Question 13 Marks
Find domain, co-domain and range for the following functions :
$(1)$ $f: \mathrm{A} \rightarrow \mathrm{N}, f(x)=x^{2}+1, \mathrm{~A}=\{x \mathrm{I}-2 \leq x<1, x \in \mathrm{Z}\}$
$(2)$ $f: \mathrm{Z} \rightarrow \mathrm{N}, f(x)=x^{2}+2, x \in \mathrm{Z}$
$(3)$ $f: \mathrm{N} \rightarrow \mathrm{N}, f(x)=4 x, x \in \mathrm{N}$
Answer$(1)$ Here, Domain $\mathrm{D}_{f}=\mathrm{A}=\{-2,-1,0\}$
Co-domain $\mathrm{B}=\mathrm{N}$
function $f(x)=x^{2}+1$
For $x=-2, f(-2)=4+1=5$
For $x=-1, f(-1)=1+1=2$
For $x=0, f(0)=0+1=1$
$\therefore$ Range of the function $\mathrm{R}_{f}=\{5,2,1\}$
$(2)$ Domain $\mathrm{D}_{f}=\mathrm{A}=\mathrm{Z}$
Co-domain B $=\mathrm{N}$
For the range of the function $f(x)=x^{2}+2$,
$ \mathrm{R}_{f} =\{\ldots . ., f(-2), f(-1), f(0), f(1), f(2), \ldots \ldots\}$
$ =\{\ldots . ., 2,3,6,11, \ldots \ldots\} $
$(3)$ Domain $\mathrm{D}_{f}=\mathrm{A}=\mathrm{N}$
Co-domain B $=\mathrm{N}$,
The range of function $f(x)-4 x$
$ \mathrm{R}_{f} =\{f(1), f(2), f(3), f(4), \ldots \ldots\}$
$ =\{4,8,12,16, \ldots \ldots\} $
View full question & answer→Question 23 Marks
If $f(x)=x^{2}-4 x+8$ then for which value of $x$, is $f(2 x)=2 f(x) ?$
Answer$\quad f(x)=x^{2}-4 x+8$
$\therefore \quad f(2 x)=(2 x)^{2}-4(2 x)+8$
$\quad=4 x^{2}-8 x+8$
Now, $f(2 x)=2 f(x)$
$\therefore \quad 4 x^{2}-8 x+8=2 x^{2}-8 x+16$
$\therefore \quad 2 x^{2}-8=0$
$\therefore \quad x^{2}-4=0$
$\therefore \quad x^{2}=4$
$\therefore \quad x=\pm 2$
$\therefore \quad$ When $x=\pm 2, f(2 x)=2 f(x) .$
View full question & answer→Question 33 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 $
View full question & answer→Question 43 Marks
If $f(x)=x^{3}+3^{x}-x^{2}-2^{x}$ then find $f(3)-6 f(2)$
Answer$ f(x) =x^{3}+3^{x}-x^{2}-2^{x} $
$f(3) =(3)^{3}+3^{3}-(3)^{2}-(2)^{3} \text { and } f(2) =2^{3}+3^{2}-2^{2}-2^{2}$
$ =27+27-9-8 =8+9-4-4$
$ =37 =9 $
$ \therefore f(3)-6 f(2) =37-6(9)$
$ =37-54$
$ =-17 $
View full question & answer→Question 53 Marks
If $f(x) \frac{x^{3}+1}{x^{2}-2 x+1}$ where $x \in \mathbf{Z}-\{1\}$ then find $f(-2), f(-1)$ and $f(0)$.
Answer$f(x) =\frac{x^{3}+1}{x^{2}-2 x+1} f(-1) =\frac{(-1)^{3}+1}{(-1)^{2}-2(-1)+1}$
$f(-2) =\frac{(-2)^{3}+1}{(-2)^{2}-2(-2)+1} ; =\frac{-1+1}{1+2+1}$
$ =\frac{-8+1}{4+4+1} =\frac{0}{4}$
$ =-\frac{7}{9} =0$
$f(0) =\frac{0^{3}+1}{0^{2}-2(0)+1}=\frac{1}{1}=1 $
View full question & answer→Question 63 Marks
If $f(x)=x^2$ and $g(x)=5 x-6$, where $x \in\{2,3,4\}$, check the equality of the functions.
