Question

Let [t] denote the greatest integer ≤ t and {t} denote the fractional part of t. The integral value of α for which the left-hand limit of the function \(f(x) = \left\lfloor 1 + x \right\rfloor + \frac{\alpha^{2\left\lfloor x \right\rfloor + \left\{ x \right\}} + \left\lfloor x \right\rfloor - 1}{2\left\lfloor x \right\rfloor + \left\{ x \right\}} \) at x=0 is equal to  \(\alpha -\frac{4}{3}\), is

Answer 1

Correct Answer: 3

\(f(x) = \left\lfloor 1 + x \right\rfloor + \frac{\alpha^{2\left\lfloor x \right\rfloor + \left\{ x \right\}} + \left\lfloor x \right\rfloor - 1}{2\left\lfloor x \right\rfloor + \left\{ x \right\}} \)

\(\lim_{{x \to 0^-}} f(x) = \alpha - \frac{4}{3}\)

\(⇒\) \(\lim_{{x \to 0^-}} \left[ 1 + \left\lfloor x \right\rfloor + \frac{\alpha^{x + \left\lfloor x \right\rfloor} + \left\lfloor x \right\rfloor - 1}{x + \left\lfloor x \right\rfloor} \right] = \alpha - \frac{4}{3}\)

\(⇒\) \(\lim_{{h \to 0^-}} \left[ 1 - 1 + \frac{\alpha^{-h - 1} - 1 - 1}{-h - 1} \right] = \alpha - \frac{4}{3}\)

\(∴\) \(\frac{\alpha^{-1} - 2}{-1} = \alpha - \frac{4}{3}\)
\(⇒\) \(3α^2 – 10α + 3 = 0\)
\(∴\) \(α = 3 \ or\  \frac{1}{3}\)
\(∵\) α in integer, hence \(α = 3\)