Case study (4 Marks)

Question 14 Marks

Consider the following equations of curves $x^2 = y$ and $y = x.$ On the basis of above information, answer the following questions.

The point$(s)$ of intersection of both the curves is $($are$).$

$(0, 0)(2, 2)$
$(0, 0)(1, 1)$
$(0, 0)(-1, -1)$
$(0, 0)(-2, -2)$

Area bounded by the curves is represented by which of the following graph?

The value of the integral $\int\limits_{1}^{0}\text{x}\ \text{dx}$ is.

$\frac{1}{4}$
$\frac{1}{3}$
$\frac{1}{2}$
$1$

The value of the integral $\int\limits_{0}^{1}\text{x}^2\ \text{dx}$ is.

$\frac{1}{4}$
$\frac{1}{3}$
$\frac{1}{2}$
$1$

The value of area bounded by the curves $x^2 = y$ and $x = y$ is.

$\frac{1}{6}\text{ sq}.\text{unit}$
$\frac{1}{3}\text{ sq}.\text{unit}$
$\frac{1}{2}\text{ sq}.\text{unit}$
${1}\text{ sq}.\text{unit}$

Answer

$(c)\ (0, 0)(-1, -1)$

We have, $x^2 = y…(i)$ and $x = y$
From $(i)$ and $(ii), x^2 = x$
$\Rightarrow x^2 - x = 0$
$\Rightarrow x(x - 1) = 0$
$\Rightarrow x = 0, 1$
From $(ii) y = 0, 1$
Required points of intersection are $(0, 0), (1, 1)$

$(a) $

$(c)\ \frac{1}{2}$

$\int\limits_{0}^{1}\text{x}\ \text{dx}=\Big[\frac{\text{x}^2}{2}\Big]^1_0=\frac{1}{2}-0=\frac{1}{2}$

$(b)\ \frac{1}{3}$

$\int\limits_{0}^{1}\text{x}^2\ \text{dx}=\Big[\frac{\text{x}^3}{3}\Big]^1_0=\frac{1}{3}-0=\frac{1}{3}$

$(a)\ \frac{1}{6}\text{ sq}.\text{unit}$

Required area $\int\limits_{0}^{1}\text{x}^\ \text{dx}-\int\limits_{0}^{1}\text{x}^2\ \text{dx}$
$=\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\text{ sq.units}$

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Question 24 Marks

Ajay cut two circular pieces of cardboard and placed one upon other as shown in figure. One of the circle represents the equation $(x - 1)^2 + y^2 = 1,$ while other circle represents the equation $x^2 + y^2 = 1.$

Based on the above information, answer the following questions.

Both the circular pieces of cardboard meet each other at

$\text{x}=1$
$\text{x}=\frac{1}{2}$
$\text{x}=\frac{1}{3}$
$\text{x}=\frac{1}{4}$

Graph of given two curves can be drawn as.

None of these

Value of $\int\limits_{0}^{\frac{1}{2}}\sqrt{1-(\text{x}-1)^2}\text{dx}$ is.

$\frac{\pi}{6}-\frac{\sqrt{3}}{8}$
$\frac{\pi}{6}+\frac{\sqrt{3}}{8}$
$\frac{\pi}{2}+\frac{\sqrt{3}}{4}$
$\frac{\pi}{2}-\frac{\sqrt{3}}{4}$

Value of $\int\limits_{\frac{1}{2}}^{1}\sqrt{1-\text{x}^2}\text{dx}$ is.

$\frac{\pi}{6}+\frac{\sqrt{3}}{4}$
$\frac{\pi}{6}+\frac{\sqrt{3}}{8}$
$\frac{\pi}{6}-\frac{\sqrt{3}}{8}$
$\frac{\pi}{2}-\frac{\sqrt{3}}{4}$

Area of hidden portion of lower circle is.

