Problem 36 - Calculus for JEE ADVANCED

Question

We have \(f(x)=\sqrt{\sin x+\cos x}+\sqrt{7 x-x^{2}-6}\) $$ =\sqrt{\sqrt{2} \sin \left(\frac{\pi}{4}+x\right)}+\sqrt{7 x-x^{2}-6} $$ \(f(x)\) is defined if $$ \text { (i) } \begin{aligned} 7 x-x^{2}-6 \geq 0 \\ \Rightarrow \quad 1 \leq x \leq 6 \end{aligned} $$ (ii) \(\sin \left(\frac{\pi}{4}+x\right) \geq 0\) $$ \begin{aligned} & x+\frac{\pi}{4} \in \ldots[-2 \pi,-\pi] \cup[0, \pi] \cup[2 \pi, 3 \pi] \ldots \\ \Rightarrow & x \in \ldots\left[-\frac{\pi}{4}, \frac{3 \pi}{4}\right] \cup\left[\frac{7 \pi}{4}, \frac{11 \pi}{4}\right] \ldots \end{aligned} $$ From (1) and (2), we get $$ x \in\left[1, \frac{3 \pi}{4}\right] \cup\left[\frac{7 \pi}{4}, 6\right] $$

Step-by-Step Solution

Answer

\(f(x)\) is defined for \(x \in \left[1, \frac{3\pi}{4}\right] \cup \left[\frac{7\pi}{4}, 6\right]\).

Step 1: Analyse first condition for domain

The function involves the expression \(7x - x^2 - 6\), which should be non-negative for the square root to be defined. Solving the inequality \(7x - x^2 - 6 \geq 0\) gives us the interval \(1 \leq x \leq 6\). This represents the values of \(x\) that satisfy this part of the function being real.

Step 2: Analyse second condition for domain

The function also involves \(\sin\left(\frac{\pi}{4} + x\right)\), which should be greater than or equal to zero. This condition implies that \(x + \frac{\pi}{4} \) should lie within the intervals where \(\sin(x) \geq 0\), giving us \(x \in \ldots\left[-\frac{\pi}{4}, \frac{3\pi}{4}\right] \cup \left[\frac{7\pi}{4}, \frac{11\pi}{4}\right] \ldots\). These intervals emerge from the periodic nature of sine over relevant intervals.

Step 3: Intersection of intervals

We find common values from both conditions to determine where both parts of \(f(x)\) are simultaneously defined. Given the intervals from Step 1 \([1, 6]\) and Step 2 \(...\left[-\frac{\pi}{4}, \frac{3\pi}{4}\right] \cup \left[\frac{7\pi}{4}, \frac{11\pi}{4}\right] \ldots\), the intersection for these ranges results in \(x \in \left[1, \frac{3\pi}{4}\right] \cup \left[\frac{7\pi}{4}, 6\right]\). This intersection provides the domain where both functions under the square roots in \(f(x)\) are real.

Key Concepts

Intervals and Inequalities

Inequalities are mathematical expressions involving the symbols \(<\), \(>\), \(\leq\), and \(\geq\). In the context of functions, these inequalities help us determine the range of values (or intervals) for which the function is valid or "defined". An interval is simply a set of numbers, typically representing all numbers between two endpoints.

When examining functions, inequalities help us define the domain, which is the set of all possible input values (\(x\)) for which the function will give a real number. For instance, with functions involving square roots or trigonometric functions, these inequalities ensure no negative under the square root or that values fall within valid trigonometric ranges.

In our example, the inequality \(7x - x^2 - 6 \geq 0\) defines valid \(x\) values leading to a range from 1 to 6, because solving the quadratic inequality helps us find which \(x\) gives non-negative outcomes. Intervals further help visualize these valid values, such as \([1, 6]\), representing all numbers \(x\) between 1 and 6, inclusive.

Square Root Function

The square root function is one of the most essential functions in mathematics and is written as \(\sqrt{x}\). The function takes the "square root" or the number that gives \(x\) when multiplied by itself. It is important to remember that for \(\sqrt{x}\) to be "real" and thus defined within our domain, \(x\) must be zero or a positive number.

In function compositions like \(\sqrt{7x - x^2 - 6}\), ensuring the expression inside remains non-negative is crucial because square roots of negative numbers are not defined in real numbers. Thus, any values of \(x\) should respect this limitation to avoid calculating the square root of a negative number (which leads to complex or imaginary results). Hence, the domain derived from \(7x - x^2 - 6 \geq 0\) helps us to specify \(x\) values where this square root function is valid, providing intervals like \([1, 6]\).

Sine Function

The sine function is a fundamental trigonometric function usually denoted as \(\sin(x)\). It relates to the y-coordinate of a point on the unit circle as the angle \(x\) wraps around. This function oscillates between -1 and 1 as \(x\) changes, influencing its domain and range when used in composed mathematical expressions.

For our case, with \(\sin\left( \frac{\pi}{4} + x \right)\), the sine element needs to be non-negative for \(\sqrt{\sin\left( \frac{\pi}{4} + x \right)}\) to be defined. Sine values are zero or positive over specific parts of its cycle, typically noted in intervals like \([-2\pi, -\pi] \cup [0, \pi] \cup [2\pi, 3\pi]\). This periodic nature of the sine function is due to its properties, repeating values over regular intervals.

Periodic Functions

Periodic functions are functions that repeat their values at regular intervals, known as their "period". The sine function is one of the best examples of a periodic function since it repeats every \(2\pi\).

This concept is crucial for understanding how trigonometric functions like sine behave. Their repeating nature means certain conditions can be satisfied multiple times within different intervals. In our exercise, by adjusting for the period, we consider the interval shifts - specifically recognizing when \(\sin\left( \frac{\pi}{4} + x \right) \geq 0\).

By combining the interval \([1, 6]\) derived from the quadratic inequality with sine’s periodic intervals, we find shared domains applicable to our function \(f(x)\). This results in a final domain where all parts of the function are correctly defined, such as \([1, \frac{3\pi}{4}] \cup [\frac{7\pi}{4}, 6]\), accounting for the periodic repetition.