Problem 36 - 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 answer is (B) \(2^{n+1} - 1\).

Step 1: Find the value of a

We need to find the minimum value of the quadratic function \(f(x) = x^2 + 4x + 5\). A quadratic function in the form \(ax^2 + bx + c\) achieves its minimum (or maximum) value at \(x = -\frac{b}{2a}\). Here, \(a = 1\), \(b = 4\), so the vertex is at \(x = -\frac{4}{2 \times 1} = -2\). Evaluating the function at \(x = -2\), we have:\[f(-2) = (-2)^2 + 4(-2) + 5 = 4 - 8 + 5 = 1.\]Thus, \(a = 1\).

Step 2: Calculate the value of b

The limit is given as \(b = \lim_{\theta \to 0} \frac{1 - \cos 2\theta}{\theta^2}\). We use the identity \(1 - \cos x = 2\sin^2\left(\frac{x}{2}\right)\), so we get:\[\lim_{\theta \to 0} \frac{1 - \cos 2\theta}{\theta^2} = \lim_{\theta \to 0} \frac{2\sin^2(\theta)}{\theta^2}.\]Factoring this, we know \(\sin \theta \approx \theta\) when \(\theta\) is small, thus:\[= \lim_{\theta \to 0} \frac{2(\theta^2)}{\theta^2} = 2.\]So, \(b = 2\).

Step 3: Find the value of the sum expression

We are asked to evaluate the expression \(\sum_{r=0}^{n} a^r \cdot b^{n-r}\). Since \(a = 1\) and \(b = 2\), it becomes:\[\sum_{r=0}^{n} 1^r \cdot 2^{n-r} = \sum_{r=0}^{n} 2^{n-r}.\]By changing the variable \(s = n - r\), the sum becomes \(\sum_{s=0}^{n} 2^s\), which is a geometric series sum. Using the formula for a geometric series, \(\sum_{s=0}^{n} x^s = \frac{x^{n+1} - 1}{x - 1}\), we get:\[= \frac{2^{n+1} - 1}{2 - 1} = 2^{n+1} - 1.\]

Step 4: Match with the given options

From Step 3, we concluded that \(\sum_{r=0}^{n} a^r \cdot b^{n-r} = 2^{n+1} - 1\). Comparing this with the provided options, we see that the correct answer is option (B) \(2^{n+1} - 1\).

Key Concepts

Quadratic Functions

A quadratic function is a polynomial of degree two and is generally expressed in the form \( ax^2 + bx + c \). This type of function is also known as a parabola when graphed, which can open upwards or downwards depending on the sign of \( a \). In our exercise, we were given the quadratic function \( f(x) = x^2 + 4x + 5 \), and we needed to find its minimum value.

• The minimum value of a quadratic function occurs at its vertex. The x-coordinate of the vertex can be found using the formula \( x = -\frac{b}{2a} \).
• For the given function, \( a = 1 \) and \( b = 4 \). Hence, the vertex lies at \( x = -\frac{4}{2 \cdot 1} = -2 \).
• By substituting \( x = -2 \) into the function, \( f(-2) = (-2)^2 + 4(-2) + 5 \), we calculated the minimum value as 1.

Thus, the minimum value of our quadratic function is 1, and we denote this as \( a = 1 \). Understanding the properties and behavior of quadratic functions is crucial, as they frequently appear in both calculus and real-world problems.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that hold true for all values within the function's domain. In calculus, these identities are very useful for simplifying expressions and solving limits, as seen in the exercise.In the original problem, we needed to evaluate the limit expression \( \lim_{\theta \to 0} \frac{1 - \cos 2\theta}{\theta^2} \). To solve this, we employed the identity:

• \( 1 - \cos x = 2\sin^2\left(\frac{x}{2}\right) \) which helps transform complex trigonometric expressions into simpler forms.
• Substituting \( x = 2\theta \), we get \( 1 - \cos 2\theta = 2\sin^2(\theta) \).
• Using \( \sin \theta \approx \theta \) when \( \theta \to 0 \), this becomes \( \lim_{\theta \to 0} \frac{2\theta^2}{\theta^2} = 2 \).

Thus, \( b = 2 \) which highlights how trigonometric identities can greatly simplify limit problems in calculus.

Geometric Series

A geometric series is a sequence where each term after the first is found by multiplying the previous term by a constant, called the common ratio. The sum of the first \( n \) terms of a geometric series with the first term \( a \) and common ratio \( r \) is calculated by the formula:\[ S_n = \frac{a(r^n - 1)}{r - 1} \]In the exercise, we faced the expression \( \sum_{r=0}^{n} a^r \cdot b^{n-r} \), which simplifies to a geometric series:

• Given \( a = 1 \) and \( b = 2 \), the expression becomes \( \sum_{r=0}^{n} 2^{n-r} \).
• By substituting \( s = n - r \), this sum transforms into \( \sum_{s=0}^{n} 2^s \).
• Applying the formula for geometric series with \( x = 2 \), it sums to \( \frac{2^{n+1} - 1}{2 - 1} = 2^{n+1} - 1 \).

Therefore, understanding geometric series can simplify complex summations, making it a powerful tool in calculus problems.