$\bigg(\frac{2\pi}{3}+\frac{\sqrt{3}}{2}\bigg)\text{ sq.units}$
$\bigg(\frac{\pi}{3}-\frac{\sqrt{3}}{8}\bigg)\text{ sq.units}$
$\bigg(\frac{\pi}{3}+\frac{\sqrt{3}}{8}\bigg)\text{ sq.units}$
$\bigg(\frac{2\pi}{3}-\frac{\sqrt{3}}{2}\bigg)\text{ sq.units}$

Answer

$(b)\ \text{x}=\frac{1}{2}$

We have, $(x - 1)^2 + y^2 = 1$
$\Rightarrow\text{y}=\sqrt{1-(\text{x}-1)^2}$
Also $\text{x}^2+\text{y}^2=1$
$\Rightarrow\text{y}=\sqrt{1-\text{x}^2}$
From $(i)$ and $(ii),$ we get
$\sqrt{1-(\text{x}-1)^2}=\sqrt{1-\text{x}^2}$
$\Rightarrow(\text{x}-1)^2=\text{x}^2$
$\Rightarrow2\text{x}=1$
$\Rightarrow\text{x}=\frac{1}{2}$

$(c)$

$(a)\ \frac{\pi}{6}-\frac{\sqrt{3}}{8}$

$\bigg[\int\limits_{0}^{\frac{1}{2}}\sqrt{1-(\text{x}-1)^2}+\frac{1}{2}\text{sin}^{-1}\Big(\frac{\text{x}-1}{1}\Big)\bigg]$
$=\frac{1}{2}\Big(\frac{1}{2}-1\Big)\sqrt{1-\frac{1}{4}}+\frac{1}{2}+\text{sin}^{-1}\Big(-\frac{1}{2}\Big)-\Big(-\frac{1}{2}\Big)$
$-\frac{1}{2}\text{sin}^{-1}$
$=\bigg[\frac{-1}{4}.\frac{\sqrt{3}}{2}-\frac{1}{2}.\frac{\pi}{6}+0+\frac{1}{2}.\frac{\pi}{2}\bigg]=\frac{\sqrt{3}}{8}-\frac{\pi}{12}+\frac{\pi}{4}$
$=\frac{\pi}{6}-\frac{\sqrt{3}}{8}$

$(c)\ \frac{\pi}{6}-\frac{\sqrt{3}}{8}$

$\int\limits_{\frac{1}{2}}^{1}\sqrt{1-\text{x}^2}\text{dx}=\bigg[{\frac{\text{x}}{2}}\sqrt{1-\text{x}^2}+\frac{1}{2}\text{sin}^{-1}\text{x}\bigg]^1_\frac{1}{2}$
$=0+\frac{1}{2}\text{sin}^{-1}(1)-\frac{1}{4}\sqrt{1-\frac{1}{4}}-\frac{1}{2}\text{sin}^{-1}\big(\frac{1}{2}\big)$
$=\frac{\pi}{4}-\frac{\sqrt{3}}{8}-\frac{\pi}{12}=\frac{\pi}{6}-\frac{\sqrt{3}}{8}$

$(d)\ \bigg(\frac{2\pi}{3}-\frac{\sqrt{3}}{2}\bigg)\text{ sq.units}$

$=2\bigg[\int\limits_{0}^{\frac{1}{2}}\sqrt{1-(\text{x}-1)^2}\text{dx}+\int\limits_{\frac{1}{2}}^{1}\sqrt{1-\text{x}^2}\text{dx}\bigg]$
$=2\bigg[\frac{\pi}{6}-\frac{\sqrt{3}}{8}+\frac{\pi}{6}-\frac{\sqrt{3}}{8}\bigg]$
$=2\bigg[\frac{\pi}{3}-\frac{\sqrt{3}}{4}\bigg]=\bigg(\frac{2\pi}{3}-\frac{\sqrt{3}}{2}\bigg)\text{ sq.units}$

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Question 34 Marks

Consider the following equation of curve $y^2 = 4x$ and straight line $x + y = 3.$
Based on the above information, answer the following questions.

The line $x + y = 3$ cuts the $x-$axis and $y-$axis respectively at.

$(0, 2), (2, 0)$
$(3, 3), (0, 0)$
$(0, 3), (3, 0)$
$(3, 0), (0, 3)$

Point(s) of intersection of two given curves is $($are$).$

$(1, -2), (-9, 6)$
$(2, 1), (-6, 9)$
$(1, 2), (9, -6)$
None of these.

Which of the following shaded portion re present the area bounded by given curves?
4. None of these

Value of the integral $\int\limits_{-6}^{2}(3-\text{y})\text{ dy}$ is

$10$
$20$
$30$
$40$

Value of area bounded by given curves is.

