1. The relation f is defined by $f(x)=\left\{\begin{matrix} x^{2}, &0\leq x\leq 3 \\ 3x, & 3\leq x\leq 10 \end{matrix}\right.$
The relation g is defined by $g(x)=\left\{\begin{matrix} x^{2}, &0\leq x\leq 2 \\ 3x, & 2\leq x\leq 10 \end{matrix}\right.$
Show that f is a function and g is not a function.
(i) f(x)
f(x)=$\displaystyle \small x^{2}$ is well defined in 0 ≤ x ≤ 3
f(x)=3x is well defined in 3 ≤ x ≤ 10
At x=3,
f(x)= $\displaystyle \small x^{2}$ = $\displaystyle \small 3^{2}$ = 9
f(x)= 3x = 3*3 = 9
∴ f is well defined at x=3
Hence f is a function.
(ii) g(x)
g(x)=$\displaystyle \small x^{2}$ is well defined in 0 ≤ x ≤ 2
g(x)=3x is well defined in 2 ≤ x ≤ 10
At x=2,
g(x)= $\displaystyle \small x^{2}$ = $\displaystyle \small 2^{2}$ = 4
g(x)= 3x = 3*2 = 6
element 2 has two different images 4 and 6.
Hence g is not a function.
2. If f (x) = $\displaystyle \small x^{2}$ , Find $\frac{f(1.1)-f(1)}{(1.1-1)}$
f (x) = $\displaystyle \small x^{2}$
$\frac{f(1.1)-f(1)}{(1.1-1)} = \frac{(1.1)^{2}-1^{2}}{0.1}=\frac{1.21-1}{0.1}
=\frac{0.21}{0.1}=2.1$
3. Find the domain of the function $f(x)=\frac{x^{2}+2x+1}{x^{2}-8x+12}$
$f(x)=\frac{x^{2}+2x+1}{x^{2}-8x+12}=\frac{(x+1)^{2}}{(x-2)(x-6)}$
The function f is not defined at x=2 and x=6
∴ Domain = R - {2,6}
4. Find the domain and the range of the real function f defined by $f(x)=\sqrt{x-1}$
f is not defined for x-1 < 0 or x < 1
∴ f is defined for x ≥ 1
Hence, Domain = [1,∞)
let $y=\sqrt{x-1}$
as x ≥ 1 ⇒ x-1 ≥ 0 ⇒ $\sqrt{x-1}$ ≥ 0
∴ Range = [0, ∞)
5. Find the domain and the range of the real function f defined by f(x)=|x-1|
It is clear that |x-1| is defined for all real numbers
∴ Domain = R
y=|x-1| can acquire only non negative values
∴ Range = $\displaystyle \small R^{+}$
6. Let $f=\begin{Bmatrix} \left ( x,\frac{x^{2}}{1+x^{2}} \right ) & : & x\epsilon R \end{Bmatrix}$ be a function from R into R. Determine the range of f.
$y=\frac{x^{2}}{1+x^{2}}$
when x=0, y=0
y is positive for all values of x. Since denominator > numerator, y < 1
∴ Range = [0,1)
7. Let f , g : R→R be defined respectively by f(x)=x+1 and g(x)=2x-3. Find f+g, f-g and $\frac{f}{g}$
f(x)=x+1, g(x)=2x-3
(f+g)(x) = f(x)+g(x) = x+1+2x-3 = 3x-2
(f-g)(x) = f(x)-g(x) = x+1-2x+3 = -x+4
$\frac{f}{g}=\frac{x+1}{2x-3}$ , 2x-3≠0
8. Let f = {(1,1), (2,3), (0,–1), (–1, –3)} be a function from Z to Z defined by f(x)=ax+b, for some integers a, b. Determine a, b.
f(x)=ax+b
(1,1) ⇒ f(1) = a*1+b=1 ⇒ a+b=1 ⇒ b=1-a...(i)
(2,3) ⇒ f(2) = a*2+b=3 ⇒ 2a+b=3
substitute (i) ⇒ 2a+(1-a)=3 ⇒ a=2
and b=1-a ⇒ b=1-2 ⇒ b=-1
∴ value of a and b are 2 and -1 respectively.
9. Let R be a relation from N to N defined by R = {(a, b) : a, b∈N and a=$\displaystyle \small b^{2}$}. Are the following true? Justify your answer in each case.
