Question

Using Rolles theorem, prove that there is at least one real root of the equation 51x^101-2323x^100-45x+1035=0 in (45^1/100, 46)​

Answer 1

$\large\underline{\sf{Solution-}}$

Given equation is

$\rm :\longmapsto\: {51x}^{101} - {2323x}^{100} - 45x + 1035 = 0$

To show that the above equation have at least one real root in the given interval, we first assume the polynomial which is integration of above equation, i.e.

Let assume that

$\rm :\longmapsto\: f(x) = \displaystyle\int\rm ({51x}^{101} - {2323x}^{100} - 45x + 1035)dx$

$\rm :\longmapsto\:f(x) = \dfrac{ {x}^{102} }{2} - \dfrac{2323 {x}^{101} }{101} - \dfrac{45 {x}^{2} }{2} +1035x + c$

and the interval is considered as

$\rm :\longmapsto\:[ {45}^{ \frac{1}{100} } , \: 46]$

Now,

Step :- 1 Every polynomial function is always continuous.

$\purple{\rm\implies \:f(x) \: is \: continuous \: on \: [ {45}^{ \frac{1}{100} } , \: 46]}$

Step :- 2

Consider,

$\rm :\longmapsto\:f(x) = \dfrac{ {x}^{102} }{2} - \dfrac{2323 {x}^{101} }{101} - \dfrac{45 {x}^{2} }{2} +1035x + c$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:f'(x) = {51x}^{101} - {2323x}^{100} - 45x + 1035$

Thus, f'(x) is a polynomial function.

$\purple{\rm\implies \:f(x) \: is \: differentiable \: on \: ({45}^{ \frac{1}{100} } , \: 46)}$

Step :- 3

$\rm :\longmapsto\:f(x) = \dfrac{ {x}^{102} }{2} - \dfrac{2323 {x}^{101} }{101} - \dfrac{45 {x}^{2} }{2} +1035x + c$

So,

$\rm :\longmapsto\:f(46) = \dfrac{ {46}^{102} }{2} - \dfrac{2323 {(46)}^{101} }{101} - \dfrac{45 {(46)}^{2} }{2} +1035(46) + c$

$\rm :\longmapsto\:f(46) = \dfrac{46 \times {46}^{101} }{2} - 23 {(46)}^{101} - 45 \times 23 \times 46 +1035(46) + c$

$\rm :\longmapsto\:f(46) =23 {(46)}^{101} - 23 {(46)}^{101} - 45.23.46 +1035(46) + c$

$\rm :\longmapsto\:f(46) = 1035 \times 46 - 1035(46) + c$

$\rm :\longmapsto\:f(46) = c$

Now, Consider

$\rm :\longmapsto\:f({45}^{ \frac{1}{100} }) = \dfrac{( {{45}^{ \frac{1}{100} })}^{102} }{2} - \dfrac{2323 {({45}^{ \frac{1}{100} })}^{101} }{101} - \dfrac{45 {({45}^{ \frac{1}{100} })}^{2} }{2} + 1035({45}^{ \frac{1}{100} }) + c$

$\rm :\longmapsto\:f({45}^{ \frac{1}{100} }) = \dfrac{( {{45}^{ \frac{51}{50} })}}{2} - 23({45}^{ \frac{101}{100} }) - \dfrac{ {({45}^{ \frac{51}{50} })}}{2} + 1035({45}^{ \frac{1}{100} }) + c$

$\rm :\longmapsto\:f({45}^{ \frac{1}{100} }) = c$

$\purple{\rm\implies \:f({45}^{ \frac{1}{100} }) = f(46) = c}$

So, Rolle's Theorem is applicable, so there exist atleast one real number c such that

$\rm :\longmapsto\:f'(x) = 0 \: on \: x \in \: ({45}^{ \frac{1}{100} }, \: 46)$

$\rm\implies \: {51x}^{101} - {2323x}^{100} - 45x + 1035 = 0 \: in \: ({45}^{ \frac{1}{100} },46)$

Hence, Proved