Kerala KEAM 5 June 2024 Solved Paper

We are asked to evaluate the following integral: $$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos^2 x}{1 + 2^{-x}} \, dx.$$Step 1: First, let's check if we can use symmetry. We notice that the limits of integration are symmetric, from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$.Consider the substitution $x = -t$, so that $dx = -dt$. The limits of integration will change as follows: when $x = -\frac{\pi}{2}$, $t = \frac{\pi}{2}$, and when $x = \frac{\pi}{2}$, $t = -\frac{\pi}{2}$. The integral becomes:$$I = \int_{\frac{\pi}{2}}^{-\frac{\pi}{2}} \frac{\cos^2(-t)}{1 + 2^{t}} (-dt).$$Since $\cos^2(-t) = \cos^2(t)$, the integral simplifies to: $$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos^2 t}{1 + 2^{t}} \, dt.$$This is exactly the same as the original integral.Step 2: Let's now add the original integral and the transformed integral. By symmetry, we have:$$2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( \frac{\cos^2 x}{1 + 2^{-x}} + \frac{\cos^2 x}{1 + 2^{x}} \right) dx.$$Notice that: $$\frac{\cos^2 x}{1 + 2^{-x}} + \frac{\cos^2 x}{1 + 2^{x}} = \cos^2 x \left( \frac{1}{1 + 2^{-x}} + \frac{1}{1 + 2^{x}} \right).$$Step 3: Simplify the sum inside the parentheses: $$\frac{1}{1 + 2^{-x}} + \frac{1}{1 + 2^{x}} = \frac{(1 + 2^x) + (1 + 2^{-x})}{(1 + 2^{-x})(1 + 2^x)} = \frac{2 + 2^x + 2^{-x}}{(1 + 2^{-x})(1 + 2^x)}.$$Using the identity $2^x + 2^{-x} = 2 \cosh(x \ln 2)$, we get: $$\frac{2 + 2^x + 2^{-x}}{(1 + 2^{-x})(1 + 2^x)} = 2.$$Step 4: The integral becomes: $$2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 2 \cos^2 x \, dx.$$Now, use the identity $\cos^2 x = \frac{1 + \cos 2x}{2}$, so the integral becomes: $$2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( 1 + \cos 2x \right) \, dx.$$Step 5: Evaluate the integral: $$2I = \left[ x + \frac{\sin 2x}{2} \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}.$$The sine term vanishes at both limits, so we are left with: $$2I = \left( \frac{\pi}{2} - \left( -\frac{\pi}{2} \right) \right) = \pi.$$Thus: $$I = \frac{\pi}{4}.$$Thus, the correct answer is option (B). Quick Tip: When faced with integrals involving symmetric limits, use substitution to exploit the symmetry and simplify the computation. In this case, symmetry allowed us to reduce the problem.