Let the function f: (0, π) → ℝ be defined by f(θ) = (sin θ + cos θ)² + (sin θ - cos θ)⁴. Suppose the

Q: Why can't we just find the minimum of the quadratic $f(t) = t^2 - t + 2$ and stop there?

That is a brilliant question! Finding the minimum of the quadratic $f(t)$ tells you the minimum *value* of the function, but it doesn't tell you *where* (at what $\theta$) that minimum occurs. Since the problem asks for the values of $\theta$ that produce local minima, you must map the minimum of the quadratic back to the original variable $\theta$ using the substitution $t = \sin 2\theta$. The quadratic gives you the 'what', but the calculus gives you the 'where'.

Q: Why do we have four critical points? I thought a quadratic only had one vertex.

You are absolutely right that a quadratic has only one vertex. However, our function is $f(\theta) = f(\sin 2\theta)$. The derivative $\frac{df}{d\theta}$ involves the derivative of $\sin 2\theta$, which is $2\cos 2\theta$. So, the critical points arise from two sources: where the quadratic's derivative is zero (the vertex) AND where the derivative of the inner function $\sin 2\theta$ is zero. This is why we get more points than just the vertex of the parabola!

Q: Is there a faster way than the first derivative test to identify the minima?

You could use the second derivative test, but it would be quite algebraically intensive here. The first derivative test is actually the most robust method for this problem. By checking the sign of the derivative around each critical point, you gain a clear, visual understanding of the function's behavior, which is less prone to calculation errors than computing the second derivative of a trigonometric expression.

Question

Let the function $f: (0, \pi) 	o \mathbb{R}$ be defined by $f(	heta) = (\sin 	heta + \cos 	heta)^2 + (\sin 	heta - \cos 	heta)^4$. Suppose the function $f$ has a local minimum at $	heta$ precisely when $	heta \in \{\lambda_1 \pi, \dots, \lambda_r \pi\}$, where $0 < \lambda_1 < \dots < \lambda_r < 1$. Then the value of $\lambda_1 + \dots + \lambda_r$ is ____.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the mathematical landscape. Today, we are going to dismantle a problem that, at first glance, looks like a tangled mess of trigonometric powers.

We have the function $f(	heta) = (\sin 	heta + \cos 	heta)^2 + (\sin 	heta - \cos 	heta)^4$.

The powers of two and four seem to suggest a long, arduous expansion. But here is the secret of JEE Advanced: the problem setter rarely wants you to do brute force. They want you to see the structure.

### Phase 1: Peeling Back the Layers

Let us look at the first term: $(\sin 	heta + \cos 	heta)^2$. If we expand this, we get $\sin^2 	heta + \cos^2 	heta + 2\sin 	heta \cos 	heta$.

We know the fundamental identity $\sin^2 	heta + \cos^2 	heta = 1$, and we know that $2\sin 	heta \cos 	heta = \sin 2	heta$. Suddenly, the first term becomes $1 + \sin 2	heta$.

Now, let us tackle the second term: $(\sin 	heta - \cos 	heta)^4$. We can view this as $((\sin 	heta - \cos 	heta)^2)^2$.

Expanding the inner square gives us $\sin^2 	heta + \cos^2 	heta - 2\sin 	heta \cos 	heta$, which simplifies to $1 - \sin 2	heta$. So, the entire second term becomes $(1 - \sin 2	heta)^2$.

### Phase 2: The Quadratic Lens

By substituting these back into our original function, we get $f(	heta) = (1 + \sin 2	heta) + (1 - \sin 2	heta)^2$.

Let us expand the square:
$$f(	heta) = 1 + \sin 2	heta + 1 - 2\sin 2	heta + \sin^2 2	heta$$

Collecting the terms, we arrive at:
$$f(	heta) = \sin^2 2	heta - \sin 2	heta + 2$$

This is the moment of clarity. If we let $t = \sin 2	heta$, our function transforms into a simple quadratic: $f(t) = t^2 - t + 2$.

We have moved from a complex trigonometric expression to a beautiful, familiar parabola. The vertex of this parabola occurs at:
$$t = -\frac{b}{2a} = -\frac{-1}{2 \cdot 1} = \frac{1}{2}$$

This tells us that the function $f(t)$ reaches its minimum when $t = 1/2$.

