Problem 64 - A Complete Resource Book in Mathematics for JEE Main

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 value of the expression is (B) \(2^{n+1}-1\).

Step 1: Find Minimum Value of Quadratic Function

We know that a quadratic function of the form \(f(x) = x^2 + bx + c\) has its minimum value at \(x = -\frac{b}{2}\). Hence for \(x^2 + 4x + 5\), substitute \(b = 4\) to find the x that minimizes the value: \(x = -\frac{4}{2} = -2\).Substitute \(x = -2\) back into the function:\(f(-2) = (-2)^2 + 4(-2) + 5\)\(= 4 - 8 + 5\)\(= 1\).Therefore, \(a = 1\).

Step 2: Calculate Limit

Apply L'Hôpital's rule to solve for the limit if direct substitution leads to an indeterminate form:\(b = \lim_{\theta \to 0} \frac{1 - \cos 2\theta}{\theta^2}\).Using the double angle identity, \(1 - \cos 2\theta = 2\sin^2 \theta\), transform the expression:\(b = \lim_{\theta \to 0} \frac{2\sin^2 \theta}{\theta^2}\)This simplifies to:\(= \lim_{\theta \to 0} \frac{2(\sin \theta)^2}{\theta^2}\).Since \(\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\),\(b = 2\cdot 1^2 = 2\).

Step 3: Evaluate the Sum

Given that \(a = 1\) and \(b = 2\), the sum is:\(\sum_{r=0}^{n} a^{r} \cdot b^{n-r} = \sum_{r=0}^{n} 1^r \cdot 2^{n-r}\)\(= \sum_{r=0}^{n} 2^{n-r}\).This simplifies using the sum of a geometric series:\( = 2^n \cdot \left(\frac{1-\left(\frac{1}{2}\right)^{n+1}}{1-\frac{1}{2}}\right)\)\(= 2^n \cdot \left(\frac{1 - \frac{1}{2^{n+1}}}{\frac{1}{2}}\right)\)\(= 2^{n+1} - 1\).

Step 4: Choose the Correct Answer

Comparing the result from Step 3 to the provided options, the correct answer is \(2^{n+1} - 1\), which corresponds to option (B).

Key Concepts

Quadratic Function

A quadratic function is generally expressed in the form \(f(x) = ax^2 + bx + c\). Understanding this function is important in many areas of mathematics and its applications, as it forms the basis of parabolic equations and can be used to model various real-world scenarios.
In our specific case, we have the quadratic expression \(x^2 + 4x + 5\). To find the minimum value, first identify the vertex of the parabola. The vertex of a parabola given by \(ax^2 + bx + c\) occurs at \(x = -\frac{b}{2a}\).
For this equation, substitute \(b = 4\) and \(a = 1\) into the formula to get \(x = -\frac{4}{2} = -2\). This x-value is where the function reaches its minimum.
Substituting \(x = -2\) into the quadratic gives us \(f(-2) = (-2)^2 + 4(-2) + 5 = 1\). Therefore, for this function, the minimum value is \(a = 1\).
Quadratic functions are important as they help us understand how to determine extreme values efficiently, which is key when solving real problems.

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool used in calculus to find limits that result in indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). It simplifies complex limit calculations by allowing us to differentiate the numerator and the denominator independently until the indeterminate form is resolved.
In this exercise, we need to compute the limit \(b = \lim_{\theta \to 0} \frac{1 - \cos 2\theta}{\theta^2}\). Initially, substituting \(\theta = 0\) directly into the expression results in \(\frac{0}{0}\), an indeterminate form. This is a perfect application for L'Hôpital's Rule.
By using the double angle identity \(1 - \cos 2\theta = 2\sin^2 \theta\), we rewrite the expression: \(b = \lim_{\theta \to 0} \frac{2\sin^2 \theta}{\theta^2}\).
Next we simplify further because we know \(\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\). Therefore, \(b = 2\cdot 1^2 = 2\). This method highlights the utility of L'Hôpital's Rule in breaking down complex limit problems into more manageable derivatives.

Geometric Series

A geometric series is a series with a constant ratio between successive terms. The sum of a finite geometric series can be calculated using the formula:\( S_n = a \frac{1 - r^{n+1}}{1 - r} \), where \(a\) is the first term of the series, \(r\) is the common ratio, and \(n\) is the number of terms.
In this problem, the sum \(\sum_{r=0}^{n} a^{r} \, b^{n-r}\) is evaluated where \(a = 1\) and \(b = 2\). Thus, it becomes \(\sum_{r=0}^{n} 1^r \cdot 2^{n-r} = \sum_{r=0}^{n} 2^{n-r}\).
Recognizing it as a geometric series with \(r = \frac{1}{2}\) and \(n+1\) terms, we apply the geometric series formula:

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

Understanding geometric series is essential as they frequently appear in finance, physics, and computer science, where summing over regularly multiplying factors is common.

Double Angle Identity

The double angle identity is a trigonometric identity used to simplify expressions when angles are doubled. These identities are key to simplifying limit problems and many trigonometric computations.
For cosine, the double angle formula is \( \cos 2\theta = 1 - 2\sin^2 \theta \), and for sine, it is \( \sin 2\theta = 2\sin \theta \cos \theta \).
In our scenario, we use \(1 - \cos 2\theta = 2\sin^2 \theta\) to transform the original limit expression \(\frac{1 - \cos 2\theta}{\theta^2}\).
By applying this identity, the expression becomes \(\frac{2\sin^2 \theta}{\theta^2}\), simplifying the limit calculation and enabling the direct application of L'Hôpital's Rule.
These formulas are not only crucial in calculus but also in understanding fields like physics and engineering, where periodic functions often take center stage.