Problem 105 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(f(x)=(x-3)(x-4)(x-4)(x-5)(x-6)\), then (A) \(f^{\prime}(x)\) has four roots (B) three roots of \(f^{\prime \prime}(x)=0\) lie in \((3,4) \cup(4,5) \cup(5,6)\) (C) the equation \(f^{\prime}(x)=0\) has only one root (D) three roots of \(f^{\prime}(x)=0\) lie in \((2,3) \cup(3,4) \cup(4,5)\)

Step-by-Step Solution

Answer

Options A and D are true.

Step 1: Understanding the function

The given function is \(f(x)=(x-3)(x-4)^2(x-5)(x-6)\). It consists of distinct and repeated linear factors. Notice that \((x-4)^2\) is repeated, making \(x=4\) a repeated root.

Step 2: Finding the first derivative

To find \(f'(x)\), apply the product rule and power rule to differentiate the function. This will involve recognizing that the derivative of a product is the sum of the derivative of each part times the other parts. Begin with \(f(x) = (x-3)(x-4)^2(x-5)(x-6)\) and differentiate.

Step 3: Applying the derivative

First, simplify the function by expanding the polynomial, or use properties of derivatives to manage each part. Here, recognize that the derivative of a function with a repeated factor will include terms like \((x-4)\) from the factor \((x-4)^2\). This symmetry allows direct identification of roots.

Step 4: Finding roots of the first derivative

Using calculus principles such as the product rule and critical points, determine that \(f'(x)=0\) when evaluated at respective critical values. By inspection, or from factor identification, the critical points appear at differentiating each factor and equating to zero.

Step 5: Evaluating the options

Option (A) states \(f^{\prime}(x)\) has four roots, which is true because each differentiated factor contributes a root except \((x-4)^2\), which contributes two distinct roots in the zero scheme at \(x=4\).

Step 6: Verification of roots in intervals

Verify Option (B), stating the four roots would be tested if three roots of \(f^{\prime\prime}(x)=0\) lie in \((3,4) \cup (4,5) \cup (5,6)\). Observe that second derivatives magnify contribution near roots, specifically in segments with change.

Step 7: Conclusion of Options

Analyze the remaining options: (C) for contradiction as it proposes only one root (untrue), and (D) stating three roots lie within specified intervals. Since observed roots are in contiguous ranges \((2,3)\), \((3,4)\), and \((4,5)\), this matches distribution.

Key Concepts

Polynomial Functions

Polynomial functions are mathematical expressions involving a sum of powers of a variable multiplied by coefficients. For example, in the polynomial function \(f(x) = (x-3)(x-4)^2(x-5)(x-6)\), each factor represents a root of the equation. In this case:

• \(x-3\), \(x-4\), \(x-5\), and \(x-6\) are the linear factors.
• The factor \((x-4)\) is squared, indicating it is a repeated root, or it has a multiplicity of 2.

Polynomial functions can have different characteristics based on their degree and the nature of their roots. These roots are points where the value of the function is zero. They are critical when finding derivative roots.
They are fundamental in calculating derivatives because roots guide us to critical points and determinable intervals of change.

Critical Points

Critical points in a function are values of \(x\) where the derivative \(f'(x)\) is zero or undefined. These points are important because they determine where the function reaches a local maximum, minimum, or inflection point. In the context of our polynomial like \(f(x) = (x-3)(x-4)^2(x-5)(x-6)\):

• The first derivative \(f'(x)\) would be zero at these points, helping identify changes in the slope.
• Analyzing these roots tells us about the behavior of \(f(x)\) between its roots.

In solving the problem, these critical points become evident by applying the rules of derivatives and examining where the derivative of a function equals zero. This provides the potential locations of maxima and minima in the graph.

Product Rule

The product rule is a fundamental technique used in calculus when differentiating products of two or more functions. If you have a function defined as a product, such as \(f(x) = u(x) \, v(x) \, w(x) \, z(x)\), the product rule integrates the derivatives of each part as follows:

• \(f'(x) = u'(x) \, v(x) \, w(x) \, z(x) + u(x) \, v'(x) \, w(x) \, z(x) + u(x) \, v(x) \, w'(x) \, z(x) + u(x) \, v(x) \, w(x) \, z'(x)\)

To solve our polynomial problem, apply the product rule to each linear factor and consider the multiple derivatives involved. This technique is powerful, especially when handling polynomials like \((x-3)(x-4)^2(x-5)(x-6)\), as it gives a structured approach to obtain \(f'(x)\). The rule ensures each part individually contributes to the derivative, essential for analysing roots.

Power Rule

The power rule is another key differentiation rule used in calculus, particularly for functions expressed as powers of \(x\) like \(x^n\). This rule states that if \(y = x^n\), then the derivative is \(dy/dx = nx^{n-1}\).

• For example, the derivative of \((x-4)^2\) is \(2(x-4)^1\).

The power rule simplifies the differentiation process significantly, allowing for easy calculation of derivatives of polynomial expressions. In the original exercise, identifying repeated factors like \( (x-4)^2 \) means using the power rule helps in determining their contribution to the overall derivative, which is crucial when these terms are part of larger polynomial expressions. Coupled with the product rule, it builds a comprehensive understanding of the polynomial's derivative structure.