Differentiation (Part-38) | S N De

 

$~9.~$ In the mean value theorem, $~f(a+h)=f(a)+hf'(a+\theta h)~~(0<\theta<1),~$ find $~\theta~$ when $~f(x)=\sqrt{x},~a=1~$ and $~h=3.$

Sol. We have, $~~f(x)=\sqrt x=x^{1/2}\rightarrow(1) \\ \therefore~f'(x)\\=\frac 12~\cdot x^{\frac 12-1}\\=\frac{1}{2\sqrt{x}}\rightarrow(2)$

Now, $~f(a+h)=f(a)+hf'(a+\theta h) \\ \text{or,}~~\sqrt{a+h}=\sqrt{a}+h \cdot \frac{1}{2\sqrt{a+\theta h}}~~[\text{By (1),(2)}] \\ \therefore~ \sqrt{1+3}=\sqrt{1}+3~\cdot \frac{1}{2\sqrt{1+\theta \times 3}} \\ \text{or,}~~\sqrt{4}=1+\frac{3}{2\sqrt{1+3\theta}}  \\ \text{or,}~~ 2-1=\frac 32 \cdot \frac{1}{\sqrt{1+3\theta}} \\ \text{or,}~~ 1 \times \frac 23= \frac{1}{\sqrt{1+3\theta}}  \\ \text{or,}~~(2/3)^2=\left(\frac{1}{\sqrt{1+3\theta}}\right)^2  \\ \text{or,}~~ \frac 49=\frac{1}{1+3\theta}\\ \text{or,}~~4(1+3\theta)=9 \\ \text{or,}~~ 4+12 \theta=9 \\ \text{or,}~~ 12\theta=9-4 \\ \text{or,}~~ \theta=\frac{5}{12}~~\text{(ans.)}$

$~10.~$ In the mean value theorem, $~~f(x+h)=f(x)+hf'(x+\theta h)~~(0<\theta<1),~~$ if $~~f(x)=e^x,~~$ then show that the value of $~\theta~$ is independent of $~x.$

Sol. We have, $~~f(x)=e^x \rightarrow(1) \\ \therefore~f'(x)=e^x\rightarrow(2)$

Now, $~~f(x+h)=f(x)+hf'(x+\theta h) \\ \text{or,}~~e^{x+h}=e^x+h~\cdot e^{x+\theta h}~~[\text{By (1),(2)}] \\ \text{or,}~~ e^x \cdot e^h-e^x=he^x \cdot e^{\theta h} \\ \text{or,}~~ e^x(e^h-1)=e^x \cdot he^{\theta h} \\ \text{or,}~~ e^h-1=he^{\theta h} \\ \text{or,}~~ e^{\theta h}=\frac{e^h-1}{h} \\ \therefore~~ \log_e(e^{\theta h})=\log_e \left(\frac{e^h-1}{h}\right) \\ \text{or,}~~ \theta h=\log_e \left(\frac{e^h-1}{h}\right) \\ \text{or,}~~ \theta=\frac 1h \log_e \left(\frac{e^h-1}{h}\right)\rightarrow(3)$

Hence, from $\,(3)\,$ it follows that the value of $~\theta~$ is independent of $~x.$

$~11.~$ Using Lagrange's mean value theorem prove that, 

$~(i)~~e^x>1+x .$

Sol. Let  $~~f(x)=\log_e(1+x) \rightarrow(1)\\ \therefore~f'(x)=\frac{1}{1+x}\rightarrow(2)$

Now, according to Lagrange's mean value theorem, 

$~~f(a+h)=f(a)+hf'(a+\theta h)\rightarrow(3)\\~~~~~ \text{where}~~[~0<\theta<1] $

Now, we put $~~a=0,~h=x~$ in $~(3)~$ so that we get,

$~~f(0+x)=f(0)+xf'(0+\theta x) \\ \text{or,}~~f(x)=f(0)+xf'(\theta x) \\ \text{or,}~~\log_e(1+x)=\log_e(1+0)\\~~~+x~\cdot \frac{1}{1+\theta x}~~[\text{By (1),(2)}] \\ \text{or,}~~ \log_e(1+x)=0+x \cdot \frac{1}{1+\theta x} \\ \therefore~ \log_e(1+x)<x \cdot 1=x ~~[\because ~\frac{1}{1+\theta x}<1~]\\ \text{so,}~~~1+x <e^x \\ \therefore~e^x> 1+x~~\text{(proved)}$

