Differentiation (Part-38) | S N De

 

 In the previous article , we have discussed Very Short Answer Type Questions of Relation and Mapping Chapter (From Qst. 1-21) of S.N. Dey Math Exercise. In this article,  we are going to start few Short Answer Type Questions (Quest. 1-12) of Relation and Mapping (Part-5) as a part of our S.N. Dey Math solution series.  So, let's start.

1. A right circular cylinder is inscribed in a sphere of radius $\,r\,$. Express the volume $\,v\,$ of the cylinder as a function of its height $\,x\,$.

Sol. Let the radius of the circular cylinder be $\,AB=R\,$ and the center of the sphere be $\,O.\,$  

The height of cylinder is $\,x\,$ (given). 

So, $\,OB=\frac x2.$  

Now, $AO^2=AB^2+OB^2 \\ \Rightarrow r^2=R^2+(x/2)^2 \\ \Rightarrow R^2=r^2-(x/2)^2.$ 

The volume of the cylinder $\,\,v=\pi R^2 x \\ \Rightarrow v=\pi(r^2-(x/2)^2)x \\ \Rightarrow v=\frac{\pi x}{4}(4r^2-x^2)$ 

So, the volume $\,v\,$ of the cylinder as a function of its height $\,x\,$ can be expressed as $\,\,v=\frac{\pi x}{4}(4r^2-x^2).$

2. A right circular cone is inscribed in a sphere of radius $\,a.$ Express the volume $\,v\,$ of the cone as a function of its slant height $\,x.$

Sol. Let the center of the sphere be $\,O.\,$ Also suppose that the radius of the cone be $\,BD=r.$

Now, for $\,\,\Delta ABD,\,\,h=AD=\sqrt{AB^2-BD^2}\\ \Rightarrow AD=\sqrt{x^2-r^2} \cdots (1)$

Also, for $\,\,\Delta BOD,\,\,OD=\sqrt{OB^2-BD^2}\\ \Rightarrow OD=\sqrt{a^2-r^2} \cdots (2)$

Now, $\,\,AD=AO+OD \\ \Rightarrow \sqrt{x^2-r^2}=a+\sqrt{a^2-r^2} \\ \text{By (1) and (2)} \\ \Rightarrow x^2-r^2=a^2+2a \sqrt{a^2-r^2}+a^2-r^2 \\ \Rightarrow 2a \sqrt{a^2-r^2}=x^2-2a^2 \\ 4a^4-4a^2r^2=x^4-4a^2x^2+4a^4 \\ \Rightarrow r^2=\frac{4a^2x^2-x^4}{4a^2}$  

Now, $\,\,h=AD=\sqrt{x^2-r^2}\\=\sqrt{x^2-\frac{4a^2x^2-x^4}{4a^2}}\\=\frac{x^2}{2a} \\ \therefore v=\frac 13 \pi r^2h \\ \Rightarrow v=\frac 13 \pi \frac{4a^2x^2-x^4}{4a^2} . \frac{x^2}{2a} \\ \Rightarrow v=\frac{\pi x^4}{24a^3}.(4a^2-x^2) \cdots(3)$ 

From (3), the result follows.

4. If $\,\,F(x)=\frac{(x-b)(x-c)}{(a-b)(a-c)}+\frac{(x-c)(x-a)}{(b-c)(b-a)}+\frac{(x-a)(x-b)}{(c-a)(c-b)},\,\,$ show that $\,\,F(0)=1.$

Sol. $\,\,F(0)=\frac{(0-b)(0-c)}{(a-b)(a-c)}+\frac{(0-c)(0-a)}{(b-c)(b-a)}+\frac{(0-a)(0-b)}{(c-a)(c-b)}\\~~~~~~~~~~=\frac{bc}{(a-b)(a-c)}+\frac{ca}{(b-c)(b-a)}+\frac{ab}{(c-a)(c-b)}\\~~~~~~~~~~=-\frac{bc(b-c)+ca(c-a)+ab(a-b)}{(a-b)(b-c)(c-a)}\\~~~~~~~~~~=-\frac{p}{(a-b)(b-c)(c-a)} \\ ~~~~~~~~~\cdots(1) \\ \text{where}\,\,\, p=bc(b-c)+ca(c-a)+ab(a-b)$

Now, we calculate $\,\,\,p=bc(b-c)+ca(c-a)+ab(a-b)\\~~~~=bc(b-c)+c^2a-ca^2+a^2b-ab^2\\~~~~=bc(b-c)-a(b^2-c^2)+a^2(b-c)\\~~~~=(b-c)\{bc-a(b+c)+a^2\}\\~~~~=(b-c)(bc-ab-ac+a^2)\\~~~~=(b-c)\{b(c-a)-a(c-a)\}\\~~~~=-(b-c)(c-a)(a-b) \cdots(2)$ 

From (1) and (2), we get $\,\,F(0)= \frac{p}{p}=1$

5.(i) If $\,\,f(x)=\frac{x-1}{x+1},\,\,$show that $\,\,\frac{f(a)-f(b)}{1+f(a)f(b)}=\frac{a-b}{1+ab}$

Sol. $\,\,f(a)-f(b)\\=\frac{a-1}{a+1}-\frac{b-1}{b+1}\\=\frac{(a-1)(b+1)-(b-1)(a+1)}{(a+1)(b+1)}\\=\frac{ab+a-b-1-(ab+b-a-1)}{(a+1)(b+1)}\\=\frac{2(a-b)}{(a+1)(b+1)} \cdots(1)$

Now, $\,\,1 +f(a)f(b)\\=1+\frac{a-1}{a+1} \frac{b-1}{b+1} \\= \frac{(a+1)(b+1)+(a-1)(b-1)}{(a+1)(b+1)}\\=\frac{ab+a+b+1+ab-a-b+1}{(a+1)(b+1)} \\=\frac{2(ab+1)}{(a+1)(b+1)} \cdots(2)$

From (1) and (2), the result follows.

