Problem 36 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(a=\min \left\\{x^{2}+4 x+5, x \in R\right\\}\) and \(b=\lim _{\theta \rightarrow 0} \frac{1-\cos 2 \theta}{\theta^{2}}\) then the value of \(\sum_{r=0}^{n} a^{r} \cdot b^{n-r}\) is (A) \(\frac{2^{n+1}-1}{4 \cdot 2^{n}}\) (B) \(2^{n+1}-1\) (C) \(\frac{2^{n+1}-1}{3 \cdot 2^{n}}\) (D) None of these

Step-by-Step Solution

Answer

The answer is (B): \(2^{n+1} - 1\).

Step 1: Determine the minimum value of the quadratic function

The function given is \(f(x) = x^2 + 4x + 5\). This is a quadratic function, and its minimum value occurs at the vertex. To find this, we use the vertex formula for quadratics: \(x = -\frac{b}{2a}\), where \(a = 1\) and \(b = 4\). So, \(x = -\frac{4}{2 \times 1} = -2\). Substituting \(x = -2\) back into the function: \(f(-2) = (-2)^2 + 4(-2) + 5 = 4 - 8 + 5 = 1\). Therefore, \(a = 1\).

Step 2: Evaluate the limit for b

We need to solve \(b = \lim_{\theta \to 0} \frac{1-\cos 2\theta}{\theta^2}\). Using the Taylor series expansion, we know \(1 - \cos(2\theta) \approx 2\theta^2\) near \(\theta = 0\). Substituting, the expression simplifies to \(\lim_{\theta \to 0} \frac{2\theta^2}{\theta^2} = \lim_{\theta \to 0} 2 = 1\). Therefore, \(b = 2\).

Step 3: Evaluate the sum expression

The expression to evaluate is \(\sum_{r=0}^{n} a^{r} \cdot b^{n-r}\). Substituting for \(a = 1\) and \(b = 2\), it becomes \(\sum_{r=0}^{n} 1^{r} \cdot 2^{n-r} = \sum_{r=0}^{n} 2^{n-r}\). This is the sum of a geometric series with common ratio \(1/2\) and \(n+1\) terms. The sum of a geometric series \(S = a(\frac{1-r^{n+1}}{1-r})\), where first term \(a = 2^n\) and \(r = 1/2\), gives \(S = 2^n(\frac{1-(1/2)^{n+1}}{1-1/2})\). Simplifying, \(S = 2^n(2-2^{-(n+1)}) = 2^{n+1} - 1\).

Step 4: Compare with options and conclude

The value obtained from the geometric series sum is \(2^{n+1} - 1\). Matching this with the given options, the correct answer is option (B): \(2^{n+1} - 1\).

Key Concepts

Limit Evaluation

The concept of limit evaluation is essential in calculus, particularly when dealing with expressions that appear indeterminate as some variable approaches a point, often zero. In this problem, we deal with the limit \(b = \lim_{\theta \to 0} \frac{1-\cos 2\theta}{\theta^2}\). This might initially look complicated, but can be simplified using the Taylor series expansion for cosine functions.

Close to zero, the Taylor series for the cosine function provides us with a handy approximation:

• \(1 - \cos 2\theta \approx 2\theta^2\)

This approximation allows us to substitute and simplify our fraction to \(\lim_{\theta \to 0} \frac{2\theta^2}{\theta^2} = \lim_{\theta \to 0} 2 = 2\).

This simple step shows how powerful limit evaluation techniques and approximations can be, turning what seems complex into a straightforward conclusion.

Quadratic Functions

Quadratic functions are a fundamental concept in mathematics, often taking the form \(f(x) = ax^2 + bx + c\). In this context, to find the minimum value of the function \(f(x) = x^2 + 4x + 5\), we use the vertex formula, a key tool for understanding the graph of a parabola. The vertex form tells us where the minimum or maximum point of the quadratic function occurs.

Here:

• \(a = 1\) (the coefficient of \(x^2\)), and
• \(b = 4\) (the coefficient of \(x\)).

To find this vertex, the formula \(x = -\frac{b}{2a}\) is applied, yielding \(x = -2\). Substituting back into the function gives us the minimum value, \(f(-2) = 1\).

Understanding the properties of quadratic functions helps us solve a wide range of mathematical problems, and here it helps determine the constant \(a = 1\) for the sum we later evaluate.

Geometric Series

A geometric series is the sum of the terms of a geometric sequence, where each term after the first is found by multiplying the previous one by a constant called the "common ratio." In this exercise, the series in question is \(\sum_{r=0}^{n} 1^{r} \cdot 2^{n-r} = \sum_{r=0}^{n} 2^{n-r}\).

The sum of such a geometric series can be calculated using the formula:

• \(S = a\left(\frac{1-r^{n+1}}{1-r}\right)\)

where \(a\) is the first term and \(r\) is the common ratio. Here, \(a = 2^n\) and the common ratio \(r = \frac{1}{2}\). This simplifies to:

• \(S = 2^n\left(2 - \frac{1}{2^{n+1}}\right)\)

Simplifying this gives \(S = 2^{n+1} - 1\), matching option (B).

Grasping the formula for geometric series calculations is invaluable for efficiently solving problems involving repeated multiplicative sequences.