3 Marks Question

Question 13 Marks

Find the domain and range of the real functions$
f(x)=\frac{|x-1|}{x-1}
$

Answer

Given, $\quad f(x)=\frac{|x-1|}{x-1}$
Domain : Clearly, $f(x)$ is defined for all $x \rightarrow R$ except $x=1$
$\therefore$ Domain of $f=R-\{1\}$
Range: Now, $f(x)=\frac{x-1}{x-1}=1$, when $x>1$
and $f(x)=\frac{-(x-1)}{x-1}=-1$, when $x<1$
$\therefore$ Range of $f=\{-1,1\}$.

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Question 23 Marks

Find the domain of the function
$
f(x)=\frac{1}{\sqrt{[x]^2-[x]-6}}
$

Answer

Given, $f(x)=\frac{1}{\sqrt{[x]^2-[x]-6}}$
Clearly, $f(x)$ is defined when $[x]^2-[x]-6>0$
$
\begin{aligned}
{[x]^2-3[x]+2[x]-6 } & >0 \\
([x]-3)([x]+2) & >0 \\
{[x]>3 \text { and }[x] } & <-2 \\
x \geq 3+1 \text { and } x & <-2 \\
x \geq 4 \text { and } x & <-2
\end{aligned}
$
$\therefore$ Domain of $f=(-\infty,-2), \cup[4, \infty)$.

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Question 33 Marks

If $f(x)=y=\frac{a x-b}{c x-a}$, then prove that $f(y)=x$.

Answer

Given,
$
f(x)=y=\frac{a x-b}{c x-a}
$
$
\begin{aligned}
\therefore \quad f(y) & =\frac{a y-b}{c y-a}=\frac{a\left(\frac{a x-b}{c x-a}\right)-b}{c\left(\frac{a x-b}{c x-a}\right)-a} \\
& =\frac{a(a x-b)-b(c x-a)}{c(a x-b)-a(c x-a)} \\
& =\frac{a^2 x-a b-b c x+a b}{a c x-b c-a c x+a^2} \\
& =\frac{a^2 x-b c x}{a^2-b c} \\
& =\frac{x\left(a^2-b c\right)}{\left(a^2-b c\right)}=x
\end{aligned}
$
$\therefore f(y)=x$
Hence Proved.

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Question 43 Marks

Let $f$ and $g$ be real functions defined by
$
f(x)=2 x+1 \text { and } g(x)=4 x-7
$
(a) For what real numbers $x, f(x)=g(x)$ ?
(b) For what real numbers $x, f(x)<g(x)$ ?

Answer

Given, $f(x)=2 x+1$ and $g(x)=4 x-7$
(a) $\because\quad$ $f(x)=g(x)$
$\therefore \quad 2 x+1=4 x-7$
$2 x=8$
$x=4$
(b) $\because$ $f(x)<g(x)$
$\therefore$ $2 x+1<4 x-7$
$2 x-4 x+1<4 x-7-4 x$
$-2 x+1<-7$
$-2 x<-7-1$
$-2 x<-8$
$\frac{-2 x}{-2}>\frac{-8}{-2}$
$x>4$

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Applied Maths 3 Marks Question

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