It is given that a=$\displaystyle \small b^{2}$
(i) (a,a) ∈ R, for all a ∈ N
(a,a) ⇒ a=$\displaystyle \small a^{2}$ is true only when a=0 or a=1
for any other value say 2∈N, 2≠$\displaystyle \small 2^{2}$
∴ (a,a) ∈ R, for all a∈N is not true
(ii) (a,b ) ∈ R, implies (b,a) ∈ R
(a,b) and (b,a) ⇒ a=$\displaystyle \small b^{2}$ and b=$\displaystyle \small a^{2}$
(9,3) ⇒ 9=$\displaystyle \small 3^{2}$ but for (3,9) ⇒ 3≠$\displaystyle \small 9^{2}$
∴(a,b ) ∈ R, implies (b,a) ∈ R is not true
(iii) (a,b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R.
(a,b) (b,c) (a,c) ⇒ a=$\displaystyle \small b^{2}$, b=$\displaystyle \small c^{2}$, a=$\displaystyle \small c^{2}$
9=$\displaystyle \small 3^{2}$, 16=$\displaystyle \small 4^{2}$ but 9≠$\displaystyle \small 4^{2}$
∴ (a,b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R is not true
10. Let A ={1,2,3,4}, B = {1,5,9,11,15,16} and f = {(1,5), (2,9), (3,1), (4,5), (2,11)} Are the following true? Justify your answer in each case.
A={1,2,3,4}, B={1,5,9,11,15,16}
A⨯B={(1,1),(1,5),(1,9),(1,11),(1,15),(1,16), (2,1),(2,5),(2,9),(2,11),(2,15),(2,16),(3,1),(3,5),(3,9),(3,11),(3,15),(3,16),(4,1),(4,5),(4,9),(4,11),(4,15),(4,16)}
(i) f is a relation from A to B
f = {(1,5), (2,9), (3,1), (4,5), (2,11)}
f ⊂ A⨯B
∴ f is a relation from A to B
(ii) f is a function from A to B.
f = {(1,5), (2,9), (3,1), (4,5), (2,11)}
element 2∈A has two images 9 and 11
∴ f is not a function.
11. Let f be the subset of Z × Z defined by f ={( ab, a + b) : a , b ∈ Z}. Is f a function from Z to Z? Justify your answer.
let a=0, b=1
ab=0 and a+b=1 ⇒ (0,1)∈f
let a=0, b=−1
ab=0 and a+b=−1 ⇒ (0,−1)∈f
element 0 has two images 1 and −1
∴ f is not a function
12. Let A = {9,10,11,12,13} and let f : A→N be defined by f (n) = the highest prime factor of n. Find the range of f.
for n=9; 3 is highest prime factor
n=10; 5 is highest prime factor
n=11; 11 is highest prime factor
n=12; 3 is highest prime factor
n=13; 13 is highest prime factor
∴ Range = {3,5,11,13}
The relation g is defined by $g(x)=\left\{\begin{matrix} x^{2}, &0\leq x\leq 2 \\ 3x, & 2\leq x\leq 10 \end{matrix}\right.$
Show that f is a function and g is not a function.
(i) f(x)
f(x)=$\displaystyle \small x^{2}$ is well defined in 0 ≤ x ≤ 3
f(x)=3x is well defined in 3 ≤ x ≤ 10
At x=3,
f(x)= $\displaystyle \small x^{2}$ = $\displaystyle \small 3^{2}$ = 9
f(x)= 3x = 3*3 = 9
∴ f is well defined at x=3
Hence f is a function.
(ii) g(x)
g(x)=$\displaystyle \small x^{2}$ is well defined in 0 ≤ x ≤ 2
g(x)=3x is well defined in 2 ≤ x ≤ 10
At x=2,
g(x)= $\displaystyle \small x^{2}$ = $\displaystyle \small 2^{2}$ = 4
g(x)= 3x = 3*2 = 6
element 2 has two different images 4 and 6.
Hence g is not a function.
2. If f (x) = $\displaystyle \small x^{2}$ , Find $\frac{f(1.1)-f(1)}{(1.1-1)}$
f (x) = $\displaystyle \small x^{2}$
$\frac{f(1.1)-f(1)}{(1.1-1)} = \frac{(1.1)^{2}-1^{2}}{0.1}=\frac{1.21-1}{0.1}
=\frac{0.21}{0.1}=2.1$
3. Find the domain of the function $f(x)=\frac{x^{2}+2x+1}{x^{2}-8x+12}$
$f(x)=\frac{x^{2}+2x+1}{x^{2}-8x+12}=\frac{(x+1)^{2}}{(x-2)(x-6)}$
The function f is not defined at x=2 and x=6
∴ Domain = R - {2,6}
4. Find the domain and the range of the real function f defined by $f(x)=\sqrt{x-1}$
f is not defined for x-1 < 0 or x < 1
∴ f is defined for x ≥ 1
Hence, Domain = [1,∞)
let $y=\sqrt{x-1}$
as x ≥ 1 ⇒ x-1 ≥ 0 ⇒ $\sqrt{x-1}$ ≥ 0
∴ Range = [0, ∞)
5. Find the domain and the range of the real function f defined by f(x)=|x-1|
It is clear that |x-1| is defined for all real numbers
∴ Domain = R
y=|x-1| can acquire only non negative values
∴ Range = $\displaystyle \small R^{+}$
6. Let $f=\begin{Bmatrix} \left ( x,\frac{x^{2}}{1+x^{2}} \right ) & : & x\epsilon R \end{Bmatrix}$ be a function from R into R. Determine the range of f.
