3 Marks Question

Question 513 Marks

Give examples of two one$-$one functions $f_1$ and $f_2$ from $R$ to $R,$ such that $f_1 + f_2 : R \rightarrow R.$ defined by $(f_1 + f_2)(x) = f_1(x) + f_2(x)$ is not one$-$one.

Answer

We know that $f_1: R \rightarrow R,$
given by $f_1(x) = x,$ and $f_2(x) = -x$ are one$-$one.
Proving $f_{1 }$ is one$-$one:
Let $f_1(x) = f_1(y)$
$\Rightarrow x = y$
So, $f_1$ is one$-$one.
Proving $f_2$ is one$-$one:
Let $f_2(x) = f_2(y)$
$\Rightarrow -x = -y$
$\Rightarrow x = y$
So, $f_2$ is one$-$one.
Proving $(f_1 + f_2)$ is not one$-$one:
Given: $(f_1 + f_2)(x) = f_1(x) + f_2(x)= x + (-x) =0$
So, for every real number $x, (f_1 + f_2)(x)=0$
So, the image of ever number in the domain is same as $0.$
Thus, $(f_1 + f_2)$ is not one$-$one.

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Question 523 Marks

Let $A = R - {3}$ and $B = R - {1}.$ Consider the function $f : A \rightarrow B$ defined by $\text{f(x)}=\frac{\text{x}-2}{\text{x}-3}.$ Show that f is one$-$one and onto and hence find $f^{-1}.$

Answer

We have, $A = R - {3}$ and $B = R - {1}.$
Consider the function $f : A \rightarrow B$ defined by
$\text{f}(\text{x})=\frac{\text{x}-2}{\text{x}-3},$
Show that $f$ is one$-$one and onto and hence find $f^{-1}.$
Let $\text{x, y}\in\text{A}$ such that $f(x) = f(y).$
Then, $\frac{\text{x}-2}{\text{x}-3}=\frac{\text{y}-2}{\text{y}-3}$
$\Rightarrow xy - 3x - 2y + 6 = xy - 2x - 3y + 6$
$\Rightarrow -x = -y$
$\Rightarrow x = y$
$\therefore f$ is one$-$one.

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Question 533 Marks

Let $R^+$ be the set of all non$-$negative real numbers. If $f : R^+ \rightarrow R^+$ and $g : R^+ \rightarrow R^+$ are defined as $f(x) = x^2$ and $\text{g(x)}=+\sqrt{\text{x}},$ find $fog$ and $gof.$ Are they equal functions?

Answer

We have, $f : R^+ \rightarrow R^+$ given by
$f(x) = x^2$
$g : R^+ \rightarrow R^+$ given by
$\text{g}(\text{x})=\sqrt{\text{x}}$
$\therefore fog(x) = f(g(x))$
$=\text{f}(\sqrt{\text{x}})=(\sqrt{\text{x}})^2=\text{x}$
Also, $gof(x) = g(f(x))$
$=\text{g}(\text{x}^2)\sqrt{\text{x}^2}=\text{x}$
Thus, $fog(x) = gof(x)$

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Question 543 Marks

Find $gof$ and $fog$ when $f : R \rightarrow R$ and $g : R \rightarrow R$ are defined by$: f(x) = 8x^3$ and $\text{g(x)}=\text{x}^\frac{1}{3}$

Answer

Given, $f : R \rightarrow R$ and $g : R \rightarrow R$
Therefore, $gof : R \rightarrow R$ and $fog : R \rightarrow R$
$f(x) = 8x^3$ and $\text{g(x)}=\text{x}^\frac{1}{3}$
$(gof)(x) = g(f(x))$
$= g(8x^3)$
$=(8\text{x}^3)^\frac{1}{3}$
$=[(2\text{x})^3]^\frac{1}{3}$
$=2\text{x}$
$(fog)(x) = f(g(x))$
$=\text{f}\Big(\text{x}^\frac{1}{3}\Big)$
$=8\Big(\text{x}^\frac{1}{3}\Big)^3$
$=8\text{x}$

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Question 553 Marks

Find fog and gof if:f(x) = x + 1, g(x) = 2x + 3

Answer

f(x) = x + 1, g(x) = 2x + 3f : R → R; g : R → R
Computing fog: Clearly, the range of g is a subset of the domain of f.
⇒ fog : R → R
(fog)(x) = f(g(x))
= f(2x + 3)
= 2x + 3 + 1
= 2x + 4
Computing gof: Clearly, the range of f is a subset of the domain of g.
⇒ fog : R → R
(gof)(x) = g(f(x))
= g(x + 1)
= 2(x + 1) + 3
= 2x + 5

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MATHS 3 Marks Question