Question

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

Answer 1

Answer:

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

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