Differentiation (Part-38) | S N De

 

In the previous article , we have discussed  Short Answer Type Questions  of Exercise-2D  . In this Chapter , we will discuss Long Answer Type Questions and its solutions.   So, let's start.                                                                                            

1. Find the domain of definitions of each of the following functions:

(i) $\,\sqrt{6-x}$

Sol. Let $\,\,y=\sqrt{6-x}.$  

Now, the function $\,y\,$ will be defined if $\,\,6-x \geq 0 \\ \Rightarrow x \leq 6.$

So, the domain of definition of $\,y\,$ is : $\,\{x \in \mathbb R : x \leq 6\}.$

1(ii) $\, \frac{x-2}{x^2-3x+2}$

Sol. Let $\,y=\frac{x-2}{x^2-3x+2}.$  

The function $\,y\,$ is defined only when $\,\,x^2-3x+2 \neq 0 \\ \Rightarrow (x-2)(x-1)\neq 0\\ \Rightarrow x \neq 1,2.$

So, the domain of definition of $\,y\,$ is : $\,\,\mathbb R-\{1,2\}$

1(iii) $\,\sqrt{2+x-x^2}$

Sol. Let $\,y=\sqrt{2+x-x^2}.$  

The function $\,y\,$ is defined only when $\, 2+x-x^2 \geq 0 \\ \Rightarrow -(x^2-x-2) \geq 0 \\ \Rightarrow x^2-2x+x-2 \leq 0 \\ \Rightarrow (x-2)(x+1)\leq 0 \cdots (1)$ 

From (1), we can conclude that $\,\, -1 \leq x \leq 2$

So, the domain of definition of $\,y\,$ is : $\,\{x: x \in \mathbb R \wedge -1\leq x \leq 2\}$

1(iv) $\sqrt{12-x-x^2}$

Sol. Let $\,y=\sqrt{12-x-x^2}.$  

The function $\,y\,$ is defined only when $\, 12-x-x^2 \geq 0 \\ \Rightarrow -(x^2+x-12) \geq 0 \\ \Rightarrow x^2+x-12 \leq 0 \\ \Rightarrow x^2+4x-3x-12 \leq 0 \\ \Rightarrow x(x+4)-3(x+4) \leq 0 \\ \Rightarrow (x+4)(x-3) \leq 0\cdots (1)$ 

From (1), we can conclude that $\,\, -4 \leq x \leq 3$

So, the domain of definition of $\,y\,$ is : $\,\{x: x \in \mathbb R \wedge -4 \leq x \leq 3\}$

1(v) $\,\frac{x^2}{1+x^2}$ 

Sol. Let $\,y=\frac{x^2}{1+x^2}.$  

The function $\,y\,$ is not defined only when $\, 1+x^2= 0, \,$ which is impossible $\forall x \in \mathbb R\,\,$ as $\,x^2 \geq 0.$ 

So, the domain of definition of $\,y\,$ is : $\,\{x: x \in \mathbb R \}$

1(vi) $\,\log_{10} x$

Sol.  Let $\,y=\log_{10} x.$

The function $\,y\,$ is defined if $x >0.$ 

So, the domain of definition of $\,y\,$ is : $\,\,\{x \in \mathbb R: x>0.\}$ 

1(vii) $\,\,f(x)=\frac{x+2}{\sqrt{x^2-x-2}}$

Sol. The function $\,\,f(x)\,$ is defined if  

 $\,\,x^2-x-2 >0 \\ \Rightarrow x^2-2x+x-2 >0 \\ \Rightarrow x(x-2)+1(x-2)>0 \\ \Rightarrow (x-2)(x+1)>0 \cdots(1)$

(1) is true if $\,x-2>0, x+1>0  \\ \text{or,}\,\, x-2<0,x+1<0 \cdots (2)$ 

From (2), we conclude (1) is true if $\,\,x>2 \,\,\text{or,}\,\, x<-1.$

So, the domain of definition of $\,f(x)\,$ is : $\,\{x: x \in \mathbb R\, \wedge \, x>2 \,\,\text{or,}\,x<-1.\}$

1(viii) $\,\,f(x)=\frac{1}{\sin x-\cos x}$

Sol. The function $\,\,f(x)\,$ is defined if  

 $\,\,\sin x-\cos x \neq 0 $ 

Now, $\,\,\sin x-\cos x =0 \\ \Rightarrow \tan x =1 \\ \Rightarrow x=n\pi+\pi/4,\,$ where $\,n\,$ is any integer.

