Differentiation (Part-38) | S N De

 

 In the previous article , we have discussed few Short Answer Type Questions (From Qst. 1-12) of S.N. Dey Math Exercise [of Relation and Mapping chapter]. In this article,  we are going to start Long Answer Type Questions of Quadratic Equations (Part-6) as a part of our S.N. Dey Math solution series.  So, let's start.

13. If $\,\,f(x)=\tan^{-1}x,\,\,$ find the relation by which $\,\,f(x),f(y),f(x+y)\,$ are connected.

Sol. $\,\,f(x)=\tan^{-1}x \\ \Rightarrow  x=\tan[f(x)] \cdots (1) \\ \therefore y=\tan[f(y)] \cdots(2)$

Now ,  replace $\,\,x\,$ with $\,x+y\,$ in (1), we get  

$x+y=\tan[f(x+y)]\\ \Rightarrow \tan[f(x)]+\tan[f(y)]=\tan[f(x+y)]\\ [\text{By (1) and (2)}]$

14. What value of $k$ makes the function below continuous?

$$a)\qquad \phi(x)~=~\begin{cases} 4 &:& x \,\,\text{is rational}\\[1ex] -4 &:& x \,\,\text{is irrational}\end{cases}$$

Find $\,\,\phi(0),\,\,\phi(2.4),\,\,\phi(\sqrt3),\,\,\phi(3.6),\,\,\phi(\pi),\,\,\phi(e),\,\phi(\sin {\pi/4})$

Sol. $\,\,\phi(0)=\phi(2.4)=\phi(3.6)=4,$  

[Since $ \,0,2.4,3.6\,\,$are rational no.s]

And $\,\,\phi(\sqrt3)=\phi(\pi)=\phi(e)=\phi(\sin {\pi/4}) [=\phi(1/\sqrt2)]=-4, $  

[Since, $\,\sqrt3,\pi,e, \frac{1}{\sqrt2}\,\,$are irrational no.s]

15. Let

$$a)\qquad f(x)~=~\begin{cases} 2x^2-3 &:& x\leq 2\\[3ex] 2x+1 &:& 2<x \leq 3 \\[3ex] \frac{1}{2x-1} &:& x>3\end{cases}$$

Find $\,\,f(0),\,f(3),\,\,f(-2),\,f(4),\,\,f(\sqrt 2),\,f(e).$ 

Sol. (i)$\,f(0)=2 \times 0^2-3 \,\,[\, \text{Since,}\,f(x)=2x^2-3,\,\,\text{when}\\ x \leq 2]\\~~~~~~~~~=-3$

(ii) $\,f(3)= 2 \times 3+1 [\, \text{Since,}\,f(x)=2x+1,\,\,\text{when}\\ 2<x \leq 3]\\~~~~~~~~~~~=7$

(iii) $\,\,f(-2)= 2 \times (-2)^2-3 \\  [\, \text{Since,}\,f(x)=2x^2-3,\,\,\text{when} \\ x \leq 2]\\~~~~~~~~~~~=5$

(iv) $\,\,f(4)= \frac{1}{2\times 4-1}\,\, [\, \text{Since,}\,f(x)=\frac{1}{2x-1},\\ \text{when}\,\,x >3]\\~~~~~~~~~~~=\frac17$

(v) $\,\,f(\sqrt 2)= 2 \times (\sqrt2)^2-3 \\ [\, \text{Since,}\,f(x)=2x^2-3,\\ \text{when}\,\,x \leq 2]\\~~~~~~~~~~~=1$

(vi) $\,\,f(-2)= 2e+1\,\, [\, \text{Since,}\,f(x)=2x+1,\,\,\text{when}\\ 2 <x < 3]$

16. Find the values of $\,x\,$ for which the following functions are undefined:  

$(i) \frac{x}{x+2} \quad (ii) \sqrt{4x-4x^2-1}\quad (iii) \frac{x}{\sin x} \\ (iv) \frac{x^2+x-6}{2x^2-x-6} \quad (v) \sqrt{x^2-4x+3}$

Sol. (i) $\,\,\frac{x}{x+2}\,$ will be undefined if $\,\,x+2=0 \Rightarrow x=-2$. 

