Differentiation (Part-38) | S N De

 

$\,8(i)\,$ If $\,y\,$ is the H.M. between $\,x\,$ and $\,z\,$, prove that 

$\,\frac{1}{y-x}+\frac{1}{y-z}=\frac 1x+\frac 1z.$

Sol. Since $\,y\,$ is the H.M. between $\,x\,$ and $\,z\,$,

$\,y=\frac{2xz}{x+z}.$

Now, $\,y-x=\frac{2zx}{x+z}-x=x\left(\frac{2z}{x+z}-1\right)=x\left(\frac{2z-x-z}{x+z}\right)\\ \Rightarrow y-x=x\left(\frac{z-x}{x+z}\right)\rightarrow(1)$

Similarly, $\,y-z=\frac{2zx}{x+z}-z=z\left(\frac{2x}{x+z}-1\right)=z\left(\frac{2x-x-z}{x+z}\right)\\ \Rightarrow y-x=z\left(\frac{x-z}{x+z}\right)\rightarrow(2)$

Hence, $\,\frac{1}{y-x}+\frac{1}{y-z}\\=\frac{x+z}{z-x}\left(\frac 1x-\frac 1z\right)\\=\frac{x+z}{z-x}\cdot \frac{z-x}{zx}\\=\frac{x+z}{xz}\\=\frac 1x+\frac 1z\,\,\text{(proved)}$

$\,8(ii)\,$ The harmonic mean of two numbers is $\,4,\,$ their arithmetic mean $\,'A'\,$ and the geometric mean $\,G\,$ satisfy the relation $\,2A+G^2=27.\,$ Find the numbers.

Sol. Let $\,x,\,y\,$ be two numbers  so that harmonic mean of these two numbers  is :

$\,\frac{2xy}{x+y}=4\,\,\text{(given)}\\ \Rightarrow 2xy=4(x+y)\rightarrow(1)$

Again, $\,A=\frac{x+y}{2}\rightarrow(2),\\ ~G=\sqrt{xy}\rightarrow(3)$

So, $\,2A+G^2=27 \\ \Rightarrow (x+y)+xy=27\,\,\text{[By (2),(3)]}\\ \Rightarrow 4(x+y)+4xy=27 \times 4\\ \Rightarrow 2xy+4xy=27\times 4\,\,[\text{By (1)}]\\ \Rightarrow 6xy=27 \times 4 \\ \Rightarrow xy=\frac{27 \times 4}{6}=18 \rightarrow(4)$

Hence from $\,(1),(4)\,$ we get,

$\,2 \times 18=4(x+y)\\ \Rightarrow x+y=\frac{36}{4}=9\rightarrow(5)$

Now, from $\,(5),(4)\,$ we get,

$\,x+\frac{18}{x}=9 \\ \Rightarrow x^2+18=9x  \\ \Rightarrow x^2-9x+18=0  \\ \Rightarrow x^2-6x-3x+18=0  \\ \Rightarrow x(x-6)-3(x-6)=0  \\ \Rightarrow (x-6)(x-3)=0  \\ \Rightarrow x=6,3.$

When $\,x=6,\,y=9-6=3\,[\text{By (5)}]$ and 

When $\,x=3,\,y=9-3=6\,[\text{By (5)}]$.

$\,\therefore \,$ The numbers are $\,6,\,3.$

$\,9.\,$ If the $\,p\,$th, $\,q\,$th and the $\,r\,$th terms of a H.P. be $\,x,y,z\,$ respectively, prove that, 

$\,yz(q-r)+zx(r-p)+xy(p-q)=0.$

Sol.  If the $\,p\,$th, $\,q\,$th and the $\,r\,$th terms of a H.P. be $\,x,y,z\,$ respectively, then

 the $\,p\,$th, $\,q\,$th and the $\,r\,$th terms of the corresponding A.P. be $\,\frac 1x,\frac 1y,\frac 1z\,$ .

Suppose that $\,a\,$ is the first term , $\,d\,$ is the common difference of that A.P. and $\,t_n\,$ is the $\,n\,$ th term of the A.P.

So, $\,t_p=\frac 1x\,\,\text{(Given)}\\ \Rightarrow a+(p-1)d=\frac 1x \rightarrow(1)$

Similarly, $\,t_q=\frac 1y \\ \Rightarrow a+(q-1)d=\frac 1y\rightarrow(2)$

and $\,t_r=\frac 1z \\ \Rightarrow a+(r-1)d=\frac 1z\rightarrow(3)$

Now, subtracting $\,(2),\,$ from $\,(1),\,$ we get,

$\,(p-q)d=\frac 1x-\frac 1y =\frac{y-x}{yx}\\ \Rightarrow d=\frac{y-x}{yx(p-q)}$

Again, subtracting $\,(3),\,$ from $\,(2),\,$ we get,

$\,(q-r)d=\frac 1y-\frac 1z =\frac{z-y}{zy}\\ \Rightarrow d=\frac{z-y}{zy(q-r)}$

Finally,  subtracting $\,(1),\,$ from $\,(3),\,$ we get,

$\,(r-p)d=\frac 1z-\frac 1x =\frac{x-z}{xz}\\ \Rightarrow d=\frac{x-z}{zx(r-p)}$

So, $\,d=\frac{y-x}{yx(p-q)}=\frac{z-y}{zy(q-r)}=\frac{x-z}{zx(r-p)}\\~~~~=\frac{(y-x)+(z-y)+(x-z)}{yx(p-q)+zy(q-r)+zx(r-p)}\,\,[*]\\~~~~=\frac{0}{yx(p-q)+zy(q-r)+zx(r-p)}\\ \Rightarrow d[yx(p-q)+zy(q-r)+zx(r-p)]=0 \\ \Rightarrow yz(q-r)+zx(r-p)+xy(p-q)=0\,\,\text{(proved)}$

Note[*] : We know if $\,x=\frac ab=\frac cd=\frac ef \\ \text{then}\,\,x=\frac{a+c+e}{b+d+f}.$

$\,10.\,$ A man travels a certain distance with a uniform speed of $\,5\,$ km/h and returns the same distance with a uniform speed of $\,4\,$ km/h. What is the average speed of the man?

Sol. Let the total distance $=x\,$ km. So, the time taken (when speed is $\,5\,$ km/h) $=\frac x5\,$ hrs. the time taken (when speed is $\,4\,$ km/h) $=\frac x4\,$ hrs.

Hence, total time taken $=\left(\frac x5+\frac x4\right)=\frac{9x}{20}\,$ hrs.

$\,\therefore \,$ Avg. speed $=\frac{\text{Total Distance}}{\text{Total time}}\\=\frac{x+x}{\frac{9x}{20}}\\=\frac{2x}{9x/20}\\=\frac{40}{9}\\=4\frac 49\,\, \text{km/h}\,\,\text{(ans.)}$