So, the domain of definition of $\,f(x)\,$ is : $\,\{x: x \in \mathbb R :\, x \neq n\pi+\pi/4\}$

1(ix) $\,\,f(x)=\frac{1}{\sqrt{(x-2)(3-x)}}$

Sol. The function $\,\,f(x)\,$ is defined if  

 $\,\,(x-2)(3-x) > 0 \cdots(1) $ 

Now,  (1) is true if $\,\,x-2 >0 \,\wedge \,\,3-x>0 \\ \Rightarrow 2<x<3$

So, the domain of definition of $\,f(x)\,$ is : $\,\{x: x \in \mathbb R \wedge \,\, 2<x<3\}$

1(x) $\,\,f(x)=\sin^{-1} (2x)$

Sol. The function $\,\,f(x)\,$ is defined if  

 $\,\,-1 \leq 2x \leq 1 \Rightarrow -\frac12 \leq x \leq \frac 12. $ 

So, the domain of definition of $\,f(x)\,$ is : $\,\{ x \in \mathbb R : \,-\frac12 \leq x \leq \frac 12\}$

1(xi) $\,f(x)=\frac{\sqrt{3x-7}}{^6\sqrt{x+1}-2}$

Sol. $\,f(x)=\frac{\sqrt{3x-7}}{^6\sqrt{x+1}-2}\,\,$ will be defined if  

$\,3x-7 \geq 0, \,\,x+1 \geq 0 ,\,\, ^6\sqrt{x+1}-2 \neq 0 \\ \Rightarrow  x \geq \frac 73,\,x \geq -1,\, x \neq 63 (=2^6-1)$

So, the required domain of definition is: $\,\{ x \in \mathbb R: \frac 73 \leq x \leq 63,\,\, 63 <x< \infty\}$ 

1(xii) $\,\,f(x)=\log_{10x} \left(\frac{2 \log_{10} (x)+2}{-x} \right)$

Sol. $\,f(x)\,$ will be defined if   

$\,\left(\frac{2 \log_{10} (x)+2}{-x} \right)>0, x>0,\,\,100x \neq 1 \\ \Rightarrow 2 \log_{10} (x)+2 < 0,\,x >0,\,\,x \neq \frac{1}{100}\\ \Rightarrow  \log_{10} (x^2) < -2,\,x >0,\,\,x \neq \frac{1}{100} \\ \Rightarrow x < \frac{1}{10},\,x >0,\,\,x \neq \frac{1}{100}.$ 

Hence, the domain of definition of $\,f(x)\,$ is : $\,\,\{x \in \mathbb R: 0 <x< \frac{1}{10},\,x \neq \frac{1}{100}\}$

2. Find the range of the following functions: 

2(i) $\,y=\sqrt{4-x^2}$

Sol. For the function $\,y\,$ we have  $\,y^2=4-x^2 \\ \Rightarrow x^2=4-y^2 \cdots (1) $

Now, since $\,\, -2 \leq x \leq 2 \\ \Rightarrow 0 \leq x^2 \leq 4 \\ \Rightarrow 0 \leq 4-y^2 \leq 4 \cdots (2)\,\,[\text{By (1)}]$

Now, from (2) we have, $\,0 \leq 4-y^2 \\ \Rightarrow y^2 \leq 4\\ \Rightarrow y \leq2 \cdots(3)$

Again from (2), we have , $\,\,4-y^2 \leq 4 \\ \Rightarrow -y^2 \leq 0 \\ \Rightarrow y^2 \geq 0 \\ \Rightarrow y \geq 0 \cdots(4)$

Hence, from (3) and (4), we have , the range of $\,y\,$is:

$\{y: y \in \mathbb R \wedge 0 \leq y \leq 2\}$

2(ii) $\,\,y=\frac{x^2}{1+x^2}$

Sol. We know , for all real values of $\,x,\, \frac{x^2}{1+x^2} <1 .$ 

Again, $\,\frac{x^2}{1+x^2} \geq 0\,\,[\text{Since}, x^2 \geq 0] \\ \Rightarrow 0 \leq y <1$

