2 Marks Questions

Question 512 Marks

Write whether f : R → R, given by $\text{f(x)}=\text{x}+\sqrt{\text{x}^2},$ is one-one, many-one, onto or into.

Answer

f : R → R given by $\text{f(x)}=\text{x}+\sqrt{\text{x}^2}$
$\because\ \text{f(x)=}\begin{cases}2\text{x};&\text{for x}>0\\0;&\text{for x}<0\end{cases}$
$\therefore$ f is many-one function.

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Question 522 Marks

Which one the following relations on A = {1, 2, 3} is a function?
f = {(1, 3), (2, 3), (3, 2)}, g = {(1, 2), (1, 3), (3, 1)}

Answer

As, each element of the domain set has unique image in the relation f = {(1, 3), (2, 3), (3, 2)}
So, f is a function.
Also, the element 1 of the domain set has two images 2 and 3 of the range set in the relation g = {(1, 2), (1, 3), (3, 1)}
So, g is not a function.

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Question 532 Marks

Let f be a function from C (set of all complex numbers) to itself given by $f(x) = x^3$. Write $f^{-1}(-1)$.

Answer

Let $f^{-1}(-1) = x$ .....(1)
$\Rightarrow f(x) = -1$
$\Rightarrow x^3 = -1$
$\Rightarrow x^3 + 1 = 0$
$\Rightarrow (x + 1)(x^2 - x + 1) = 0$
[Using the identity: $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$]
$\Rightarrow\ (\text{x}+1)(\text{x}+\omega)(\text{x}+\omega^2)=0,$ where $\omega=\frac{1\pm\text{i}\sqrt{3}}{2}$
$\Rightarrow\ \text{x}=-1,-\omega,-\omega^2$ $(\text{as x}\in\text{C})$
$\Rightarrow\ \text{f}^{-1}(-1)=\{-1,-\omega,-\omega^2\}$ [from 1]

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Question 542 Marks

If $f(x) = 4 - (x - 7)^3$, then write $f^{-1}(x)$.

Answer

We have, $f(x) = 4 - (x - 7)^3$ Let $y = 4 - (x - 7)^3$
$\Rightarrow\ (\text{x} - 7)^3 = 4 - \text{y}$
$\Rightarrow\ \text{x}-7=\sqrt[3]{4-\text{y}}$
$\Rightarrow\ \text{x}=7+\sqrt[3]{4-\text{y}}$
$\Rightarrow\ \text{f}^{-1}(\text{y})=7+\sqrt[3]{4-\text{y}}$
$\therefore\ \text{f}^{-1}(\text{x})=7+\sqrt[3]{4-\text{x}}$

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Question 552 Marks

Which of the following functions from A to B are one-one and onto?
$f_3 = \{(a, x), (b, x), (c, z), (d, z)\}; A = \{a, b, c, d,\}, B = \{x, y, z\}$

Answer

$ f_3=\{(a, x),(b, x),(c, z),(d, z)\}$
$ A=\{a, b, c, d\}, B=\{x, y, z\}$
Since, $f_3(a)=x=f_3(b)$ and $f_3(c)=z=f_3(d)$
$\therefore f_3$ in not one-one.
Again, $\mathrm{y} \in \mathrm{B}$ in not the image of any of the element of A .
$\therefore f_3$ in not on to.

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Question 562 Marks

Write the domain of the real function f defined by $\text{f(x)}=\sqrt{25-\text{x}^2}.$

Answer

We have, $\text{f(x)}=\sqrt{25-\text{x}^2}$ The function is defined only when $25-\text{x}^2\geq0$$\text{x}^2-25\leq0$
$(\text{x}+5)(\text{x}-5)\leq0$
$\text{x}\in[-5,5]$
Therefore, the domain of the given function is [-5, 5].

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Question 572 Marks

Let A = {x ∈ R : -4 ≤ x ≤ 4 and x ≠ 0} and f : A → R be defined by $\text{f(x)}=\frac{|\text{x}|}{\text{x}}.$ Write the range of f.

Answer

We have, $\text{A}=\{\text{x}\in\text{R}:-4\leq\text{x}\leq4\text{ and x }\neq0\}$
f : A → R defined by $\text{f(x)}=\frac{|\text{x}|}{\text{x}}$
Clearly, $\text{f(x)}=\begin{cases}1;&\text{x}>0\\-1;&\text{x}<0\end{cases}$
$\therefore$ Range of f = {-1, 1}

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Question 582 Marks

What is the range of the function $\text{f(x)}=\frac{|\text{x}-1|}{\text{x}-1}?$

Answer

$\text{f(x)}=\frac{|\text{x}-1|}{\text{x}-1}=\frac{\pm(\text{x}-1)}{\text{x}-1}=\pm1$Range of f = {-1, 1}

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Question 592 Marks

If f : A → A, g : A → A are two bijections, then prove that:
fog is a surjection.

Answer

Given: A → A, g : A → A are two bijections. Then, fog : A → A Surjectivity of fog: let z be an element in the co-domain of fog (A).Now, $\text{z}\in\text{A}$ (co-domain of f) and f is a surjection.
So, z = f(y), where $\text{y}\in\text{A}$ (domain of f) .....(1)
Now, $\text{y}\in\text{A}$ (co-domain of g) and g is a surjection.
So, y = g(x), where $\text{x}\in\text{A}$ (domain of g) .....(2)
From (1) and (2),
z = f(y) = f(g(x)) = (fog)(x), where $\text{x}\in\text{A}$ (domain of fog)
So, fog is a surjection.

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Question 602 Marks

Let $\text{f}:\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\rightarrow\ \text{R}$ be a function defined by f(x) = cos[x]. write range (f).

Answer

$\text{f}:\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\rightarrow\ \text{R}$ given by f(x) = cos[x]$\because\ \cos\text{x}$ in position in $\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)$
$\therefore$ cos[x] will be $\{1, \cos1, \cos2\}$
$\therefore$ Range of $\text{f}=\{1, \cos1, \cos2\}$

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Question 612 Marks

Let $f, g : R \rightarrow R$ be defined by $f(x) = 2x + 1$ and $g(x) = x^2 - 2$ for all $x \in R$, respectively. Then, find gof.

Answer

We have,
$f, g : R \rightarrow R$ are defined by $f(x) = 2x + 1 $ and$ g(x) = x^2 - 2$ for all $x \in R$, respectively
Now,
$gof(x) = g(f(x))$
$= g(2x + 1)$
$= (2x + 1)^2 - 2$
$= 4x^2 + 4x + 1 - 2$
$= 4x^2 + 4x - 1$

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MATHS 2 Marks Questions