1. Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range.
(i) {(2,1), (5,1), (8,1), (11,1), (14,1), (17,1)}
It is a function because 2,5,8,11,14,17 have unique images.
Domain = {2,5,8,11,14,17}
Range= {1}
(ii) {(2,1), (4,2), (6,3), (8,4), (10,5), (12,6), (14,7)}
It is a function because 2,4,6,8,10,12,14 have unique images.
Domain= {2,4,6,8,10,12,14}
Range= {1,2,3,4,5,6,7}
(iii) {(1,3), (1,5), (2,5)}
It is not a function because element 1 has two images 3 and 5.
2. Find the domain and range of the following real functions:
(i) f(x) = – |x|
We know that $f(x)=\left\{\begin{matrix} x & if & x\geq 0\\ -x & if & x< 0 \end{matrix}\right.$
∴$f(x)=-|x|=\left\{\begin{matrix} -x & if & x\geq 0\\ x & if & x< 0 \end{matrix}\right.$
Domain= R
Range= (∞,0]
(ii) $f(x)=\sqrt{9-x^{2}}$
f is not defined for 9-$\displaystyle \small x^{2}$≤ 0 ⇒ $\displaystyle \small x^{2}$≥ 9 ⇒ x ≥ 3 or x ≤ -3
But for each real number x lying between -3 and 3, f(x) is unique.
∴ Domain= {x: -3≤x≤3} or [-3,3]
Again, $y=\sqrt{9-x^{2}}$ ⇒ $y^{2}=9-x^{2}$ ⇒ $x=\sqrt{9-y^{2}}$
Here, x is not defined for 9-$\displaystyle \small y^{2}$<0 ⇒ $\displaystyle \small y^{2}$>9 ⇒ y>3 or y<-3
But y cannot be negative. For any value of x such that -3≤x≤3, the value of f(x) lie between 0 and 3.
∴ Range= {y: 0≤y≤3} or [0,3]
3. A function f is defined by f(x) = 2x –5. Write down the values of
(i) f (0) = 2*0-5 = -5
(ii) f (7) = 2*7-5 = 9
(iii) f (–3) =2*(-3)-5 = -11
4. The function ‘t ’ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by $t(C)=\frac{9C}{5}+32$ . Find
(i) $t(0)=\frac{9*0}{5}+32$ = 32
(ii) $t(28)=\frac{9*28}{5}+32=\frac{252+160}{5}=\frac{412}{5}=82.4$
(iii) $t(-10)=\frac{9*(-10)}{5}+32=-18+32=14$
(iv) The value of C, when t (C) = 212
$212=\frac{9C}{5}+32$
$\frac{9C}{5}=212-32=180$
$C=\frac{180*5}{9}=20*5=100$
5. Find the range of each of the following functions.
(i) f (x) = 2–3x, x ∈ R, x > 0
y=2-3x ⇒ $x=\frac{2-y}{3}$
x>0 ⇒ 2-y >0 ⇒ y<2
∴ Range = (-∞,2)
(ii) f (x) = $\displaystyle \small x^{2}$+ 2, x is a real number
y=$\displaystyle \small x^{2}$+2 ⇒ $\displaystyle \small x^{2}$= y-2 ⇒$x=\sqrt{y-2}$
$\displaystyle \small x^{2}$≥ 0 ⇒ y-2 ≥ 0 ⇒ y ≥ 2
∴ Range = [2,∞)
(iii) f (x) = x, x is a real number
y = x = real number
∴ Range = R
(i) {(2,1), (5,1), (8,1), (11,1), (14,1), (17,1)}
It is a function because 2,5,8,11,14,17 have unique images.
Domain = {2,5,8,11,14,17}
Range= {1}
(ii) {(2,1), (4,2), (6,3), (8,4), (10,5), (12,6), (14,7)}
It is a function because 2,4,6,8,10,12,14 have unique images.
Domain= {2,4,6,8,10,12,14}
Range= {1,2,3,4,5,6,7}
(iii) {(1,3), (1,5), (2,5)}
It is not a function because element 1 has two images 3 and 5.
2. Find the domain and range of the following real functions:
(i) f(x) = – |x|
We know that $f(x)=\left\{\begin{matrix} x & if & x\geq 0\\ -x & if & x< 0 \end{matrix}\right.$
∴$f(x)=-|x|=\left\{\begin{matrix} -x & if & x\geq 0\\ x & if & x< 0 \end{matrix}\right.$
Domain= R
Range= (∞,0]
(ii) $f(x)=\sqrt{9-x^{2}}$
f is not defined for 9-$\displaystyle \small x^{2}$≤ 0 ⇒ $\displaystyle \small x^{2}$≥ 9 ⇒ x ≥ 3 or x ≤ -3
But for each real number x lying between -3 and 3, f(x) is unique.
∴ Domain= {x: -3≤x≤3} or [-3,3]
Again, $y=\sqrt{9-x^{2}}$ ⇒ $y^{2}=9-x^{2}$ ⇒ $x=\sqrt{9-y^{2}}$
Here, x is not defined for 9-$\displaystyle \small y^{2}$<0 ⇒ $\displaystyle \small y^{2}$>9 ⇒ y>3 or y<-3
But y cannot be negative. For any value of x such that -3≤x≤3, the value of f(x) lie between 0 and 3.
∴ Range= {y: 0≤y≤3} or [0,3]
3. A function f is defined by f(x) = 2x –5. Write down the values of
(i) f (0) = 2*0-5 = -5
(ii) f (7) = 2*7-5 = 9
(iii) f (–3) =2*(-3)-5 = -11
4. The function ‘t ’ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by $t(C)=\frac{9C}{5}+32$ . Find
(i) $t(0)=\frac{9*0}{5}+32$ = 32
(ii) $t(28)=\frac{9*28}{5}+32=\frac{252+160}{5}=\frac{412}{5}=82.4$
(iii) $t(-10)=\frac{9*(-10)}{5}+32=-18+32=14$
(iv) The value of C, when t (C) = 212
$212=\frac{9C}{5}+32$
$\frac{9C}{5}=212-32=180$
$C=\frac{180*5}{9}=20*5=100$
5. Find the range of each of the following functions.
(i) f (x) = 2–3x, x ∈ R, x > 0
y=2-3x ⇒ $x=\frac{2-y}{3}$
x>0 ⇒ 2-y >0 ⇒ y<2
∴ Range = (-∞,2)
(ii) f (x) = $\displaystyle \small x^{2}$+ 2, x is a real number
y=$\displaystyle \small x^{2}$+2 ⇒ $\displaystyle \small x^{2}$= y-2 ⇒$x=\sqrt{y-2}$
$\displaystyle \small x^{2}$≥ 0 ⇒ y-2 ≥ 0 ⇒ y ≥ 2
∴ Range = [2,∞)
(iii) f (x) = x, x is a real number
y = x = real number
∴ Range = R
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