Problem 64 - 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

Option (B) \(2^{n+1}-1\).

Step 1: Identify the function to minimize

We need to find the minimum value of the function \(f(x) = x^2 + 4x + 5\). This is a quadratic function, and its minimum value can be found using the formula for the vertex of a parabola \(x = -\frac{b}{2a}\), where \(a\) and \(b\) are the coefficients of \(x^2\) and \(x\) respectively.

Step 2: Find the vertex of the quadratic function

The quadratic function is \(f(x) = x^2 + 4x + 5\). Here, \(a = 1\) and \(b = 4\). Using the vertex formula \(x = -\frac{4}{2 \times 1} = -2\).

Step 3: Calculate the minimum value

Substitute \(x = -2\) back into the function \(f(x) = x^2 + 4x + 5\) to get the minimum value: \(f(-2) = (-2)^2 + 4(-2) + 5 = 4 - 8 + 5 = 1\). Thus, \(a = 1\).

Step 4: Evaluate the limit for b

We need to find \(b = \lim_{\theta \to 0} \frac{1 - \cos 2\theta}{\theta^2}\). Use the trigonometric identity \(1 - \cos x = 2\sin^2\left(\frac{x}{2}\right)\) to transform the expression: \(1 - \cos 2\theta = 2\sin^2 \theta\). Thus, the limit becomes \(b = \lim_{\theta \to 0} \frac{2\sin^2 \theta}{\theta^2}\).

Step 5: Simplify the limit expression

Use the small-angle approximation \(\sin \theta \approx \theta\) as \(\theta \to 0\). Therefore, \(\sin^2 \theta \approx \theta^2\), which gives \(b = \lim_{\theta \to 0} \frac{2\theta^2}{\theta^2} = 2\).

Step 6: Substitute a and b into the series

With \(a = 1\) and \(b = 2\), substitute into the series \(\sum_{r=0}^{n} a^r \cdot b^{n-r}\). This simplifies to \(\sum_{r=0}^{n} 1^r \cdot 2^{n-r} = \sum_{r=0}^{n} 2^{n-r}\).

Step 7: Evaluate the series

\(\sum_{r=0}^{n} 2^{n-r} = 2^n + 2^{n-1} + 2^{n-2} + \ldots + 2^0\) is a geometric series with \(n+1\) terms. The sum of a geometric series is given by \(S_{n+1} = \frac{a(r^{n+1} - 1)}{r - 1}\), where \(a = 1\) and \(r = 2\), resulting in \(S_{n+1} = \frac{2^{n+1} - 1}{1} = 2^{n+1} - 1\).

Step 8: Select the correct option

The value of the series \(\sum_{r=0}^{n} a^r \cdot b^{n-r}\) is \(2^{n+1} - 1\), which matches option (B).

Key Concepts

Limit evaluation

Understanding how to evaluate limits is a crucial part of calculus. This is the process of finding the value a function approaches as the input approaches a particular point. In the context of this exercise, the limit evaluation is focused on finding the limit of the expression \( \lim_{\theta \to 0} \frac{1 - \cos 2\theta}{\theta^2} \).

A commonly used technique involves trigonometric identities, like the identity \( 1 - \cos x = 2\sin^2\left(\frac{x}{2}\right) \), which helps in transforming the expression into a more workable form. By substituting \( x = 2\theta \) into this identity, we get \( 1 - \cos 2\theta = 2\sin^2 \theta \). Therefore, the limit becomes \( \lim_{\theta \to 0} \frac{2\sin^2 \theta}{\theta^2} \).

At this point, you can use the small angle approximation \( \sin \theta \approx \theta \), which is valid when \( \theta \to 0 \). By applying this approximation, \( \sin^2 \theta \approx \theta^2 \), making the limit simplify to \( \lim_{\theta \to 0} \frac{2\theta^2}{\theta^2} = 2 \). This means as \( \theta \) approaches 0, the expression approaches the value 2.

Quadratic minimization

Quadratic minimization is the process of finding the minimum value of a quadratic function. For any quadratic function of the form \( f(x) = ax^2 + bx + c \), the function's minimum value lies at the vertex of its parabola.

The vertex form of a quadratic equation can be calculated using the formula \( x = -\frac{b}{2a} \), where \( a \) and \( b \) are the coefficients of \( x^2 \) and \( x \) respectively.

In this exercise, the quadratic function is \( f(x) = x^2 + 4x + 5 \). With \( a = 1 \) and \( b = 4 \), applying the vertex formula yields \( x = -\frac{4}{2 \times 1} = -2 \).

Substituting \( x = -2 \) back into the original function gives \( f(-2) = (-2)^2 + 4(-2) + 5 = 1 \). Thus, the minimum value of the quadratic function, \( a \), is 1.

• This process involves finding a single calculation based on a formula which can often simplify what may look like a complicated problem.
• It's vital in optimization problems to understand precisely how finding the vertex gives you critical values. This allows for planning and analysis in real-world applications involving minimization and maximization.

Geometric series

A geometric series is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the 'common ratio'. Geometric series appear frequently in calculus, physics, and finance due to their predictable and summable pattern.

In the problem at hand, the summation \( \sum_{r=0}^{n} a^r \cdot b^{n-r} \) simplifies to a geometric series when \( a = 1 \) and \( b = 2 \). The formula for the sum of a geometric series is: \[ S = \frac{a(r^{n+1} - 1)}{r - 1} \] where \( a \) is the first term and \( r \) is the common ratio.

In this exercise, we see that the sequence turns into \( \sum_{r=0}^{n} 2^{n-r} \), meaning each term is a power of 2, with the common ratio \( r = 2 \). As a result, we calculate the sum with \( a = 1 \) to find that

• The series sum is \( 2^n + 2^{n-1} + 2^{n-2} + \ldots + 2^0 \).
• This indicates \( S_{n+1} = \frac{2^{n+1} - 1}{1} = 2^{n+1} - 1 \).

Understanding geometric series is not only fundamental for solving mathematical problems but also essential for various applications like calculating interest in finance or analyzing exponential growth in scientific models.