Question

For real x, the expression \[y=\dfrac{(x-a)(x-b)}{(x-c)}\] will assume all real values provided
(Given that a>b)

Answer 1

HINT: - For an expression to assume all real values, its denominator must not be zero or 0 for any value of x. To achieve this, we must try to prevent the denominator from becoming zero by getting it cancelled by the numerator.


Complete step-by-step answer:

As mentioned in the hint, one must try to cancel the denominator that is (x-c) in this case.
We will try to get this (x-c) cancelled from the numerator by forming a factor in the numerator particularly when the value of x becomes c as otherwise the denominator will become zero. Now, for this either of the two, ‘a’ or ‘b’ must be equal to c so that the denominator gets cancelled with the numerator.
Along with this, as it is clearly mentioned in the question that a>b, therefore, option (c) is ruled out.
As both the options, (a) and (b) have \[a\ge b\] , hence the correct answer to this question is option (d).

NOTE: -The students can make a mistake when checking the domain function and also what domain is given to check in the question.
Other than this, the fact that for an expression to assume all real values, its denominator must not be zero or 0 for any value of x should be known. To achieve this, we must try to prevent the denominator from becoming zero by getting it cancelled by the numerator.