(ii)Let  $\,f(x)=\sqrt{4x-4x^2-1}\,$ will be undefined if $\,\,4x-4x^2-1<0.$  

Now, $\,\,4x-4x^2-1\\=-(4x^2-4x+1)\\=-(2x-1)^2 <0 \quad \forall x \in \mathbb R-\{1/2\}$

So, $\,\,f(x)\,$ will be undefined for all values of $\, x \in \mathbb R$ except for $\,\,x=\frac12.$

(iii) $\,\frac{x}{\sin x}\,$ will be undefined if $\,\sin x=0 \Rightarrow x=n\pi, \, n \in \mathbb Z$

(iv) $\,\frac{x^2+x-6}{2x^2-x-6}\,$ will be undefined if $\,2x^2-x-6=0\\  \Rightarrow (2x+3)(x-2)=0 \\ \Rightarrow x=2,-\frac32$

(v) $\,\,\sqrt{x^2-4x+3}\,$  will be undefined if $\, x^2-4x+3 <0 \\ \Rightarrow (x-3)(x-1)<0 \cdots(1)$ 

From (1)  to be true, either $\,x-3>0 ,x-1<0\,\,$ or $\,\,x-3<0,x-1>0 .$   

So, either $\,\,x>3,x<1\,\,$ or, $\,\,x<3,\, x>1 \Rightarrow 1<x<3$  

So, $\,\,1<x<3\,\,[\text{Since,}\, x>3 \wedge x<1 \,\text{is not possible}]$

17. If the maximum and minimum values of $\,\,f(x)=a+b \sin x,\,\,$ be $\,7\,$ and $\,1\,$ respectively, find the value of $\,\,f(\pi/6).$

Sol. Case- 1. $\,\,b >0. \\ -1\leq \sin x \leq 1 \\ \Rightarrow -b \leq b \sin x \leq b \\ \Rightarrow a-b \leq a+b \sin x \leq a+b \\ \Rightarrow a-b \leq f(x) \leq a+b$

So, the max. value of $\,f(x)\,$ is $\,(a+b)\,$ and  the min. value of $\,f(x)\,$ is $\,(a-b).\,$

Hence, $\,a+b=7,\,a-b=1 \Rightarrow a=4,\, b=3. \\ \therefore f(x)=4+3\sin x \\ \therefore f(\pi/4)=4+3.\sin{\pi /6}=4+\frac32=\frac{11}{2}$

Case-2: $b<0.\,\,$ In this case, max.$\,f(x)=a-b,\,\,$ min. $\,f(x)=a+b.$

Hence, $\,\,a-b=7,\,\,a+b=1\\ \Rightarrow a=4,b=-3. \\ \therefore f(x)=4-3\sin x \\ \therefore f(\pi/6)=4-3\sin{\pi/6}=4-\frac32=\frac52$

18. If $\,\,f(x)=ax^2+bx+x,\,\,f(x+1)=f(x)+x+1\,\,$ is an identity , find the values of $\,\,a,\,b.$

Sol. $\,f(x+1)=f(x)+x+1 \\ \Rightarrow a(x+1)^2+b(x+1)+c \\=ax^2+bx+c+x+1 \\ \Rightarrow 2ax+a+b=x+1 \cdots(1)$

Hence , (1) will be an identity if $\,\,2a=1 \cdots(2) \\ a+b=1 \cdots (3)$ 

So, solving (2) and (3), we get $\,\,a=\frac12,\,b=\frac12$

19. If $\,f(x)=x^2+ax+b,\,f(1)=1,\,f(2)=2,\,\,$ find the value of $\,\,f(3).$

Sol. $\,\,f(1)=1 \\ \Rightarrow 1+a+b=1 \,\,[\text{Since,}f(x)=x^2+ax+b]\\ \Rightarrow a+b=0 \cdots(1) \\ f(2)=2 \\ \Rightarrow 4+2a+b=2 \\ \Rightarrow 2a+b=-2 \cdots(2)$ 

Solving (1) and (2), we get $\,\,a=-2,b=2 \\ \therefore f(x)=x^2-2x+2 \\ \therefore f(3)=3^2-2 \times 3+2=5$

 20. If the function $\,f: \mathbb R \to \mathbb R\,\,$ is given by $\,\,f(x)=x\,\,\forall x\in \mathbb R$ and the function $\,\,g: \mathbb R -\{0\} \to \mathbb R\,\,$ is given by $\,g(x)=\frac1x, \forall x \in \mathbb R-\{0\} ,\,$ then find the functions $\,f+g\,$ and $\,f-g.\,$

Sol. For $\,\,f+g:$

For $\,f+g,$ the domain of definition is the intersection of domain of definitions of $\,f\,$ and $\,g\,$ which is $\mathbb R \cap \mathbb R-\{0\}=\mathbb R-\{0\}.$

Now, $\,\,(f+g)(x)=f(x)+g(x)=x+\frac1x. $  

Hence , $ (f+g): \mathbb R-\{0\} \to \mathbb R$ 

 is defined by $\,(f+g)(x)=x+\frac1x$

For $\,\,f-g:$

For $\,f-g,$ the domain of definition is the intersection of domain of definitions of $\,f\,$ and $\,g\,$ which is $\mathbb R \cap \mathbb R-\{0\}=\mathbb R-\{0\}.$