So, So, the range of $\,\,y\,$ is : $\,\{y \in \mathbb R: 0 \leq y \leq 1\}$

(iii) $\,\,y=\sin x\,\,[0 \leq x \leq \pi]$

Sol. $\,\, 0 \leq x \leq \pi \Rightarrow 0 \leq \sin x \leq 1$

So, the range of $\,\,y\,$ is : $\,\{y \in \mathbb R: 0 \leq y \leq 1\}$

2(iv) $\,y=\frac{x}{x^2-5x+9}$

Sol.  $\,y=\frac{x}{x^2-5x+9} \\ \Rightarrow yx^2-(5y+1)x+9y=0 \cdots(1)$

Since $\,\,x \in \mathbb R,\,$  discriminant $(D) \geq 0\\  \Rightarrow \{-(5y+1)\}^2-4.y.(9y) \geq 0  \\ \Rightarrow 25y^2+10y+1-36y^2 \geq 0 \\ \Rightarrow -11y^2+10y+1 \geq 0 \\ \Rightarrow (1-y)(1+11y) \geq 0 \\ \Rightarrow -\frac{1}{11} \leq y \leq 1 $

So, the range of $\,y:\, \{ y \in \mathbb R: -\frac{1}{11} \leq y \leq 1\}$

2(v) $\,\,y=\frac{3x-5}{x^2-1},\,\, x \neq 1$

Sol. $\,\,y=\frac{3x-5}{x^2-1},\,\, x \neq 1 \\ \Rightarrow yx^2-3x+(5-y)=0.$

Since $\,\,x \in \mathbb R,\,$  discriminant $(D) \geq 0\\ \Rightarrow (-3)^2 -4y(5-y) \geq 0 \\ \Rightarrow 9-20y+4y^2 \geq 0 \\ \Rightarrow (2y-1)(2y-9) \geq 0 \\ \Rightarrow y \leq \frac12,\, \text{or}\,\, y \geq \frac92$

2(vi)$\,\,y=\sin x + \cos x$

Sol. $\,\,y=\sin x + \cos x \\ \Rightarrow y=\sqrt{2}(\frac{1}{\sqrt2}\sin x+ \frac{1}{\sqrt2} \cos x)\\~~~~~~~=\sqrt2(\sin x \cos \pi/4+\cos x \sin \pi/4)\\~~~~~~~=\sqrt2 \sin(x+\pi/4)$

Since $\,\,-1 \leq \sin(x +\pi/4) \leq 1\\ \Rightarrow -\sqrt2 \leq f(x)\leq \sqrt2$

Hence, the range of $\,f(x)\,$ is : $\,[-\sqrt2,\sqrt2]$

2(vii) $\,y=\frac{1}{3-\cos 2x}$

Sol. we know , $\,-1 \leq \cos 2x \leq 1 \\ \Rightarrow -1 \leq -\cos 2x \leq 1 \\ \Rightarrow 3-1 \leq 3-\cos 2x \leq 3+1 \\ \Rightarrow 2 \leq 3-\cos 2x\leq 4 \\ \Rightarrow \frac12 \geq y \geq \frac 14 \\ \Rightarrow \frac 14 \leq y \leq \frac 12$

Hence, the range of $\,y\,$ is : $\{y \in \mathbb R \wedge \frac 14 \leq y \leq \frac 12\}$

2(viii) $\,\,y=\tan x\,\,[-\frac{\pi}{2} <x<\frac{\pi}{2}] $

Sol. Since , $-\frac{\pi}{2} <x<\frac{\pi}{2} \\ \Rightarrow -\infty <\tan x <\infty$

So, the range of $\,y\,$ is : $\{y: y \in \mathbb R\}$

3. If $\,\,f(x)\,\,$is a quadratic function and $\,f(1)=5,\,f(-1)=11,\,f(2)=8,\,\,$ find $\,f(-2).$

Sol. Let $\,f(x)=ax^2+bx+c \\ \Rightarrow f(1)=a+b+c=5\cdots(1) \\ f(-1)=a-b+c=11 \cdots(2) \\ f(2)=4a+2b+c=8 \cdots(3)$