$56\text{ sq.units}$
$\frac{63}{5}\text{ sq. units}$
$\frac{64}{3}\text{ sq. units}$
$31\text{ sq.units}$

Answer

$(d) (3, 0), (0, 3)$

Line $x + y = 3$ cuts the $x-$axis and $y-$axis at $(3, 0)$ and $(0, 3)$ respectively.
$[$Since, at $x-$axis, $y = 0$ and at $y-$axis, $x = 0]$

$(c) (1, 2), (9, -6)$

We have $y^2 = 4x$ and $x + y = 3$
From $(i)$ and $(ii),$ we have $y^2 = 4(3 - y)$
$\Rightarrow y^2 + 4y - 12 = 0$
$\Rightarrow y^2 + 6y - 2y - 12 = 0$
$\Rightarrow y(y + 6) - 2 (y + 6) = 0$
$\Rightarrow (y + 6)(y - 2) = 0$
$\Rightarrow y = 2, y = -6$
From $(ii), x = 3 - 2 = 1$
or $x = 3 + 6 = 9$
$\therefore$ Required points of intersection are $(1, 2), (9, -6)$

$(b) $

$(d)\ 40$

$\int\limits_{-6}^{2}(3-\text{y})\text{dy}=\bigg[3\text{y}-\frac{\text{y}^2}{2}\bigg]^2_{-6}$
$=\bigg[6-\frac{4}{2}-\bigg[{(-6)}-\frac{{(-6)}^2}{2}\bigg]\bigg]=4+36=40$

$(c)\ \frac{64}{3}\text{ sq. units}$

Required area $=\int\limits_{-6}^{2}(3-\text{y})\text{dy}-\int\limits_{-6}^{2}\frac{\text{y}^2}{4}\text{dy}$
$=40-\frac{1}{4}\Big[\frac{\text{y}^3}{3}\Big]^2_{-6}$
$=40-\frac{1}{4}\Big[\frac{8}{3}-\frac{(-6)}{3}^3\Big]$
$=40-\frac{2}{3}-\frac{216}{12}$
$=\frac{480-8-216}{12}$
$=\frac{256}{12}=\frac{64}{3}\text{ sq.units}$

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Question 44 Marks

A mirror in the shape of an ellipse represented by $\frac{\text{x}^2}{9}+-\frac{\text{y}^2}{4}=1$ was hanging on the wall. Arun and his sister were playing with ball inside the house, even their mother refused to do so. All of sudden, ball hit the mirror and got a scratch in the shape of line represented by $\frac{\text{x}}{3}+\frac{\text{y}}{2}=1$

Based on the above information, answer the following questions.

Point(s) of intersection of ellipse and scratch (straight line) is (are).

(0, 2), (3, 0)
(2, 0), (3, 0)
(2, 3), (0, 0)
(0, 3), (3, 0)

Area of smaller region bounded by the ellipse and line is represented by.

The value of $\frac{2}{3}\int\limits_{0}^{3}\sqrt{9-\text{x}^2}\text{dx}$ is.
1. $\frac{\pi}{2}$
2. $\pi$
3. $\frac{3\pi}{2}$
4. $\frac{\pi}{4}$

The value of $2\int\limits_{0}^{3}\bigg(1-\frac{\text{x}}{3}\bigg)\text{dx}$ is.
1. 0
2. 1
3. 2
4. 3

Area of the smaller region bounded by the mirror and scratch is.

$3\Big(\frac{\pi}{2}+1\Big)\text{ sq.units}$
$\Big(\frac{\pi}{2}+1\Big)\text{ sq.units}$
$\Big(\frac{\pi}{2}-1\Big)\text{ sq.units}$
$3\Big(\frac{\pi}{2}-1\Big)\text{ sq.units}$

Answer

(a) (0, 2), (3, 0)

Solution:
Points (0, 2) and (3, 0) pass through both line and ellipse.

