Differentiation (Part-38) | S N De

 

Differentiation (Part-31) | S N De

5. If the function $~~f(x)=x^3+ax^2-bx+4~$ defined in $~~-2 \leq x \leq 2~~$ satisfies Rolle's theorem when $~-2 <c<2~$ where $~c=\frac 13(1+\sqrt{13})~$, then find the values of $~a~$ and $~b.$

Sol.  We know if  $~x~$ is a real valued variable defined in the closed interval $a \leq x \leq b,~$ then $~f(x)~$ is 

$~(i)~$ continuous in $~a \leq x \leq b,~~(ii)~~$ differentiable in $~a<x<b~~(iii)~~f(a)=f(b),~~$ then there exists a point $~c \in (a ,b)~$ such that $~f'(c)=0.$

Now, from the context of the given problem we can say  since $~x~$ is a real valued variable defined in the closed interval $-2 \leq x \leq 2,~$ then $~f(x)~$ is 

$~(i)~$ continuous in $~ -2 \leq x \leq 2,~~(ii)~~$ differentiable in $~ -2<x<2~~(iii)~~f(2)=f(-2),~~$ then there exists a point $~c \in (-2 ,2)~$ such that $~f'(c)=0.$

Now, $~~f(x)=x^3+ax^2-bx+4 \\ \text{or,}~~f'(x)=3x^2+2ax-b \\ \therefore~f'(c)=3c^2+2ac-b \\ \Rightarrow \left[f'(c)\right]_{c=\frac{1+\sqrt{13}}{3}}=3 \times \frac 19(1+\sqrt{13})^2\\~~~+\frac{2a}{3}(1+\sqrt{13})-b \\ \text{or,}~~0=\frac 13(1+13+2\sqrt{13})\\~~~+\frac{2a}{3}(1+\sqrt{13})-b \\ \text{or,}~~b=\frac{14+2\sqrt{13}}{3}+\frac{2a}{3}(1+\sqrt{13})\rightarrow(1)$

Now, $~~f(2)=f(-2) \\ \text{or,}~~2^3+a \times 2^2-b \times 2+4\\~~~=(-2)^3+a \times (-3)^2-b \times (-2)+4 \\ \text{or,}~~ 8+4a-2b=-8+4a+2b \\ \text{or,}~~ 8+8=2b+2b \\ \text{or,}~~4b=16 \\ \text{or,}~~ b=\frac{16}{4}=4 \rightarrow(2) $

Hence, by $~(1)~$ and $~(2)~$ we get,

$~~4=\frac{14+2\sqrt{13}}{3}+\frac{2a}{3}(1+\sqrt{13}) \\ \text{or,}~~4-\left(\frac{14+2\sqrt{13}}{3}\right)=\frac{2a}{3}(1+\sqrt{13}) \\ \text{or,}~~ \frac{12-14-2\sqrt{13}}{3}=\frac{2a}{3}(1+\sqrt{13}) \\ \text{or,}~~ -2-2\sqrt{13}=2a(1+\sqrt{13}) \\ \text{or,}~~ -2(1+\sqrt{13})=2a(1+\sqrt{13}) \\ \therefore a=-1 \rightarrow(3)$

Hence, from $\,(2)\,$ and $\,(3)\,$, we can get, $~~a=-1,~~b=4.$

Senior Secondary School Mathematics for Class 12 Paperback

6. Using Rolle's theorem, find the point on the curve 

$~(a)~~y=x(x-4),~~x\in[0,4]\\~(b)~f(x)=(1-\cos x)~\text{in}~ 0\leq x \leq 2\pi~$

where the tangent is parallel to $~x-$axis.

