Problem 124 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(f(x)=\sin x+\frac{1}{2} \cos 2 x\). Then (A) \(\min\) \(x \in\left[0, \frac{\pi}{2}\right] f(x)<\frac{4}{3}\) (B) \(\min _{x \in\left[0, \frac{\pi}{2}\right] f(x)>\frac{3}{4}}\) (C) \(\min _{x \in\left[0, \frac{\pi}{2}\right] f(x)>\frac{2}{3}}\) (D) \(\min _{x \in\left[0, \frac{\pi}{2}\right]} f(x)<\frac{3}{2}\)

Step-by-Step Solution

Answer

Options (A) and (D) are true.

Step 1: Write Down the Function

Let \( f(x) = \sin x + \frac{1}{2} \cos 2x \). We'll analyze the function to understand its behavior over the interval \( [0, \frac{\pi}{2}] \).

Step 2: Convert the Expression

Using the identity \( \cos 2x = 2\cos^2 x - 1 \), rewrite \( \cos 2x \) in terms of \( x \). Thus, \( f(x) = \sin x + \frac{1}{2}(2\cos^2 x - 1) = \sin x + \cos^2 x - \frac{1}{2} \).

Step 3: Identify Critical Points

Find the derivative: \( f'(x) = \cos x - 2\cos x \sin x = \cos x (1 - 2\sin x) \). Set \( f'(x) = 0 \): \( \cos x = 0 \) or \( \sin x = \frac{1}{2} \). In \( [0, \frac{\pi}{2}] \), the solution is \( x = \frac{\pi}{6} \).

Step 4: Evaluate Function at Critical Points and Bounds

Calculate \( f(0) = \frac{1}{2}, f(\frac{\pi}{6}) = \frac{1}{2} + \frac{3}{4} - \frac{1}{2} = \frac{3}{4}, \) and \( f(\frac{\pi}{2}) = 1 + 0 - \frac{1}{2} = \frac{1}{2} \).

Step 5: Determine Minimum Value

Comparing values: \( f(0) = \frac{1}{2}, f(\frac{\pi}{6}) = \frac{3}{4}, f(\frac{\pi}{2}) = \frac{1}{2} \). Thus, the minimum value is \( \frac{1}{2} \).

Step 6: Compare Min Value with Given Options

Check each option against \( \min f(x) = \frac{1}{2} \):- (A) True since \( \frac{1}{2} < \frac{4}{3} \)- (B) False since \( \min > \frac{3}{4} \) is not true- (C) False since \( \min > \frac{2}{3} \) is not true- (D) True since \( \frac{1}{2} < \frac{3}{2} \).

Key Concepts

Derivative

Understanding derivatives is a cornerstone of calculus and essential in finding how a function behaves. In simple terms, the derivative of a function at any point measures how the function's value changes as its input changes. It's like discovering the slope of the function at a given point. This is particularly useful with trigonometric functions to analyze their slope.

For the given function, which is a combination of trigonometric terms \(f(x) = \sin x + \frac{1}{2} \cos 2x\), finding the derivative involves some familiarity with basic trigonometric derivatives. The derivative of \(\sin x\) is \(\cos x\), and the chain rule applies when dealing with more complex parts like \(\cos 2x\).

When we differentiate \(f(x)\), using the identity of \(\cos 2x = 2\cos^2 x - 1\), it helps simplify the derivative to \(f'(x) = \cos x (1 - 2\sin x)\). Zeroing in on this function's derivative helps determine critical points, which are key to understanding how and when the function's input influences its behavior drastically.

Critical Points

Critical points are where the derivative of a function is zero or undefined. These points are significant in calculus because they often represent local maxima, minima, or points of inflection, where the function changes its direction or curvature.

For \(f(x) = \sin x + \frac{1}{2} \cos 2x\), setting the derivative equal to zero and solving for \(x\) helps us find critical points.

• Set \(\cos x (1 - 2\sin x) = 0\) to find critical values.
• This gives two possibilities: \(\cos x = 0\) or \(\sin x = \frac{1}{2}\).

On the interval \([0, \frac{\pi}{2}]\), only \(\sin x = \frac{1}{2}\) is valid, leading to a critical point at \(x = \frac{\pi}{6}\).

Evaluating \(f\) at critical points and endpoints verifies their role in finding extremal values like minimum or maximum values within the interval. These critical points help ensure we fully understand the behavior of the function across a specific domain, especially when combined with trigonometric functions.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are always true for every value of the variable where both sides are defined. They are extremely useful in simplifying expressions and solving trigonometric equations.

In the problem, the key identity used is the double angle formula: \(\cos 2x = 2\cos^2 x - 1\). This identity allows us to express \(\cos 2x\) in terms of a simpler form involving \(\cos x\). By substituting this into the original function, \(f(x) = \sin x + \frac{1}{2}(2\cos^2 x - 1)\), we simplify it to \(f(x) = \sin x + \cos^2 x - \frac{1}{2}\).

This step ensures that finding critical points and evaluating the derivative is more straightforward. It highlights the power of trigonometric identities in reducing complex expressions and helps efficiently address the trigonometric problems by breaking them into simpler, more manageable parts.

• Using identities can reveal the hidden structures within functions.
• They are essential tools in both simplifying problems and solving them effectively.

Understanding these identities and applying them to manipulate and simplify expressions is critical, especially in calculus, where such transformations are often needed.