Solution: $\begin{array}{ll} \mathrm{f}(\mathrm{x})=|\mathrm{x}|+\mathrm{x} \quad & \text { (given }) \\ \mathrm{g}(\mathrm{x})=|\mathrm{x}|-\mathrm{x} \quad & \text { (given) } \end{array}$...
Solution: $\mathrm{f}(\mathrm{x})=\frac{x-1}{x-2} \quad \text { (as given) }$ One-One function Suppose $\mathrm{p}, \mathrm{q}$ be two arbitrary elements in $\mathrm{R}$ Therefore, $f(p)-f(q)$...
Solution: $\mathrm{f}(\mathrm{x})=4 \mathrm{x} 2+12 \mathrm{x}+15 \quad \text { (as given) }$ $\mathrm{f}(\mathrm{x})$ is invertible if $\mathrm{f}(\mathrm{x})$ is a bijection (i.e one-one onto...
Solution: $f(x)=9 x_{2}+6 x-5 \text { (as given) }$ $\mathrm{f}(\mathrm{x})$ is invertible if $\mathrm{f}(\mathrm{x})$ is a bijection (i.e one-one onto function) One-One function Suppose p,q be two...
Solution: $\mathrm{f}(\mathrm{x})=\frac{4 x}{3 x+4} \quad$ (as given) One-One function Suppose $\mathrm{p}, \mathrm{q}$ be two arbitrary elements in $\mathrm{R}$ Therefore, $f(p)=f(q)$...
Solution: $\mathrm{f}(\mathrm{x})=\frac{4 x+3}{6 x-4} \quad \text { (given) }$ $\mathrm{f}(\mathrm{x})$ is invertible if $\mathrm{f}(\mathrm{x})$ is a bijection (i.e one-one onto function) One-One...
Solution: $\mathrm{f}(\mathrm{x})=\frac{4 x+3}{6 x-4}, \mathrm{x} \neq \frac{2}{3} \text { (given) }$ We need to Show: fof $(x)=x$ for all $x \neq \frac{2}{3}$ It is known that fof $(x)=f(f(x))$...
Solution: $\mathrm{f}(\mathrm{x})=\frac{1}{2}(3 \mathrm{x}+1) \quad$ (given) $\mathrm{f}(\mathrm{x})$ is invertible if $\mathrm{f}(\mathrm{x})$ is a bijection (i.e one-one onto function) One-One...
Solution: $\mathrm{f}(\mathrm{x})=3 \mathrm{x}-4 \quad$ (as given) $\mathrm{f}(\mathrm{x})$ is invertible if $\mathrm{f}(\mathrm{x})$ is a bijection (i.e one-one onto function) One-One function...
Solution: $\mathrm{f}(\mathrm{x})=2 \mathrm{x}+3$ (as given) $\mathrm{f}(\mathrm{x})$ is invertible if $\mathrm{f}(\mathrm{x})$ is a bijection (i.e one-one onto function) One-One function Suppose...
Solution: A function is invertible if it is a bijection. (i.e. One-One Onto function) One-One function $\mathrm{f}=\{(2,7),(3,9),(4,11),(5,13)\}$ It is observed that different elements of A have...
Solution: We need to state: Whether $\mathrm{f}$ is one-one Given that: $f=\{(1,4),(2,5),(3,6)\}$ Here the function is defined from $A \rightarrow B$ For a function to be one-one if the images of...
Solution: We need to find: the function $g: R \rightarrow R: g \circ f=f$ o $g=I_{g}$ Formula used: (i) $g$ o $f=g(f(x))$ (ii) f o $g=f(g(x))$ Given that: $f: \mathbb{R} \rightarrow \mathbb{R}:...
Solution: We need to find: $g$ o f and f o $\mathrm{g}$ Formula used: (i) f o $\mathrm{g}=\mathrm{f}(\mathrm{g}(\mathrm{x}))$ (ii) $\mathrm{g} \circ \mathrm{f}=\mathrm{g}(\mathrm{f}(\mathrm{x}))$...
Solution: We need to find: f of Formula used: $f o f=f(f(x))$ Given that: (i) $f=\{(1,4),(2,1)(3,3),(4,2)\}$ We have, $\text { fof }(1)=f(f(1))=f(4)=2$ $\begin{array}{l}...
Solution: We need to find: g of Formula used: $g$ o $f=g(f(x))$ Given that: (i) $f=\{(1,2),(3,5),(4,1)\}$ (ii) $g=\{(1,3),(2,3),(5,1)\}$ We have, $\begin{array}{l}...
Solution: We need to find: $f\{f(x)\}$ Formula used: (i) $f \circ f=f(f(x))$ Given that: (i) $f: R \rightarrow R: f(x)=3 x+2$ We have, $\begin{array}{l} f\{f(x)\}=f(f(x))=f(3 x+2) \\ \text { fo }...
