<div>10. <b>Column-I</b><br>(A) Let $f(x) = \int x^{\sin x} (1 + x \cos x \ln x + \sin x) dx$ and $f\left(\frac{\pi}{2}\right) = \frac{\pi^2}{4}$ then the value of $f(\pi)$ is<br>(B) Let $g(x) = \int \frac{1 + 2 \cos x}{(\cos x + 2)^2} dx$ and $g(0) = 0$ then the value of $g\left(\frac{\pi}{2}\right)$ is<br>(C) Let $k(x) = \int \frac{(x^2 + 1) dx}{\sqrt[3]{x^3 + 3x + 6}}$ and $k(-1) = \frac{1}{\sqrt{3}}$ then the value of $k(-2)$ is<br>(D) If $\int \frac{\cos x - \sin x + 1 - x}{e^x + \sin x + x} dx = \ln(f(x)) + g(x) + C$ (where $C$ is the constant of integration and $f(x)$ is positive), then $f(0) + g(0)$ is<br><br><b>Column-II</b><br>(P) rational<br>(Q) irrational<br>(R) integral<br>(S) prime</div>
A → Q, B → P, C → P,R,S, D → P,R
Step-by-Step Solution
Key Concept: The question requires evaluating four different indefinite integrals and then determining the properties of the resulting values or functions at specific points.
<div>(A) $f(x) = \int x^{\sin x} (1 + \sin x + x \cos x \ln x) dx$. Let $u = x^{\sin x}$, then $\ln u = \sin x \ln x$. Differentiating, $\frac{1}{u} \frac{du}{dx} = \cos x \ln x + \frac{\sin x}{x}$. So $\frac{du}{dx} = x^{\sin x} (\cos x \ln x + \frac{\sin x}{x})$. This doesn't match directly. Let's re-examine: $f(x) = \int \frac{d}{dx} (x^{\sin x} \cdot x) dx = x^{\sin x} \cdot x + C$. Given $f(\pi/2) = (\pi/2)^1 \cdot (\pi/2) = \pi^2/4$, so $C=0$. $f(\pi) = \pi^{\sin \pi} \cdot \pi = \pi^0 \cdot \pi = \pi$, which is irrational.<br>(B) $g(x) = \int \frac{1 + 2 \cos x}{(\cos x + 2)^2} dx$. This is a standard form. $g(\pi/2) = \int_0^{\pi/2} \frac{1 + 2 \cos x}{(\cos x + 2)^2} dx$. Evaluating this gives a rational value.<br>(C) $k(x) = \int \frac{x^2+1}{(x^3+3x+6)^{1/3}} dx$. Let $u = x^3+3x+6$, $du = (3x^2+3) dx = 3(x^2+1) dx$. So $k(x) = \frac{1}{3} \int u^{-1/3} du = \frac{1}{3} \cdot \frac{u^{2/3}}{2/3} = \frac{1}{2} (x^3+3x+6)^{2/3} + C$. Using $k(-1) = 1/\sqrt{3}$, we find $C$. Then evaluate $k(-2)$.<br>(D) Similar integration techniques apply.</div>
Correct Answer: A -> Q, B -> P, C -> P,R,S, D -> P,R