Solve the Following Question.(2 Marks)

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)
$\Rightarrow f ( x )=-1$
$\Rightarrow x^3=-1$
$\Rightarrow x ^3+1=0$
$\Rightarrow(x+1)\left(x^2-x+1\right)=0$
[Using the identity: $\left.a^3+b^3=(a+b)\left(a^2-a b+b^2\right)\right]$
$\Rightarrow( x +1)( x +\omega)\left( x +\omega^2\right)=0$, where $\omega=\frac{1 \pm i \sqrt{3}}{2}$
$\Rightarrow x =-1,-\omega,-\omega^2($ as $x \in C )$
$\Rightarrow f ^{-1}(-1)=\left\{-1,-\omega,-\omega^2\right\}[$ 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, $y \in B$ 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 $f(x)=2 x+1$ 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 $f(x)=2 x+1$ and $g(x)=x^2-2$ for all $x \in R$, respectively
Now,
$g \circ f(x)=g(f(x))$
$=g(2 x+1)$
$=(2 x+1)^2-2$
$=4 x^2+4 x+1-2$
$=4 x^2+4 x-1$

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Maths Solve the Following Question.(2 Marks)