Problem 24 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If \(\lim _{x \rightarrow 2} \frac{\tan (x-2)\left\\{x^{2}+(k-2) x-2 k\right\\}}{x^{2}-4 x+4}=5\), then \(\mathrm{k}\) is equal to: \(\quad\) [Online April 11, 2014] (a) 0 (b) 1 (c) 2 (d) 3

Step-by-Step Solution

Answer

The value of \(k\) that satisfies the limit is 3.

Step 1: Identify the Type of Indeterminate Form

The given limit \[\lim _{x \rightarrow 2} \frac{\tan (x-2)\left\{x^{2}+(k-2) x-2k\right\}}{x^{2}-4 x+4}=5\]needs to be solved. When plugging \(x = 2\) directly into the denominator \(x^{2}-4x+4\), we get \(0\), thus indicating a potential indeterminate form of type \(\frac{0}{0}\). Check for indeterminate form in the numerator as well.

Step 2: Simplify and Factor the Denominator

The denominator can be rewritten and factored as:\[x^{2} - 4x + 4 = (x-2)^{2}\]which confirms the indeterminate form \(\frac{0}{0}\) since both numerator and denominator involve \((x-2)\).

Step 3: Analyze and Factorize the Numerator

Rewrite the numerator\[\tan(x-2)\left(x^{2} + (k-2)x - 2k\right)\].The non-tangent part resembles a polynomial that likely contains a factor of \((x-2)\). Note that directly \(x=2\), the expression \(x^{2} + (k-2)x - 2k\) must equal zero for true indeterminate form. Substitute \(x=2\) into the polynomial:\[(2)^2 + (k-2)\cdot2 - 2k = 0\]Solve this to find some restriction on \(k\).

Step 4: Set Equation Equal to Zero for Factorization

Substitute \(x=2\) into the polynomial:\[4 + 2(k-2) - 2k = 0\]Simplify it to solve for \(k\):\[4 + 2k - 4 - 2k = 0\Rightarrow 0 = 0\].This doesn't give a solution; meaning instead check factor compatibility with the given limit value.

Step 5: Determine Polynomial Value Through Limit Set-up

For the given expression to not be indeterminate further apart from the \((x-2)\) term:The non-tangent polynomial \(x^{2} + (k-2)x - 2k\) should result in such a factor at \(x=2\) for the numerator to remove \((x-2)\):Use the limit provided,\[\lim_{x \to 2} \frac{\tan(x-2)(x-2)u}{(x-2)^2} = 5\]where \(u\) simplifies to a constant as \(x\to2\), specifically analyze through polynomial division of \((x-2)\) and set up against 5.

Step 6: Solve and Compare Limit Operations

The cancellation should result in \[ \lim_{x\rightarrow 2} \frac{\tan(x-2)((x-2)v)}{(x-2)^2} = \lim_{x\rightarrow 2} \frac{ (v) \tan(x-2)}{(x-2)} \]using the derivative of \(\tan(x)\) where \(u \rightarrow 1\) as equivalent to \(\cos(x)^{-2} \approx 1\); realize \((\tan(x)\to(x))\):When simplified here to equal actual values, \(v = 5\) directly impacts value formation. Simplification implication occurs once equals 5.

Key Concepts

Indeterminate forms

When dealing with limits, we often encounter situations called indeterminate forms. These forms arise when the limit yields an expression like \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), or similar expressions that do not provide direct information about the limit. In our exercise, substituting \(x = 2\) into the expression results in the indeterminate form \(\frac{0}{0}\). This is because both the numerator and the denominator become zero when \(x\) approaches the value of 2. Identifying such forms is crucial because they indicate the need for further simplification or manipulation to determine the limit.

• Recognition of indeterminate forms is the first step in solving many limit problems.
• Techniques such as factoring, L'Hôpital's rule, or algebraic manipulation are often used to resolve these forms.

Carefully examining the initial expression reveals where these forms can occur and prepares us to employ strategies to resolve them.

Factoring polynomials

Factoring polynomials is a particularly powerful tool in simplifying expressions to resolve indeterminate forms. In the given exercise, the denominator \(x^2 - 4x + 4\) can be factored. Factoring transforms it into \((x-2)^2\), revealing a common factor with the numerator. This shows both numerator and denominator share the factor \((x-2)\), leading to the indeterminate \(\frac{0}{0}\) form.

• Factoring helps in canceling out common terms in the numerator and denominator.
• It simplifies expressions, allowing us to evaluate limits more directly.

Replacing the polynomial in the numerator with known values helps determine if \(k\) must adjust to allow consistent simplification. Factoring also helps in identifying any restrictions on variable values that prevent division by zero.

Limits and continuity

The concept of limits is fundamental in understanding how functions behave as their inputs approach certain values. Specifically, in this exercise, we are tasked with evaluating the limit as \(x\) approaches 2. The presence of a trigonometric function \(\tan(x-2)\) complicates things, but using the fact that the polynomial \(x^2 + (k-2)x - 2k\) must also tend to zero helps us resolve it. This illustrates the principle of continuity, where smooth behavior of a function is implied around a point except at specific singularities.

• Continuity tells us a function doesn't suddenly "pop" to an unexpected value.
• In this problem, solving for \(k\) ensures that the function remains continuous around \(x = 2\).

Solving limits often involves examining the behavior near the point of interest, leveraging known results and algebraic simplifications to achieve precise answers.