Problem 71 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(A=\sin ^{2} x+\cos ^{4} x\) then, for all real values of \(x\), (A) \(\frac{13}{16} \leq A \leq 1\) (B) \(1 \leq A \leq 2\) (C) \(\frac{3}{4} \leq A \leq \frac{13}{16}\) (D) \(\frac{3}{4} \leq A \leq 1\)

Step-by-Step Solution

Answer

Option (A) \(\frac{13}{16} \leq A \leq 1\) is correct.

Step 1: Expression Analysis

We start analyzing the expression \(A = \sin^2 x + \cos^4 x\). We know that \(\sin^2 x + \cos^2 x = 1\) for any angle \(x\). The square and fourth powers of the trigonometric functions limit the value of \(A\).

Step 2: Substitute and Simplify

Let \(y = \cos^2 x\). Then \(\sin^2 x = 1 - y\). So, we have \(A = (1-y) + y^2\). Simplifying gives \(A = y^2 - y + 1\).

Step 3: Derivative and Critical Points

Find the derivative \(A' = 2y - 1\). Setting this to zero gives the critical point: \(y = \frac{1}{2}\).

Step 4: Evaluate at Critical and Boundary Values

Evaluate \(A\) at the critical point \(\frac{1}{2}\), and at the boundary values \(y = 0\) and \(y = 1\). For \(y = 0\), \(A = 1\); for \(y = 1\), \(A = 1\); and for \(y = \frac{1}{2}\), \(A = \frac{13}{16}\).

Step 5: Conclusion from Calculations

These evaluations show that \(A\) achieves its minimum value of \(\frac{13}{16}\) at \(y = \frac{1}{2}\) and a maximum value of 1 at both \(y = 0\) and \(y = 1\). Hence, \(\frac{13}{16} \leq A \leq 1\).

Key Concepts

Function Analysis

When analyzing functions, we focus on understanding how they behave under different inputs. In this exercise, we start with the given expression \(A = \sin^2 x + \cos^4 x\). Our task is to investigate all real values of \(x\) that this expression can produce. Function analysis helps us explore key characteristics like trends, limits, and possible values.
Let's consider \(\sin^2 x\) and \(\cos^4 x\) specifically:

• \(\sin^2 x\) represents the square of the sine function, which always falls between 0 and 1 for real values of \(x\).
• Likewise, \(\cos^4 x\), being a fourth power, also ranges from 0 to 1.

This basic insight guides us to see that \(A\) is composed of these bounded terms. The use of trigonometric identities such as \(\sin^2 x + \cos^2 x = 1\) is essential in simplifying and transforming functions to uncover their net impact. With these transformations, we express \(A\) to mean more predictable terms for further analysis.

Critical Points

Critical points are vital in our analysis because they indicate where important changes occur in a function. With the equation \(A = y^2 - y + 1\), derived from defining \(y = \cos^2 x\), finding critical points involves differentiating this expression. The derivative, \(A' = 2y - 1\), shows the slope of \(A\) with respect to \(y\).
To find where the slope is zero, we set the derivative equal to zero:

• \(2y - 1 = 0\)
• Solve for \(y\) to find: \(y = \frac{1}{2}\)

This \(y = \frac{1}{2}\) is a critical point because it's where the rate of increase or decrease in the function pauses. Critical points assist us in recognizing local minima, maxima, and potential inflection points, which can deeply impact the interpretation of a function's graph. Therefore, they indicate where we should delve deeper into the function’s behavior.

Derivative Application

The derivative is an essential tool for analyzing the behavior of functions, particularly in determining rates of change. In our problem with \(A = y^2 - y + 1\), we used the derivative \(A' = 2y - 1\) to find critical points. Derivatives enable us to apply rules of calculus to predict and calculate the behavior of functions beyond their basic definitions.
To apply the derivative effectively:

• We evaluate \(A\) at critical and boundary values: \(y = \frac{1}{2}, 0, 1\).
• Substitute these values into \(A\) => \(A(0) = 1\), \(A(1) = 1\), and \(A(\frac{1}{2}) = \frac{13}{16}\).

The derivative shows not only where changes occur but also the nature (minimum or maximum) of these points. By applying the derivative correctly, we establish that the function achieves a minimum value of \(\frac{13}{16}\) at \(y = \frac{1}{2}\) and a maximum of 1, solidifying that \(\frac{13}{16} \leq A \leq 1\). This application underscores the importance of calculus for understanding nuances in function behavior.