2 Marks Questions

Question 512 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, $\text{y}\in\text{B}$ in not the image of any of the element of $A.$
$\therefore f_3$ in not on to.

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Question 522 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 532 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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Question 542 Marks

If $f : R \rightarrow R$ defined by $f(x) = 3x - 4$ is invertible, then write $f^{-1}(x).$

Answer

Let $f^{-1}(x) = y .....(1)$
$\Rightarrow f(y) = x$
$\Rightarrow 3y - 4 = x$
$\Rightarrow 3y = x + 4$
$\Rightarrow\ \text{y}=\frac{\text{x}+4}{3}$
$\Rightarrow\ \text{f}^{-1}(\text{x})=\frac{\text{x}+4}{3} [$from $(1)]$

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Question 552 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 562 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 572 Marks

Which of the following functions from $A$ to $B$ are one$-$one and onto? $f_1 = \{(1, 3), (2, 5), (3, 7)\}; A = \{1, 2, 3\}, B = \{3, 5, 7\}$

Answer

$f_1 = \{(1, 3), (2, 5), (3, 7)\}$
$A = \{1, 2, 3\}, B = \{3, 5, 7\}$
We can earily observe that in $f_1$ every element of $A$ has different image from $B.$
$\therefore f_1$ in not one$-$one.
Also, each element of $B$ is the image of some element of $A.$
$\therefore f_1$ in not on to.

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Question 582 Marks

If $f : R \rightarrow R$ is defined by $f(x) = x^2,$ find $f^{-1}(-25).$

Answer

$f : R \rightarrow R$ defined by $f(x) = x^2 $
$\therefore f^{-1}(x^2) = x$
$\Rightarrow\ \text{f}^{-1}(-25)=\phi$ $[\because\ \sqrt{-25}\notin\text{R}]$

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Question 592 Marks

Let $A = \{a, b, c, d\}$ and $f : A \rightarrow A$ be given by $f = \{(a, b), (b, d), (c, a), (d, c)\}.$ Write $f ^{-1}.$

Answer

We have,
$A = \{a, b, c, d\}$ and $f : A \rightarrow A$ be given by
$f = \{(a, b), (b, d), (c, a), (d, c)\}$
Since, the elements of a function when interchanged gives inverse function.
Therefore, $f^{-1} = \{(b, a), (d, b), (a, c), (c, d)\}$

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Question 602 Marks

Let $\text{f}:\text{R}-\Big\{-\frac{3}{5}\Big\}\rightarrow\ \text{R}$ be a function defined as $\text{f(x)}=\frac{2\text{x}}{5\text{x}+3}.$ Write $f^{-1}:$ Range of $\text{f}\rightarrow\ \text{R}-\Big\{-\frac{3}{5}\Big\}.$

Answer

Let $f^{-1}(x) = y ......(1)$
$\Rightarrow f(y) = x$
$\Rightarrow\ \frac{2\text{y}}{5\text{y}+3}=3\text{x}$
$\Rightarrow 2y = 5xy + 3x $
$\Rightarrow 2y - 5xy = 3x $
$\Rightarrow y(2 - 5x) = 3x $
$\Rightarrow\ \text{y}=\frac{3\text{x}}{2-5\text{x}}$
$\Rightarrow\ \text{f}^{-1}(\text{x})=\frac{3\text{x}}{2-5\text{x}}[$ from $1]$

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Question 612 Marks

Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. State whether f is one-one or not.

Answer

f = {(1, 4), (2, 5), (3, 6)}
Here, different elements of the domain have different images in the co-domain.
So, f is one-one.

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MATHS 2 Marks Questions