### Phase 3: The Calculus Bridge

We need to find where the derivative $f'(	heta)$ is zero. Using the chain rule on our function $f(	heta) = \sin^2 2	heta - \sin 2	heta + 2$, we get:

$$f'(	heta) = \frac{d}{d	heta}(\sin^2 2	heta - \sin 2	heta + 2)$$

$$f'(	heta) = 2\sin 2	heta \cdot \cos 2	heta \cdot 2 - \cos 2	heta \cdot 2$$

$$f'(	heta) = 4\sin 2	heta \cos 2	heta - 2\cos 2	heta$$

$$f'(	heta) = 2\cos 2	heta (2\sin 2	heta - 1)$$

Setting $f'(	heta) = 0$ gives us two conditions: $\cos 2	heta = 0$ or $\sin 2	heta = 1/2$.

### Phase 4: The Final Verdict

For $\cos 2	heta = 0$ in the interval $	heta \in (0, \pi)$, we have $2	heta \in (0, 2\pi)$, so $2	heta = \pi/2, 3\pi/2$, which means $	heta = \pi/4, 3\pi/4$.

For $\sin 2	heta = 1/2$, we have $2	heta = \pi/6, 5\pi/6$, which means $	heta = \pi/12, 5\pi/12$.

Now, we perform the sign analysis. By checking the sign of $f'(	heta) = 2\cos 2	heta (2\sin 2	heta - 1)$, we find that at $	heta = \pi/12$ and $	heta = 5\pi/12$, the derivative changes from negative to positive.

These are our local minima! The other points, $\pi/4$ and $3\pi/4$, turn out to be local maxima.

Thus, our $\lambda_1 = 1/12$ and $\lambda_2 = 5/12$. The sum is:
$$\lambda_1 + \lambda_2 = \frac{1}{12} + \frac{5}{12} = \frac{6}{12} = 0.5$$

**The final result is 0.5.**

Enter Numerical Value:

Visualized Solution

$Analyze f (θ)$

$Expand First Term$

$Expand Second Term$

$Combine and Simplify$

$Quadratic Substitution$

$Minimum of Parabola$

$Derivative f^{'} (θ)$

$Critical Points Setup$

$Solve Case 1$

$Solve Case 2$

$First Derivative Test$

$Sign Analysis at π /12$

$Sign Analysis at 5 π /12$

$Identify λ_{1} and λ_{2}$

$Final Sum$

The Sigma Insight: Maxima and Minima

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

Let the set of all positive values of $λ$ , for which the point of local minimum of the function $(1 + x (λ^{2} - x^{2}))$ satisfies $\frac{x ^{2} + x + 2}{x ^{2} + 5 x + 6} < 0$ , be $(α, β)$ . Then $α^{2} + β^{2}$ is equal to _________

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:

Let $f (x) = sin^{3} x + λ sin^{2} x, - π /2 < x < π /2$ . Find the intervals in which $λ$ should lie in order that $f (x)$ has exactly one minimum and exactly one maximum.

The function $f (x) = \frac{x}{2} + \frac{2}{x}$ has a local minimum at

Let $f (x) = \frac{s i n π x}{x ^{2}}, x > 0$ . Let $x_{1} < x_{2} < x_{3} < \dots < x_{n} < \dots$ be all the points of local maximum of $f$ and $y_{1} < y_{2} < y_{3} < \dots < y_{n} < \dots$ be all the points of local minimum of $f$ . Then which of the following options is/are correct ?

Let $f (x) = ∣ (x - 1) (x^{2} - 2 x - 3) ∣ + x - 3, x \in R$ . If m and M are respectively the number of points of local minimum and local maximum of f in the interval (0, 4), then m + M is equal to ______

Let $f (x) = x^{2} + \frac{1}{x ^{2}}$ and $g (x) = x - \frac{1}{x}, x \in R - {- 1, 0, 1}$ . If $h (x) = \frac{f ( x )}{g ( x )}$ , then the local minimum value of $h (x)$ is :

Let $g (x) = f (x) + f (1 - x)$ and $f^{''} (x) > 0, x \in (0, 1)$ . If $g$ is decreasing in the interval $(0, α)$ and increasing in the interval $(α, 1)$ , then $tan^{- 1} 2 α + tan^{- 1} (\frac{1}{α}) + tan^{- 1} (\frac{α + 1}{α})$ is equal to

Let $f : R \to R$ be a function defined by $f (x) = ∣∣ x + 2∣ - 2∣ x ∣∣$ . If $m$ is the number of points of local minima and $n$ is the number of points of local maxima of $f$ , then $m + n$ is

Enter Numerical Value:

Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Expand First Term

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Expand Second Term

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Combine and Simplify

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Quadratic Substitution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Minimum of Parabola

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Derivative f′(θ)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Critical Points Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solve Case 1

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solve Case 2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } First Derivative Test

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Sign Analysis at π/12

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Sign Analysis at 5π/12

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Identify λ1​ and λ2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Sum