If A = sin² x + cos⁴ x, then for all real x: (A) 13/16 ≤ A ≤ 1 (B) 1 ≤ A ≤ 2 (C) 3/4 ≤ A ≤ 13/16 (D)

Q: Wait, why can't we just use $\sin^2 x + \cos^2 x = 1$ to simplify this?

That identity is powerful, but it only works when the powers match. Here, you have $\cos^4 x$, not $\cos^2 x$. You must convert the $\sin^2 x$ term to $\cos^2 x$ first to unify the expression.

Q: Why is the domain of $t$ restricted to $[0, 1]$?

Excellent question! Since $t = \cos^2 x$, and we know that for any real $x$, $-1 \le \cos x \le 1$, squaring this value forces $t$ to be between $0$ and $1$. Ignoring this domain restriction is the most common trap in JEE problems!

Q: How do we know the vertex is the minimum?

Because the coefficient of $t^2$ is positive ($1$), the parabola opens upwards. The vertex is the lowest point. Since the vertex $t = 1/2$ falls within our domain $[0, 1]$, the minimum value of the function is indeed at the vertex.

Question

If $A = \sin^2 x + \cos^4 x$, then for all real $x$:

Options:

(A) $13/16 \le A \le 1$

(B) $1 \le A \le 2$

(C) $3/4 \le A \le 13/16$

(D) $3/4 \le A \le 1$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, aspiring engineers! Today, we are going to dissect a problem that perfectly captures the spirit of JEE Advanced: it looks like a simple trigonometric expression, but it is actually a masterclass in algebraic manipulation and domain analysis.

Let us dive into the expression $A = \sin^2 x + \cos^4 x$. Our goal is to find the range of $A$ for all real $x$.

The first thing that should jump out at you is the lack of harmony. We have a $\sin^2 x$ and a $\cos^4 x$. In the world of mathematics, when you see a mix of ratios, your first instinct should be unification.

We know the fundamental identity $\sin^2 x = 1 - \cos^2 x$. By substituting this, we transform our expression into:

$$A = (1 - \cos^2 x) + \cos^4 x$$

Rearranging this gives us:

$$A = \cos^4 x - \cos^2 x + 1$$

Now, we have a beautiful, unified polynomial in terms of $\cos^2 x$.

### The Algebraic Bridge: Stepping into the Quadratic Realm

Trigonometry can sometimes obscure the underlying algebraic structure. To clear the fog, let us introduce a dummy variable.

Let $t = \cos^2 x$. Suddenly, our expression becomes a simple quadratic function:

$$f(t) = t^2 - t + 1$$

This is the moment where many students rush forward, but pause! We must define the domain of $t$.

Since $x$ is any real number, we know that $-1 \le \cos x \le 1$. When we square this, the range of $\cos^2 x$ becomes $0 \le \cos^2 x \le 1$.

Therefore, our variable $t$ is strictly confined to the interval $[0, 1]$. This is the trap! If you treat $t$ as any real number, you will miss the boundaries and fail the problem.

### The Parabolic Landscape: Finding the Extremes

Now, we are looking at the parabola $f(t) = t^2 - t + 1$ within the domain $t \in [0, 1]$. To find the minimum, we look for the vertex.

Completing the square is our most reliable tool here. We take the coefficient of $t$, which is $-1$, halve it to get $-1/2$, and square it to get $1/4$.

We rewrite the function as:

$$f(t) = (t^2 - t + 1/4) - 1/4 + 1$$

This simplifies to:

$$f(t) = (t - 1/2)^2 + 3/4$$

This vertex form is a revelation. Since any real number squared is non-negative, $(t - 1/2)^2 \ge 0$. The minimum value of this square is $0$, which occurs at $t = 1/2$.

Since $1/2$ is within our domain $[0, 1]$, the minimum value of our function is $3/4$.

### The Final Boundary Check

We have the minimum, but what about the maximum? For a parabola on a closed interval, the maximum must occur at one of the boundaries.

Let us test the endpoints of our domain $t \in [0, 1]$:

$$f(0) = 0^2 - 0 + 1 = 1$$
$$f(1) = 1^2 - 1 + 1 = 1$$

Both boundaries yield the same value of $1$. Thus, the maximum value of $A$ is $1$.

We have successfully navigated the landscape: the minimum is $3/4$ and the maximum is $1$. The **range of $A$ is $[3/4, 1]$**.

This problem teaches us that the most complex-looking trigonometric expressions are often just quadratic functions in disguise. Keep your eyes on the domain, trust your algebraic tools, and you will conquer any JEE problem that comes your way!

Animated Solution for Mathematics - Trigonometry: If $A = sin^{2} x + cos^{4} x$ , then for all real $x$ :

Select Answer:

Visualized Solution

Analyze the Expression

Unifying the Trigonometric Ratios

Substituting the Identity

Algebraic Substitution

Establishing the Domain

Visualizing the Quadratic

Completing the Square

Perfect Square Form

Finding the Minimum Value

Calculating the Minimum

Checking the Boundaries

Final Range of $A$

The Sigma Insight: Trigonometric Ratios and Identities

Analyzing the Setup

The Algebraic Bridge

The Parabolic Landscape

The Final Boundary Check

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Animated Solution for Mathematics - Trigonometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If A=sin2x+cos4x, then for all real x: