Differentiation (Part-38) | S N De

 In the previous article , we have discussed Long Answer Type Questions (Qst. 1-6) of Relation and Mapping Chapter  of S.N. Dey Math Exercise. In this article,  we are going to complete the rest of Long Answer Type Questions (Quest. 7-12) of Relation and Mapping  as a part of our S.N. Dey Math solution series.  So, let's start.

 7. Find the domain of definition of $\,\,(i) f(x)=\sqrt{x+1}+\sqrt{4-x}$

Sol.$\,f(x)\,$ will be defined if $\,x+1 \geq 0 \, \text{and}\,\, 4-x \geq 0 \\ \therefore -1 \leq x \leq 4.$

So, domain of definition of $f(x)=\{x \in \mathbb R: -1 \leq x \leq4\}$

7. Find the domain of definition of $\,\,(ii)\,\, \phi(x)=\cos^{-1} \left(\frac{x-4}{3}\right)+\log(5-x)$

Sol.$\,\,\,\,\phi(x)\,$ will be defined if $\,5-x > 0 \, \text{and}\,\, -1 \leq \frac{x-4}{3} \leq 1 \\ \Rightarrow  x <5 \,\, \wedge \,\, 1 \leq x \leq 7 \\ \therefore 1 \leq x <5$

So, domain of definition of $\phi(x)=\{x \in \mathbb R: 1 \leq x <5\}$   

8. The function $\,\,f(x)\,$ is defined in $\,\,0 \leq x \leq 1;\,\,$  

Find the domain of definition of $\,(i)\,f(2x-1)\, (ii)\,f(x^2)$

   

Sol. of  8(i)  The function $\,f(x)\,$ is defined in   

$\,\,0 \leq x \leq 1 ;$ and so $\,f(2x-1)\,$ is defined in  

 $\,\,0 \leq 2x-1 \leq 1 \Rightarrow \frac 12 \leq x \leq 1$

So, the domain of definition of $\,f(x)\,$ is : $\{ x \in \mathbb R:\frac 12 \leq x \leq 1 \}$

sol. of  8(ii) The function $\,f(x^2)\,$ will be defined if $\,0 \leq x^2 \leq 1 \Rightarrow -1 \leq x \leq 1$

So, the domain of definition of $\,f(x)\,=\{x \in \mathbb R:-1 \leq x \leq 1 \}$                                                                                   

9. If two real functions $\,\,f(x)\,$ and $\,\,\phi(x)\,\,$ are defined respectively, by $\,\,f(x)=\sqrt{x-2}\,\,$ and  $\,\,\phi(x)=x+3,\,$

 then find each of the following functions: $(i)\,\frac 1f \,\,(ii)\frac{1}{\phi}\,\,(iii)\, f +\phi\,\,(iv) \, f \phi \,\,(v)\, \frac{f}{\phi}$

Sol. (i) $\frac 1f(x)=\frac{1}{f(x)}=\frac{1}{\sqrt{x-2}}\,\,$ and it will be defined if $\,x-2>0.$

Therefore, $\,\frac 1f(x)=\frac{1}{\sqrt{x-2}},\,\, 2 <x<\infty$

(ii) $\,\,\frac{1}{\phi}(x)=\frac {1}{\phi(x)}=\frac{1}{x+3}\,\,$ and it will be defined if $\,\,x+3 \neq 0,\,\,$ 

Now, $\,\,\frac{1}{\phi}(x)=\frac{1}{x+3},\,\, x \in \mathbb R-\{-3\}$

(iii)$\,\,(f +\phi)(x)=f(x)+\phi(x)=\sqrt{x-2}+x+3\,\,$ and it will be defined if $\,x-2 \geq 0.$

Now, $(f+\phi)(x)=\sqrt{x-2}+x+3,\,\,2 \leq x < \infty$

(iv) $\,\, (f \phi)(x)=f(x)\phi(x)=(\sqrt{x-2})(x+3)$ and it will be defined if $\,\,$ if   