Solution: We need to find: f o f Formula used: (i) f o $f=f(f(x))$ Given that: (i) $f: R \rightarrow R: f(x)=\left(3-x^{3}\right)^{1 / 3}$ We have, $\begin{array}{l} \text { fo }...
Solution: We need to prove: $g$ o $f \neq f$ o $g$ Formula used: (i) f o $\mathrm{g}=\mathrm{f}(\mathrm{g}(\mathrm{x}))$ (ii) $g$ of $=g(f(x))$ Given that: (i) $f: R \rightarrow R: f(x)=x^{2}$ (ii)...
Solution: We need to prove: function is many-one and into It is given that: $f: R \rightarrow R: f(x)=\left\{\begin{array}{c}1, \text { if } x \text { is rational } \\ -1, \text { if } x \text { is...
Solution: We need to find: $\mathrm{f}^{-1}$ It is given that: $f: R \rightarrow R: f(x)=10 x+3$ We have, $f(x)=10 x+3$ Suppose $f(x)=y$ such that $y \in R$ $\begin{array}{l} \Rightarrow y=10 x+3 \\...
Solution: We need to find: $\mathrm{f}^{-1}$ It is given that: $f: R \rightarrow R: f(x)=\frac{2 x-7}{4}$ We have, $f(x)=\frac{2 x-7}{4}$ Suppose $f(x)=y$ such that $y \in R$ $\begin{array}{l}...
Solution: We need to prove: function is many-one into It is given that: $f: R \rightarrow R: f(x)=1+x^{2}$ We have, $f(x)=1+x^{2}$ For, $f\left(x_{1}\right)=f\left(x_{2}\right)$ $\begin{array}{l}...
Solution: We need to prove: function is one-one and onto It is given that: $f: R_{0} \rightarrow R_{0}: f(x)=\frac{1}{x}$ We have, $f(x)=\frac{1}{x}$ For, $f\left(x_{1}\right)=f\left(x_{2}\right)$...
Solution: We need to prove: function is one-one and into It is given that: $f: Z \rightarrow Z: f(x)=x^{3}$ Solution: We have, $f(x)=x^{3}$ For, $f\left(x_{1}\right)=f\left(x_{2}\right)$...
Solution: We need to prove: function is neither one-one nor onto It is given that: $f: R \rightarrow R: f(x)=x^{4}$ We have, $f(x)=x^{4}$ For, $f\left(x_{1}\right)=f\left(x_{2}\right)$...
Solution: We need to prove: function is one-one and into It is given that: $f: \mathbb{N} \rightarrow N: f(x)=x^{2}$ Solution: We have, $f(x)=x^{2}$ For, $f\left(x_{1}\right)=f\left(x_{2}\right)$...
Solution: We need to prove: function is neither one-one nor onto It is given that: $f: R \rightarrow R: f(x)=x^{2}$ Solution: We have, $f(x)=x^{2}$ For, $f\left(x_{1}\right)=f\left(x_{2}\right)$...
Solution: We need to prove: function is one-one and into It is given that: $f: N \rightarrow N: f(x)=3 x$ We have, $f(x)=3 x$ For, $f\left(x_{1}\right)=f\left(x_{2}\right)$ $\begin{array}{l}...
Solution: We need to prove: function is one-one and onto It is given that: f: $R \rightarrow R: f(x)=2 x$ We have, $f(x)=2 x$ For, $f\left(x_{1}\right)=f\left(x_{2}\right)$ $\begin{array}{l}...
Solution: (i) $g \circ f$ We need to find: $\mathrm{g}$ o $\mathrm{f}$ Formula used: $g$ o $f=g(f(x))$ It is given that: (i) $f: \mathbb{R} \rightarrow R: f(x)=\left(x^{2}+3 x+1\right)$ (ii) $g: R...
Solution: We need to find: $f$ o $g . g \circ f,(f \circ g)(2)$ and $(g \circ f)(-3)$ Formula used: (i) $f$ o $g=f(g(x))$ (ii) $g$ o $f=g(f(x))$ It is given that: (i) $f: R \rightarrow R:...
Solution: We need to find: $(f \circ g)\left(\frac{-3}{2}\right)+(g \circ f)\left(\frac{4}{3}\right)$ Formula used: (i) $f$ o $g=f(g(x))$ (ii) $g \circ f=g(f(x))$ It is given that: (i) $f$ is a...
Solution: We need to show: $h \circ(g \circ f)=(h \circ g)$ of Formula used: (i) fo $g=f(g(x))$ (ii) $\mathrm{g} \circ \mathrm{f}=\mathrm{g}(\mathrm{f}(\mathrm{x}))$ It is given that: (i) $f:...
