{-63-1}{3}=-\frac{64}{3}$
$f(3)=(3)^3-2(3)+\frac{1}{3}=21+\frac{1}{3}=\frac{63+1}{3}=\frac{64}{3}$
Therefore, $f(3)+f(-3)=\frac{64}{3}+\left(-\frac{64}{3}\right)=\frac{64}{3}-\frac{64}{3}=0$
View full question & answer→Question 162 Marks
For a real function $f(x)=6 x^3-5^x+15$, find the value of $f(0)$.
Answer$f(x)=6 x^3-5^x+15$
$\therefore f(0)=6 \times 0-5^0+15=0-1+15=14$
$\therefore f(0)=14$
View full question & answer→Question 172 Marks
$f: Z-\{2\} \rightarrow Z, f(x)=\frac{x^2+x-6}{x-2}$, State the type of the function.
Answer$f: Z-\{2\} \rightarrow Z$
$\therefore$ Domain $=\{\ldots-3,-2,0,1,2,3,4, \ldots\}$
$f(x)=\frac{x^2+x-6}{x-2}$,
$\therefore f(-3)=\frac{9-3-6}{-3-2}=\frac{0}{-5}=0$
$f(-2)=\frac{4-2-6}{-2-2}=1$
$f(-1)=\frac{1-1-6}{-1-2}=2$
$f(0)=\frac{-6}{-2}=3$
$f(1)=\frac{1+1-6}{1-2}=\frac{-4}{-1}=4$
$f(3)=\frac{9+3-6}{3-2}=6$
$f(4)=\frac{16+4-6}{4-2}=\frac{14}{2}=7$
Here, for two different elements of domain, their images are not same. Therefore, the given function is one - one function.
View full question & answer→Question 182 Marks
If $f(x)=\frac{2 x-4}{x+7}$ is a real function then for which value of $x$ the image is zero?
Answer$f(x)=\frac{2 x-4}{x+7}$
Image $=0$. Therefore, put $f(x)=0$
$0=\frac{2 x-4}{x+7}$
$\therefore 0=2 x-4 \therefore 4=2 x$
Therefore, $x=2$, the image is zero.
View full question & answer→Question 192 Marks
If $f: R \rightarrow R$ and $f(x)=x^2+2 x-1$, then state the type of function $f$.
Answer$f: R \rightarrow R$
$f(x)=x^2+2 x-1$
Domain $=\{\ldots . .-2,-1,0,1, \ldots\}$
Co - domain $=\{\ldots-1,-2,-1,2, \ldots\}$.
Here, for two different elements $-2$ and $0$ of domain, their images are same in co - domain. Therefore function $f$ is many - one function.
View full question & answer→Question 202 Marks
If $f(x)=\frac{x^2(x+1)^2}{4}$ is a real function then find the value of $f(3)-f(2)$.
Answer$f(x)=\frac{x^2(x+1)^2}{4}$
$\therefore f(2)=\frac{2^2(2+1)^2}{4}=\frac{4(3)^2}{4}=9$
$f(3)=\frac{3^2(3+1)^2}{4}=\frac{9(4)^2}{4}=9 \times 4=36$
Therefore, $f(3)-f(2)=36-9=27$
View full question & answer→Question 212 Marks
Give definition of a function.
AnswerIf $A$ and $B$ are any two non-empty sets and each element of set $A$ is related with one and only one element of set $B$ by some rule, relation or correspondence, then It is called a function from set $A$ to $B$ and is denoted by $f: A \rightarrow B$.
View full question & answer→Question 222 Marks
If $\mathrm{f}(\mathrm{x})=\frac{x(x+1)}{2}$ for which value of $\mathrm{x}, \mathrm{f}(-\mathrm{x})=\mathrm{f}(3) ?$
View full question & answer→Question 232 Marks
If $f(x) = ax + b$ and $f(0) = 7, f(2) = 17$; obtain the form of the function.
