The set of all real values of λ for which the function f(x) = (1 - cos²x) ⋅ (λ + sin x), x ∈ (-π/2,

Q: Why do we ignore the endpoints of the interval?

The interval is open, meaning the function is not defined at the endpoints, so we only care about what happens inside. We are looking for extrema within the open interval $(-\pi/2, \pi/2)$.

Q: Why can't $\lambda$ be zero?

If $\lambda=0$, the second critical point becomes $\sin x = 0$, which implies $x=0$. This is the same as the first critical point. We need two distinct points to have both a maxima and a minima.

Q: How do we know the derivative changes sign?

Since $\sin x$ is monotonic in the interval, the linear factor $2\lambda + 3\sin x$ will change sign as it crosses its root, ensuring that the critical point is indeed an extremum.

Question

The set of all real values of $\lambda$ for which the function $f(x) = (1 - \cos^2 x) \cdot (\lambda + \sin x)$, $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ has exactly one maxima and exactly one minima, is:

Options:

(A) $\left(-\frac{3}{2}, \frac{3}{2}\right)$

(B) $\left(-\frac{1}{2}, \frac{1}{2}\right)$

(C) $\left(-\frac{3}{2}, \frac{3}{2}\right) - \{0\}$

(D) $\left(-\frac{3}{2}, \frac{3}{2}\right) - \{0\}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with analyzing the function $f(x) = (1 - \cos^2 x)(\lambda + \sin x)$ on the interval $x \in (-\pi/2, \pi/2)$. Our goal is to determine the values of $\lambda$ such that the function possesses exactly one local maximum and one local minimum.

Using the fundamental trigonometric identity $\sin^2 x + \cos^2 x = 1$, we can simplify the expression. The term $(1 - \cos^2 x)$ becomes $\sin^2 x$.

Thus, the function simplifies to:
$$f(x) = \sin^2 x (\lambda + \sin x) = \lambda \sin^2 x + \sin^3 x$$

### The Calculus Engine

To identify the extrema, we must find the critical points by calculating the derivative $f'(x)$ and setting it to zero. Applying the chain rule:
$$f'(x) = 2\lambda \sin x \cos x + 3\sin^2 x \cos x$$

We can factorize the derivative to simplify the search for roots:
$$f'(x) = \sin x \cos x (2\lambda + 3\sin x)$$

Setting $f'(x) = 0$ yields three potential conditions for critical points.

### The Detective Work

We analyze the three factors derived from $f'(x) = 0$:

1. $\sin x = 0$: Within the interval $(-\pi/2, \pi/2)$, this yields the critical point $x = 0$.

2. $\cos x = 0$: Within the open interval $(-\pi/2, \pi/2)$, $\cos x$ is never zero. Thus, this provides no solutions.

3. $2\lambda + 3\sin x = 0$: This implies $\sin x = -\frac{2\lambda}{3}$.

For the function to have exactly one maxima and one minima, we require exactly two distinct critical points where the derivative changes sign. Since $x=0$ is already a critical point, the equation $\sin x = -\frac{2\lambda}{3}$ must provide exactly one additional valid root within the interval.

### The Constraint

The range of $\sin x$ for $x \in (-\pi/2, \pi/2)$ is $(-1, 1)$. Therefore, the value $-\frac{2\lambda}{3}$ must satisfy:
$$-1 < -\frac{2\lambda}{3} < 1$$

Solving this inequality for $\lambda$:
$$-\frac{3}{2} < \lambda < \frac{3}{2}$$

We must also ensure that the root $\sin x = -\frac{2\lambda}{3}$ is distinct from the root $\sin x = 0$. If $-\frac{2\lambda}{3} = 0$, then $\lambda = 0$, which would result in only one critical point. Therefore, we must exclude $\lambda = 0$.

### Final Result

Combining these conditions, the set of all real values for $\lambda$ is:
$$\lambda \in \left( -\frac{3}{2}, 0 \right) \cup \left( 0, \frac{3}{2} \right)$$

**The final range for $\lambda$ is $\lambda \in (-\frac{3}{2}, \frac{3}{2}) \setminus \{0\}$.**

Animated Solution for Mathematics - Differentiation: The set of all real values of $λ$ for which the function $f (x) = (1 - cos^{2} x) \cdot (λ + sin x)$ , $x \in (- \frac{π}{2}, \frac{π}{2})$ has exactly one maxima and exactly one minima, is:

Select Answer:

Visualized Solution

The Function and the Interval

Simplifying the Function

Substituting the Identity

Expanding the Function

Finding Critical Points

Differentiating the Function

Factorizing the Derivative

Analyzing the First Root

Analyzing the Second Root

Analyzing the Third Root

Conditions for Exactly Two Extrema

Bounding the Sine Function

Solving the Inequality

The Distinctness Condition

Final Range of $λ$

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Calculus Engine

The Detective Work

The Constraint

Final Result

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