Let $f : [ - 1,2 ] \rightarrow [ 0 , \infty )$ be a continuous function such that $f ( x ) = f ( 1 - x )$ for all $x \in [ - 1,2 ]$. Let $R _ { 1 } = \int _ { - 1 } ^ { 2 } x f ( x ) d x$, and $R _ { 2 }$ be the area of the region bounded by $y = f ( x ) , x = - 1 , x = 2$, and the $x$-axis. Then (A) $R _ { 1 } = 2 R _ { 2 }$ (B) $R _ { 1 } = 3 R _ { 2 }$ (C) $2 R _ { 1 } = R _ { 2 }$ (D) $3 R _ { 1 } = R _ { 2 }$
For a point $P$ in the plane, let $d_1(P)$ and $d_2(P)$ be the distances of the point $P$ from the lines $x - y = 0$ and $x + y = 0$ respectively. The area of the region $R$ consisting of all points $P$ lying in the first quadrant of the plane and satisfying $2 \leq d_1(P) + d_2(P) \leq 4$, is
A farmer $F _ { 1 }$ has a land in the shape of a triangle with vertices at $P ( 0,0 ) , Q ( 1,1 )$ and $R ( 2,0 )$. From this land, a neighbouring farmer $F _ { 2 }$ takes away the region which lies between the side $P Q$ and a curve of the form $y = x ^ { n } ( n > 1 )$. If the area of the region taken away by the farmer $F _ { 2 }$ is exactly $30 \%$ of the area of $\triangle P Q R$, then the value of $n$ is $\_\_\_\_$.
The area of the region $$\left\{ (x,y) : 0 \leq x \leq \frac{9}{4}, \quad 0 \leq y \leq 1, \quad x \geq 3y, \quad x + y \geq 2 \right\}$$ is (A) $\frac{11}{32}$ (B) $\frac{35}{96}$ (C) $\frac{37}{96}$ (D) $\frac{13}{32}$
Consider the functions $f , g : \mathbb { R } \rightarrow \mathbb { R }$ defined by $$f ( x ) = x ^ { 2 } + \frac { 5 } { 12 } \quad \text { and } \quad g ( x ) = \begin{cases} 2 \left( 1 - \frac { 4 | x | } { 3 } \right) , & | x | \leq \frac { 3 } { 4 } \\ 0 , & | x | > \frac { 3 } { 4 } \end{cases}$$ If $\alpha$ is the area of the region $$\left\{ ( x , y ) \in \mathbb { R } \times \mathbb { R } : | x | \leq \frac { 3 } { 4 } , 0 \leq y \leq \min \{ f ( x ) , g ( x ) \} \right\}$$ then the value of $9 \alpha$ is $\_\_\_\_$ .
Let $f : [ 0,1 ] \rightarrow [ 0,1 ]$ be the function defined by $f ( x ) = \frac { x ^ { 3 } } { 3 } - x ^ { 2 } + \frac { 5 } { 9 } x + \frac { 17 } { 36 }$. Consider the square region $S = [ 0,1 ] \times [ 0,1 ]$. Let $G = \{ ( x , y ) \in S : y > f ( x ) \}$ be called the green region and $R = \{ ( x , y ) \in S : y < f ( x ) \}$ be called the red region. Let $L _ { h } = \{ ( x , h ) \in S : x \in [ 0,1 ] \}$ be the horizontal line drawn at a height $h \in [ 0,1 ]$. Then which of the following statements is(are) true? (A) There exists an $h \in \left[ \frac { 1 } { 4 } , \frac { 2 } { 3 } \right]$ such that the area of the green region above the line $L _ { h }$ equals the area of the green region below the line $L _ { h }$ (B) There exists an $h \in \left[ \frac { 1 } { 4 } , \frac { 2 } { 3 } \right]$ such that the area of the red region above the line $L _ { h }$ equals the area of the red region below the line $L _ { h }$ (C) There exists an $h \in \left[ \frac { 1 } { 4 } , \frac { 2 } { 3 } \right]$ such that the area of the green region above the line $L _ { h }$ equals the area of the red region below the line $L _ { h }$ (D) There exists an $h \in \left[ \frac { 1 } { 4 } , \frac { 2 } { 3 } \right]$ such that the area of the red region above the line $L _ { h }$ equals the area of the green region below the line $L _ { h }$