Answer$a = 5, b = 7, f(x) = 5x + 7$
View full question & answer→Question 242 Marks
If $f(x) = ax + k$, for which value of k, $f(1) + f( - 1) = 4?$
View full question & answer→Question 252 Marks
If $f(x)=2 x^{2}-4 x+8$, for which value of $x, f(x)=$$24 ?$
View full question & answer→Question 262 Marks
If $f(x)=\frac{3 x+4}{2 x-1}$, for which value of $x, f(x)=7 ?$
View full question & answer→Question 272 Marks
If $f(x)=\frac{3 x^{2}-1}{2 x-1}$ find $f(2)+3, f(0)$
View full question & answer→Question 282 Marks
If $\mathrm{f}(\mathrm{x})=x^{3}-x$ and $\mathrm{g}(\mathrm{x})=x^{2}+3 x+5 .$ Find $\mathrm{f}(2)+$
View full question & answer→Question 292 Marks
If $\mathrm{f}(\mathrm{x})=x^{3}-x$ and $\mathrm{g}(\mathrm{x})=x^{2}+3 x+5 .$ Find $\mathrm{f}(3)$ $2 g(1)$
View full question & answer→Question 302 Marks
If $f(x)=3+\frac{2}{5+\frac{1}{x}}$, find $f(2)-f(1)$
View full question & answer→Question 312 Marks
If $\mathrm{f}(\mathrm{x})=x^{2}-2^{x}+x^{x}$ then find $\mathrm{f}(2)+\mathrm{f}(0)+\mathrm{f}(-2)$
View full question & answer→Question 322 Marks
Find $\frac{f(0)+f(1)}{f(-1)}$ for $f(x)=\frac{2 x^{2}-7 x+3}{x+2}$
View full question & answer→Question 332 Marks
If $\mathrm{f}(\mathrm{x})=\frac{x^{2}-3 x+1}{2 x+1}$ and $\mathrm{g}(\mathrm{x})=3 x^{2}-x-1$, then find $f(2)+3 . g(1)$
View full question & answer→Question 342 Marks
If $f(x)=\frac{x^{2}-3 x+5}{x+1}$ then find $f(3), f(-2)$ and $f\left(\frac{1}{2}\right)$
Answer$f(3)=\frac{5}{4}, f(-2)=-15, f\left(\frac{1}{2}\right)=\frac{5}{2}$
View full question & answer→Question 352 Marks
If $\mathrm{f}(\mathrm{x})=\frac{x^{2}-7 x+5}{x-2} x \in Z-\{2\}$, then find $\mathrm{f}(0), \mathrm{f}(1)$ and $\mathrm{f}(- 2).$
Answer$f(0)=-\frac{5}{2}, f(1)=1, f(-2)=-\frac{23}{4}$
View full question & answer→Question 362 Marks
$\mathrm{P}=\{0,1,2,3\}, \mathrm{Q}=\left\{0,1, \frac{1}{2}, \frac{1}{3}\right\} ; f(x)=\frac{x-1}{x+1} x \in A .$ Can this be called a function?
Answer$f(x)=\frac{x-1}{x+1}$ cannot be called a function
View full question & answer→Question 372 Marks
$A=\{1,2,3\}, B=\{2,5,8\} ; f(x)=3 x-1, x \in A$. Is this a function?
Answer$f(x)=3 x-1$ is a function.
View full question & answer→Question 382 Marks
For a function $f(x) = 9x + 2$, its range $= \{2, -7, 11\}$; then find the domain.
AnswerDomain $= \{0, -1, 1\}$
View full question & answer→Question 392 Marks
Co-domain of a function $f(x) =7x-3$ is $\{4, 32, 67\}$, then find the domain.
AnswerDomain$=\{1, 5, 10\}$
View full question & answer→Question 402 Marks
If $f: N \rightarrow R, f(x)=\sqrt{2 x^{2}}+1 ; x \in N$ and $x<5$, then find range of the function.
AnswerRange $=\{\sqrt{3}, 3, \sqrt{19}, \sqrt{33}\}$
View full question & answer→Question 412 Marks
If $f: A \rightarrow B, A=\{1,2\}, B=\{1,4,7,10\} ; f(x)=3 x-2$ and $g: A \rightarrow B ; A=\{1,2\}, B=\{1,4,9,16\} ; g(x)=x^{2}$ check the equality of functions.
Answer$f$ and $g$ are equal functions
View full question & answer→Question 422 Marks
If $\mathrm{f}(\mathrm{x})=\frac{1}{x}+x^{2}$, find the value of $f\left(\frac{1}{2}\right)-f\left(-\frac{1}{2}\right)$.
View full question & answer→Question 432 Marks
If $g(x)=3 x^{2}-2 x+1$ and $h(x)=x(x-1)$; the domain set $=\{-1,0,2\}$, then state whether $g$ and $h$ are equal functions or not.
Answer$\mathrm{g}$ and $\mathrm{h}$ are not equal functions
View full question & answer→Question 442 Marks
If the cost of producing $\times$ units is $C(x)=\frac{x}{5}+7 x+$ 100 and total cost is $Rs. 950$, find the value of $x .$
View full question & answer→Question 452 Marks
$\mathrm{f}: \mathrm{Z}-\{3\} \rightarrow \mathrm{Z}, \mathrm{f}(\mathrm{x})=\frac{x^{2}+x-12}{x-3} ; x \in Z-3$ Determine the type of the function.
View full question & answer→Question 462 Marks
If for $\mathrm{f}(\mathrm{x})=5 \mathrm{x}-4 ; R_{f}=\{-14,-4,6,16\}$, find the domain of the function.
Answer$D_{f}=\{-2,0,2,4\}$
View full question & answer→Question 472 Marks
If $\mathrm{h}(\mathrm{x})=\frac{x^{2}-25}{x-5}$ and domain $=\{2,4,6\}$, find $R_{h}$.
View full question & answer→Question 482 Marks
$f ( x )= x (2 x -7)$ where $x \in R$. if $f ( x )=15$ then find the value of $x$.
Answer$f(x)=15$
$\therefore x(2 x-7)=15$
$\therefore 2 x^2-7 x-15=0$
$\therefore 2 x^2+3 x-10 x-15=0$
$\therefore x(2 x+3)-5(2 x+3)=0$
$\therefore(2 x+3)(x-5)=0$
$\therefore(2 x+3)=0 \text { or }(x-5)=0$
$\therefore x=-\frac{3}{2} \text { or } x=5$
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