Question

For how many integral values of x inequality,
3x^2 - 7x+8/x^2+1
≤2 holds true?
​

Answer 1

Step-by-step explanation:

We have,

$\frac{3 {x}^{2} - 7x + 8 }{ {x}^{2} + 1 } \leqslant 2$

$\implies\frac{3 {x}^{2} - 7x + 8 }{ {x}^{2} + 1 } - 2 \leqslant 0$

$\implies\frac{3 {x}^{2} - 7x + 8 - 2( {x }^{2} + 1) }{ {x}^{2} + 1 } \leqslant 0$

$\implies\frac{3 {x}^{2} - 7x + 8 - 2 {x }^{2} - 2 }{ {x}^{2} + 1 } \leqslant 0$

$\implies\frac{ {x}^{2} - 7x + 6 }{ {x}^{2} + 1 } \leqslant 0$

$\implies\frac{ {x}^{2} - 6x - x+ 6 }{ {x}^{2} + 1 } \leqslant 0$

$\implies\frac{ x(x - 6) -1 (x - 6 ) }{ {x}^{2} + 1 } \leqslant 0$

$\implies\frac{( x - 1)(x - 6) }{ {x}^{2} + 1 } \leqslant 0$

$\implies \: x \in \: [1 , 6]$

So, integral solutions are

$\tt \purple{x \in \{1 ,2 ,3 ,4,5 ,6 \}}$

Answer 2

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

Given inequality is

$\rm :\longmapsto\:\dfrac{ 3{x}^{2} - 7x + 8 }{ {x}^{2} + 1} \leqslant 2$

$\rm :\longmapsto\:\dfrac{ 3{x}^{2} - 7x + 8 }{ {x}^{2} + 1} - 2 \leqslant 0$

$\rm :\longmapsto\:\dfrac{ 3{x}^{2} - 7x + 8 - 2( {x}^{2} + 1)}{ {x}^{2} + 1} \leqslant 0$

$\rm :\longmapsto\:\dfrac{ 3{x}^{2} - 7x + 8 - 2{x}^{2} - 2}{ {x}^{2} + 1} \leqslant 0$

$\rm :\longmapsto\:\dfrac{ {x}^{2} - 7x + 6}{ {x}^{2} + 1} \leqslant 0$

As,

$\rm :\longmapsto\: {x}^{2} + 1 > 0 \: \forall \: x \: \in \: Z$

So,

$\rm :\longmapsto\: {x}^{2} - 7x + 6 \leqslant 0$

$\rm :\longmapsto\: {x}^{2} - 6x - x+ 6 \leqslant 0$

$\rm :\longmapsto\:x(x - 6) - 1(x - 6) \leqslant 0$

$\rm :\longmapsto\:(x - 6)(x - 1) \leqslant 0$

$\bf\implies \:1 \leqslant x \leqslant 6$

As, it is given that x can assume only integral values,

So,

$\bf\implies \:x = \{1,2,3,4,5,6 \}$

$\bf\implies \:x \: can \: assume \: 6 \: integral \: values$

Additional Information :-

If a < b, then

$\red{ \boxed{ \sf{ \:(x - a)(x - b) < 0 \: \implies \: a < x < b}}}$

$\red{ \boxed{ \sf{ \:(x - a)(x - b) \leqslant 0 \: \implies \: a \leqslant x \leqslant b}}}$

$\red{ \boxed{ \sf{ \:(x - a)(x - b) > 0 \: \implies \: x < a \: \: or \: \: x > b}}}$

$\red{ \boxed{ \sf{ \:(x - a)(x - b) \geqslant 0 \: \implies \: x \leqslant a \: \: or \: \: x \geqslant b}}}$