Question

Discuss continuity of function f(x) at x=1 where

$\text{f(x)} = \begin{cases} \dfrac{x-1}{e^{\left(\dfrac{1}{x-1}\right)}+1}} &, \text{if}\: x \neq 1\\\\0 &, \text{if}\: x = 1\end$

Answer 1

Answer:

$discuss \: the \: continuity \: of \: the \: function \: f(x) \: defined \: by :$

$fx = \frac{log(sec {}^{2}x) }{x \: sinx}$

$for \: x \: ≠ \: 0$

$= e \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: for \: x \: = 0$

$at \: x \: = 0$

$if \: the \: discontinuity \: is \: removable \: redifine \: the \: function$

$so , \: that \: it \: becomes \: continous$

Answer 2

Answer:

$\sf{\underline{\boxed{\green{\large{\bold{ Question}}}}}}$

$solve : \sf\sqrt { \dfrac{x}{x-1}} + \sqrt { \dfrac{ 1- x }{ x }} =2 \dfrac{1}{6}$

⠀⠀

$\sf{\underline{\boxed{\green{\large{\bold{ Solution}}}}}} </p><p>$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\sf\implies let \: \sqrt{ \dfrac{x}{x-1}}</p><p>$

$\sf\implies and\: \sqrt{\dfrac{ 1 - x} {x }} = \dfrac{1}{y}</p><p>$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Given equation reduces to

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\sf\implies y + \dfrac{1}{y} = \dfrac{13}{6}$

$\sf\implies \dfrac{ y^2 + 1 } { y } = \dfrac{13}{6}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\sf\implies 6 ( y^2 + 1 ) = 13y$

$\sf\implies 6y^2 + 6 =$

$\sf\implies 6y^2 - 13y + 6 = 0$

Fctorisation by middle term split

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$</p><p>\sf\implies 6y^2 -9y - 4y + 6 = 0$

$\sf\implies 3y ( 2y - 3 ) -2 ( 2y - 3 ) =$

$\sf\implies ( 3y - 2 ) ( 2y - 3 ) = 0</p><p>⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

\sf\implies 3y - 2 = 0⟹3y−2=0

\sf \implies 3y = 2⟹3y=2

\sf\implies y = \dfrac{2}{3}⟹y=

3

2

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\sf{\underline{\boxed{\blue{\large{\bold{ y = \dfrac{2}{3} }}}}}}$

$\sf\implies 2y - 3 = 0$

$\sf \implies 2y = 3</p><p>$

$\sf\implies y = \dfrac{3}{2}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\sf{\underline{\boxed{\blue{\large{\bold{ y = \dfrac{3}{2} }}}}}} </p><p>⠀⠀⠀⠀⠀⠀$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\sf{\underline{\boxed{\purple{\large{\bold{ y = \dfrac{3}{2} or \dfrac{2}{3} }}}}}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

_________________________

$\sf\implies when\: y = \dfrac{3}{2}$

$\sf\implies then \sqrt{\dfrac{x}{1 - x }} = \dfrac{3}{2}$

Squaring both the sides

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\sf\implies \dfrac{x}{1 - x } = \dfrac{9}{4}$

$\sf\implies 4x = 9 - 9x$

$\sf\implies 9x + 4x = 9$

$\sf\implies 13x = 9$

$\sf\implies x =\dfrac{9}{13}$

$</p><p>\sf{\underline{\boxed{\blue{\large{\bold{ x = \dfrac{9}{13 } }}}}}} </p><p>⠀⠀⠀$

$\sf\implies when\: y = \dfrac{2}{3}$

$\sf\implies then \sqrt{\dfrac{x}{1-x }} = \dfrac{2}{3}</p><p>$

Squaring both the sides ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\sf\implies \dfrac{x}{1 - x } = \dfrac{4}{9}$

$\sf\implies 9x = 4 - 4x$

$\sf\implies 9x + 4x$

$</p><p>\sf\implies 13x$

$\sf\implies x =\dfrac{4}{13}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\sf{\underline{\boxed{\blue{\large{\bold{ x = \dfrac{4}{13 } }}}}}}$

$\sf{\underline{\boxed{\purple{\large{\bold{ x = \dfrac{9}{13} or \dfrac{4}{13} }}}}}} </p><p>$