Answer$f(x)=x^2, g(x)=5 x-6$, Domain $=\{2,3,4\}$ For $f(x)=x^2$, Range $R_f=\{f(2), f(3), f(4)\}=\{4,9,16\}$
For $g(x)=5 x-6$,
Range $R_f=\{g(2), g(3), g(4)\}$
$\therefore g(2)=5(2)-6=10-6=4$
$g(3)=5(3)-6=15-6=9$
$g(4)=5(4)-6=20-6=14$
$\therefore R_f=\{4,9,14\}$
Domains of the functions $f(x)$ and $g(x)$ are same but their range are not same.
Hence, the given functions are unequal functions.
View full question & answer→Question 73 Marks
If the domain of a function $f: A \rightarrow N \cup\{0\} ; f(x)=\sqrt{ x ^2- 1 6 }$ is $A=\{4,5\}$, then find its range.
Answer$f(x)=\sqrt{x^2-16}$; Domain $=\{4,5\}$
$\therefore$ Range $R _{ f }=\{ f ( x ) \mid x \in$ Domain $\}=\{ f (4), f (5)\}$
Now, $f(x)=\sqrt{x^2-16}$
$\therefore f(4)=\sqrt{(4)^2-16}=\sqrt{16-16}=0$
$f(5)=\sqrt{(5)^2-16}=\sqrt{25-16}=\sqrt{9}=3(\because x$ is positive. $)$
Hence, Range $R_f=\{0,3\}$
View full question & answer→Question 83 Marks
If $f : A \rightarrow M , A =\{ x \mid x \in N , 1 \leq x \leq 5\}$ and $M=\{x \mid x \in N, 1 \leq x \leq 20\}$ and $f(x)=x^2+1$. then find the range of $f$.
Answer$f: A \rightarrow M$,
$A =\{ x \mid x \in N , 1 \leq x \leq 5\}=\{1,2,3,4$,
$M=\{x \mid x \in N, 1 \leq x \leq 20\}=\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20\}$
$f(x)=x^2+1$
$\therefore$ Range of $f R_f=\{f(x) \mid x \in A\}=\{f(1), f(2), f(3), f(4)\}$
Now, $f(x)=x^2+1$.
$\therefore f(1)=(1)^2+1=1+1=2$
$f(2)=(2)^2+1=4+1=5$
$f(3)=(3)^2+1=9+1=10$
$f(4)=(4)^2+1=16+1=17$
Hence, Range of $f R _{ f }=\{2,5,10,17\}$
View full question & answer→Question 93 Marks
For the real function $f(x)=2 x^2-5 x+4$, find the value of $x$, for which $f(3 x)-3 f(x)+5=0$.
Answer$f(x)=2 x^2-5 x+4$
Now, $f(3 x)-3 f(x)+5=0$
$\therefore 2(3 x)^2-5(3 x)+4-\left\{3\left(2 x^2-5 x+5\right)\right\}+5=0$
$\therefore 2\left(9 x^2\right)-15 x+4-6 x^2+15 x+4-12+5=0$
$\therefore 18 x^2-6 x^2-15 x+15 x+4-12+5=0$
$\therefore 12 x^2-3=0$
$\therefore 12 x^2=3$
$\therefore x^2=\frac{3}{12}=\frac{1}{4}=\left( \pm \frac{1}{2}\right)^2$
$\therefore x= \pm \frac{1}{2}$
Hence, when $x= \pm \frac{1}{2}$ then $f(3 x)-3 f(x)+5=0$.
View full question & answer→Question 103 Marks
For the function $f: A \rightarrow B, f(x)=4 x-3 ; R_f=\{9,13,17,25\}$ then find $D_f$.
Answer$f: A \rightarrow B, f(x)=4 x-3, R_f=\{9,13,17,25\}$
$f(x)=9 \therefore 9=4 x-3, \therefore 12=4 x \therefore x=3$
$f(x)=13 \therefore 13=4 x-3 \therefore 16=4 x \therefore x=4$
$f(x)=17 \therefore 17=4 x-3 \therefore 20=4 x \therefore x=5$
$f(x)=25 \therefore 25=4 x-3 \therefore 28=4 x \therefore x=7$
Hence, $D_f=\{3,4,5,7\}$
View full question & answer→Question 113 Marks
If: $R-\{0\} \rightarrow R, f(x)=\frac{1}{x}\left(1+\frac{1}{x}\right)-1$; than find the value of $f(-1), f(-2)$ and $f\left(\frac{1}{2}\right)$.
