Question

please solve this question friends jaldi solve kar do Mera ??????

THIS IS MODULUS INEQUALITY CLASS 11TH
PLS SOLVE MY QUESTION
Q5)
ANSWERS ARE ALSO GIVEN IN THE SIDE PLS SOLVE GUYS PLEASE PLEASE PLEASE

Answer 1

6Final Answer : x < -2 or x > -1

For Modulus see pic 2 ,
One condition is solved here , other is in pic.

For Mod,
we know if
|f(x) | < 3 :
then -3 < f(x) < 3 .

We know that,
$\frac{ {x}^{2} - 3x - 1 }{ {x}^{2} + x + 1} < 3 \\ = > {x}^{2} - 3x - 1 < 3( {x}^{2} + x + 1) \\ = > {x}^{2} - 3x - 1 < 3 {x}^{2} + 3x + 3 \\ = > 2 {x}^{2} + 6x + 4 > 0 \\ = > {x}^{2} + 3x + 2 > 0 \\ = > {x}^{2} + 2x + x + 2 > 0 \\ = > (x + 2)(x + 1) > 0$

Now,
(x^2-3x -1 )/(x^2+x+1) > -3

=> (x^2-3x-1)>-3x^2-3x-3
=> 4x^2 +0x + 2 > 0
=> 4x^2 +2 > 0
This is always positive because x^2 >=0
So, this will not affect our solution.

Now, doing this on number line :
see pic.
Assigning corresponding signs in number line.
We get ,
x < -2 and x > -1

See pic for Number line.