GUJCET 2011 Mathematics Question Paper with Answer and Solution

(A) The area $A$ of the region bounded by the curve $y = f(x)$,the $x$-axis,and the lines $x = a$ and $x = b$ is given by the formula:
$A = \int_{a}^{b} |f(x)| \, dx$
Given the curve $xy = 4$,we have $y = \frac{4}{x}$.
The region is bounded by $x = 1$ and $x = 3$.
Therefore,the area $A$ is:
$A = \int_{1}^{3} \frac{4}{x} \, dx$
$A = 4 \int_{1}^{3} \frac{1}{x} \, dx$
$A = 4 [\ln |x|]_{1}^{3}$
$A = 4 (\ln 3 - \ln 1)$
Since $\ln 1 = 0$,we get:
$A = 4 \ln 3$
Thus,the area is $4 \log 3$ sq. units.

(C) The area $A$ is given by the integral $\int_{-1}^{2} |\sin(\pi x)| \, dx$.
Since $\sin(\pi x)$ changes sign at $x = 0$ and $x = 1$,we split the integral:
$A = \int_{-1}^{0} |\sin(\pi x)| \, dx + \int_{0}^{1} |\sin(\pi x)| \, dx + \int_{1}^{2} |\sin(\pi x)| \, dx$.
For $x \in [-1, 0]$,$\sin(\pi x) \leq 0$,so $|\sin(\pi x)| = -\sin(\pi x)$.
For $x \in [0, 1]$,$\sin(\pi x) \geq 0$,so $|\sin(\pi x)| = \sin(\pi x)$.
For $x \in [1, 2]$,$\sin(\pi x) \leq 0$,so $|\sin(\pi x)| = -\sin(\pi x)$.
$A = \int_{-1}^{0} -\sin(\pi x) \, dx + \int_{0}^{1} \sin(\pi x) \, dx + \int_{1}^{2} -\sin(\pi x) \, dx$.
Evaluating these:
$\int -\sin(\pi x) \, dx = \frac{\cos(\pi x)}{\pi}$.
$A = [\frac{\cos(\pi x)}{\pi}]_{-1}^{0} + [-\frac{\cos(\pi x)}{\pi}]_{0}^{1} + [\frac{\cos(\pi x)}{\pi}]_{1}^{2}$.
$A = (\frac{1}{\pi} - \frac{-1}{\pi}) + (-(\frac{-1}{\pi} - \frac{1}{\pi})) + (\frac{1}{\pi} - \frac{-1}{\pi}) = \frac{2}{\pi} + \frac{2}{\pi} + \frac{2}{\pi} = \frac{6}{\pi}$.
Thus,the correct option is $C$.