Solution:
$\frac{2}{3}\int\limits_{0}^{3}\sqrt{9-\text{x}^2}\text{dx}=\frac{2}{3}\int\limits_{0}^{3}\sqrt{(3)^2-\text{x}^2}\text{dx}$
$=\frac{2}{3}\bigg[\frac{1}{2}\text{x}\sqrt{9-\text{x}^2}+\frac{9}{2}\text{sin}^{-1} \frac{\text{x}}{3}\bigg]^3_0$
$=\frac{2}{3}\bigg[\frac{3}{2}\sqrt{0}+\frac{9}{2}\text{sin}^{-1}(1)-\frac{1}{2}(0)-\frac{9}{2}\text{sin}^{-1}(0)\bigg]$
$=\frac{2}{3}\bigg[\frac{9}{2}.\frac{\pi}{2}\bigg]=\frac{3\pi}{2}$

(d) 3

Solution:
$2\int\limits_{0}^{3}\Big(1-\frac{\text{x}}{3}\Big)\text{dx}=2\bigg[\text{x}-\frac{\text{x}^2}{6}\bigg]^3_0$
$=2\Big(3-\frac{9}{6}-0-0\Big)=2\times\frac{3}{2}=3$

(d) $3\Big(\frac{\pi}{2}-1\Big)\text{ sq.units}$

Solution:
Area of smaller region bounded by the mirror and scratch
$=\frac{2}{3}\int\limits_{0}^{3}\sqrt{9-\text{x}^2}\text{dx}-2\int\limits_{0}^{3}\Big(1-\frac{\text{x}}{3}\Big)\text{dx}$
$=\frac{3\pi}{2}-3=3\Big(\frac{\pi}{2}-1\Big)\text{ sq.units}$

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Question 54 Marks

Graphs of two function $\text{f}(\text{x})=\text{sin}\text{ x}$ and $\text{(g)}\text{x}=\text{cos}\text{ x}$ is given below:

Based on the above information, answer the following questions.

In $(0, \pi)$, the curves $\text{f}(\text{x})=\text{sin}\text{ x}$ and $\text{g}\text{ (x)}=\text{cos}\text{ x}$ at $\text{x}=$
1. $\frac{\pi}{2}$
2. $\frac{\pi}{3}$
3. $\frac{\pi}{4}$
4. ${\pi}$
Value of $\int\limits_{0}^{\frac{\pi}{4}}\text{sin}\text{ x}\text{ dx}$ is.
1. $1-\frac{1}{\sqrt{2}}$
2. $1+\frac{1}{\sqrt{2}}$
3. $2-\frac{1}{\sqrt{2}}$
4. $2+\frac{1}{\sqrt{2}}$

Value of $\int\limits_\frac{\pi}{4}^{\frac{\pi}{2}}\text{cos}\text{ x}\text{ dx}$ is.
1. $1+\frac{1}{\sqrt{2}}$
2. $1-\frac{1}{\sqrt{2}}$
3. $2-\sqrt{2}$
4. $2+\sqrt{2}$
Value of $\int\limits_{0}^{\pi}\text{sin}\text{ x}\text{ dx}$ is.

0
1
2
-2

Value of $\int\limits_{0}^\frac{\pi}{2}\text{sin}\text{ x}\text{ dx}$ is.

Answer

Solution:
for point of intersection, we have
$\text{sin}\text{ x}=\text{cos}\text{ x}$
$\Rightarrow\frac{\text{sin}\text{ x}}{\text{cos}\text{ x}}=1$
$\Rightarrow\text{tan}\text{ x}=1$
$\Rightarrow\text{x}=\frac{\pi}{4}$

(a) $1-\frac{1}{\sqrt{2}}$

Solution:
$\int\limits_{0}^{\frac{\pi}{4}}\text{sin}\text{ x}\text{ dx}=\big[-\text{cos x}\big]^\frac{\pi}{4}_0=-\text{cos}\frac{\pi}{4}+\text{cos}0$
$=1-\frac{1}{\sqrt{2}}$

(b) $1-\frac{1}{\sqrt{2}}$

Solution:
$\int\limits_{\frac{\pi}{2}}^{\frac{\pi}{4}}\text{cos}\text{ x}\text{ dx}=\big[\text{sin x}\big]^\frac{\pi}{2}_\frac{\pi}{4}=\text{sin}\frac{\pi}{2}-\text{sin}\frac{\pi}{4}$
$=1-\frac{1}{\sqrt{2}}$

