Key Concepts and Formulas

• Inverse Cosine Property: $\cos^{-1}(-A) = \pi - \cos^{-1}(A)$
• Complementary Angle Property: $\cos^{-1}(y) + \sin^{-1}(y) = \frac{\pi}{2}$, which implies $\cos^{-1}(y) = \frac{\pi}{2} - \sin^{-1}(y)$.
• Principal Value Branch of $\sin^{-1}(x)$: For $x \in [-1, 1]$, $\sin^{-1}(\sin x) = x$ if $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.
• Definition of Absolute Value: $|x| = x$ for $x \ge 0$ and $|x| = -x$ for $x < 0$.
• Differentiability of Piecewise Functions: To check differentiability at a point where the function definition changes (e.g., $x=0$), we compare the Left Hand Derivative (LHD) and Right Hand Derivative (RHD).
• Monotonicity of a Function's Derivative: A function $g(x)$ is increasing if its derivative $g'(x) > 0$ and decreasing if $g'(x) < 0$. In this problem, we are analyzing the monotonicity of $f'(x)$, so we need to examine its derivative, $f''(x)$.

Step-by-Step Solution

Step 1: Simplify the expression for $f(x)$ using inverse trigonometric identities.

Given the function: $f(x) = x\cos^{-1}(-\sin|x|), \quad x \in \left[ { - {\pi \over 2},{\pi \over 2}} \right]$

We apply the property $\cos^{-1}(-A) = \pi - \cos^{-1}(A)$ to the term $\cos^{-1}(-\sin|x|)$. Here, $A = \sin|x|$. $f(x) = x \left( \pi - \cos^{-1}(\sin|x|) \right)$

Next, we simplify $\cos^{-1}(\sin|x|)$. We use the identity $\cos^{-1}(y) = \frac{\pi}{2} - \sin^{-1}(y)$. So, $\cos^{-1}(\sin|x|) = \frac{\pi}{2} - \sin^{-1}(\sin|x|)$.

Now, we simplify $\sin^{-1}(\sin|x|)$. Since $x \in \left[ { - {\pi \over 2},{\pi \over 2}} \right]$, it implies that $|x| \in \left[ {0,{\pi \over 2}} \right]$. The interval $\left[ {0,{\pi \over 2}} \right]$ lies entirely within the principal value branch of $\sin^{-1}(y)$, which is $\left[ { - {\pi \over 2},{\pi \over 2}} \right]$. Therefore, for $|x| \in \left[ {0,{\pi \over 2}} \right]$, we have $\sin^{-1}(\sin|x|) = |x|$.

Substituting this back: $\cos^{-1}(\sin|x|) = \frac{\pi}{2} - |x|$

Now, substitute this entire expression back into the formula for $f(x)$: $f(x) = x \left( \pi - \left( \frac{\pi}{2} - |x| \right) \right)$ $f(x) = x \left( \pi - \frac{\pi}{2} + |x| \right)$ $f(x) = x \left( \frac{\pi}{2} + |x| \right)$ $f(x) = \frac{\pi}{2}x + x|x|$

Step 2: Express $f(x)$ as a piecewise function.

The term $|x|$ needs to be defined based on the sign of $x$.

• If $x \ge 0$, then $|x| = x$. $f(x) = \frac{\pi}{2}x + x(x) = \frac{\pi}{2}x + x^2$
• If $x < 0$, then $|x| = -x$. $f(x) = \frac{\pi}{2}x + x(-x) = \frac{\pi}{2}x - x^2$

Combining these, $f(x)$ can be written as: $f(x) = \begin{cases} \frac{\pi}{2}x - x^2, & x < 0 \\ \frac{\pi}{2}x + x^2, & x \ge 0 \end{cases}$

Step 3: Calculate the first derivative, $f'(x)$.

We differentiate each piece of the function with respect to $x$:

• For $x < 0$: $f'(x) = \frac{d}{dx} \left( \frac{\pi}{2}x - x^2 \right) = \frac{\pi}{2} - 2x$
• For $x > 0$: $f'(x) = \frac{d}{dx} \left( \frac{\pi}{2}x + x^2 \right) = \frac{\pi}{2} + 2x$

Now, we need to check the differentiability at $x=0$. The Left Hand Derivative (LHD) at $x=0$: $f'(0^-) = \lim_{x \to 0^-} f'(x) = \lim_{x \to 0^-} \left(\frac{\pi}{2} - 2x\right) = \frac{\pi}{2}$ The Right Hand Derivative (RHD) at $x=0$: $f'(0^+) = \lim_{x \to 0^+} f'(x) = \lim_{x \to 0^+} \left(\frac{\pi}{2} + 2x\right) = \frac{\pi}{2}$ Since LHD = RHD, $f'(0)$ exists and $f'(0) = \frac{\pi}{2}$.

Therefore, we can write $f'(x)$ as: $f'(x) = \begin{cases} \frac{\pi}{2} - 2x, & x < 0 \\ \frac{\pi}{2} + 2x, & x \ge 0 \end{cases}$

Step 4: Calculate the second derivative, $f''(x)$.

We differentiate each piece of the function $f'(x)$ with respect to $x$:

• For $x < 0$: $f''(x) = \frac{d}{dx} \left( \frac{\pi}{2} - 2x \right) = -2$
• For $x > 0$: $f''(x) = \frac{d}{dx} \left( \frac{\pi}{2} + 2x \right) = 2$

Thus, $f''(x) = \begin{cases} -2, & x < 0 \\ 2, & x > 0 \end{cases}$

Step 5: Analyze the monotonicity of $f'(x)$.

Since $f''(x) = -2 < 0$ for $x \in \left( -{\pi \over 2}, 0 \right)$, $f'(x)$ is decreasing in $\left( -{\pi \over 2}, 0 \right)$. Since $f''(x) = 2 > 0$ for $x \in \left( 0, {\pi \over 2} \right)$, $f'(x)$ is increasing in $\left( 0, {\pi \over 2} \right)$.

Therefore, $f'$ is decreasing in $\left( { - {\pi \over 2},0} \right)$ and increasing in $\left( {0,{\pi \over 2}} \right)$.

Common Mistakes & Tips

• Remember to consider the principal value branch of the inverse trigonometric functions when simplifying expressions.
• When dealing with absolute values, always split the function into piecewise definitions to handle different cases.
• Pay close attention to differentiability at points where the function definition changes. Check LHD and RHD.

Summary

We simplified the expression for $f(x)$ using inverse trigonometric identities and absolute value properties. We then expressed $f(x)$ as a piecewise function and calculated its first derivative, $f'(x)$, also as a piecewise function. After confirming differentiability at $x=0$, we found the second derivative, $f''(x)$. Finally, we analyzed the sign of $f''(x)$ to determine the intervals where $f'(x)$ is increasing or decreasing. We found that $f'(x)$ is decreasing in $\left( -{\pi \over 2}, 0 \right)$ and increasing in $\left( 0, {\pi \over 2} \right)$.

Final Answer

The final answer is \boxed{ƒ' is decreasing in $\left( { - {\pi \over 2},0} \right)$ and increasing in $\left( {0,{\pi \over 2}} \right)$}, which corresponds to option (A).