$~11.~$ Using Lagrange's mean value theorem prove that, 

$~(ii)~~\log(1+x)<x~~~~(\text{when}~x>0).$

Sol. Let  $~~f(x)=\log_e(1+x) \rightarrow(1)\\ \therefore~f'(x)=\frac{1}{1+x}\rightarrow(2)$

Clearly, $~f(x)~$ is continuous in $~[0,x]~,$ differentiable in $~(0,x)~.$ So, according to Lagrange's mean value theorem,  there exists at least one value of $~x~$ say $~x=c \in (0,x)~$ such that 

$~~f'(c)=\frac{f(x)-f(0)}{x-0} \\ \text{or,}~~ f'(c)=\frac{f(x)-0}{x}\\~~[\because~f(0)=\log_e(1+0)=0] \\ \therefore ~\frac{1}{1+c}=\frac{\log_e(1+x)}{x} \rightarrow(3)~~~~[\text{By (1),(2)}] $

Now, $~0<c<x \\ \text{so,}~~c>0 \\ \therefore c+1>0+1=1 \\ \Rightarrow \frac{1}{1+c}<1\rightarrow(4)$

Hence, from $\,(3)\,$ and $\,(4)\,$ we get,

$~~\frac{\log_e(1+x)}{x}=\frac{1}{1+c}<1 \\ \therefore~\log_e(1+x)<x~~~~(\text{when}~x>0)~~\text{(proved)}$

$~11.~$ Using Lagrange's mean value theorem prove that, 

$~(iii)~~b^n-a^n<nb^{n-1}(b-a)~~\text{when}~~b>a.$

Sol. Sol. Let  $~~f(x)=x^n~ \rightarrow(1)~~~(a \leq x \leq b)\\ \therefore~f'(x)=nx^{n-1}\rightarrow(2)$

Clearly, $~f(x)~$ is continuous in $~[a,b]~,$ differentiable in $~(a,b)~.$ So, according to Lagrange's mean value theorem,  there exists at least one value of $~x~$ say $~x=c \in (a,b)~$ such that 

$~~f'(c)=\frac{f(b)-f(a)}{b-a} \\ \text{or,}~~ nc^{n-1}=\frac{b^n-a^n}{b-a}~~[\text{By (1),(2)}]\\ \therefore (b^n-a^n)=nc^{n-1}(b-a) \rightarrow(3)$

Now, $~~\because a<c<b \\ \therefore c<b \\ \text{so,}~~c^{n-1}<b^{n-1} \rightarrow(4)$

So, from $~(3),(4)~$ we get,

$~~b^n-a^n<nb^{n-1}(b-a)~~\text{when}~~b>a~~\text{(proved)}$

$~11.~$ Using Lagrange's mean value theorem prove that, 

$~(iv)~~(b-a)\sec^2a<\tan b-\tan b\\~~~~<(b-a)\sec^2b~~\text{when}~ ~0<a<b<\pi/2.$

Sol. Let  $~~f(x)=\tan x~ \rightarrow(1)~~~[~x \in (0,\pi/2)]\\ \therefore~f'(x)=\sec^2x\rightarrow(2)$

Clearly, $~f(x)~$ is continuous and  differentiable in $~(0,\pi/2)~.$ So, according to Lagrange's mean value theorem,  there exists at least one value of $~x~$ say $~x=c \in (0, \pi/2)~$ such that 

$~~f'(c)=\frac{f(b)-f(a)}{b-a} \\ \text{or,}~~ \sec^2c=\frac{\tan b-\tan a}{b-a}~\rightarrow(3)\\ ~~~~~~~~~~[\text{By (1),(2)}] $

Now, $~~\because a<c<b \\ \therefore \sec^2a<\sec^2c<\sec^2b\rightarrow(4)$

So, from $~(3),(4)~$ we get,

$~~\sec^2a<\frac{\tan b-\tan a}{b-a}<\sec^2b \\ \therefore~(b-a)\sec^2a<\tan b-\tan a \\~~~<(b-a)\sec^2b~\\~~~\text{when}~~0<a<b<\pi/2~~\text{ (proved)}$