Question

Passage-II: Consider the following functions ({x} = x - max{a ∈ Z: a ≤ x})

$$f(x) = \begin{cases} \ln(1+|x|+(b-2)|x+1|)+atan^{-1}x+2, & -2 < x < 0 \\ b, & x = 0 \\ \frac{e^{2x} - e^{\ln(4^{\frac{x}{2}})} + (c+2)\sin^{-1}\sqrt{\{x\}}}{x^2} & 0 < x < 2 \end{cases} \text{and}$$ $$g(x) = \begin{cases} \frac{\ln(1+2x)}{x}, & -\frac{1}{2} < x < 0 \\ 2\cos x, & x = 0 \\ \frac{e^{2x} - 1}{x}, & 0 < x < 1 \\ \frac{x}{e^2 - 1}, & x \geq 1 \end{cases}$$

Now, based on above information, answer the following questions

Q.(19) Let m be number of points where g is dis-continuous and n be number of points where g is non-differentiable, then m + n is equal to

Answer 1

3

Answer 2

Let's analyze the function g(x) to determine the number of points where it is discontinuous (m) and non-differentiable (n).

Function g(x):

$$g(x) = \begin{cases} \frac{\ln(1+2x)}{x}, & -\frac{1}{2} < x < 0 \\ 2\cos x, & x = 0 \\ \frac{e^{2x} - 1}{x}, & 0 < x < 1 \\ \frac{x}{e^2 - 1}, & x \geq 1 \end{cases}$$

Continuity Check:

We need to check continuity at the points where the definition of g(x) changes, which are x = 0 and x = 1.

• At x = 0:

  • Left-hand limit: $\lim_{x \to 0^-} g(x) = \lim_{x \to 0^-} \frac{\ln(1+2x)}{x}$. Using L'Hôpital's rule or the standard limit $\lim_{u \to 0} \frac{\ln(1+u)}{u} = 1$, we have $\lim_{x \to 0^-} \frac{\ln(1+2x)}{x} = 2$.
  • Function value at x = 0: $g(0) = 2\cos(0) = 2$.
  • Right-hand limit: $\lim_{x \to 0^+} g(x) = \lim_{x \to 0^+} \frac{e^{2x} - 1}{x}$. Using L'Hôpital's rule or the standard limit $\lim_{u \to 0} \frac{e^u - 1}{u} = 1$, we have $\lim_{x \to 0^+} \frac{e^{2x} - 1}{x} = 2$.

  Since $\lim_{x \to 0^-} g(x) = g(0) = \lim_{x \to 0^+} g(x) = 2$, g(x) is continuous at x = 0.

• At x = 1:

  • Left-hand limit: $\lim_{x \to 1^-} g(x) = \lim_{x \to 1^-} \frac{e^{2x} - 1}{x} = \frac{e^2 - 1}{1} = e^2 - 1$.
  • Function value at x = 1: $g(1) = \frac{1}{e^2 - 1}$.
  • Right-hand limit: $\lim_{x \to 1^+} g(x) = \lim_{x \to 1^+} \frac{x}{e^2 - 1} = \frac{1}{e^2 - 1}$.

  Since $\lim_{x \to 1^-} g(x) = e^2 - 1$ and $\lim_{x \to 1^+} g(x) = g(1) = \frac{1}{e^2 - 1}$, and $e^2 - 1 \neq \frac{1}{e^2 - 1}$, g(x) is discontinuous at x = 1.

Therefore, the number of points where g(x) is discontinuous is m = 1.

Differentiability Check:

• At x = 0:

  We need to check if the left and right derivatives at x = 0 are equal.

  • Left derivative: For $x < 0$, $g(x) = \frac{\ln(1+2x)}{x}$. Using the quotient rule, $g'(x) = \frac{\frac{2}{1+2x} \cdot x - \ln(1+2x)}{x^2} = \frac{2x - (1+2x)\ln(1+2x)}{x^2(1+2x)}$. Then $g'(0^-) = \lim_{x \to 0^-} \frac{2x - (1+2x)\ln(1+2x)}{x^2(1+2x)}$. Using Taylor series or L'Hôpital's rule (twice), we find $g'(0^-) = -2$.

  • Right derivative: For $x > 0$, $g(x) = \frac{e^{2x} - 1}{x}$. Using the quotient rule, $g'(x) = \frac{2e^{2x} \cdot x - (e^{2x} - 1)}{x^2} = \frac{2xe^{2x} - e^{2x} + 1}{x^2}$. Then $g'(0^+) = \lim_{x \to 0^+} \frac{2xe^{2x} - e^{2x} + 1}{x^2}$. Using Taylor series or L'Hôpital's rule (twice), we find $g'(0^+) = 2$.

  Since $g'(0^-) = -2$ and $g'(0^+) = 2$, the left and right derivatives are not equal, so g(x) is not differentiable at x = 0.

• At x = 1:

  Since g(x) is discontinuous at x = 1, it is not differentiable at x = 1.

Therefore, the number of points where g(x) is non-differentiable is n = 2 (at x = 0 and x = 1).

Finally, m + n = 1 + 2 = 3.

Thus, m + n = 3.