For the function f(x) = (cos x) - x + 1, x ∈ R, between the following two statements (S1) f(x) = 0 f

Q: Wait, why can't we just solve $\cos x = x - 1$ by taking the inverse cosine? It seems like a standard algebra problem.

That is a classic trap! You cannot isolate $x$ because it appears both inside the trigonometric function and as a linear term. This is a 'transcendental equation.' In JEE Advanced, whenever you see a mix of algebraic and trigonometric terms, stop trying to solve for $x$ and start analyzing the function's behavior using calculus.

Q: Why is the function 'strictly decreasing' if the derivative can be zero at some points?

Excellent observation! The derivative $f'(x) = -(\sin x + 1)$ becomes zero only when $\sin x = -1$. This happens at discrete points like $x = \frac{3\pi}{2}, \frac{7\pi}{2}$, etc. Because these points are isolated and the function does not 'turn around' (it doesn't change from decreasing to increasing), the function remains strictly decreasing over the entire domain. It never 'rests' long enough to violate the strictly decreasing property.

Q: How do I know for sure there isn't a second root hidden somewhere?

That is the beauty of monotonicity! If a function is strictly decreasing, it is a one-to-one function (injective). It can cross the x-axis at most once. Since we proved it is strictly decreasing and we found a sign change between $0$ and $\pi$ using the Intermediate Value Theorem, the existence of exactly one root is mathematically guaranteed.

Question

For the function $f(x) = (\cos x) - x + 1, x \in \mathbb{R}$, between the following two statements (S1) $f(x) = 0$ for only one value of $x$ in $[0, \pi]$. (S2) $f(x)$ is decreasing in $[0, \frac{\pi}{2}]$ and increasing in $[\frac{\pi}{2}, \pi]$.

Options:

(A) Both (S1) and (S2) are correct.

(B) Both (S1) and (S2) are incorrect.

(C) Only (S2) is correct.

(D) Only (S1) is correct.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the edge of a vast mathematical landscape, looking at the function $f(x) = \cos x - x + 1$. At first glance, it looks innocent, but in the world of JEE Advanced, appearances can be deceiving.

You might be tempted to set $f(x) = 0$ and solve for $x$ directly, or rearrange it to $\cos x = x - 1$. However, you are staring at a transcendental equation where $x$ cannot be isolated using standard algebraic manipulation.

This is where the true power of calculus comes to your rescue. We do not need to find the exact value of $x$ to understand the behavior of the function; we simply need to analyze its rate of change.

### The Slope of Reality

To understand how this function moves, we must look at its derivative, $f'(x)$. Differentiating $f(x) = \cos x - x + 1$ with respect to $x$ yields:

$$f'(x) = -\sin x - 1$$

We can factor out the negative sign to express this as:

$$f'(x) = -(\sin x + 1)$$

Now, consider the properties of the sine function. For any real number $x$, the value of $\sin x$ is trapped within the interval $[-1, 1]$. Adding $1$ to this inequality gives:

$$0 \le \sin x + 1 \le 2$$

Because there is a negative sign outside the parenthesis, we find that $f'(x) \le 0$ for all real $x$. This is a profound realization: the function is never increasing. It is always sliding downwards, like a ball rolling down a hill that never levels off for long.

### Debunking the Monotonicity Myth

Statement (S2) claims that the function is decreasing in $[0, \frac{\pi}{2}]$ and increasing in $[\frac{\pi}{2}, \pi]$. Our derivative analysis has shattered this claim.

We proved that $f'(x) \le 0$ everywhere. A function that is always decreasing cannot suddenly decide to increase.

It is physically impossible for the curve to turn back up. Therefore, Statement (S2) is **fundamentally incorrect**; it is a trap designed to test your understanding of the implications of a negative derivative.

### The Hunt for the Root

Finally, let's tackle Statement (S1), which asks if the function crosses the x-axis exactly once in $[0, \pi]$. To answer this, we use the **Intermediate Value Theorem**.

First, we check the boundaries:

$$f(0) = \cos(0) - 0 + 1 = 1 + 1 = 2$$

$$f(\pi) = \cos(\pi) - \pi + 1 = -1 - \pi + 1 = -\pi$$

The function starts at a height of $2$ (above the x-axis) and ends at $-\pi \approx -3.14$ (below the x-axis). Because the function is continuous, it must cross the x-axis at least once.

Is it exactly once? Yes. Because we proved the function is strictly decreasing, it can only cross the x-axis once and cannot turn back to cross it again.

Thus, Statement (S1) is **absolutely correct**. You have successfully navigated the trap, analyzed the derivative, and used the Intermediate Value Theorem to prove the uniqueness of the root.

Animated Solution for Mathematics - Differentiation: For the function $f (x) = (cos x) - x + 1, x \in R$ , between the following two statements (S1) $f (x) = 0$ for only one value of $x$ in $[0, π]$ . (S2) $f (x)$ is decreasing in $[0, \frac{π}{2}]$ and increasing in $[\frac{π}{2}, π]$ .

Select Answer:

Visualized Solution

Defining the Function $f (x)$

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

Analyzing the Sign of $f^{'} (x)$

Verifying Statement (S2)

Evaluating $f (x)$ at $x = 0$

Evaluating $f (x)$ at $x = π$

Applying Intermediate Value Theorem

Uniqueness of the Root

Final Conclusion

The Sigma Insight: Monotonicity

Analyzing the Setup

The Slope of Reality

Debunking the Monotonicity Myth

The Hunt for the Root

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