Question

The function $f(x) = \begin{cases} (\frac{6}{5})^{\frac{tan6x}{tan5x}}, & \text{if } 0 < x < \frac{\pi}{2} \\ b+2, & \text{if } x = \frac{\pi}{2} \\ (1 + |cosx|)^{(\frac{a|tanx|}{b})}, & \text{if } \frac{\pi}{2} < x < \pi \end{cases}$

Determine the values of 'a' & 'b', if f is continuous at x = $\pi/2$.

Answer 1

a=0 and b=-1

Answer 2

For the function $f(x)$ to be continuous at $x = \pi/2$, the following condition must hold: $\lim_{x \to (\pi/2)^-} f(x) = \lim_{x \to (\pi/2)^+} f(x) = f(\pi/2)$

1. Evaluate the left-hand limit (LHL): For $0 < x < \pi/2$, $f(x) = (\frac{6}{5})^{\frac{tan6x}{tan5x}}$. Let $x = \frac{\pi}{2} - h$, where $h \to 0^+$ as $x \to (\pi/2)^-$. The exponent becomes: $\frac{\tan(6(\frac{\pi}{2} - h))}{\tan(5(\frac{\pi}{2} - h))} = \frac{\tan(3\pi - 6h)}{\tan(\frac{5\pi}{2} - 5h)} = \frac{\tan(-6h)}{\tan(\frac{\pi}{2} - 5h)} = \frac{-\tan(6h)}{\cot(5h)}$ $= \frac{-\tan(6h)}{1/\tan(5h)} = -\tan(6h)\tan(5h)$ As $h \to 0^+$, $\tan(kh) \approx kh$. So, the limit of the exponent is: $\lim_{h \to 0^+} -\tan(6h)\tan(5h) = \lim_{h \to 0^+} -(6h)(5h) = \lim_{h \to 0^+} -30h^2 = 0$ Therefore, the LHL is: $\lim_{x \to (\pi/2)^-} f(x) = \left(\frac{6}{5}\right)^0 = 1$

2. Evaluate the right-hand limit (RHL): For $\pi/2 < x < \pi$, $f(x) = (1 + |cosx|)^{\frac{a|tanx|}{b}}$. Let $x = \frac{\pi}{2} + h$, where $h \to 0^+$ as $x \to (\pi/2)^+$. For $x \in (\pi/2, \pi)$, $cosx < 0$ and $tanx < 0$. So, $|cosx| = -cosx = -cos(\frac{\pi}{2} + h) = -(-sin(h)) = sin(h)$. And $|tanx| = -tanx = -tan(\frac{\pi}{2} + h) = -(-cot(h)) = cot(h)$. The limit is of the indeterminate form $1^\infty$. We use the formula $\lim_{x \to c} (1+g(x))^{h(x)} = e^{\lim_{x \to c} g(x)h(x)}$. The limit of the exponent product is: $\lim_{h \to 0^+} |cosx| \cdot \frac{a|tanx|}{b} = \lim_{h \to 0^+} sin(h) \cdot \frac{a \cdot cot(h)}{b}$ $= \frac{a}{b} \lim_{h \to 0^+} sin(h) \cdot \frac{cos(h)}{sin(h)} = \frac{a}{b} \lim_{h \to 0^+} cos(h) = \frac{a}{b} \cdot 1 = \frac{a}{b}$ Therefore, the RHL is: $\lim_{x \to (\pi/2)^+} f(x) = e^{a/b}$

3. Evaluate the function value at $x=\pi/2$: $f(\pi/2) = b+2$.

4. Equate the limits and the function value for continuity: $1 = e^{a/b} = b+2$ From the first equality, $1 = e^{a/b}$: $\frac{a}{b} = \ln(1) = 0$ This implies $a=0$, provided $b \neq 0$.

   Now, substitute $e^{a/b} = 1$ into the second equality: $1 = b+2$ $b = 1 - 2 = -1$ Since $b = -1 \neq 0$, our condition for $a=0$ is satisfied.

   Thus, the values are $a=0$ and $b=-1$.