Question

If sgn (x3 – x2 – 6x) = 1, then number of integral value(s) of x satisfying the equation is (if x (–3, 5))

Answer 1

Given : sgn(x³ - x² - 6x) = 1  , x ∈ (–3, 5)  

To find : number of integral value(s) of x satisfying the equation

Step-by-step explanation:

The sign of a real number, also called sgn or signum, is -1 for a negative numbers (i.e., one with a minus sign "), 0 for the number zero, or +1 for a positive numbers (i.e., one with a plus sign").

sgn(x) = 1  if  x > 0

              = 0  if x  = 0

            -1   if x < 0

x³ - x² - 6x

= x(x² - x  - 6)

= x(x - 3)(x + 2)

x  = 0 , 3  , - 2   => x³ - x² - 6x = 0

=> sgn(x³ - x² - 6x) = 0  for  x = 0 , 3  , - 2

x ∈ (–3, 5)  

x = - 1  =>  x(x - 3)(x + 2)  =  4  > 0 => sgn(x³ - x² - 6x) = 1  for x = -1

x = 1  =>   x(x - 3)(x + 2)  =  -6  < 0 => sgn(x³ - x² - 6x) = -1  for x = 1

x = 2  =>   x(x - 3)(x + 2)  =  -4  < 0 => sgn(x³ - x² - 6x) = -1  for x = 2

x = 4  =>  x(x - 3)(x + 2)  =  24  > 0 => sgn(x³ - x² - 6x) = 1  for x = 4

x = - 1  & x  = 4

are integral values satisfying   sgn(x³ - x² - 6x) = 1

two integral values satisfying the equation

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Answer 2

If sgn (x³ – x² – 6x) = 1, then number of integral value(s) of x satisfying the equation is 2

Step-by-step explanation:

We know that

$\sgn f(x)=\frac{|f(x)|}{f(x)}$

And

$|f(x)|=f(x)$ for all x for which $f(x)\ge0$

$|f(x)|=-f(x)$ for all x for which $f(x)<0$

Here

$f(x)=x^3-x^2-6x$

$\implies f(x)=x(x^2-x-6)$

$\implies f(x)=x(x^2-3x+2x-6)$

$\implies f(x)=x[x(x-3)+2(x-3)]$

$\implies f(x)=x(x+2)(x-3)$

Now,

$f(x)<0$

$\implies x(x+2)(x-3)<0$

$\implies x\in (-\infty,-2)\cup(0,3)$

And

$f(x)\ge0$

$\implies x(x+2)(x-3)\ge0$

$\implies x\in [-2,0]\cup[3,\infty)$

Now given in the question that

$x\in (-3,5)$

i.e. $x^3-x^2-6x$ will be negative for $(-3,-2)\cup(0,3)$ and positive or equal to zero for $[-2,0]\cup[3,5)$

$f(x)$ will be strictly positive for $(-2,0)\cup(3,5)$

Now,

$\text{sgn} (x^3-x^2-6x)=\frac{|x^3-x^2-6x|}{x^3-x^2-6x}$

$\text{sgn} (x^3-x^2-6x)$ will be 1 only when $|x^3-x^2-6x|=x^3-x^2-6x$

Which will be possible for all x where $x^3-x^2-6x>0$ which is for $x\in(-2,0)\cup(3,5)$

Therefore, the integral values of x will be -1 and 4

The number of integral values  = 2

Hope this answer is helpful.

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