Question

The area of the region
$R = {(x, y) \in ℝ^{2} : 0 \leq x \leq 1, 0 \leq y \leq 1 and \frac{1}{4} \leq x y \leq \frac{1}{2}}$
is _____________ (rounded off to two decimal places).

Answer: $\underset{_____________}{0.25}$

Solution :-

Set R is the collection of all those points which are in intersection of three regions given below

\begin{array}{clcc} Region (1) & 0 \leq x \leq 1 & ⟹ & x \geq 0 and x \leq 1 \\ Region (2) & 0 \leq y \leq 1 & ⟹ & y \geq 0 and y \leq 1 \\ Region (3) & \frac{1}{4} \leq x y \leq \frac{1}{2} & ⟹ & y \geq \frac{1}{4 x} and y \leq \frac{1}{2 x} \end{array}

For Region(3) :

if $x = 1$ then $\frac{1}{4} \leq y \leq \frac{1}{2}$ and if $y = 1$ then $\frac{1}{4} \leq x \leq \frac{1}{2}$

if $x = \frac{1}{4}$ then $y = 1$ and if $y = \frac{1}{4}$ then $x = 1$

if $x = \frac{1}{2}$ then $\frac{1}{2} \leq y \leq 1$ and if $y = \frac{1}{2}$ then $\frac{1}{2} \leq x \leq 1$

Therefore, if $x \in [\frac{1}{4}, \frac{1}{2}]$ then $y \in [\frac{1}{4 x}, 1]$ and if $x \in [\frac{1}{2}, 1]$ then $y \in [\frac{1}{4 x}, \frac{1}{2 x}]$

Hence, intersection points of those three regions are $(1, \frac{1}{4}), (1, \frac{1}{2}), (\frac{1}{4}, 1), (\frac{1}{2}, 1), (\frac{1}{2}, \frac{1}{2}) .$

Now, the area of the region

\begin{array}{llllllllll} R & = & \int_{\frac{1}{4}}^{\frac{1}{2}} \int_{\frac{1}{4 x}}^{1} d y d x + \int_{\frac{1}{2}}^{1} \int_{\frac{1}{4 x}}^{\frac{1}{2 x}} d y d x \\ = & \int_{\frac{1}{4}}^{\frac{1}{2}} {[y]}_{\frac{1}{4 x}}^{1} d x + \int_{\frac{1}{2}}^{1} {[y]}_{\frac{1}{4 x}}^{\frac{1}{2 x}} d x \\ = & \int_{\frac{1}{4}}^{\frac{1}{2}} (1 - \frac{1}{4 x}) d x + \int_{\frac{1}{2}}^{1} (\frac{1}{2 x} - \frac{1}{4 x}) d x \\ = & {[x - \frac{1}{4} log x]}_{\frac{1}{4}}^{\frac{1}{2}} + \int_{\frac{1}{2}}^{1} \frac{1}{4 x} d x \\ = & [(\frac{1}{2} - \frac{1}{4} log \frac{1}{2}) - (\frac{1}{4} - \frac{1}{4} log \frac{1}{4})] + {[\frac{1}{4} log x]}_{\frac{1}{2}}^{1} \\ = & [\frac{1}{2} - \frac{1}{4} log \frac{1}{2} - \frac{1}{4} + \frac{1}{4} log \frac{1}{4}] + [\frac{1}{4} log 1 - \frac{1}{4} log \frac{1}{2}] \\ = & [\frac{1}{4} - \frac{2}{4} log \frac{1}{2} + \frac{1}{4} log {(\frac{1}{2})}^{2} + 0] (∵ log 1 = 0) \\ = & \frac{1}{4} - \frac{1}{2} log \frac{1}{2} + \frac{1}{4} \cdot 2 log \frac{1}{2} (∵ log {(x)}^{n} = n log x) \\ = & \frac{1}{4} - \frac{1}{2} log \frac{1}{2} + \frac{1}{2} log \frac{1}{2} \\ = & \frac{1}{4} = 0.25 . \end{array}

Hence, The area of the region R, rounded off to two decimal places is 0.25.

So, the correct answer is $\to$ $\underset{_____________}{0.25}$

#Integral Calculus #IIT JAM PYQ 2024 #Mathematics #IIT JAM PYQs

Question

The area of the region R={(x,y)∈ℝ2:0≤x≤1,0≤y≤1and14≤x⁢y≤12} is _____________ (rounded off to two decimal places).

Solution :-

The area of the region
$R = {(x, y) \in ℝ^{2} : 0 \leq x \leq 1, 0 \leq y \leq 1 and \frac{1}{4} \leq x y \leq \frac{1}{2}}$
is _____________ (rounded off to two decimal places).