 $x-2 \geq 0. \\ \therefore (f\phi)(x)=(\sqrt{x-2})(x+3),\,\,\, 2 \leq x < \infty$  

(v) $\,\,\, \frac {f}{\phi}(x)=\frac{f(x)}{\phi(x)}=\frac{\sqrt{x-2}}{x+3}\quad$

will be defined if $\,\, x-2 \geq 0.$

So, $\,\,\frac{f}{\phi}(x)=\frac{\sqrt{x-2}}{x+3},\,\,2 \leq x < \infty.$

10. Find the domain of definition of the function defined by $\,f(x)=\frac{1}{\log_e(2-x)}+\sqrt{x+3}.$

Sol. The aforementioned function will be defined if  

 $\,\,2-x \neq 1,\,2-x >0 \,\wedge \, x+3 \geq 0. \\ \Rightarrow -3 \leq x <1,\,\, 1 <x<2.$

So, the required the domain of definition is: $\{x \in \mathbb R : -3 \leq x <1,\,1 <x<2.\}$

11. Find the domain of definition of each of the following functions :  

$\,\,(i)\,\,f(x)=\frac{1}{\sqrt{x+|x|}}\\(ii)\,\,g(x)=\frac{1}{\sqrt{x-|x|}}\\ (iii)\,\,h(x)=\sqrt{x+[x]} \\ (iv)\,\,\phi(x)=\sqrt{x-[x]}$

Sol. (i) $\,\,f(x)\,$ will be defined if $\,\,x+|x| >0.$    

Now, $\,\, \forall x, |x|>0\,\,$and so the function will be defined if $\,\,x >0.$

So, the required domain of def. is : $\,\, x \in \mathbb R : 0 <x< \infty$

Sol. (ii) $\,\,g(x)\,$ will be defined if $\,\,x-|x| >0.$    

Now, we know, $\,\, \forall x \in \mathbb R, |x| \not <x\,\,$and so $\,\,x-|x| >0,\,$ is impossible.

So, the required domain of def. is : $\,\, \phi,\,\,\,\text{Null Set.}$

Sol. (iii) $\,\,h(x)\,$ will be defined if $\,\,x+[x] \geq 0.$    

Now, we know, $\,\, \forall x \in \mathbb R, x +[x] \geq 0\,\,$ if $\,\,x \geq 0.$    

So, the required domain of def. is : $\,\, \{x : \in \mathbb R: x \geq 0\}$

Sol. (iv) $\,\, \phi(x)\,$ will be defined if $\,\,x-[x] \geq 0.$    

Now, we know, $\,\, \forall x \in \mathbb R, [x] \leq x\,\,$ if $\,\,x \geq 0.$    

So, the required domain of def. is : $\,\, \mathbb R$

12. Find the range of each of the following functions : 

 $\,\,(i) \,f(x)=2-|x-2| \\ (ii)\,f(x)=\frac{1}{\sqrt{x+[x]}}\,\,(iii)\,h(x)=\frac{|x-3|}{x-3}$

Sol. (i) We know, $\,\, 0 \leq |x-2| < \infty \\ \Rightarrow -\infty < -|x-2| \leq 0 \\ \Rightarrow -\infty <2-|x-2| \leq 2$

So, the range of the function is : $\,\, (-\infty,2].$

Sol. (ii)  The function will be defined if $\,\,x+[x] >0,\,$ which is possible when $\,x>0.$

Now, $\,x>0 \Rightarrow 0 < x+[x] < \infty, \\ ~~~~~~~~~~\Rightarrow 0 < \frac{1}{x+[x]} < \infty$

So, the range of the function is : $\,\,(0,\infty)$

Sol .(iii) Let 

$\qquad f(x)=|x-3|~=~\begin{cases} x-3 &:& x\geq 3\\[3ex] 3-x &:& x < 3 \end{cases} \\ \therefore \qquad h(x)=~\begin{cases}1 &:& |x-3|\geq 0\\[3ex] -1 &:& |x-3| < 0 \end{cases}$

So, the range of the function: $\,\{-1,1\}.$

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