Question

Let \(-\frac{\pi}{6}<\theta<-\frac{\pi}{12} .\) Suppose \(\alpha_{1}\) and \(\beta_{1}\) are the roots of the equation \(x^{2}-2 x \sec \theta+1=0\) and \(\alpha_{2}\) and \(\beta_{2}\) are the roots of the equation \(x^{2}+2 x \tan \theta-1=0\). If \(\alpha_{1}>\beta_{1}\) and \(\alpha_{2}>\beta_{2}\), then \(\alpha_{1}+\beta_{2}\) equals (A) \(2(\sec \theta-\tan \theta)\) (B) \(2 \sec \theta\) (C) \(-2 \tan \theta\) (D) 0

Short answer

0

Step-by-step solution

Identify the Quadratic Equations

Begin by recognizing that the task requires us to deal with two quadratic equations,\[x^{2}-2x\sec\theta+1=0\]and\[x^{2}+2x\tan\theta-1=0.\]We are given that \(\alpha_{1}\) and \(\beta_{1}\) are the roots of the first quadratic equation and \(\alpha_{2}\) and \(\beta_{2}\) are the roots of the second. Also, it is provided that \(\alpha_{1} > \beta_{1}\) and \(\alpha_{2} > \beta_{2}\), meaning that \(\alpha_{1}\) and \(\alpha_{2}\) are the larger roots of their respective quadratic equations.

Use Vieta's Formulas

Recall Vieta's formulas which state that for a quadratic equation \(ax^2 + bx + c = 0\) with roots \(r_1\) and \(r_2\), the sum of the roots is given by \(r_1 + r_2 = -\frac{b}{a}\) and the product is given by \(r_1 \times r_2 = \frac{c}{a}\). Apply this to both equations:For \(x^{2}-2x\sec\theta+1=0\), the sum of the roots \(\alpha_{1}+\beta_{1} = 2\sec\theta\),For \(x^{2}+2x\tan\theta-1=0\), the sum of the roots \(\alpha_{2}+\beta_{2} = -2\tan\theta\).

Calculate \(\alpha_{1} + \beta_{2}\)

Notice that we need to find \(\alpha_{1} + \beta_{2}\), not the sum of the roots of either equation. However, we can express \(\beta_{2}\) in terms of \(\alpha_{2}\) using the sum of the roots for the second equation: \(\beta_{2} = -2\tan\theta - \alpha_{2}\). Now, using the sum for the first equation \(\alpha_{1} + \beta_{1} = 2\sec\theta\), and replacing \(\beta_{1}\) with \(\beta_{1} = 2\sec\theta - \alpha_{1}\), we deduce that \(\alpha_{1} + \beta_{2} = \alpha_{1} + (-2\tan\theta - \alpha_{2})\).

Utilize the Range of \(\theta\)

The values of secant and tangent for the range \(-\frac{\pi}{6}<\theta< -\frac{\pi}{12}\) are negative because secant is reciprocal of cosine and tangent is sine over cosine, both of which are negative in the fourth quadrant (where our \(\theta\) lies). So \(\sec\theta\) and \(\tan\theta\) are negative, meaning \(2\sec\theta\) and \(2\tan\theta\) are also negative.

Determine the Correct Option Based on the Calculations

Combining the results from steps 3 and 4, replace \(\alpha_{1}\) with \(2\sec\theta - \beta_{1}\) and \(\alpha_{2}\) with \(-2\tan\theta - \beta_{2}\) in the expression for \(\alpha_{1} + \beta_{2}\):\[\alpha_{1} + \beta_{2} = (2\sec\theta - \beta_{1}) + (-2\tan\theta - (-2\tan\theta - \beta_{2}))\]Simplify the expression:\[\alpha_{1} + \beta_{2} = 2\sec\theta - \beta_{1} + \beta_{2}\]Now, since \(\beta_{1} = 2\sec\theta - \alpha_{1}\) and \(\beta_{2} = -2\tan\theta - \alpha_{2}\), and considering that secant and tangent are negative in the given range, the only possible option that will result in a cancellation that yields zero is option (D) zero. This is because the sum of \(\beta_{1}\) and \(-\beta_{2}\) should be equal to the sum of \(\alpha_{1}\) and \(-\alpha_{2}\), leaving out the constants secant and tangent.

Key concepts

Vieta's Formulas

Vieta's formulas provide a direct link between the coefficients of a polynomial equation and the sums and products of its roots. These formulas are especially useful when dealing with quadratic equations, which are equations of the second degree. For the quadratic equation \(ax^2 + bx + c = 0\) with roots \(r_1\) and \(r_2\), Vieta's formulas tell us that the sum of the roots is \(r_1 + r_2 = -\frac{b}{a}\) and the product of the roots is \(r_1 \times r_2 = \frac{c}{a}\).

Understanding Vieta's formulas is crucial for solving many mathematical problems, particularly when working with the JEE Advanced Mathematics syllabus. These formulas not only simplify the calculation of the roots but also aid in solving problems without actually finding the roots, which can save time during a high-stakes exam like the JEE Advanced. It's also important for students to recognize that Vieta's formulas can be extended to higher degree polynomials, with each formula providing relationships between coefficients and combinations of roots.

Trigonometric Functions

Trigonometric functions are fundamental in understanding the behavior of angles and their relationship to triangles. These functions include sine, cosine, tangent, cotangent, secant, and cosecant. In the context of the JEE Advanced Mathematics, trigonometry plays a vital role across numerous topics such as algebra, calculus, and geometry.

In solving the given exercise, the functions secant (sec) and tangent (tan) are employed, which are derived from the basic sine and cosine functions. The secant of an angle is the reciprocal of its cosine, while the tangent is the sine divided by the cosine. These functions are periodic and have specific values at standard angles. It's essential for students to be comfortable with these functions' values, especially at common angles like \(\frac{\pi}{6}\), \(\frac{\pi}{4}\), and \(\frac{\pi}{3}\), and to understand their signs in different quadrants to solve trigonometric equations effectively.

JEE Advanced Mathematics

The JEE (Joint Entrance Examination) Advanced is known for its challenging and diverse set of mathematics problems, which require a deep understanding of various mathematical concepts and their applications. It tests students on topics such as algebra, calculus, vectors, and complex numbers, so a strong grasp of each area is imperative.

In the context of this quadratic equations problem, the examination is testing the student's ability to merge their understanding of algebra (through Vieta's formulas) with trigonometry. The JEE Advanced tends to present problems that integrate multiple mathematical disciplines, encouraging students to have a comprehensive mastery over the subject. Such questions also evaluate analytical skills and problem-solving strategies, which are core to success in the examination. It's essential for aspirants to practice problems that not only involve straightforward applications of formulas but also challenge them to connect various mathematical concepts in creative ways.