Question

Let

$$f(x) = \begin{cases} a + \frac{[tan(x)]}{x}, & x > 0 \\ 3, & x = 0 \\ b + [\frac{x-tanx}{x^3}], & x < 0 \end{cases}$$

If f(x) is continuous at x = 0, then find the value of $\sum_{r=0}^{\infty} (\frac{a}{b})^r$.

(Note : [k] denotes the largest integer less than or equal to k)

• A. 4
• B. 3
• C. 1
• D. 2

Answer 1

Answer: (A)

4

Answer 2

For the function $f(x)$ to be continuous at $x=0$, the following condition must be met: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$ We are given $f(0) = 3$.

1. Evaluate the limit from the right ($x \to 0^+$): For $x > 0$, $f(x) = a + \frac{[tan(x)]}{x}$. As $x \to 0^+$, $x$ is a small positive number. For $x \in (0, 1)$, the greatest integer function $[x] = 0$. Thus, $\tan[x] = \tan(0) = 0$. $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \left(a + \frac{[tan(x)]}{x}\right) = \lim_{x \to 0^+} \left(a + \frac{0}{x}\right) = a$ For continuity, $a = f(0) = 3$.

2. Evaluate the limit from the left ($x \to 0^-$): For $x < 0$, $f(x) = b + \left[\frac{x-\tan x}{x^3}\right]$. We need to evaluate the limit of the argument of the greatest integer function: $\lim_{x \to 0^-} \frac{x-\tan x}{x^3}$. Using the Taylor series expansion of $\tan x$ around $x=0$: $\tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + O(x^7)$ Substituting this into the expression: $x - \tan x = x - \left(x + \frac{x^3}{3} + \frac{2x^5}{15} + O(x^7)\right) = -\frac{x^3}{3} - \frac{2x^5}{15} + O(x^7)$ Now, divide by $x^3$: $\frac{x - \tan x}{x^3} = \frac{-\frac{x^3}{3} - \frac{2x^5}{15} + O(x^7)}{x^3} = -\frac{1}{3} - \frac{2x^2}{15} + O(x^4)$ As $x \to 0^-$, $x^2$ is a small positive quantity. Therefore, $-\frac{2x^2}{15}$ is a small negative quantity. This implies that $\frac{x - \tan x}{x^3}$ approaches $-\frac{1}{3}$ from values slightly less than $-\frac{1}{3}$. Let $y = \frac{x - \tan x}{x^3}$. As $x \to 0^-$, $y \approx -\frac{1}{3} - (\text{small positive number})$. For example, $y$ could be approximately $-0.33333 - 0.00001 = -0.33334$. The greatest integer of such a number is: $\left[\frac{x - \tan x}{x^3}\right] = \left[-\frac{1}{3} - \frac{2x^2}{15} + O(x^4)\right] = -1$ So, the limit from the left is: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \left(b + \left[\frac{x-\tan x}{x^3}\right]\right) = b + (-1) = b-1$ For continuity, $b-1 = f(0) = 3$. This gives $b = 4$.

3. Calculate the sum of the geometric series: We have found $a=3$ and $b=4$. We need to find the value of $\sum_{r=0}^{\infty} (\frac{a}{b})^r$. Substituting the values of $a$ and $b$: $\sum_{r=0}^{\infty} \left(\frac{3}{4}\right)^r$ This is an infinite geometric series with the first term $A = (\frac{3}{4})^0 = 1$ and the common ratio $R = \frac{3}{4}$. Since $|R| = |\frac{3}{4}| < 1$, the series converges. The sum of an infinite geometric series is given by $\frac{A}{1-R}$. $\text{Sum} = \frac{1}{1 - \frac{3}{4}} = \frac{1}{\frac{1}{4}} = 4$