Question

Find the smallest positive integral value satisfying :-

\frac{(x - 1)( {x }^{2} - 9)}{x(x + 2)} < 0x(x+2)(x−1)(x2−9)​<0
​

Answer 1

we have to find the smallest positive integral value satisfying

$\frac{(x-1)(x^2-9)}{x(x+2)}$ < 0

solution : let f(x) = (x - 1)(x² - 9)/x(x + 2) < 0

⇒{(x - 1)(x - 3)(x + 3)}/{x(x + 2) < 0

here (x - 1) = 0 ⇒x = 1

x - 3 = 0 ⇒x = 3

x + 3 = 0⇒x = -3

x = 0

x + 2 = 0 ⇒x = -2

putting -3, -2, 0, 1 and 3 in number lines as shown in figure.

case 1 : x > 3 , f(x) > 0

case 2 : 1 < x < 3 , f(x) < 0

case 3 : 0 < x < 1 , f(x) > 0

case 4 : 0 < x < -2 , f(x) < 0

case 5 : -2 < x < -3 , f(x) > 0

case 6 : x < -3 , f(x) < 0

but we have given f(x) = (x - 1)(x² - 9)/x(x + 2) < 0

so, intervals 1 < x < 3, 0 < x < -2 and x < -3 are the solutions.

we have to find smallest positive integral value of x. this can be found in the interval (1, 3) i.e., 2

Therefore the smallest positive integral value is 2.

Answer 2

Step-by-step explanation:

we have to find the smallest positive integral value satisfying

\frac{(x-1)(x^2-9)}{x(x+2)}x(x+2)(x−1)(x2−9) < 0

solution : let f(x) = (x - 1)(x² - 9)/x(x + 2) < 0

⇒{(x - 1)(x - 3)(x + 3)}/{x(x + 2) < 0

here (x - 1) = 0 ⇒x = 1

x - 3 = 0 ⇒x = 3

x + 3 = 0⇒x = -3

x = 0

x + 2 = 0 ⇒x = -2

putting -3, -2, 0, 1 and 3 in number lines as shown in figure.

case 1 : x > 3 , f(x) > 0

case 2 : 1 < x < 3 , f(x) < 0

case 3 : 0 < x < 1 , f(x) > 0

case 4 : 0 < x < -2 , f(x) < 0

case 5 : -2 < x < -3 , f(x) > 0

case 6 : x < -3 , f(x) < 0

but we have given f(x) = (x - 1)(x² - 9)/x(x + 2) < 0

so, intervals 1 < x < 3, 0 < x < -2 and x < -3 are the solutions.

we have to find smallest positive integral value of x. this can be found in the interval (1, 3) i.e., 2

Therefore the smallest positive integral value is 2.