Let $n \geq 2$ be a natural number and $f : [ 0,1 ] \rightarrow \mathbb { R }$ be the function defined by $$f ( x ) = \begin{cases} n ( 1 - 2 n x ) & \text { if } 0 \leq x \leq \frac { 1 } { 2 n } \\ 2 n ( 2 n x - 1 ) & \text { if } \frac { 1 } { 2 n } \leq x \leq \frac { 3 } { 4 n } \\ 4 n ( 1 - n x ) & \text { if } \frac { 3 } { 4 n } \leq x \leq \frac { 1 } { n } \\ \frac { n } { n - 1 } ( n x - 1 ) & \text { if } \frac { 1 } { n } \leq x \leq 1 \end{cases}$$ If $n$ is such that the area of the region bounded by the curves $x = 0 , x = 1 , y = 0$ and $y = f ( x )$ is 4, then the maximum value of the function $f$ is
Let $S = \left\{ ( x , y ) \in \mathbb { R } \times \mathbb { R } : x \geq 0 , y \geq 0 , y ^ { 2 } \leq 4 x , y ^ { 2 } \leq 12 - 2 x \right.$ and $\left. 3 y + \sqrt { 8 } x \leq 5 \sqrt { 8 } \right\}$. If the area of the region $S$ is $\alpha \sqrt { 2 }$, then $\alpha$ is equal to (A) $\frac { 17 } { 2 }$ (B) $\frac { 17 } { 3 }$ (C) $\frac { 17 } { 4 }$ (D) $\frac { 17 } { 5 }$
Let $\mathbb { R }$ denote the set of all real numbers. Then the area of the region $$\left\{ ( x , y ) \in \mathbb { R } \times \mathbb { R } : x > 0 , y > \frac { 1 } { x } , 5 x - 4 y - 1 > 0,4 x + 4 y - 17 < 0 \right\}$$ is
The area bounded between the parabolas $x^{2} = \frac{y}{4}$ and $x^{2} = 9y$, and the straight line $y = 2$ is (1) $20\sqrt{2}$ (2) $\frac{10\sqrt{2}}{3}$ (3) $\frac{20\sqrt{2}}{3}$ (4) $10\sqrt{2}$
The area (in square units) bounded by the curves $y = \sqrt{x}$, $2y - x + 3 = 0$, $X$-axis and lying in the first quadrant is (1) 18 sq. units (2) $\frac{27}{4}$ sq. units (3) 9 sq. units (4) 36 sq. units
The area (in sq. units) of the region described by $\{(x, y) : y^2 \leq 2x \text{ and } y \geq 4x - 1\}$ is: (1) $\frac{7}{32}$ (2) $\frac{5}{64}$ (3) $\frac{15}{64}$ (4) $\frac{9}{32}$
The area (in sq. units) of the region described by $\left\{ ( x , y ) : y ^ { 2 } \leq 2 x \text{ and } y \geq 4 x - 1 \right\}$ is (1) $\frac { 9 } { 32 }$ sq. units (2) $\frac { 7 } { 32 }$ sq. units (3) $\frac { 5 } { 64 }$ sq. units (4) $\frac { 15 } { 64 }$ sq. units
The area (in sq. units) of the region $\{(x, y) : y^2 \geq 2x$ and $x^2 + y^2 \leq 4x, x \geq 0, y \geq 0\}$ is: (1) $\pi - \frac{4\sqrt{2}}{3}$ (2) $\pi - \frac{8}{3}$ (3) $\pi - \frac{4}{3}$ (4) $\frac{\pi}{2} - \frac{2\sqrt{2}}{3}$
The area (in sq. units) of the region $\{(x,y): y^2 \geq 2x$ and $x^2 + y^2 \leq 4x, x \geq 0, y \geq 0\}$ is: (1) $\pi - \frac{4\sqrt{2}}{3}$ (2) $\pi - \frac{8}{3}$ (3) $\pi - \frac{4}{3}$ (4) $\frac{\pi}{2} - \frac{2\sqrt{2}}{3}$
The area (in sq. units) of the region $\{(x, y) : x \geq 0,\ x + y \leq 3,\ x^2 \leq 4y$ and $y \leq 1 + \sqrt{x}\}$ is (1) $\dfrac{59}{12}$ sq. units (2) $\dfrac{3}{2}$ sq. units (3) $\dfrac{7}{3}$ sq. units (4) $\dfrac{5}{2}$ sq. units