Key Concepts and Formulas

• Definite Integral Property for Symmetric Limits: For a continuous function $f(x)$ on $[-a, a]$, $\int_{-a}^a f(x) dx = \int_0^a [f(x) + f(-x)] dx$.
• Trigonometric Identity: $\cos(-\theta) = \cos(\theta)$.
• Basic Integral: $\int \cos(kx) dx = \frac{\sin(kx)}{k} + C$.
• Limit of $\frac{\sin x}{x}$: $\lim_{x \to 0} \frac{\sin x}{x} = 1$.

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Step-by-Step Solution

Step 1: Understand the Problem and Identify the Integrand We are asked to find a value of $\alpha$ given the definite integral: $I = \int\limits_{-1}^1 \frac{\cos \alpha x}{1+3^x} d x$ The value of the integral is given as $I = \frac{2}{\pi}$. Let the integrand be $f(x) = \frac{\cos \alpha x}{1+3^x}$.

Step 2: Apply the Definite Integral Property for Symmetric Limits The limits of integration are from $-1$ to $1$, which are symmetric. We can apply the property $\int_{-a}^a f(x) dx = \int_0^a [f(x) + f(-x)] dx$. Here, $a=1$. We need to find $f(-x)$: $f(-x) = \frac{\cos(\alpha (-x))}{1+3^{-x}}$ Using the property $\cos(-\theta) = \cos(\theta)$, we get $\cos(\alpha (-x)) = \cos(\alpha x)$. So, $f(-x) = \frac{\cos \alpha x}{1+3^{-x}}$ Now, we apply the property: $I = \int_0^1 [f(x) + f(-x)] dx = \int_0^1 \left( \frac{\cos \alpha x}{1+3^x} + \frac{\cos \alpha x}{1+3^{-x}} \right) dx$

Step 3: Simplify the Integrand Let's simplify the sum of $f(x)$ and $f(-x)$: $f(x) + f(-x) = \frac{\cos \alpha x}{1+3^x} + \frac{\cos \alpha x}{1+3^{-x}}$ Factor out $\cos \alpha x$: $= \cos \alpha x \left( \frac{1}{1+3^x} + \frac{1}{1+3^{-x}} \right)$ We know that $3^{-x} = \frac{1}{3^x}$. Substitute this into the expression: $= \cos \alpha x \left( \frac{1}{1+3^x} + \frac{1}{1+\frac{1}{3^x}} \right)$ To simplify the second term's denominator: $1+\frac{1}{3^x} = \frac{3^x+1}{3^x}$. So, the expression becomes: $= \cos \alpha x \left( \frac{1}{1+3^x} + \frac{1}{\frac{3^x+1}{3^x}} \right)$ $= \cos \alpha x \left( \frac{1}{1+3^x} + \frac{3^x}{1+3^x} \right)$ The terms inside the parenthesis have a common denominator: $= \cos \alpha x \left( \frac{1+3^x}{1+3^x} \right)$ $= \cos \alpha x \cdot 1 = \cos \alpha x$ Thus, the integral simplifies to: $I = \int_0^1 \cos \alpha x \, dx$

Step 4: Evaluate the Simplified Integral Now, we integrate $\cos \alpha x$ with respect to $x$: $I = \left[ \frac{\sin \alpha x}{\alpha} \right]_0^1$ Evaluate the definite integral by applying the limits: $I = \frac{\sin(\alpha \cdot 1)}{\alpha} - \frac{\sin(\alpha \cdot 0)}{\alpha}$ $I = \frac{\sin \alpha}{\alpha} - \frac{\sin 0}{\alpha}$ Since $\sin 0 = 0$: $I = \frac{\sin \alpha}{\alpha}$

Step 5: Equate and Solve for $\alpha$ We are given that $I = \frac{2}{\pi}$. Therefore, we have the equation: $\frac{\sin \alpha}{\alpha} = \frac{2}{\pi}$ This is a transcendental equation. For multiple-choice questions in exams like JEE, the solution for $\alpha$ is usually one of the given options. Let's test the options.

• Option (A): $\alpha = \frac{\pi}{2}$ Substitute $\alpha = \frac{\pi}{2}$ into the equation: $\frac{\sin(\pi/2)}{\pi/2} = \frac{1}{\pi/2} = \frac{2}{\pi}$ This matches the given value of the integral.

• Option (B): $\alpha = \frac{\pi}{4}$ $\frac{\sin(\pi/4)}{\pi/4} = \frac{1/\sqrt{2}}{\pi/4} = \frac{4}{\sqrt{2}\pi} = \frac{2\sqrt{2}}{\pi} \neq \frac{2}{\pi}$

• Option (C): $\alpha = \frac{\pi}{3}$ $\frac{\sin(\pi/3)}{\pi/3} = \frac{\sqrt{3}/2}{\pi/3} = \frac{3\sqrt{3}}{2\pi} \neq \frac{2}{\pi}$

• Option (D): $\alpha = \frac{\pi}{6}$ $\frac{\sin(\pi/6)}{\pi/6} = \frac{1/2}{\pi/6} = \frac{3}{\pi} \neq \frac{2}{\pi}$ Only $\alpha = \frac{\pi}{2}$ satisfies the equation.

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Common Mistakes & Tips

• Incorrect Application of Properties: Ensure the integrand is well-defined and continuous over the interval $[-a, a]$ before applying the symmetric limit property.
• Algebraic Errors: Be meticulous when simplifying fractions, especially those involving exponential terms like $3^{-x}$.
• Integration Errors: Forgetting to divide by the coefficient of $x$ (i.e., $\alpha$) when integrating $\cos(\alpha x)$ is a common oversight.
• Solving Transcendental Equations: For exams, expect that the solution to equations like $\frac{\sin \alpha}{\alpha} = C$ will be among the provided options and can be verified by substitution.

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Summary

The problem involves evaluating a definite integral with symmetric limits. By applying the property $\int_{-a}^a f(x) dx = \int_0^a [f(x) + f(-x)] dx$, the integrand $\frac{\cos \alpha x}{1+3^x}$ was simplified to $\cos \alpha x$. The resulting integral, $\int_0^1 \cos \alpha x \, dx$, evaluated to $\frac{\sin \alpha}{\alpha}$. Equating this to the given value $\frac{2}{\pi}$ led to the equation $\frac{\sin \alpha}{\alpha} = \frac{2}{\pi}$. Testing the given options, we found that $\alpha = \frac{\pi}{2}$ satisfies this equation.

The final answer is $\boxed{\frac{\pi}{2}}$ which corresponds to option (A).