Answer$f(x)=\frac{1}{x}\left(1+\frac{1}{x}\right)-1$
$\therefore f(-1)=\frac{1}{(-1)}\left(1+\frac{1}{(-1)}\right)-1=-(1-1)-1=-0-1=-1$
$f(-2)=\frac{1}{(-2)}\left[1+\frac{1}{(-2)}\right]-1=-\frac{1}{2}\left[1-\frac{1}{2}\right]-1=-\frac{1}{2}\left(\frac{1}{2}\right)-1=-\frac{1}{4}-1=\frac{-1-4}{4}=\frac{-5}{4}$
$f\left(\frac{1}{2}\right)=\frac{1}{\frac{1}{2}}\left[1+\frac{1}{2}\right]-1$
$=2(1+2)-1=2(3)-1=6-1=5$
Hance, $f(-1)=-1, f(-2)=-\frac{5}{4}, f\left(\frac{1}{2}\right)=5$
View full question & answer→Question 123 Marks
Obtain domain, co - domain and range for $f: P \rightarrow Q, P=\left\{-\frac{1}{2}, 1, \frac{1}{2}, \frac{3}{2}\right\} ; Q=\left\{-\frac{1}{5}, 1, \frac{1}{3^3}, 3\right\}, f(x)=\frac{ x }{2- x }$
Answer
- $f: P \rightarrow Q, P=\left\{-\frac{1}{2^{\prime}}, 1, \frac{1}{2}, \frac{3}{2}\right\} ; Q=\left\{-\frac{1}{5}, 1, \frac{1}{3^2}, 3\right\}, f(x)=\frac{x}{2-x^{\prime}}$
- Domain: $D _{ f }=\left\{-\frac{1}{2}, 1, \frac{1}{2}, \frac{3}{2}\right\}$
- Co - domain : $\left\{-\frac{1}{5}, 1, \frac{1}{3}, 3\right\}$
- Range : $R_f=\left\{f(x) \mid x \in D_f\right\}=\left\{f\left(-\frac{1}{2}\right), f(1), f\left(\frac{1}{2}\right), f\left(\frac{3}{2}\right)\right\}$
- Now, $f(x)=\frac{x}{2-x}$
- $\therefore f\left(-\frac{1}{2}\right)=\frac{-\frac{1}{2}}{2-\left(-\frac{1}{2}\right)}=-\frac{1}{2\left(2+\frac{1}{2)}\right.}=-\frac{1}{2 x \frac{5}{2}}=-\frac{1}{5}$
- $f(1)=\frac{1}{2-1}=1$
- $f\left(\frac{1}{2}\right)=\frac{\frac{1}{2}}{2-\frac{1}{2}}=\frac{1}{2\left(\frac{3}{2}\right)}=\frac{1}{3}$
- $f\left(\frac{3}{2}\right)=\frac{\frac{3}{2}}{2-\frac{3}{2}} \frac{3}{2\left(\frac{3}{2}\right)}=3$
- $\therefore$ Range $R_f=\left\{-\frac{1}{5}, 1, \frac{1}{3}, 3\right\}$
View full question & answer→Question 133 Marks
If $f: A \rightarrow N, A=\{500,1000,1300,1400\}, f(x)=\sqrt{ 5 6 0 0 - 4 x }, x \in R$ then find the value of $f(x)$ for $x=1000$. Also, for which value of $x, f(x)=20$ ?
Answer$f(x)=\sqrt{5600-4 x}$
Putting $x=1000$,
$f(4)=\sqrt{5600-4(1000)}=\sqrt{5600-4000}=\sqrt{1600}=40$
$f(x)=20$
$\therefore 20=\sqrt{5600-4 x}$
$\therefore(20)^2=5600-4 x$
$\therefore 4 x=5600-400$
$\therefore 4 x=5200$
$\therefore x=1300$
Hence, when $x=1300$, then $f(x)=20$.
