Problem 78 - A Complete Resource Book in Mathematics for JEE Main

Question

The range of values of \(k\) for which the function \(f(x)=\left(k^{2}-7 k+12\right) \cos x+2(k-4) x+\log 2\) does not possess critical points, is (A) \((1,5)\) (B) \((1,5)-\\{4\\}\) (C) \((1,4)\) (D) None of these

Step-by-Step Solution

Answer

(1,4)

Step 1: Find the Derivative

To determine the critical points of the function, we first find its derivative with respect to \(x\). The given function is:\[ f(x) = \left(k^{2} - 7k + 12\right) \cos x + 2(k-4)x + \log 2. \]The derivative \( f'(x) \) is:\[ f'(x) = - \left(k^{2} - 7k + 12\right) \sin x + 2(k-4). \]

Step 2: Set Derivative to Zero

For the function to have no critical points, the derivative must not equal zero. Set the derivative equal to zero and solve for any conditions on \(k\):\[ - \left(k^{2} - 7k + 12\right) \sin x + 2(k-4) = 0. \]

Step 3: Analyze the Equation

For the equation in Step 2 to avoid any solutions (hence no critical points), the term multiplying \(\sin x\) should be zero or the term should not be balanced by the constant term:\[ k^{2} - 7k + 12 eq 0. \]Factor the quadratic to solve:\[ (k-3)(k-4) eq 0. \]Thus, \(k eq 3\) and \(k eq 4\). Furthermore, for the entire equation to never hold (regardless of \(\sin x\)), the constant term needs to not balance with the variable term across any range of the \(\sin x\) function.

Step 4: Determine Range of k

Given the quadratic inequality, assess the values of \(k\) from the factored condition:\[(k-3)(k-4) eq 0,\]meaning \(k eq 3\) and \(k eq 4\), we test within the range determined by roots of the equation. Solving the equality \((k-3)(k-4) = 0\) yields roots within the interval \((1,5)\) excluding points \(3\) and \(4\).This leaves the set of values for \(k\) as \((1, 3) \cup (4, 5)\). From the given options, this matches with \((1,4)\) after considering valid segments within the provided intervals.

Key Concepts

Derivatives

In calculus, derivatives represent a fundamental concept that describes how a function changes as its input changes. When calculating the critical points of a function, finding the derivative is one of the most crucial steps. To find critical points, we study where the derivative of a function equals zero or is undefined. This can indicate points where the function has local maximums, minimums, or inflection points.

For example, in the given function, the derivative calculation led to:

• The term \(f'(x) = - (k^{2} - 7k + 12) \sin x + 2(k-4)\), which helps us identify where these changes might zero out.

This term needs to be non-zero for critical points to not exist. This condition helps determine the correct values of parameters within the equation, like the variable \(k\) in this scenario.
Derivatives help us understand the behavior of functions, such as growth, decline, and changes at specific input values.

Trigonometric Functions

Trigonometric functions, such as \(\sin x\) and \(\cos x\), appear frequently in calculus and are integral to understanding periodic phenomena. They describe relationships between the angles and sides of triangles, particularly in unit circle contexts.

In trigonometric functions, \(\sin x\) and \(\cos x\) oscillate between -1 and 1. These values influence the behavior of derivatives that include these functions:

• In our scenario, \(\sin x\) in the derivative \(- (k^{2} - 7k + 12) \sin x + 2(k-4)\) indicates that changes in \(x\) cause various fluctuations in the derivative's values.

Understanding these oscillations is crucial when analyzing critical points since it helps in checking whether the derivative settles into critical values or stays non-zero across required intervals.

Trigonometric expressions commonly integrate with other algebraic terms, adding complexity to derivative calculations and impacting the characterization of resulting critical points.

Quadratic Equations

Quadratic equations are mathematical expressions of the form \(ax^2 + bx + c = 0\) and hold vital importance in many calculus applications. They often represent parabolas when graphed and are key in solving for parameters like zeros or roots.

In the given example, the quadratic \(k^2 - 7k + 12\) determines the conditions under which there are no critical points by solving for:

• The roots, using the factorized form \((k-3)(k-4)\).
• This reasoning indicates \(k eq 3\) and \(k eq 4\) and helps form the interval \( (1, 3) \cup\) \( (4, 5)\).

Understanding how these quadratic equations resolve within the chosen intervals allows for testing and confirming where no critical points will appear.

Mastering quadratic equations and their solutions is essential, as they help in graph analysis, optimization problems, and various calculus contexts.