Question

Let $f(u) = \frac{1}{ln|u|-1}$, $u=\frac{x+e}{x+1}$. Number of values of x in domain for which f(u) is discontinuous is -

• A. 2
• B. 1
• C. 3
• D. 0

Answer 1

Answer: (C)

3

Answer 2

The function $f(u)$ is discontinuous at $u$ values where it is not defined, which are $u=0$, $u=e$, and $u=-e$. The question asks for the number of values of x such that $f(u(x))$ is discontinuous because $u(x)$ is one of these values. We find the values of x for which $u(x) = 0$, $u(x) = e$, and $u(x) = -e$.

Solving $u(x) = \frac{x+e}{x+1} = 0$ gives $x=-e$.

Solving $u(x) = \frac{x+e}{x+1} = e$ gives $x=0$.

Solving $u(x) = \frac{x+e}{x+1} = -e$ gives $x=\frac{-2e}{1+e}$.

These three values of x ($-e$, $0$, $\frac{-2e}{1+e}$) are distinct, and for each of these values, $u(x)$ is defined. Thus, there are 3 values of x for which $u(x)$ is a point of discontinuity for $f(u)$.