Let f(x) = x cos⁻¹(sin(-|x|)), x ∈ (-π/2, π/2), then which of the following is true ? (A) f'(0) = -π

Q: Wait, why can't we just use the chain rule directly on $f(x) = x \cos^{-1}(\sin(-|x|))$?

You absolutely could, but it would be a mathematical nightmare! Differentiating $\cos^{-1}(\sin(-|x|))$ directly involves the derivative of $\cos^{-1}(u)$, which is $-\frac{1}{\sqrt{1-u^2}}$, and then applying the chain rule repeatedly. By simplifying the expression first using trigonometric identities, we reduce the function to a simple polynomial form, making the derivative trivial to compute. Always look for simplification before differentiation!

Q: Is $f(x)$ differentiable at $x=0$?

Yes, it is! To check differentiability at $x=0$, we must ensure the Left Hand Derivative (LHD) equals the Right Hand Derivative (RHD). As we found, $f'(0^-) = \frac{\pi}{2}$ and $f'(0^+) = \frac{\pi}{2}$. Since they match, the function is smooth at the origin, and the derivative exists.

Q: Why does the derivative change behavior at $x=0$?

The behavior changes because of the absolute value function $|x|$. The absolute value function is defined differently for positive and negative inputs, which creates a 'kink' in the expression of the function. Even though the function is differentiable at $x=0$, the formula for the derivative is different on either side of zero, leading to different slopes for $f'(x)$.

Question

Let $f(x) = x \cos^{-1}(\sin(-|x|)), x \in (-\frac{\pi}{2}, \frac{\pi}{2})$, then which of the following is true ?

Options:

(A) $f'(0) = -\frac{\pi}{2}$

(B) $f'$ is decreasing in $(-\frac{\pi}{2}, 0)$ and increasing in $(0, \frac{\pi}{2})$

(C) $f$ is not differentiable at $x = 0$

(D) $f'$ is increasing in $(-\frac{\pi}{2}, 0)$ and decreasing in $(0, \frac{\pi}{2})$

Sigma Percentile · Accepted Answer

### The Art of Unmasking: A Journey Through $f(x)$

Welcome, fellow traveler on the road to JEE Advanced. Today, we are going to dismantle a function that, at first glance, seems designed to confuse.

We are looking at $f(x) = x \cos^{-1}(\sin(-|x|))$ for $x \in (-\frac{\pi}{2}, \frac{\pi}{2})$. It looks like a tangled mess of inverse trigonometry and absolute values, but mathematics is not about brute force; it is about elegance.

### Phase 1: The Unmasking

First, we stare down the innermost part: $\sin(-|x|)$. We know that the sine function is odd, meaning $\sin(-	heta) = -\sin	heta$.

So, our inner term becomes $-\sin|x|$. Now our function is $f(x) = x \cos^{-1}(-\sin|x|)$.

This is better, but that negative sign inside the $\cos^{-1}$ is still a nuisance. We reach into our toolkit and pull out the identity:

$$\cos^{-1}(-z) = \pi - \cos^{-1}(z)$$

By setting $z = \sin|x|$, we transform our function into $f(x) = x [\pi - \cos^{-1}(\sin|x|)]$. We are making progress!

### Phase 2: The Transformation

We still have a $\cos^{-1}$ term. Let us convert it to $\sin^{-1}$ using the complementary identity:

$$\cos^{-1}(z) = \frac{\pi}{2} - \sin^{-1}(z)$$

Substituting this in, we get:

$$f(x) = x \left[ \pi - \left( \frac{\pi}{2} - \sin^{-1}(\sin|x|) ight) ight]$$

Simplifying the constants, $\pi - \frac{\pi}{2}$ gives us $\frac{\pi}{2}$, and the double negative makes the $\sin^{-1}$ term positive. Thus:

$$f(x) = x \left( \frac{\pi}{2} + \sin^{-1}(\sin|x|) ight)$$

Now, consider the domain $x \in (-\frac{\pi}{2}, \frac{\pi}{2})$. In this range, $|x|$ is always between $0$ and $\frac{\pi}{2}$. Within this principal domain, $\sin^{-1}(\sin|x|) = |x|$.

Our function has collapsed into the elegant form:

$$f(x) = x \left( \frac{\pi}{2} + |x| ight)$$

### Phase 3: The Crossroads at $x=0$

Because of the absolute value, we must respect the two worlds of $x$.

When $x \ge 0$, $|x| = x$, so:

$$f(x) = x\left(\frac{\pi}{2} + xight) = \frac{\pi x}{2} + x^2$$

When $x < 0$, $|x| = -x$, so:

$$f(x) = x\left(\frac{\pi}{2} - xight) = \frac{\pi x}{2} - x^2$$

Now, we differentiate. For $x > 0$, $f'(x) = \frac{\pi}{2} + 2x$. For $x < 0$, $f'(x) = \frac{\pi}{2} - 2x$.

Checking the point $x=0$, both the left-hand and right-hand derivatives yield $\frac{\pi}{2}$. The function is differentiable, and the slope is continuous!

### Phase 4: The Final Verdict

Finally, let us look at the monotonicity of $f'(x)$.

In the interval $(-\frac{\pi}{2}, 0)$, $f'(x) = \frac{\pi}{2} - 2x$. The derivative of this slope is $-2$, which is negative, meaning $f'(x)$ is **decreasing**.

In the interval $(0, \frac{\pi}{2})$, $f'(x) = \frac{\pi}{2} + 2x$. The derivative of this slope is $2$, which is positive, meaning $f'(x)$ is **increasing**.

We have solved the puzzle. The derivative $f'$ is decreasing in $(-\frac{\pi}{2}, 0)$ and increasing in $(0, \frac{\pi}{2})$. Mathematics is not just about finding the answer; it is about seeing the structure beneath the chaos.

Animated Solution for Mathematics - Differentiation: Let $f (x) = x cos^{- 1} (sin (- ∣ x ∣)), x \in (- \frac{π}{2}, \frac{π}{2})$ , then which of the following is true ?

Select Answer:

Visualized Solution

Understanding the Function $f (x)$

Simplifying the Inner Term $sin (- ∣ x ∣)$

Applying Inverse Cosine Property

Converting $cos^{- 1}$ to $sin^{- 1}$

Simplifying the Expression

Evaluating $sin^{- 1} (sin ∣ x ∣)$

Defining $f (x)$ Piecewise

Finding the Derivative $f^{'} (x)$

Checking Differentiability at $x = 0$

Analyzing Monotonicity of $f^{'} (x)$

Final Conclusion

The Sigma Insight: Monotonicity

The Art of Unmasking

Phase 1

Phase 2

Phase 3

Phase 4

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Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=xcos−1(sin(−∣x∣)),x∈(−2π​,2π​), then which of the following is true ?

Select Answer:

Visualized Solution

Understanding the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Simplifying the Inner Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sin(−∣x∣)