Solution:
$\int\limits_{0}^{\pi}\text{sin}\text{ x}\text{ dx}=\big[-\text{cos x}\big]^{\pi}_0=\big[-\text{cos}{\pi}+\text{cos}0\big]=2$

(b) 1

Solution:
$\int\limits_{0}^\frac{\pi}{2}\text{sin}\text{ x}\text{ dx}=\big[-\text{cos x}\big]^\frac{\pi}{2}_0=\Big[-\text{cos}\frac{\pi}{2}+\text{cos}0\Big]$
$=0+1+1$

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Question 64 Marks

In a classroom, teacher explains the properties of a particular curve by saying that this particular curve has beautiful up and downs. It starts at 1 and heads down until $\pi$ radian, and then heads up again and closely related to sine function and both follow each other, exactly $\frac{\pi}{2}$ radians apart as shown in figure.

Based on the above information, answer the following questions.

Name the curve, about which teacher explained in the classroom.

Cosine
Sine
Tangent
Cotangent

Area of curve explained in the passage from 0 to $\frac{\pi}{2}$ is.

$\frac{1}{3}\text{ sq.unit}$
$\frac{1}{2}\text{ sq.unit}$
${1}\text{ sq.unit}$
${2}\text{ sq.units}$

Area of curve discussed in classroom from $\frac{\pi}{2}$ to $\frac{3\pi}{2}$ is.

-2 sq. units
2 sq. units
3 sq. units
-3 sq. units

Area of curve discussed in classroom from $\frac{3\pi}{2}$ to $2\pi$ is.

1 sq. unit
2 sq. units
3 sq. units
4 sq. units

Area of explained curve from 0 to $2\pi$ is.

1 sq. unit
2 sq. units
3 sq. units
4 sq. units

Answer

(a) Cosine

Solution:
Here, teacher explained about cosine curve.

Solution:
Required area $=\int\limits_{0}^{\frac{\pi}{2}}\text{cos x dx}$
$=\big[\text{sin}\text{x}\big]^\frac{\pi}{2}_0=\text{sin}\frac{\pi}{2}-\text{sin}0=1-0=1\text{ sq.unit}$

(b) 2 sq. units

Solution:
Required area $=\begin{vmatrix}\int\limits_\frac{\pi}{2}^{\frac{3\pi}{2}}&\cos\text{x dx}\end{vmatrix}=\begin{vmatrix}\big[\sin\text{x}\big]^{\frac{3\pi}{2}}_{\frac{\pi}{2}}\end{vmatrix}$
$=\begin{vmatrix}\sin{\frac{3\pi}{2}}-\text{sin}{\frac{\pi}{2}}\end{vmatrix}$
$=\begin{vmatrix}-1-1\end{vmatrix}=|-2|$
$=2\text{ sq.units}$ [Since, area can't be negative]

(a) 1 sq. unit

Solution:
Required area $=\int\limits_{\frac{3\pi}{2}}^{2\pi}\text{cos x dx}=\big[\text{sin}\text{ x}\big]^{2\pi}_\frac{3\pi}{2}$
$=\text{sin } 2\pi-\text{sin}\frac{3\pi}{2}$
$=0(-1)=1\text{ sq.unit}$

(d) 4 sq. units

Solution:
Required area $=\int\limits_{0}^{\frac{\pi}{2}}\text{cos x dx}+=\begin{vmatrix}\int\limits_\frac{\pi}{2}^{\frac{3\pi}{2}}&\cos\text{x dx}\end{vmatrix}+\int\limits_{\frac{3\pi}{2}}^{2\pi}\text{cos x dx}$
$=1+2+1=4\text{ sq.units}$

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Question 74 Marks

Consider the following equations of curves $\text{y}=\cos\text{x},\text{y}=\text{x}+1$ and y = 0.
On the basis of above information, answer the following questions.

The curves $\text{y}=\cos\text{x}$ and y = x + 1 meet at:

(1, 0)
(0, 1)
(1, 1)
(0, 0)

$\text{y}=\cos\text{x}$ meet the x-axis at:

$\Big(\frac{-\pi}{2},0\Big)$
$\Big(\frac{\pi}{2},0\Big)$
Both (a) and (b).
None of these.

