Question

Let \((x, y)\) be such that $$ \sin ^{-1}(a x)+\cos ^{-1}(y)+\cos ^{-1}(b x y)=\frac{\pi}{2} $$ Match the statements in Column I with statements in Column II and indicate your answer by darkening the appropriate bubbles in the \(4 \times 4\) matrix given in the ORS. \(\begin{array}{ll}\text { Column I } & \text { Column II }\end{array}\) (A) If \(a=1\) and \(b=0\), then \((x, y)\) (p) lies on the circle \(x^{2}+y^{2}=1\) (B) If \(a=1\) and \(b=1\), then \((x, y)\) (q) lies on \(\left(x^{2}-1\right)\left(y^{2}-1\right)=0\) (C) If \(a=1\) and \(b=2\), then \((x, y)\) (r) lies on \(y=x\) (D) If \(a=2\) and \(b=2\), then \((x, y)\) (s) lies on \(\left(4 x^{2}-1\right)\left(y^{2}-1\right)=0\)

Short answer

(A) matches with (p), (B) matches with (q), (C) does not match with any option as it does not imply y=x, and (D) matches with (s).

Step-by-step solution

Apply the given conditions A (a=1 and b=0) to the original equation

Substitute the values of a and b into the original equation: \(\sin^{-1}(ax) + \cos^{-1}(y) + \cos^{-1}(bxy) = \frac{\pi}{2}\) to get \(\sin^{-1}(x) + \cos^{-1}(y) = \frac{\pi}{2}\). Using the fact that \(\sin^{-1}(x) = \frac{\pi}{2} - \cos^{-1}(x)\), we can rewrite the equation as \(\cos^{-1}(y) + \frac{\pi}{2} - \cos^{-1}(x) = \frac{\pi}{2}\), which simplifies to \(\cos^{-1}(x) = \cos^{-1}(y)\). Since cosine is a decreasing function, this implies \(x = y\), which satisfies the circle equation \(x^2 + y^2 = 1\). Therefore, (A) matches with (p).

Apply the given conditions B (a=1 and b=1) to the original equation

Substitute the values a=1 and b=1 into the original equation to get \(\sin^{-1}(x) + \cos^{-1}(y) + \cos^{-1}(xy) = \frac{\pi}{2}\). There are no immediate simplifications, but recognizing that \(\sin^{-1}(x)\), \(\cos^{-1}(y)\), and \(\cos^{-1}(xy)\) are all in the range \([0,\pi]\), the only way their sum can be \(\frac{\pi}{2}\) is if one of them is \(\frac{\pi}{2}\), and the other two are zero. This implies either \(x=0\), or \(y=0\), or \(xy=0\), which all satisfy \(\left(x^{2}-1\right)\left(y^{2}-1\right)=0\). Therefore, (B) matches with (q).

Apply the given conditions C (a=1 and b=2) to the original equation

Substitute a=1 and b=2 into the original equation to get \(\sin^{-1}(x) + \cos^{-1}(y) + \cos^{-1}(2xy) = \frac{\pi}{2}\). For the sum of these terms to equal \(\frac{\pi}{2}\), and since \(\cos^{-1}(2xy)\) will be undefined for \(|2xy| > 1\), the only possibility is \(2xy = 0\) which implies either \(x=0\) or \(y=0\). However, this doesn't particularly suggest \(y=x\). Therefore, (C) does not match with (r).

Apply the given conditions D (a=2 and b=2) to the original equation

Substitute a=2 and b=2 into the original equation to get \(\sin^{-1}(2x) + \cos^{-1}(y) + \cos^{-1}(2xy) = \frac{\pi}{2}\). This time, the \(\sin^{-1}(2x)\) term imposes that \(2x\) must be within \([-1, 1]\), therefore \(x\) must be within \([-\frac{1}{2}, \frac{1}{2}]\). This means the term can be 0 or \(\frac{\pi}{2}\). Similarly, \(\cos^{-1}(2xy)\) implies that \(2xy\) must be within \([-1, 1]\), resulting in the same range for \(y\). This satisfies \(\left(4 x^{2}-1\right)\left(y^{2}-1\right)=0\), hence (D) matches with (s).

Key concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are the inverses of the standard trigonometric functions, allowing us to determine the angle that corresponds to a given trigonometric ratio. For instance, \( \sin^{-1}(x) \) is the inverse sine function or arcsine, which returns the angle whose sine is \( x \). A basic understanding of these functions includes knowing their principal values, which lie within specific ranges: for arcsine \( \sin^{-1}(x) \) and arctangent \( \tan^{-1}(x) \) the range is \( [-\frac{\pi}{2}, \frac{\pi}{2}] \) and for arccosine \( \cos^{-1}(x) \) it is \( [0, \pi] \).

When solving JEE Advanced problems that involve inverse trigonometric functions, it's important to be aware of these ranges, as well as the corresponding graphs and properties. For example, knowing that \( \sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} \) can guide you in simplifying complex expressions. Moreover, you'll often have to leverage identities such as \( \sin^{-1}(x) = \frac{\pi}{2} - \cos^{-1}(x) \) to streamline equations and find solutions to tricky questions involving variables and parameters.

When it comes to exercises like the one in our example, the intelligent application of inverse trigonometric properties becomes a tool for deducing relationships between variables. In this context, the solutions emerge from a mix of algebraic manipulation and conceptual understanding of these functions.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for all values of the included variables. These identities are crucial tools for simplifying and solving trigonometric expressions in JEE Advanced Mathematics problems. They encompass basic relationships like \( \sin^2(x) + \cos^2(x) = 1 \) as well as sum and difference formulas, double and half-angle formulas, and product-to-sum formulas.

It is vital for students to memorize these identities and learn how to apply them flexibly. In the context of our example, one would realize that the correct application of trigonometric identities can simplify equations significantly, which is often the first step in solving the problem. A deeper understanding of the identities allows students to recognize opportunities for their use in converting complex equations into solvable ones. For example, recognizing that the sum of the inverse trigonometric functions equals a specific value might suggest the strategic use of a complementary angle identity to facilitate a solution. Incorporating trigonometric identities into solutions increases efficiency and accuracy in problem-solving, which is paramount for competitive exams like the JEE Advanced.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, merges algebra and geometry to describe the position of points on a plane using numerical coordinates. It's a crucial area of mathematics for the JEE Advanced exam, as it forms the basis for understanding and solving problems related to shapes, sizes, relative positions, and properties of space.

In solving JEE Advanced problems, one must be proficient in equations that define various geometric entities such as lines, circles, and conics. For instance, the equation \(x^{2} + y^{2} = r^{2}\) represents a circle with a radius \(r\) centered at the origin. Familiarity with these equations and figures is essential. In our sample problem, understanding that the condition \(x = y\) can describe the line\(y = x\) or, with further constraints, the equation of a circle, illustrates how coordinate geometry can provide geometric interpretations to algebraic results.

Moreover, interpreting the products of equations like \((x^{2}-1)(y^{2}-1)=0\) involves recognizing that the equation represents the axes or the rectangle hyperbolas depending on the context. To build a solid foundation in coordinate geometry, students should focus on mastering different forms of equations and their geometric interpretations, transformations, and loci. These are instrumental in visualizing the solution to a problem beyond the abstract manipulation of symbols and equations.