NCERT Solutions · Class 11 Mathematics Relations and Functions

Test for a function. A relation is a function when no first element is repeated — each element of the domain has exactly one image.(i) \(\displaystyle \{(2,1),(5,1),(8,1),(11,1),(14,1),(17,1)\} \)The first elements \(\displaystyle 2,5,8,11,14,17 \) are all different, so each has a unique image.It is a function, with domain \(\displaystyle \{2,5,8,11,14,17\} \) and range \(\displaystyle \{1\} \). (Repetition of the image $\displaystyle 1$ is allowed; only repeated first elements would spoil it.)(ii) \(\displaystyle \{(2,1),(4,2),(6,3),(8,4),(10,5),(12,6),(14,7)\} \)The first elements \(\displaystyle 2,4,6,8,10,12,14 \) are all distinct.It is a function, with domain \(\displaystyle \{2,4,6,8,10,12,14\} \) and range \(\displaystyle \{1,2,3,4,5,6,7\} \).(iii) \(\displaystyle \{(1,3),(1,5),(2,5)\} \)Here \(\displaystyle 1 \) appears as a first element twice, paired with two different images \(\displaystyle 3 \) and \(\displaystyle 5 \).It is not a function.(i) function: domain \(\displaystyle \{2,5,8,11,14,17\} \), range \(\displaystyle \{1\} \); (ii) function: domain \(\displaystyle \{2,4,6,8,10,12,14\} \), range \(\displaystyle \{1,2,3,4,5,6,7\} \); (iii) not a function, since $\displaystyle 1$ has two images $\displaystyle 3$ and 5.

Domain = inputs that make the formula a real number; range = the values it then takes.(i) \(\displaystyle f(x)=-|x| \)The modulus \(\displaystyle |x| \) is defined for every real \(\displaystyle x \), so the domain is \(\displaystyle \mathbf{R} \).Since \(\displaystyle |x| \ge 0 \) for all \(\displaystyle x \), we get \(\displaystyle -|x| \le 0 \). Every non-positive value is attained: for \(\displaystyle y \le 0 \), taking \(\displaystyle x=-y \ (\ge 0) \) gives \(\displaystyle f(x)=-|{-y}|=y \).So the range is \(\displaystyle (-\infty,\,0] \).(ii) \(\displaystyle f(x)=\sqrt{9-x^{2}} \)A real square root needs a non-negative radicand: \[9-x^{2}\ge 0 \;\Longrightarrow\; x^{2}\le 9 \;\Longrightarrow\; -3\le x\le 3 . \] So the domain is \(\displaystyle [-3,\,3] \).For \(\displaystyle x \) in \(\displaystyle [-3,3] \), \(\displaystyle x^{2} \) runs over \(\displaystyle [0,9] \), so \(\displaystyle 9-x^{2} \) runs over \(\displaystyle [0,9] \) and \(\displaystyle \sqrt{9-x^{2}} \) runs over \(\displaystyle [0,3] \). The endpoints are reached at \(\displaystyle x=\pm 3 \) (value $\displaystyle 0$) and \(\displaystyle x=0 \) (value $\displaystyle 3$).So the range is \(\displaystyle [0,\,3] \).(i) Domain \(\displaystyle =\mathbf{R} \), range \(\displaystyle =(-\infty,0] \). (ii) Domain \(\displaystyle =[-3,3] \), range \(\displaystyle =[0,3] \).

Direct substitution. Replace \(\displaystyle x \) by the given number in \(\displaystyle f(x)=2x-5 \).(i) \(\displaystyle f(0)=2(0)-5=-5 \)(ii) \(\displaystyle f(7)=2(7)-5=14-5=9 \)(iii) \(\displaystyle f(-3)=2(-3)-5=-6-5=-11 \)\(\displaystyle f(0)=-5,\ f(7)=9,\ f(-3)=-11 \).

Direct substitution, then solve backwards for part (iv). The rule is \(\displaystyle t(\mathrm{C})=\dfrac{9\mathrm{C}}{5}+32 \).(i) \(\displaystyle t(0)=\dfrac{9(0)}{5}+32=0+32=32 \)(ii) \(\displaystyle t(28)=\dfrac{9(28)}{5}+32=\dfrac{252}{5}+32=50.4+32=82.4 \)(iii) \(\displaystyle t(-10)=\dfrac{9(-10)}{5}+32=-18+32=14 \)(iv) Now the output is given and the input is wanted, so solve the equation \(\displaystyle t(\mathrm{C})=212 \): \[\frac{9\mathrm{C}}{5}+32=212 \;\Longrightarrow\; \frac{9\mathrm{C}}{5}=180 \;\Longrightarrow\; 9\mathrm{C}=900 \;\Longrightarrow\; \mathrm{C}=100 . \]\(\displaystyle t(0)=32,\ t(28)=82.4,\ t(-10)=14 \), and \(\displaystyle t(\mathrm{C})=212 \) when \(\displaystyle \mathrm{C}=100 \).

Track the output as the input runs over its domain.(i) \(\displaystyle f(x)=2-3x,\ x\in\mathbf{R},\ x>0 \)\[x>0 \;\Longrightarrow\; 3x>0 \;\Longrightarrow\; -3x<0 \;\Longrightarrow\; 2-3x<2 . \] Every value below $\displaystyle 2$ occurs: given \(\displaystyle y<2 \), the number \(\displaystyle x=\dfrac{2-y}{3} \) is positive and gives \(\displaystyle f(x)=y \). The value $\displaystyle 2$ itself would need \(\displaystyle x=0 \), which is excluded.Range \(\displaystyle =(-\infty,\,2) \).(ii) \(\displaystyle f(x)=x^{2}+2,\ x\in\mathbf{R} \)Since \(\displaystyle x^{2}\ge 0 \) for every real \(\displaystyle x \), we get \(\displaystyle x^{2}+2\ge 2 \), with equality at \(\displaystyle x=0 \). For any \(\displaystyle y\ge 2 \), \(\displaystyle x=\sqrt{y-2} \) gives \(\displaystyle f(x)=y \).Range \(\displaystyle =[2,\,\infty) \).(iii) \(\displaystyle f(x)=x,\ x\in\mathbf{R} \)This is the identity function; each real \(\displaystyle y \) is the image of \(\displaystyle x=y \).Range \(\displaystyle =\mathbf{R} \).(i) \(\displaystyle (-\infty,2) \) (ii) \(\displaystyle [2,\infty) \) (iii) \(\displaystyle \mathbf{R} \).