$y=\frac{x^{2}}{1+x^{2}}$
when x=0, y=0
y is positive for all values of x. Since denominator > numerator, y < 1
∴ Range = [0,1)
7. Let f , g : R→R be defined respectively by f(x)=x+1 and g(x)=2x-3. Find f+g, f-g and $\frac{f}{g}$
f(x)=x+1, g(x)=2x-3
(f+g)(x) = f(x)+g(x) = x+1+2x-3 = 3x-2
(f-g)(x) = f(x)-g(x) = x+1-2x+3 = -x+4
$\frac{f}{g}=\frac{x+1}{2x-3}$ , 2x-3≠0
8. Let f = {(1,1), (2,3), (0,–1), (–1, –3)} be a function from Z to Z defined by f(x)=ax+b, for some integers a, b. Determine a, b.
f(x)=ax+b
(1,1) ⇒ f(1) = a*1+b=1 ⇒ a+b=1 ⇒ b=1-a...(i)
(2,3) ⇒ f(2) = a*2+b=3 ⇒ 2a+b=3
substitute (i) ⇒ 2a+(1-a)=3 ⇒ a=2
and b=1-a ⇒ b=1-2 ⇒ b=-1
∴ value of a and b are 2 and -1 respectively.
9. Let R be a relation from N to N defined by R = {(a, b) : a, b∈N and a=$\displaystyle \small b^{2}$}. Are the following true? Justify your answer in each case.
It is given that a=$\displaystyle \small b^{2}$
(i) (a,a) ∈ R, for all a ∈ N
(a,a) ⇒ a=$\displaystyle \small a^{2}$ is true only when a=0 or a=1
for any other value say 2∈N, 2≠$\displaystyle \small 2^{2}$
∴ (a,a) ∈ R, for all a∈N is not true
(ii) (a,b ) ∈ R, implies (b,a) ∈ R
(a,b) and (b,a) ⇒ a=$\displaystyle \small b^{2}$ and b=$\displaystyle \small a^{2}$
(9,3) ⇒ 9=$\displaystyle \small 3^{2}$ but for (3,9) ⇒ 3≠$\displaystyle \small 9^{2}$
∴(a,b ) ∈ R, implies (b,a) ∈ R is not true
(iii) (a,b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R.
(a,b) (b,c) (a,c) ⇒ a=$\displaystyle \small b^{2}$, b=$\displaystyle \small c^{2}$, a=$\displaystyle \small c^{2}$
9=$\displaystyle \small 3^{2}$, 16=$\displaystyle \small 4^{2}$ but 9≠$\displaystyle \small 4^{2}$
∴ (a,b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R is not true
10. Let A ={1,2,3,4}, B = {1,5,9,11,15,16} and f = {(1,5), (2,9), (3,1), (4,5), (2,11)} Are the following true? Justify your answer in each case.
A={1,2,3,4}, B={1,5,9,11,15,16}
A⨯B={(1,1),(1,5),(1,9),(1,11),(1,15),(1,16), (2,1),(2,5),(2,9),(2,11),(2,15),(2,16),(3,1),(3,5),(3,9),(3,11),(3,15),(3,16),(4,1),(4,5),(4,9),(4,11),(4,15),(4,16)}
(i) f is a relation from A to B
f = {(1,5), (2,9), (3,1), (4,5), (2,11)}
f ⊂ A⨯B
∴ f is a relation from A to B
(ii) f is a function from A to B.
f = {(1,5), (2,9), (3,1), (4,5), (2,11)}
element 2∈A has two images 9 and 11
∴ f is not a function.
11. Let f be the subset of Z × Z defined by f ={( ab, a + b) : a , b ∈ Z}. Is f a function from Z to Z? Justify your answer.
let a=0, b=1
ab=0 and a+b=1 ⇒ (0,1)∈f
let a=0, b=−1
ab=0 and a+b=−1 ⇒ (0,−1)∈f
element 0 has two images 1 and −1
∴ f is not a function
12. Let A = {9,10,11,12,13} and let f : A→N be defined by f (n) = the highest prime factor of n. Find the range of f.
for n=9; 3 is highest prime factor
n=10; 5 is highest prime factor
n=11; 11 is highest prime factor
n=12; 3 is highest prime factor
n=13; 13 is highest prime factor
∴ Range = {3,5,11,13}
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