Sol.$~(a)~$ We have , $~f(x)=y=x(x-4)=x^2-4x.$

Now, $~(i)~f(x)~$ is continuous on $~[0,4],~~~(ii)~f'(x)=2x-4~$ exist $~~\forall x\in(0,4)~~(iii)~f(0)=0=f(4).$

Hence, there exists a point $~~c \in(0,4)~$ such that 

$~~f'(c)=0. \\ \therefore~2c-4=0 \\ \text{or,}~~ 2c=4 \\ \text{or,}~~c=\frac 42=2.$

At $~c=2,\\~~f(c)\\=f(2)\\=2^2-4\times 2\\=4-8\\=-4.$

So, the point on the curve where the tangent is parallel to $~x-$axis is $~~(2,-4).$

Sol. $~(b)~$ We have , $~f(x)=y=(1-\cos x),~~\text{in}~~0\leq x \leq 2\pi$

Now, $~(i)~f(x)~$ is continuous on $~[0,2\pi],~~~(ii)~f'(x)=\sin x~$ exists $~~\forall x\in(0,2\pi)~~(iii)~f(0)=0=f(2\pi).$

Hence, by Rolle's theorem, there exists a point $~~c \in(0,2\pi)~$ such that 

$~~f'(c)=0. \\ \therefore~\sin c=0=\sin\pi \\ \text{or,}~~ c=\pi~~[\because c \in (0,2\pi)].$

At $~c=\pi,\\~~f(c)\\=f(\pi)\\=1-\cos\pi\\=1-(-1)\\=1+1\\=2.$

So, the point on the curve where the tangent is parallel to $~x-$axis is $~~(\pi,2).$

7. At what point is the tangent to the curve $~~f(x)=\log x~$ parallel to the chord joining the points $~A(1,0)~$ and $~B(e,1)~ ?$

Sol. We have, $~f(x)=\log x \\ \therefore~f'(x)=\frac 1x\rightarrow(1)$

Now, the slope of the chord joining the points $~A(1,0)~$ and $~B(e,1)~$ is given by $~\frac{1-0}{e-1}=\frac{1}{e-1}.$

Let at some point with abscissa $\,x=c\,$,  the tangent to the curve $~~f(x)=\log x~$ parallel to the given chord, we have $~~\frac 1c=\frac{1}{e-1}~~[\text{By (1)}] \\ \therefore c=e-1.$ 

Hence, the tangent to the given curve will be parallel to the given chord at the point with abscissa $~c=e-1.$

8. In the mean value theorem, $~f(b)-f(a)=(b-a)f'(c)~, (a<c<b);~$ if $~~f(x)=x^3-3x-1~$ and $~a=-\frac{11}{7},~b=\frac{13}{7},~$ find the value of $~c.$

Sol.  At first we calculate 

$~~~~~f(a)\\=f(-11/7)\\=(-\frac{11}{7})^3-3\times (-\frac{11}{7})-1\\=-\frac{1331}{343}+\frac{33}{7}-1\\=\frac{-1331+1617-343}{343}\\=\frac{-57}{343}\rightarrow(1)$

Again, $~~~f(b)\\=f(13/7)\\=(\frac{13}{7})^3-3 \times (\frac{13}{7})-1\\=\frac{2197}{343}-\frac{39}{7}-1\\=\frac{2197-1911-343}{343}\\=\frac{-57}{343}\rightarrow(2)$

Now, $~f(x)=x^3-3x-1 \\ \therefore~f'(x)=3x^2-3=3(x^2-1) \\ \text{so that}~~f'(c)=3(c^2-1)\rightarrow(3)$

Hence, from $~(1),~(2),~(3)~$ we get,

$~~~f(b)-f(a)=(b-a)f'(c) \\ \text{or,}~~f(13/7)-f(-11/7)\\~~~~~~=\left[\frac{13}{7}-(-\frac{11}{7})\right]f'(c) \\ \text{or,}~~ \frac{-57}{343}-\left(\frac{-57}{343}\right) =(\frac{13}{7}+\frac{11}{7})f'(c)\\ \text{or,}~~ 0=f'(c) \\ \text{or,}~~ 3(c^2-1)=0 \\ \text{or,}~~c^2-1=0  \\ \text{or,}~~ c^2=1  \\ \text{or,}~~ c=\pm\sqrt{1}=\pm 1~~\text{(ans.)} $