3 Marks Question

Question 513 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 523 Marks

Let $f(x) = x^2 + x + 1$ and $g(x) = sinx$. Show that fog ≠ gof.

Answer

$(fog)(x) = f(g(x))$
$f(sinx) = \sin^2x + sinx + 1$
and, (gof)(x) = g(f(x))
$= g(x^2 + x + 1)$
$= sinx^2 + x + 1$
Therefore, fog ≠ gof.

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Question 533 Marks

Find fog and gof if$:f(x) = x^2, g(x) = cosx$

Answer

$f(x) = x^2, g(x) = cosx$
Domain of f and Domain of g = R
Range of $\text{f}=(0,\infty)$
Range of g = (-1, 1)
$\therefore$ Range of f $\subset$ domain of g ⇒ gof exist
Range of g $\subset$ domain of f ⇒ fog exist
Now,
$gof(x) = g(f(x)) = g(x^2) = cosx^2$
And
$fog(x) = f(f(x)) = f(cosx) = \cos^2x$

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Question 543 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 553 Marks

Classify the following functions as injection, surjection or bijection:
$f : Z \rightarrow Z$ given by $f(x) = x^2$

Answer

$f: Z \rightarrow Z$ given by $f(x)=x^2$
Injection test: Let $x$ and $y$ be any two elements in the domain $(Z)$, such that $f(x)=f(y) . f(x)=f(y) x^2=y^2 x= \pm y$ So, f is not an injection.
Surjection test: Let $y$ be any element in the co-domain $(Z)$, such that $f(x)=y$ for some element $x$ in $Z$ (domain). $f(x)=$ $y x^2=y x= \pm \sqrt{y}$ which may not be in $Z$.
For example, if $y=3$,
$\text{x}=\pm\sqrt{3}$ is not in Z.
So, f is not a bijection.

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MATHS 3 Marks Question