5(ii) If $\,\,f(x)=\frac{1}{1-x},\,\,$ show that $\,\, f[f\{f(x)\}]=x.$ 

Sol. $\,\,f(f(x))=\frac{1}{1-\frac{1}{1-x}}=\frac{1-x}{-x}=\frac{x-1}{x}$

Now, $\,\,f[f\{f(x)\}]=f[\frac{x-1}{x}]\\~~~~~~~~~~~~~~~~~~~~=\frac{1}{1-\frac{x-1}{x}}\\~~~~~~~~~~~~~~~~~~~~=x$

Hence , the result follows. 

6. If $\,\,y=f(x)=\frac{ax-b}{bx-a},\,\,$ prove that $\,\,f(x).f(1/x)\,$ is independent of $\,x.$

Sol. $\,\,f(1/x)=\frac{a/x-b}{b/x-a}=\frac{a-bx}{b-ax}=\frac{bx-a}{ax-b}$

Now, $\,\,f(x).f(1/x)\\=\frac{ax-b}{bx-a} \times \frac{bx-a}{ax-b}\\=1 \cdots (1)$ 

From (1), the result follows.

7. If $\,\,y=f(x)=\frac{px+q}{rx-p},\,\,$ show that $\,\,x=f(y)$ 

Sol. We have ,  $\,\,y=\frac{px+q}{rx-p} \\ \Rightarrow y(rx-p)=px+q \\ \Rightarrow x(ry)-py=px+q \\ \Rightarrow x(ry-p)=py+q \\ \Rightarrow x= \frac{py+q}{ry-p}=f(y)$

8. If $\,\,y=f(x)=\frac{3x-5}{2x-m},\,\,f(y)=x,\,$ find $\,m.\,$

Sol. Since $\,\,y=f(x)=\frac{3x-5}{2x-m}, \\ \therefore f(y)=\frac{3y-5}{2y-m} =x \\ \Rightarrow \frac{3(\frac{3x-5}{2x-m})-5}{2(\frac{3x-5}{2x-m})-m}=x \\ \Rightarrow \frac{9x-15-10x+5m}{6x-10-2mx+m^2}=x \\ \Rightarrow -x+5m-15=6x^2 -10x-2mx^2+m^2x \\ \Rightarrow (6-2m)x^2+(m^2-9)x+(15-5m)=0 \cdots(1)$ 

Now, the equation (1) will be satisfied for any values of $\, x,\,$  

if $\,6-2m=0,\,m^2-9=0,\,\,15-5m=0 \\ \Rightarrow m=3.$

Hence follows the result. 

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9. If $\,\,y=f(x)=\frac{x-3}{2x+1}\,\,$ and $\,z=f(y),\,$ express $\,z\,$ as a function of $\,x.\,$

Sol. We have, $\,z=f(y)=\frac{y-3}{2y+1} \\ \Rightarrow z=\frac{\frac{x-3}{2x+1}-3}{2. \frac{x-3}{2x+1}+1} =\frac{(x-3)-3(2x+1)}{2(x-3)+(2x+1)} \\ \Rightarrow z=...=-\frac{5x+6}{4x-5}$

10. If $\,\,F(x)=\frac{4x-5}{3x-4},\,\,$ prove that $\,\,F\{F(x)\}=x.$

Sol. $\,\,F\{F(x)\}=F\{\frac{4x-5}{3x-4}\}\\~~~~~~~~~~~~~~~~~= \frac{4\frac{4x-5}{3x-4}-5}{3\frac{4x-5}{3x-4}-4}\\~~~~~~~~~~~~~~~~~= \frac{4(4x-5)-5(3x-4)}{3(4x-5)-4(3x-4)}\\~~~~~~~~~~~~~~~~~=\frac{16x-20-15x+20}{12x-15-12x+16}=x$ 

11. If $\,\,f(x)=(a-x^n)^{1/n},\,$ where $a>0\,\,$ and $\,n\,$ is a positive integer, show that $\,\,f(f(x))=x$

Sol. We have, $\,\,f(f(x))\\=f((a-x^n)^{1/n})\\=(a-((a-x^n)^{1/n})^n)^{1/n}\\=(a-a+x^n)^{1/n}\\=(x^n)^{1/n}\\=x$

12(i). If $\,\,\phi(x)=\frac{1-x}{1+x},\,\,$ prove that $\,\phi(\phi(x))=x$

Sol. $\,\phi(\phi(x))\\=\phi(\frac{1-x}{1+x})\\=\frac{1-\frac{1-x}{1+x}}{1+\frac{1-x}{1+x}}\\=\frac{(1+x)-(1-x)}{(1+x)+(1-x)}\\=\frac{2x}{2}\\=x$

12(ii). If $\,\,\phi(x)=\frac{1-x}{1+x},\,\,$ prove that $\,\phi(\phi(\cot \theta))=\cot\theta$

Sol. $\,\phi(\phi(\cot\theta))\\=\phi(\frac{1-\cot\theta}{1+\cot\theta})\\=\frac{1-\frac{1-\cot\theta}{1+\cot\theta}}{1+\frac{1-\cot\theta}{1+\cot\theta}}\\=\frac{(1+\cot\theta)-(1-\cot\theta)}{(1+\cot\theta)+(1-\cot\theta)}\\=\frac{2 \cot \theta}{2}=\cot\theta$

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