Q.1. If b>a then the equation (x-a)(x-b)+1=0 has both roots in (a,b).But how?
To understand why the equation (x-a)(x-b)+1=0 has both roots in the interval (a, b) when b>a, we need to analyze the equation and its properties.
We start by expanding the equation:
(x-a)(x-b) + 1 = 0
x^2 - bx - ax + ab + 1 = 0
x^2 - (a + b)x + ab + 1 = 0
Now, let's consider the discriminant of this quadratic equation. The discriminant (D) is given by:
D = (b^2 - 4ac)
In our case, the coefficients are:
a = -(a + b)
b = 1
c = ab + 1
Substituting these values into the discriminant formula, we have:
D = (-(a + b))^2 - 4(1)(ab + 1)
= (a + b)^2 - 4(ab + 1)
= a^2 + 2ab + b^2 - 4ab - 4
= a^2 - 2ab + b^2 - 4
= (a - b)^2 - 4
Since b > a, we can say that (a - b)^2 is negative, and therefore, (a - b)^2 - 4 is also negative. This means that the discriminant (D) of the quadratic equation is negative.
Now, let's consider the nature of the roots using the discriminant:
If D < 0, then the quadratic equation has no real roots.
If D = 0, then the quadratic equation has one real root.
If D > 0, then the quadratic equation has two distinct real roots.
Since the discriminant (D) of our equation is negative, we can conclude that the equation (x-a)(x-b)+1=0 has no real roots. However, we need to keep in mind that the question's statement claims that both roots lie in the interval (a, b).
Hence, there seems to be a contradiction between the information provided and the nature of the equation.
Please double-check the given equation or provide any additional information if necessary for a more accurate analysis.