Value of the integral $\int\limits_{-1}^{0}(\text{x}+1)\text{dx}$ is:

$\frac{1}{2}$
$\frac{2}{3}$
$\frac{3}{4}$
$\frac{1}{3}$

Value of the integral $\int\limits_{0}^{\frac{\pi}{2}}\cos\text{x dx}$ is:

0
-1
2
1

Area bounded by the given curves is:

$\frac{1}{2}\text{ sq}.\text{units}$
$\frac{3}{2}\text{ sq}.\text{units}$
$\frac{3}{4}\text{ sq}.\text{units}$
$\frac{1}{4}\text{ sq}.\text{units}$

Answer

(b) (0, 1)

Solution:
Curves $\text{y}=\cos\text{x}$ and y = x + 1 meet at point C(0, 1).

Solution:
Curve $\text{y}=\cos\text{x}$ meet the x-axis at $\text{A}'\Big(\frac{-\pi}{2},0\Big)$ and $\text{A}\Big(\frac{\pi}{2},0\Big).$

(a) $\frac{1}{2}$

Solution:
$\int\limits_{-1}^{0}(\text{x}+1)\text{dx}=\Big[\frac{\text{x}^2}{2}+\text{x}\Big]_{-1}^0=0-\Big(\frac{1}{2}-1\Big)=\frac{1}{2}$

(d) 1

Solution:
$\int\limits_{0}^{\frac{\pi}{2}}\cos\text{x dx}=[\sin\text{x}]^{\frac{\pi}{2}}_0=\sin\frac{\pi}{2}-\sin0=1$

(b) $\frac{3}{2}\text{ sq}.\text{units}$

Solution:
Required area $=\int\limits_{-1}^{0}(\text{x}+1)\text{dx}+\int\limits_{0}^{\frac{\pi}{2}}\cos\text{x dx}$
$=\frac{1}{2}+1=\frac{3}{2}\text{ sq}.\text{units}$

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Question 84 Marks

Consider the curve $x^2 + y^2 = 16$ and line $y = x$ in the first quadrant.
Based on the above information, answer the following questions.

Point of intersection of both the given curves is.

$(0, 4)$
$(0, 2\sqrt{2})$
$(2\sqrt{ 2},2\sqrt{2})$
$(2,\sqrt{2},4)$

Which of the following shaded portion represent the area bounded by given two curves?

None of these

The value of the integral $\int\limits_{0}^{2\sqrt{2}}\text{x}\text{dx}$ is.

The value of the integral $\int\limits_{2\sqrt{2}}^{0}\sqrt{16-\text{x}^2}\text{ dx}$ is.

$2(\pi-2)$
$2(\pi-8)$
$4(\pi-2)$
$4(\pi+2)$

Area bounded by the two given curves is.

$3\pi\ \text{ sq.units}$
$\frac{\pi}{2}\ \text{ sq.units}$
$\pi\ \text{ sq.units}$
$2\pi\ \text{ sq.units}$

Answer

We have $x^2 + y^2 = 16$ and $y = x$
From $(i)$ and $(ii) \text{2x}=16$
$\Rightarrow\text{x}^2=8$
$\Rightarrow\text{x}=2\sqrt{2}$
$(\therefore x$ lies in first quadrant$)$
$\therefore$ Point of intersection of $(i)$ and $(ii)$ in first quadrant $(2\sqrt{2},(2\sqrt{2})$

$(b)$

The shaded region which represent the area bounded by two given curves in first quadrant is shown below.

$(d) 4$

$\int\limits_{0}^{2\sqrt{2}}\text{x}\text{dx}=\Big[\frac{\text{x}^2}{2}\Big]^2_0=\frac{(2\sqrt{2})}{2}=\frac{8}{2}=4$

(a) $2(\pi-2)$

$\int\limits_{2\sqrt{2}}^{4}\sqrt{16-\text{x}^2}$
$\text{ dx}=\Big[\frac{\text{x}}{2}\sqrt{16-\text{x}^2}+\frac{16}{2}\text{sin}^{-1}\Big(\frac{\text{x}}{4}\Big)\Big]^4_{2\sqrt{2}}$
$=8\text{sin}^{-1}(1)-4-8\text{sin}^{-1}\Big(\frac{1}{\sqrt{2}}\Big)$
$=8\Big(\frac{\pi}{2}\Big)-4-8\Big(\frac{\pi}{4}\Big)$
$=4\pi-4-2\pi=2\pi-4=2(\pi-2)$

