Problem 144 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Column-I Column-II I. If \(f(x)=\) (A) \(R-[-\sqrt{3}, \sqrt{3}]\) \(\left\\{\begin{array}{c}4 x-x^{3}+\ln \left(a^{2}-3 a+3\right), 0 \leq x<3 \\\ x-18, & x \geq 3\end{array}\right.\) \(=\) has a local minima at \(x=3\), than \(a\) belongs to II. If the function \(f(x)=\) (B) \((0, \infty)\) \(\left(\frac{\sqrt{a+1}}{a-1}-1\right) x^{3}-x+\ln (a-1)\) is strictly decreasing \(\forall x \in R\) then \(a\) belongs to III. If the function \(f(x)=x^{3}+\) (C) \([1,2]\) \(a x^{2}+a^{2} x+2 \sin ^{2} x\) is strictly increasing \(\forall x \in R\), then \(a\) belongs to IV. The function \(f(x)=\left|e^{a x}-e^{-u x}\right|\), (D) \((3, \infty)\) \(a>0\) is strictly increasing in the interval

Step-by-Step Solution

Answer

I-A, II-D, III-C, IV-B.

Step 1: Analyze Condition I

Given that \(f(x) = 4x - x^3 + \ln(a^2 - 3a + 3)\) has a local minimum at \(x=3\), find the derivative \(f'(x) = 4 - 3x^2\). Set \(f'(x)=0\) for local extrema, giving \(3x^2 = 4 \), resulting in \(x = \pm\sqrt{\frac{4}{3}}\). So, \(a^2 - 3a + 3\) must ensure the critical point is \(x=3\), which affects \(a\) as \(x=3\) must be evaluable; thus \(a \) belongs to \(R - [-\sqrt{3}, \sqrt{3}]\).

Step 2: Analyze Condition II

For \(f(x) = \left(\frac{\sqrt{a+1}}{a-1} - 1\right)x^3 - x + \ln(a-1)\) to be strictly decreasing, \(f'(x)\) must be negative. Thus, \(f'(x)=3\left(\frac{\sqrt{a+1}}{a-1} - 1\right)x^2 - 1 < 0\). Solving, \(\sqrt{a+1} < a-1\). Solving this inequality gives \(a > 3\). So, \(a\) belongs to \((3, \infty)\).

Step 3: Analyze Condition III

For \(f(x) = x^3 + ax^2 + a^2x + 2\sin^2x\) to be strictly increasing, its derivative \(f'(x)=3x^2 + 2ax + a^2 + 2\sin(2x)\) must be non-negative for all \(x\). We deduce \(3x^2 + 2ax + a^2\geq 0\). For this quadratic to be always non-negative, the discriminant must satisfy \(4a^2 - 12a^2 \leq 0\), implying \(a \in [1,2]\).

Step 4: Analyze Condition IV

Consider \(f(x)=|e^{ax} - e^{-ux}|\) where \(a>0\). It is strictly increasing on an interval if \(a > u\). Given that \(a > 0\), this essentially puts \(a\) in \((0, \infty)\). However, to relate this to the given column options, recognize that the analysis of intervals depends on the specific setup of the question.

Key Concepts

Local Minima

In mathematical functions, a local minimum is a point where a function reaches a value smaller than the functions' values around it. You can think of it as a small "valley" in the landscape of a graph. To find local minima, we use the derivative of the function. The derivative, which is denoted as \( f'(x) \), helps us find where the slope of the function is zero or changes direction. These points are called critical points.

For instance, consider the function \( f(x) = 4x - x^3 + \ln(a^2 - 3a + 3) \). To find if there's a local minimum at \( x = 3 \), we first calculate its derivative: \( f'(x) = 4 - 3x^2 \). Setting \( f'(x) = 0 \) gives us \( x = \pm \sqrt{\frac{4}{3}} \). This indicates possible local minima or maxima. At \( x = 3 \), the value of \( a \) makes the critical point evaluable, excluding \( a \) from the range \([-\sqrt{3}, \sqrt{3}]\).

• Local Minima Exploration: Analyze each critical point with the second derivative, \( f''(x) \), or the first derivative test to determine if it truly is a minimum.
• Choosing the Right \( a \): Ensure that the logarithmic term \( \ln(a^2 - 3a + 3) \) is defined and real for \( x = 3 \).

Strictly Increasing Functions

A function is considered strictly increasing when its outputs become larger as its inputs grow. This means for any two numbers \( x_1 \) and \( x_2 \) where \( x_1 < x_2 \), the function value is \( f(x_1) < f(x_2) \). Mathematically, this is checked by ensuring that the derivative \( f'(x) \) is positive for all values of \( x \).

For example, if we want the function \( f(x) = x^3 + ax^2 + a^2x + 2\sin^2x \) to be strictly increasing, its derivative must be non-negative. The derivative \( f'(x) = 3x^2 + 2ax + a^2 + 2\sin(2x) \) must be checked across all \( x \). The challenge is ensuring this quadratic is non-negative, requiring the discriminant to satisfy \( 4a^2 - 12a^2 \leq 0 \). Hence, \( a \) should be in \([1,2]\).

• Derivative Analysis: The condition of \( f'(x) \) being positive ensures a strictly increasing function.
• Quadratic Positivity: The quadratic part must not change sign, achieved by examining its discriminant.

Strictly Decreasing Functions

A function is strictly decreasing if its outputs get smaller as the inputs grow. For numbers \( x_1 \) and \( x_2 \) with \( x_1 < x_2 \), a strictly decreasing function ensures \( f(x_1) > f(x_2) \). We verify this by ensuring \( f'(x) \), the derivative, is negative for all \( x \).

Consider the function \( f(x) = \left(\frac{\sqrt{a+1}}{a-1} - 1\right)x^3 - x + \ln(a-1) \). For it to be strictly decreasing, \( f'(x) = 3\left(\frac{\sqrt{a+1}}{a-1} - 1\right)x^2 - 1 \) must be less than zero. Solving \( \sqrt{a+1} < a-1 \) gives us \( a > 3 \). So, \( a \) would belong to \((3, \infty)\).

• Ensuring Negativity: The derivative must strictly remain below zero, signifying a decreasing function.
• Inequality Solving: Solve the inequalities derived from the derivative to determine valid ranges for \( a \).