Solving (1),(2) and (3), we have $\,\,a=2,\,b=-3,\,c=6$ and so  

$f(x)=2x^2-3x+6 \\ \Rightarrow f(-2)=2(-2)^2-3(-2)+6=20$

4. Let $\,P(x)\,$ be a quadratic function and $\,\,P(l)=a,\,\,P(m)=b,\,\,P(n)=c,\,\,$ find $\,P(x).\,$

Sol. Without any loss of generality, let  

$\,P(x)=A(x-l)(x-m)+\\B(x-m)(x-n)+C(x-n)(x-l) \cdots (1) \\ \Rightarrow P(l)=B(l-m)(l-n) \\ \Rightarrow a=B(l-m)(l-n) \\ \Rightarrow B=\frac{a}{(l-m)(l-n)}$

Again, from (1) we get, 

$P(m)=C(m-n)(m-l) \\ \Rightarrow b=C(m-n)(m-l) \\ \Rightarrow C=\frac{b}{(m-n)(m-l)}$

Similarly , from (1) gives $\,P(n)=A(n-l)(n-m) \\ \Rightarrow c=A(n-l)(n-m) \\ \Rightarrow A=\frac{c}{(n-l)(n-m)}$

Putting the values of $\,\, A,B,C\,$ we get from (1),

$\,P(x)=\frac{c}{(n-l)(n-m)}. (x-l)(x-m)\\+ \frac{a}{(l-m)(l-n)}. (x-m)(x-n)\\+\frac{b}{(m-n)(m-l)}.(x-n)(x-l)$

5(i). If $\,\,F(x)=a/x+b+cx,\,\,F(1)=5,\,F(-2)=2, \\ F(-1)=-3,\,\,$ find the value of $\,\,F(-3).$

Sol. $\,F(x)=a/x+b+cx \\ \Rightarrow F(1)=a+b+c=5=5 \cdots (1)\\ F(-2)=-a/2+b-2c=2 \cdots(2)\\ F(-1)=-a+b-c=-3 \cdots(3)$

Hence , solving (1),(2) and (3), we get $\,\,a=6,\,b=1,\,c=-2.$

So, $\,F(x)=6/x+1-2c \\ \Rightarrow F(-3)=6/(-3)+1-2(-3)\\~~~~~~~~~~~~~~~~=-2+1+6\\~~~~~~~~~~~~~~~~=5.$

5(ii). If $\,\,f(n+1)=\frac{2f(n)+1}{2},\,n=1,2,3,..,f(1)=2,$ then find the value of $\,f(101).$

Sol. $\,\,f(1)=2 \\ \therefore f(2)=\frac{2f(1)+1}{2}\\~~~~~~~~~~~=\frac{2.2+1}{2} \\ \therefore f(3)=\frac{2f(2)+1}{2}\\~~~~~~~~~~~=\frac{2.2+1+1}{2} \\ \therefore f(4)=\frac{2f(3)+1}{2}\\~~~~~~~~~~~=\frac{2.2+1+1+1}{2} \\ .................. \\ f(101)=\frac{2.2+1+1+1+......+1}{2}\\~~~~~~~~~~~~=\frac{2.2+100}{2}\\~~~~~~~~~~~~=52$

6. The taxi fare is Rs. $\,3\,$ for $\,1\,$ Km or less from starting point and Rs. $\,1.20\,$ per km or any fraction thereof, for additional distance. If the fare be Rs. $\,y\,$ for a total journey of $\,x\,$ km, express $\,y\,$ as function of $\,x\,$.

Sol. We notice that taxi fare is Rs. 3 for travelling $\,x\,$ km. where $\,0 \leq x \leq 1.$ But after crossing $\,1\,$ km. distance for each additional distance , one has to pay Rs. $\,1.20\,$ per km or any fraction thereof. So, if we consider after crossing $\,1\,$ km., one has traversed additional $\,p\,$ km. then it will be charged $=3+1.20p\,\,$ rs. 

So, in short we can say if the fare be Rs. $\,y\,$ for a total journey of $\,x\,$ km. then $\,\,y=3+1.20p\,\,\text{when}\,\,p<x \leq (p+1), \\ \text{where}\,\,p=0\,\text{or a positive integer.}$  

 

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