(d) $2\pi\text{ sq.units}$

Required area $=$ Area $\text{(OLA)} +$ Area $\text{(BAL)}$
$\int\limits_{0}^{2\sqrt{2}}\text{x}\text{dx}+\int\limits_{2\sqrt{2}}^{4}\sqrt{16-\text{x}^2}\text{dx}$
$4+2(\pi-2)=2\pi\text{ sq.units}$

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Question 94 Marks

Location of three houses of a society is represented by the points A(-1, 0), B(1, 3) and C(3, 2) as shown in figure.

Based on the above information, answer the following questions

Equation of line AB is.
1. $\text{y}=\frac{3}{2}(\text{x}+1)$
2. $\text{y}=\frac{3}{2}(\text{x}-1)$
3. $\text{y}=\frac{1}{2}(\text{x}+1)$
4. $\text{y}=\frac{1}{2}(\text{x}-1)$
Equation of line BC is.
1. $\text{y}=\frac{1}{2}\text{x}-\frac{7}{2}$
2. $\text{y}=\frac{3}{2}\text{x}-\frac{7}{2}$
3. $\text{y}=\frac{-1}{2}\text{x}+\frac{7}{2}$
4. $\text{y}=\frac{3}{2}\text{x}+\frac{7}{2}$
Area of region ABCD is.

2 sq. units
4 sq. units
6 sq. units
8 sq. units

Area of $\triangle\text{ADC}$ is,

4 sq. units
8 sq. units
16 sq. units
32 sq. units

Area of $\triangle\text{ABC}$ is.

3 sq. units
4 sq. units
5 sq. units
6 sq. units

Answer

(a) $\text{y}=\frac{3}{2}(\text{x}+1)$

Solution:
Equation of line AB is.$\text{y}-0=\frac{3-0}{1+1}(\text{x}+1)\Rightarrow\text{y}=\frac{3}{2}(\text{x}+1)$

Solution:
Equation of line BC is $\text{y}-3=\frac{2-3}{3-1}(\text{x}+1)$
$\Rightarrow\text{y}=-\frac{1}{2}\text{x}+\frac{1}{2}+3\Rightarrow\text{y}=\frac{-1}{2}\text{x}+\frac{7}{2}$

(d) 8 sq. units

Solution:
Area of region ABCD = Area of $\triangle\text{ABE }+$ Area of region BCDE
$=\int\limits_{-1}^{1}\frac{3}{2}(\text{x}+1)\text{dx}+\int\limits_{1}^{3}\bigg(\frac{-1}{2}\text{x}+\frac{7}{2}\bigg)\text{dx}$
$=\frac{3}{2}\bigg[\frac{\text{x}^2}{2}+\text{x}\bigg]^1_{-1}+\bigg[\frac{\text{-x}^2}{4}+\frac{7}{2}\text{x}\bigg]^3_{1}$
$\frac{3}{2}\bigg[\frac{1}{2}+1-\frac{1}{2}+1\bigg]+\bigg[\frac{-9}{4}+\frac{21}{2}+\frac{1}{4}-\frac{7}{2}\bigg]$
$=3+5=8\text{ sq.units}$

(a) 4 sq. units

Solution:
Equation of line AC is $\text{y}-0=\frac{2-0}{3+1}(\text{x}+1)$
$\Rightarrow\text{y}=\frac{1}{2}(\text{x}+1)$
$\therefore\text{Area of }\triangle\text{ADC}=\int\limits_{-1}^{3}\frac{1}{2}(\text{x}+1)\text{dx}=\bigg[\frac{\text{x}^2}{4}+\frac{1}{2}\text{x}\bigg]^3_{-1}$
$=\frac{9}{4}+\frac{3}{2}-\frac{1}{4}+\frac{1}{2}=4\text{ sq.units}$

(b) 4 sq. units

Solution:
Area of $\triangle\text{ABC }=$ Area of region ABCD - Area of $\triangle\text{ACD}=8-4=4\text{ sq.units}$

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MATHS Case study (4 Marks)

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