View full question & answer→Question 143 Marks
If $f(x)=15 x^2-4 x^2+x+10, x \in R$; obtain the value of $\frac{f(2)}{f(1)}$.
Answer$f(x)=15 x^2-4 x^2+x+10$
$\therefore \frac{f(2)}{f(1)}=\frac{15(2)^3-4(2)^2+2+10}{15(1)^3-4(1)^2+1+10}=\frac{(15 \times 8)-(4 \times 4)+12}{15-4+11}=\frac{120-16+12}{22}=\frac{116}{22}=\frac{58}{11}$
View full question & answer→Question 153 Marks
If $f(x)=2 x^2+\frac{1}{x^{\prime}} x \in R$; Obtain the value of $f(3)+f(-3)$.
Answer$f(x)=2 x^2+\frac{1}{x}$
$\therefore f(3)+f(-3)=2(3)^2+\frac{1}{3}+2(-3)^2+\frac{1}{(-3)}$
$=(2 \times 9)+\frac{1}{3}+(2 x 9)-\frac{1}{3}=18+18=36$
View full question & answer→Question 163 Marks
If $f(x)=\frac{2 x+3}{5 x+2}, x \in R-\left\{-\frac{2}{5}\right\}$; then find the value of $f(2) . f\left(\frac{1}{2}\right)$.
Answer$f(x)=\frac{2 x+3}{5 x+2}$
$\therefore f (2) . f \left(\frac{1}{2}\right)=\frac{2(2)+3}{5(2)+2} \times \frac{2\left(\frac{1}{2}\right)+3}{5\left(\frac{1}{2}\right)+2}$
$=\frac{4+3}{10+2} \times \frac{1+3}{\frac{5}{2}+2}$
$=\frac{7}{12} \times \frac{\frac{4}{9}}{2}=\frac{7}{12} \times \frac{4 \times 2}{9}=\frac{7 \times 2}{3 \times 9}=\frac{14}{27}$
View full question & answer→Question 173 Marks
If $f(x)=x(3 x-2), g(x)=x^3$ and $x \in\{0,1,2\}$; then prove that $f$ and $g$ are equal functions.
Answer$f(x)=x(3 x-2), x \in\{0,1,2\}...........(1)$
$\therefore$ Range $R_f=\{f(0), f(1), f(2)\}$
Now, $f(0)=0(3 \times 0-2)=0$,
$f(1)=1(3 \times 1-2)=1$,
$f(2)=2(3 \times 2-2)=2(6-2)=2 \times 4=8$
$\therefore$ Range $R _{ f }=\{0,1,8\}..........(2)$
$g(x)=x^3, x \in\{0,1,2\}.........(3)$
Now, $g(0)=0 . g(1)=1, g(2)=8$
$\therefore$ Range $R_f=\{0,1,8\}..........(4)$
From $(1)$ and $(3)$ the domain of $f$ and $g$ is equal.
From $(2)$ and $(4)$ the range of $f$ and $g$ is equal.
Hence, $f$ and $g$ are equal functions.
View full question & answer→Question 183 Marks
If $k: R \rightarrow R, k(x)=x^2+3 x-12$; then determine the type of the function $k$.
Answer$k : R \rightarrow R$
$\therefore$ Domain $=\{\ldots,-3,-2,-1,0,1,2, \ldots\}$
$K ( x )= x ^2+3 x -12$
$\therefore k (-3)=(-3)^2+3(-3)-12=-12$
$K(-2)=(-2)^2+3(-2)-12=4-6-12=-14$
$K(-1)=(-1)^2+3(-1)-12=1-3-12=-14$
$K(0)=0+0-12=-12$
$K(1)=1+3-12=-8$
$K(2)=4+12-12=4$
$\therefore \text { Co- domain }=\{\ldots,-12,-14,-14,-12,-8,4, \ldots\}$
Here, for the elements $-3$ and $0,-2$ and $-1$ of domain, their images are same. Therefore function $k$ is many - one function.
View full question & answer→Question 193 Marks
For $f : A \rightarrow B , A =\{10,20,30\} ; B=(18,48,98,128,148\} ; f ( x )=5 x -2$; obtain domain, co-domain and range.
