Question

Find number of integral roots of the inequation

${x}^{6} - {x}^{5} + {x}^{4} - {x}^{2} + x - 1 < 0$


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Answer 1

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

Given inequality is

$\rm \: {x}^{6} - {x}^{5} + {x}^{4} - {x}^{2} + x - 1 < 0$

$\rm \: {x}^{5}(x - 1) + {x}^{2}({x}^{2} - 1) + (x - 1) < 0$

$\rm \: {x}^{5}(x - 1) + {x}^{2}(x - 1)(x + 1) + (x - 1) < 0$

$\rm \: (x - 1)\bigg[ {x}^{5} + {x}^{2}(x + 1) + 1\bigg] < 0$

$\rm \: (x - 1)\bigg[ {x}^{5} + {x}^{3} + {x}^{2} + 1\bigg] < 0$

$\rm \: (x - 1)\bigg[ {x}^{3}({x}^{2} + 1) + ({x}^{2} + 1)\bigg] < 0$

$\rm \: (x - 1)( {x}^{2} + 1)( {x}^{3} + 1) < 0$

$\rm \: (x - 1)( {x}^{2} + 1)(x+ 1)( {x}^{2} - x + 1) < 0$

Now,

$\boxed{ \rm{ \: {x}^{2} + 1 > 0 \: \forall \: x \: \in \: R}} - - - (1)$

and

$\boxed{ \rm{ \: {x}^{2} - x + 1 > 0 \: \forall \: x \: \in \: R \: }} - - - (2)$

[ Reason :- If a quadratic expression f(x) = ax² + bx + c such that a > 0 and Discriminant, D = b² - 4ac < 0, then f(x) > 0 ]

[ For, x² - x + 1, a = 1 > 0 and Discriminant, D = 1 - 4 = - 3 < 0 ]

So, above inequality can be rewritten as

$\rm \: (x - 1)(x + 1) < 0$

$\rm\implies \: - 1 < x < 1$

$\rm\implies \:x \: \in \: ( - 1,1)$

But we have to find the integral solution.

So,

$\rm\implies \:x \: = 0 \: is \: only \: integral \: solution$

So,

$\rm\implies \:Number \: of \: integral \: solution \: = \: 1$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{ |x| < y\rm\implies \: - y < x < y}\\ \\ \bigstar \: \bf{ |x| \leqslant y\rm\implies \: - y \leqslant x \leqslant y}\\ \\ \bigstar \: \bf{ |x| > y\rm\implies \: x < - y \: or \: x > y} \: \\ \\ \bigstar \: \bf{ |x| \geqslant y\rm\implies \:x \leqslant - y \: or \: x \geqslant y}\\ \\ \bigstar \: \bf{ |x - a| < y\rm\implies \:a - y < x < a + y}\\ \\ \bigstar \: \bf{ |x - a| \leqslant y\rm\implies \:a - y \leqslant x \leqslant a + y}\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}$