Question

please answer .....;;;;;;;;;;;;;

Answer 1

Sorry to say, i am not sure about the solution of this question..........If it is wronf pardon me

Answer 2

Any value whether Positive or Negative inside Modulus function becomes Positive. That is IxI = x and I-xI = x

⇒ the value inside the modulus can be positive or negative but it finally becomes positive

⇒ here the expression inside the modulus is (x - 1)

from the definition of modulus function the expression (x - 1) can be positive or negative.

let us consider the case when (x - 1) is positive

⇒ the given equation becomes : x² + x - 1 = 1

⇒ x² + x - 2 = 0

⇒ x² + 2x -x - 2 = 0

⇒ x = - 2 or x  = 1

now let us consider the case when (x - 1) is negative

⇒ the given equation becomes x² -(x - 1) = 1

⇒ x² - x + 1 = 1

⇒ x² - x = 0

⇒ x(x - 1) = 0

⇒ x = 0 or x = 1

so the possible solutions of the equation  x² + Ix - 1I = 1 are :

x = -2 (or) x = 0 (or) x = 1

these are possible solutions but not actual solutions which satisfy the equation.

let us verify whether they are actually satisfying the equation or not

substituting x = -2 in the given question we get :

⇒ (-2)² + I-2 - 1I = 1

we know that anything inside modulus function finally becomes positive

⇒ 4 + 3 = 1

⇒ 7 = 1 but we know that 7 cannot be equal to 1

so -2 is not the solution the given question.

substituting x = 0 in the equation we get :

⇒ 0 + I-1I = 1

we know that I-1I = 1

⇒ 1 = 1

so the solution x = 0 is an actual solution to the given question.

substituting x = 1 in the equation we get :

⇒ 1² + 0 = 1

⇒ 1 = 1

so the solution x = 1 is an actual solution to the given question.

⇒ the number of solutions of x² + Ix - 1I = 1 are 2 and they are 0 and 1