Answer$F: A \rightarrow B, A=\{10,20,30\} B=(18,48,98,128,148\} f(x)=5 x-2$
Domain : $A=\{10,20.30\}$
Co-domain : $B=\{18,48,98,128,148\}$
Range: $R_f=\{f(x) \mid x \in A\}=\{f(10), f(20), f(30)\}$
Now, $f(10)=(5 \times 10)-2=50-2=48$
$f(20)=(5 \times 20)-2=100-2=98$
$f(30)=(5 \times 30)-2=(150-2)=148$
$\therefore$ Range $R _{ f }=\{48,98,148\}$
View full question & answer→Question 203 Marks
The demand function of a commodity $d=100+\frac{500}{p}$ If price per unit is $Rs. 25$ , then find its demand.
View full question & answer→Question 213 Marks
If $f(x)=3 x^{3}-2 x+1$, for which value of $x . f(2 x)=$ $f(0) ?$
Answer$x=0$ $OR$ $x=\frac{1}{3}$
View full question & answer→Question 223 Marks
Find $f(0), f(2)$ and $f(-2)$ for $f(x)=5 x^{2}+3 x-1$
Answer$f(0)=-1, f(2)=25, f(-2)=13$
View full question & answer→Question 233 Marks
If $f(x)=\frac{x^{2}-7 x+6}{x-1}, x \in Z-\{1\}$ and $g(x)=x-6, x \in Z$ $-\{1\}$. Check the equality of functions $f$ and $g$.
View full question & answer→Question 243 Marks
Check the equality of the following functions:
$f(x)=\frac{x^{2}+10 x+21}{x} x \in Z-\{-3\}$
$g(x)=x+7, x \in Z-\{-3\}$
View full question & answer→Question 253 Marks
If $f: R \rightarrow R, f(x)=2 x+7, g: Z \rightarrow$ $Z, g(x)=x^{2}+x-1$ whether, $f$ and $g$ are equal or not?
View full question & answer→Question 263 Marks
If $\mathrm{f}:(0,1,2\} \rightarrow \mathrm{Z}, \mathrm{f}(\mathrm{x})=5 \mathrm{x}-7$, find, domain, co-domain and range of the function.
AnswerDomain $=\{0,1,2\}$, Co-domain $=\{Z\}$, Range $=\{-7,-2,3\}$
View full question & answer→Question 273 Marks
$\mathrm{f}: A \rightarrow N$, $f(x)=3 x^{2}+1$, $\mathrm{~A} = x \mid -1 < x \leq$ $3,x\in Z\mid.$ State the domain, co-domain and range of the function.
AnswerDomain $=\{0,1,2,3\}$, Co-domain $=\{N\}$, Range $=\{1,4,13,$, $28\}$
View full question & answer→Question 283 Marks
If $\mathrm{f}:\{1,2,3,4\} \rightarrow R, f(x)=\frac{x+5}{x^{2}+1}$ then find domain, co-domain and range of the function.
AnswerDomain $=\{1,2,3,4\} ;$ Co-domain $\{R\} ;$ Range $=\{3,$, $\left.\frac{7}{5}, \frac{4}{5}, \frac{9}{17}\right\}$
View full question & answer→Question 293 Marks
Find domain, co-domain and range for $f:\{-2,-1,0,1\}$ $\rightarrow Z, f(x)=4 x^{2}-x+3$
AnswerDomain $=\{-2,-1,0,1\} ;$ Co-domain $=\{$ Z $\} ;$ Range $=\{21,8,3,$, 6)
View full question & answer→Question 303 Marks
If $f:\{1,2,3\} \rightarrow z ; f(x)=4 x+5$, find the domain, codomain and range of the function.
AnswerDomain $=\{1,2,3\}$, Co-domain $=\{Z\}$, Range $=\{9,13,17\}$
View full question & answer→Question 313 Marks
If $\mathrm{f}(\mathrm{x})=x^{2}+3 x-2$, for which value of $\mathrm{x}, \mathrm{f}(\mathrm{x})=\mathrm{f}(0) ?$
View full question & answer→Question 323 Marks
If $g(x)=\frac{x(x-1)}{x+1}$ for which value of $x, g(x)=0 ?$
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