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 $\therefore f(2+\sqrt{3})=-1$
Show that the function is many – one into.
Show that the function is many – one into.
Find (i)
(ii)
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...
Find the domain and range of the real function, defined by . Show that is many – one.
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}...
Which of the following relations are functions? Give reasons. In case of a function, find its domain and range.
(i)
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)...
Which of the following relations are functions? Give reasons. In case of a function, find its domain and range.
(i)
(ii)
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...
Find the domain and range of the function
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 $\mathrm{f}(\mathrm{x})=1$ $\Rightarrow$...
Show that the function f: , defined by is both one – one and onto.
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 \\...
Prove that the function is one – one but not onto.
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,...
Show that the function is neither one – one nor onto.
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...
Show that the function
(i) is one – one into
(ii) is one – one into
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...
Show that the function
(i) is one – one into.
(ii) is many – one into
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...
Let and . Show that each one of and is one one but is not one – one.
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...
Show that the function is one – one and onto.
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....
Show that the function is many – one and into.
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...
Show that the function is many – one into.
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...
Let be defined by
Find (i) (ii) .
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 \\...
Let be defined by
Find (i) (ii)
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...
Give an example of a function which is
(i) neither one – one nor onto
(ii) onto but not one – one.
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...
Find the quadratic polynomial, sum of whose zeroes is 0 and their product is . Hence, find the zeroes of the polynomial.
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...
Find the zeroes of the quadratic polynomial and verify the relation between the zeroes and the coefficients.
$$ \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}...