Now, $\,\,(f-g)(x)=f(x)-g(x)=x-\frac1x. $  

Hence , $ (f-g): \mathbb R-\{0\} \to \mathbb R$ 

 is defined by $\,(f-g)(x)=x-\frac1x$

21. If two real functions $\,f\,$ and $\,g\,$ are defined respectively by $\,\,f(x)=\sqrt{x+1}\,$ and $\,\,g(x)=\sqrt{x-1},\,\,$ then find the values of $\,\,\frac{f}{g}(1).$

Sol.$\,\,\frac fg(1)=\frac{f(1)}{g(1)}=\frac{\sqrt{1+1}}{\sqrt{1-1}},\,\,$ which is undefined. 

Again, $\,\,\frac{g}{f}(1)=\frac{g(1)}{f(1)}=\frac{\sqrt{1-1}}{\sqrt{1+1}}=\frac{0}{\sqrt2}=0$

Hence,  $\,\,f(x)=\sqrt{x+1}\,$ will be defined if $\,\,x+1 \geq 0 \Rightarrow x \geq -1.$

$\,\,g(x)=\sqrt{x-1}\,\,$ will be defined if $\,\,x-1 \geq 0 \Rightarrow x \geq 1.$

So, domain of definition of $\,\,f\,$is: $\,D_f=\{x \in \mathbb R: x \geq -1\}\\~~~~~~=[-1,\infty).$ 

And, domain of definition of $\,\,g\,$is: $\,D_g=\{x \in \mathbb R: x \geq +1\}\\~~~~~~=[1,\infty).$ 

For $\,\,\frac fg:$ 

The domain of definition of $\,\,\frac fg:\\ D_f \cap D_g -\{x: g(x)=0\}\\=[-1,\infty)\cap [1,\infty)-\{x: \sqrt{x-1} =0\}\\=[1, \infty)-\{1\}\\=(1, \infty)$  

For $\,\,\frac gf:$ 

The domain of definition of $\,\,\frac gf:\\ D_f \cap D_g -\{x: f(x)=0\}\\=[1,\infty)-\{x: \sqrt{x+1} =0\}\\=[1, \infty)-\{-1\}\\=[1, \infty)$ 

22. Find the range of the function $\,\,f(x)=3-|x-3|.$ 

Sol.  We know, $\,\, 0 \leq |x-3| < \infty \\ \Rightarrow -\infty < -|x-3| \leq 0 \\ \Rightarrow -\infty +3 <3-|x-3| \leq 3+0 \\ \Rightarrow -\infty < 3-|x-3| \leq 3 \\ \Rightarrow -\infty < f(x) \leq 3.$

Hence, the range of $\,\,f(x)\,$ is given by : $\,(-\infty,3].$

23.Find $\,\,f \circ g\, \circ\,h,\, $ if $\,f(x)=\frac{x}{x+1},\,g(x)=x^{10}\,$ and $\,h(x)=x+3.$

Sol. $\,\,(f \circ g\, \circ\,h)(x)\\=f[g\{h(x)\}] \\=f[g(x+3)]\,\,[\text{As}\,\,\,h(x)=x+3]\\=f[(x+3)^{10}]\,\,[\text{As}\,\,\,g(x)=x^{10}]\\=\frac{(x+3)^{10}}{(x+3)^{10}+1} \,\,[\text{As}\,\,\,f(x)=\frac{x}{x+1}]$

24. If $\,\,f(x)=\frac{4^x}{4^x+2},\,\,$ then show that $\,\,f(x)+f(1-x)=1.\,\,$  

Hence, prove that $\,f(\frac{1}{1997})+f(\frac{2}{1997})+.....+f(\frac{1996}{1997})=998$

Sol. $\,\,f(x)+f(1-x)\\=\frac{4^x}{4^x+2}+\frac{4^{x-1}}{4^{x-1}+2}\\=\frac{4^x}{4^x+2}+\frac{4}{4+2 \times 4^x}\\=\frac{4^x}{4^x+2}+\frac{2}{2+4^x}\\=\frac{4^x+2}{4^x+2}\\=1.$

Now,$\,f(\frac{1}{1997})+f(\frac{2}{1997})+.....+f(\frac{1996}{1997})\\=[f(\frac{1}{1997})+f(\frac{1996}{1997})]+[f(\frac{2}{1997})+f(\frac{1995}{1997})]+.... \\\text{total}\,(\frac{1996}{2}) \, \text{pairs}\\=[f(\frac{1}{1997})+f(1-\frac{1}{1997})]+[f(\frac{2}{1997})+f(1-\frac{2}{1997})]+....\\(998 \,\,\text{pairs})\\=1+1+.....(998 \,\,\text{pairs})\\=998$

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