Question

The largest value of the non-negative integer a for which $$ \lim _{x \rightarrow 1}\left\\{\frac{-a x+\sin (x-1)+a}{x+\sin (x-1)-1}\right\\}^{1-\sqrt{2}}=\frac{1}{4} $$

Short answer

The largest non-negative integer value for a for which the limit equals 1/4 is 0.

Step-by-step solution

Express the limit in terms of y

Let's make a change of variable by letting y = x - 1. Then, as x approaches 1, y approaches 0. The expression within the limit becomes: \[\lim_{y \to 0}\left\{\frac{-a (y+1)+\sin(y)+a}{y+\sin(y)}\right\}^{1-\sqrt{2}}.\] This simplifies the expression by eliminating the ‘x - 1’ terms in the sine functions.

Simplify the numerator and denominator

Simplify the limit expression by distributing -a and canceling +a in the numerator: \[\lim_{y \to 0}\left\{\frac{-ay - a + a + \sin(y)+a}{y+\sin(y)}\right\}^{1-\sqrt{2}} \] Becomes \[\lim_{y \to 0}\left\{\frac{-ay + \sin(y)}{y+\sin(y)}\right\}^{1-\sqrt{2}}.\]

Examine the base of the exponential expression as y approaches 0

Since \(\lim_{y \to 0} \sin(y) / y = 1\), we can substitute this limit in to obtain: \[\lim_{y \to 0}\left\{\frac{-ay +y}{y+y}\right\}^{1-\sqrt{2}} \] Simplifying within the curly brackets gives: \[\lim_{y \to 0}\left\{\frac{y(1-a)}{2y}\right\}^{1-\sqrt{2}}\] which reduces to \[\lim_{y \to 0}\left(\frac{1-a}{2}\right)^{1-\sqrt{2}}.\]

Identify the base of the exponent for the limit to be 1/4

For the entire expression to be constant (1/4), the base must also be constant and equal to the fourth root of 1/4, i.e., \(\left(\frac{1-a}{2}\right)^{1-\sqrt{2}} = \frac{1}{4}\). Raising both sides to the power of \(\frac{1}{1-\sqrt{2}}\) gives \(\frac{1-a}{2} = \left(\frac{1}{4}\right)^{\frac{1}{1-\sqrt{2}}}.\) Solving this equation will give us the required value of a.

Solve for a

Raising 1/4 to the power of \(\frac{1}{1-\sqrt{2}}\) yields 2 (since \(\left(\frac{1}{2}\right)^{2(1-\sqrt{2})}=\frac{1}{4}\)). Therefore, \(\frac{1-a}{2} = 2\), which leads to 1 - a = 4. Solving for a, we find that a = -3. However, since we are looking for non-negative integers, the maximal non-negative integer less than -3 does not exist. Thus, the value of a must be 0 to prevent the expression from diverging.

Key concepts

Limit Evaluation

Limit evaluation is a fundamental concept in calculus which involves finding the value that a function approaches as the input approaches some value. For example, in the given JEE Advanced problem, we are required to evaluate the limit as x approaches 1. This is done by changing the variable to y which approaches 0 as x approaches 1, simplifying our calculations.

Additionally, when evaluating limits, it's critical to identify forms that are indeterminate, like 0/0, and use strategies like L'Hôpital's rule, factoring, or algebraic manipulation to resolve them. In the case of our problem, we utilize the well-known limit \(\lim_{y \to 0} \sin(y) / y = 1\), which simplifies the expression greatly. Understanding these techniques is essential for solving limit problems in the JEE Advanced mathematics section.

Exponential Expressions

Exponential expressions feature prominently in JEE Advanced mathematics, especially within limit problems. They are expressed in the form \(b^{n}\), where 'b' is the base and 'n' is the exponent. The behavior of these expressions is pivotal when evaluating limits.

When we encounter an exponential expression in a limit, as in our textbook example, we need to note that the limit of the exponent can affect the overall limit of the expression. If the base is a constant and the exponent approaches 0, the expression approaches 1. Conversely, if the exponent goes to infinity, the expression may go to infinity, 0, or remain bounded, depending on the base. As shown in our exercise, finding the relevant base is crucial for solving the problem correctly.

Trigonometric Limits

Trigonometric limits are another crucial aspect of JEE Advanced problems involving limits. These typically include functions like \(\sin x\), \(\cos x\), and \(\tan x\) which have certain limits that students are expected to know. For instance, the limit \(\lim_{x \to 0}\sin x / x = 1\) is a standard result used to simplify trigonometric limits.

It's important for students to be acquainted with these special limits and how to apply them in problems. Sine and cosine limits, for instance, help to solve indeterminate forms by acting as a bridge to simplifying the functions to a point where standard limit rules apply. Our given exercise showcases the use of such a trigonometric limit to proceed with the solution.

JEE Advanced Mathematics

JEE Advanced mathematics covers a wide array of topics including algebra, calculus, trigonometry, and geometry, all requiring a deep understanding and the ability to apply concepts in challenging problems. The syllabus aims to test analytical skills and the application of concepts. With problems like the one presented, JEE Advanced pushes students to use a combination of techniques to find solutions.

Effective preparation involves not just understanding individual topics, but also recognizing how different concepts can come together within a single problem. The limit problem involving exponential expressions and a trigonometric function is a poignant example where students are tested on multiple fronts. Mastery of the subject is signified not simply by rote learning, but by the ability to manipulate and adapt known formulas and theorems to new situations.