Let $[t]$ denote the greatest integer less than or equal to $t$. If the function $f(x) = \begin{cases} b^2 \sin \left( \frac{\pi}{2} \left[ \frac{\pi}{2} (\cos x + \sin x) \cos x \right] \right), & x < 0 \\ \frac{\sin x - \frac{1}{2} \sin 2x}{x^3}, & x > 0 \\ a, & x = 0 \end{cases}$ is continuous at $x = 0$, then $a^2 + b^2$ is equal to
Function Continuity Analysis at x = 0
The function $f(x)$ is defined piecewise: $ f(x) = \begin{cases} b^2 \sin \left( \frac{\pi}{2} \left[ \frac{\pi}{2} (\cos x + \sin x) \cos x \right] \right), & x < 0 \\ \frac{\sin x - \frac{1}{2} \sin 2x}{x^3}, & x > 0 \\ a, & x = 0 \end{cases} $ For the function to be continuous at $x = 0$, the following condition must hold: $ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) $ We know that $f(0) = a$. We need to evaluate the left-hand and right-hand limits.
Left-Hand Limit Calculation ($x \to 0^-$)
For $x < 0$, $f(x) = b^2 \sin \left( \frac{\pi}{2} \left[ \frac{\pi}{2} (\cos x + \sin x) \cos x \right] \right)$. Let $g(x) = \frac{\pi}{2} (\cos x + \sin x) \cos x$. As $x \to 0$, $g(x) \to \frac{\pi}{2} (\cos 0 + \sin 0) \cos 0 = \frac{\pi}{2} (1+0)(1) = \frac{\pi}{2}$. To evaluate the greatest integer function $[g(x)]$, we examine the behavior of $g(x)$ near $x=0$. The derivative is $g'(x) = \frac{\pi}{2} (\cos(2x) - \sin(2x))$. At $x=0$, $g'(0) = \frac{\pi}{2} > 0$. Since $g'(0)$ is positive, $g(x)$ is increasing at $x=0$. Therefore, for $x < 0$ and $x$ sufficiently close to $0$, we have $g(x) < \frac{\pi}{2}$. Using Taylor expansion around $x=0$: $g(x) \approx \frac{\pi}{2} + \frac{\pi}{2}x$. For $x \to 0^-$, $g(x)$ is slightly less than $\frac{\pi}{2}$. Numerical checks and analysis show that for $x$ close enough to $0^-$, $1 < g(x) < \frac{\pi}{2}$. Thus, the greatest integer $[g(x)] = 1$. The left-hand limit is: $ \lim_{x \to 0^-} f(x) = b^2 \sin \left( \frac{\pi}{2} [g(x)] \right) = b^2 \sin \left( \frac{\pi}{2} \times 1 \right) = b^2 \sin \left( \frac{\pi}{2} \right) = b^2 \times 1 = b^2 $
Right-Hand Limit Calculation ($x \to 0^+$)
For $x > 0$, $f(x) = \frac{\sin x - \frac{1}{2} \sin 2x}{x^3}$. We use Taylor series expansions around $x=0$: $ \sin x = x - \frac{x^3}{6} + O(x^5) $ $ \sin 2x = 2x - \frac{(2x)^3}{6} + O(x^5) = 2x - \frac{4x^3}{3} + O(x^5) $ Substitute these into the expression for $f(x)$: $ f(x) = \frac{\left(x - \frac{x^3}{6} + O(x^5)\right) - \frac{1}{2}\left(2x - \frac{4x^3}{3} + O(x^5)\right)}{x^3} $ $ f(x) = \frac{x - \frac{x^3}{6} - x + \frac{2x^3}{3} + O(x^5)}{x^3} $ $ f(x) = \frac{\left(-\frac{1}{6} + \frac{4}{6}\right)x^3 + O(x^5)}{x^3} = \frac{\frac{3}{6}x^3 + O(x^5)}{x^3} $ $ f(x) = \frac{\frac{1}{2}x^3 + O(x^5)}{x^3} = \frac{1}{2} + O(x^2) $ The right-hand limit is: $ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \left(\frac{1}{2} + O(x^2)\right) = \frac{1}{2} $
Applying Continuity Condition and Final Calculation
For continuity at $x=0$, we equate the limits and $f(0)$: $ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) $ $ b^2 = \frac{1}{2} = a $ This gives us $a = \frac{1}{2}$ and $b^2 = \frac{1}{2}$. We need to find $a^2 + b^2$: $ a^2 + b^2 = \left(\frac{1}{2}\right)^2 + \frac{1}{2} = \frac{1}{4} + \frac{1}{2} = \frac{1}{4} + \frac{2}{4} = \frac{3}{4} $
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Consider the following two statements :
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Then,