Solution: We need to find: $g: Z \rightarrow Z: g$ o $f=I_{Z}$ Formula used: (i) f o $\mathrm{g}=\mathrm{f}(\mathrm{g}(\mathrm{x}))$ (ii) $\mathrm{g}$ of $=g(f(x))$ It is given that: (i) $g: Z...
Solution: We need to prove: $(f \circ g)=I_{R}=(g \circ f)$. Formula used: (i) f o $g=f(g(x))$ (ii) $\mathrm{g}$ of $=\mathrm{g}(\mathrm{f}(\mathrm{x}))$ It is given that: (i) $f: R \rightarrow R:...
Solution: We need to find: formula for $h \circ(g \circ f)$ We need to prove: Show that [h o $(g \circ f)] \sqrt{\frac{\pi}{4}}=0$ Formula used: f of $=f(f(x))$ It is given that: (i) $f: \mathbb{R}...
Solution: We need to prove: fo $f=f$ Formula used: $f$ o $f=f(f(x))$ It is given that: (i) f: $\mathbf{R} \rightarrow \mathrm{R}: \mathrm{f}(x)=|\mathrm{x}|$ Solution: We have, fo...
Solution: (i) $\mathrm{g} \circ \mathrm{g}$ We need to find: $\mathrm{g} \circ \mathrm{g}$ Formula used: $\mathrm{g}$ o $\mathrm{g}=\mathrm{g}(\mathrm{g}(\mathrm{x}))$ It is given that: (i) $g: R...
Solution: (i) f o f We need to find: $f$ o $f$ Formula used: $f$ o $f=f(f(x))$ It is given that: (i) $f: R \rightarrow R: f(x)=(2 x+1)$ Solution: We have, $\begin{array}{l} \text { fof }=f(f(x))=f(2...
Solution: (i) g of We need to find: $g \circ f$. Formula used: $g$ o $f=g(f(x))$ It is given that: (i) $f: R \rightarrow R: f(x)=(2 x+1)$ (ii) $g: R \rightarrow R: g(x)=\left(x^{2}-2\right)$...
Solution: We need to prove: $(g \circ f) \neq(f \circ g)$ Formula used: (i) $\mathrm{g}$ o $\mathrm{f}=\mathrm{g}(\mathrm{f}(\mathrm{x}))$ (ii) $f$ o $g=f(g(x))$ It is given that: (i) $f: R...
Solution: (i) $\mathrm{g} \circ \mathrm{f}$ We need to find: $g$ of Formula used: $g$ o $f=g(f(x))$ It is given that: $f=\{(3,1),(9,3),(12,4)\}$ and $g=\{(1,3),(3,3),(4,9),(5,9)\}$ Solution: We...
Solution: (i) f of We need to find: $f$ o $f$ Formula used: $f$ f $f=f(f(x))$ It is given that: $f=\{(1,4),(2,1),(3,3),(4,2)\}$ Solution: We have, $\begin{array}{l}...
Solution: (i) $\mathrm{g} \circ \mathrm{f}$ We need to find: $g$ of Formula used: $g$ o $f=g(f(x))$ It is given that: $f=\{(1,4),(2,1),(3,3),(4,2)\}$ and $g=\{(1,3),(2,1)$ $(3,2),(4,4)\}$ Solution:...
Solution: (i) $\mathrm{f}(\pi)$ Here, $x=\Pi$, which is irrational $f(\pi)=-1$ (ii) $f(2+\sqrt{3})$ Here, $x=2+\sqrt{3}$, which is irrational $\therefore f(2+\sqrt{3})=-1$
Solution: (i) $f\left(\frac{1}{2}\right)$ Here, $x=1 / 2$, which is rational $\therefore f(1 / 2)=1$ (ii) $\mathrm{f}(\sqrt{2})$ Here, $x=\sqrt{2}$, which is irrational $\therefore...
Solution: For domain $\left(1+x^{2}\right) \neq 0$ $\begin{array}{l} \Rightarrow x^{2} \neq-1 \\ \Rightarrow \operatorname{dom}(f)=R \end{array}$ For the range of $\mathrm{x}$ : $\begin{array}{l}...
Solution: (i) $h = {\{(a, b), (b, c), (c, b), (d, c)}\}$ Here, each of the first set element has different image in second set. $\therefore h$ is a function whose domain = {a, b, c, d} and range (h)...
Solution: For a relation to be a function each element of first set should have different image in the second set(Range) (i) f = {( - 1, 2), (1, 8), (2, 11), (3, 14)} Here, each of the first set...
Solution: As the function $f(x)$ can accept any values as per the given domain $R$, so, the domain of the function $f(x)=x^{2}+1$ is $R$ The minimum value of $\mathrm{f}(\mathrm{x})=1$ $\Rightarrow$...
Solution: $\begin{array}{l} f(n)=\left\{\begin{array}{l} \frac{1}{2}(n-1), \text { when } n \text { is odd } \\ -\frac{1}{2} n, \text { when } n \text { is even } \end{array}\right. \\ f(1)=0 \\...
