Question

Let [x] denotes the greatest integer function and f(x) be defined as

\frac{[x+1]}{(\exp{(x + 2)\ln4)}) 4 -16} ; x<2\\ \frac{4^x-16}{1-\cos(x-2)} \\ a\cdot\frac{1-\cos(x-2)}{(x-2) \tan(x-2)} ; x>2 \end{cases}$$ Find the value of a for which f(x) may be continuous at x = 2.

• A. 1/60

Answer 1

Answer: (A)

1/60

Answer 2

For continuity at x=2, the Left-Hand Limit (LHL) must equal the Right-Hand Limit (RHL).

1. Left-Hand Limit (LHL): $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} \frac{[x+1]}{4^{x+2} - 16} = \frac{2}{4^{4} - 16} = \frac{2}{256 - 16} = \frac{2}{240} = \frac{1}{120}$

2. Right-Hand Limit (RHL): $\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} a \cdot \frac{1-\cos(x-2)}{(x-2) \tan(x-2)}$ Let $h = x-2$. As $x \to 2^+$, $h \to 0^+$. $\lim_{h \to 0^+} a \cdot \frac{1-\cos h}{h \tan h} = a \cdot \lim_{h \to 0^+} \frac{1-\cos h}{h^2} \cdot \frac{h}{\tan h} = a \cdot \frac{1}{2} \cdot 1 = \frac{a}{2}$

3. Equate LHL and RHL for continuity: $\frac{1}{120} = \frac{a}{2}$ $a = \frac{2}{120} = \frac{1}{60}$

Therefore, the value of a for which f(x) may be continuous at x = 2 is $\frac{1}{60}$.