Solution: In the range of $\mathrm{N} \mathrm{f}(\mathrm{x})$ is monotonically increasing. $\therefore f(n)=n^{2}+n+1$ is one one. But Range of $f(n)=[0.75, \infty) \neq N($ codomain $)$ Thus,...
Solution: $\begin{array}{l} f(x)=\sin x \\ y=\sin x \end{array}$ Here, the lines cut the curve in two equal valued points of $y$, so, the function $f(x)=\sin x$ is not one - one. Range of...
Solution: (i) $f: N \rightarrow N: f(x)=x^{3}$ is one - one into. $f(x)=x^{3}$ As the function $f(x)$ is monotonically increasing from the domain $N \rightarrow N$ $\therefore f(x)$ is one -one...
Solution: (i) $f: N \rightarrow N: f(x)=x^{2}$ is one - one into. As the function $f(x)$ is monotonically increasing from the domain $N \rightarrow N$ $\therefore f(x)$ is one -one Range of...
Solution: $f: \left[0, \frac{\pi}{2}\right] \rightarrow \mathrm{R}$ for given function $\mathrm{f}(\mathrm{x})=\sin$ Recalling the graph for $\sin \mathrm{x}$, we realise that for any two values on...
Solution: We need to show that $f: R \rightarrow R$ given by $f(x)=x s$ is one-one and onto. A function which is onto has every element of co-domain mapped to the at least one element of Domain....
Solution: We need to show that f: $\mathrm{R} \rightarrow \mathrm{R}$ given by $\mathrm{f}(\mathrm{x})=\mathrm{x} 4$ is many-one into. A function which is not onto is into. A function where more...
Solution: We need to show that $f: R \rightarrow R$ given by $f(x) = 1 + x^2$ is many-one into. A function which is not onto is into. A function where more than one element in Set A maps to one...
Solution: (i) $\mathrm{f}(-1)$ $x=-1$, it is satisfying the condition $-2 \leq x \leq 3$ Therefore, $f(x)=x_{2}-2$ $\begin{aligned} \begin{aligned} \therefore \mathrm{f}(-1) =(-1)_{2}-2 \\ =1-2 \\...
Solution: (i) $\mathrm{f}(2)$ $x=2$, it is satisfying the condition $-2 \leq x \leq 3$ Therefore, $f(x)=x_{2}-2$ $\begin{aligned} \therefore \mathrm{f}(2) =22-2 \\ =4-2 \\ =2 \\ \therefore...
Solution: (i) Neither one-one nor onto $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ given by $\mathrm{f}(\mathrm{x})=|\mathrm{x}|=\left\{\begin{array}{l}\mathrm{x}, \text { if } \mathrm{x} \geq 0...
Solution: (i) One-One but not Onto $\mathrm{f}: \mathrm{N} \rightarrow \mathrm{N}$ be a mapping given by $\mathrm{f}(\mathrm{x})=\mathrm{x} 2$ For one-one $\begin{array}{l} f(x)=f(y) \\ x_{2}=y z \\...
Solution: (i) Into Function: It is is a function where there is atleast one element is Set B who is not the image of any element in set A. For Example: $f(x) = 2x - 1$ from the set of Integers to...
Solution: (i)Bijective function: It is, also known as one-one onto function and is a function where for every element of set A, there is exactly one image in set B, such that no element is set B is...
Solution: (i) Injective function: It is, also known as one-one function and is a type of function where every element in set A has an image in set B. Hence, f: A → B is one-one or injection function...
Let $f(x)=x^{3}-4 x^{2}-7 x+10$ Since 1 and $-2$ are the zeroes of $f(x)$, it follows that each one of $(x-1)$ and $(x+2)$ is a factor of $f(x)$ Consequently, $(x-1)(x+2)=\left(x^{2}+x-2\right)$ is...
The given polynomial is $p(x)=\left(x^{3}-2 x^{2}-5 x+6\right)$ $$ \begin{aligned} &\therefore \mathrm{p}(3)=\left(3^{3}-2 \times 3^{2}-5 \times 3+6\right)=(27-18-15+6)=0 \\...
Let $\alpha$ and $\beta$ be the zeroes of the required polynomial $f(x)$. Then $(\alpha+\beta)=0$ and $\alpha \beta=-1$ $\therefore f(x)=x^{2}-(\alpha+\beta) \mathrm{x}+\alpha \beta$ $\Rightarrow...
$$ \begin{aligned} &\mathrm{f}(\mathrm{x})=8 \mathrm{x}^{2}-4 \\ &\text { It can be written as } 8 \mathrm{x}^{2}+0 \mathrm{x}-4 \\ &=4\left\{(\sqrt{2} x)^{2}-(1)^{2}\right\} \\ &=4(\sqrt{2}...