Maths Case study (4 Marks)

1. (c) $(\sqrt{3},1),(-\sqrt{3,}-1)$

Solution:

We have x² + y² = 16 and y = 4 and $\text{x}=\sqrt{3\text{y}}$

From (i) and (ii) we get

3y² + y² = 4

⇒ 4y² = 4 

⇒ y² = 1

 $\Rightarrow\text{y}=\pm1$

From (ii) $\text{x}=\sqrt{3}.-\sqrt{3}$

$\therefore$ Point of intersection of pizza and edge of knife

2. (a) 

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

Solution:

Required area $\int\limits_{0}^{\sqrt{3}}\frac{\text{x}}{\sqrt{3}}\text{dx}+\int\limits_{\sqrt{3}}^{0}\sqrt{4-\text{x}^2}\text{dx}$

$=\frac{1}{\sqrt{3}}\Big[\frac{\text{x}^2}{2}\Big]^{\sqrt{3}}_0+\bigg[\frac{\text{x}}{2}\sqrt{4-\text{x}^2}+\frac{4}{2}\text{sin}^{-1}\Big(\frac{\text{x}}{2}\Big)\bigg]^2_{\sqrt{3}}$

$=\frac{1}{\sqrt{3}}\Big[\frac{3}{2}-0\Big]+\bigg[2\text{sin}^{-1}(1)-\Big(\frac{\sqrt{3}}{2}+\text{2sin}^{-1}\frac{\sqrt{3}}{2}\Big)\bigg]$

$=\frac{\sqrt{3}}{2}+\frac{2\pi}{2}-\frac{\sqrt{3}}{2}-\frac{2\pi}{3}=\frac{\pi}{3}\text{ sq.units}$

4. (a) $\pi\text{ sq.units}$

Solution:

We have, x² + y² = 4

⇒ (x - 0)² + (y - 0)² = (2)²

$\therefore$ Radius = 2

 Area of $\frac{1}{4}$ the slice of pizza $=\frac{1}{4}\pi(2)^2=\pi\text{ sq.units}$

5. (d) $4\pi\text{ sq.units}$

Solution:

Area of whole pizza $\pi(2)^2=4\pi\text{ sq.units}$

1. (c) (0, 0)(-1, -1)

Solution:

We have, x² = y…(i) and x = y

From (i) and (ii), x² = x

⇒ x² - x = 0

⇒ x(x - 1) = 0

⇒ x = 0, 1

From (ii) y = 0, 1

Required points of intersection are (0, 0), (1, 1)

2. (a) 

3. (c) $\frac{1}{2}$

Solution:

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

4. (b) $\frac{1}{3}$

Solution:

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

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

Solution:

Required area $\int\limits_{0}^{1}\text{x}^\ \text{dx}-\int\limits_{0}^{1}\text{x}^2\ \text{dx}$

$=\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\text{ sq.units}$

1. (b) $\text{x}=\frac{1}{2}$

Solution:

We have, (x - 1)² + y² = 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

$\sqrt{1-(\text{x}-1)^2}=\sqrt{1-\text{x}^2}$

$\Rightarrow(\text{x}-1)^2=\text{x}^2$

$\Rightarrow2\text{x}=1$

$\Rightarrow\text{x}=\frac{1}{2}$

2. (c) 

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

Solution:

$\bigg[\int\limits_{0}^{\frac{1}{2}}\sqrt{1-(\text{x}-1)^2}+\frac{1}{2}\text{sin}^{-1}\Big(\frac{\text{x}-1}{1}\Big)\bigg]$

$=\frac{1}{2}\Big(\frac{1}{2}-1\Big)\sqrt{1-\frac{1}{4}}+\frac{1}{2}+\text{sin}^{-1}\Big(-\frac{1}{2}\Big)-\Big(-\frac{1}{2}\Big)$

$-\frac{1}{2}\text{sin}^{-1}$

$=\bigg[\frac{-1}{4}.\frac{\sqrt{3}}{2}-\frac{1}{2}.\frac{\pi}{6}+0+\frac{1}{2}.\frac{\pi}{2}\bigg]=\frac{\sqrt{3}}{8}-\frac{\pi}{12}+\frac{\pi}{4}$

$=\frac{\pi}{6}-\frac{\sqrt{3}}{8}$

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

$\int\limits_{\frac{1}{2}}^{1}\sqrt{1-\text{x}^2}\text{dx}=\bigg[{\frac{\text{x}}{2}}\sqrt{1-\text{x}^2}+\frac{1}{2}\text{sin}^{-1}\text{x}\bigg]^1_\frac{1}{2}$

$=0+\frac{1}{2}\text{sin}^{-1}(1)-\frac{1}{4}\sqrt{1-\frac{1}{4}}-\frac{1}{2}\text{sin}^{-1}\big(\frac{1}{2}\big)$

$=\frac{\pi}{4}-\frac{\sqrt{3}}{8}-\frac{\pi}{12}=\frac{\pi}{6}-\frac{\sqrt{3}}{8}$

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

Solution:

$=2\bigg[\int\limits_{0}^{\frac{1}{2}}\sqrt{1-(\text{x}-1)^2}\text{dx}+\int\limits_{\frac{1}{2}}^{1}\sqrt{1-\text{x}^2}\text{dx}\bigg]$

$=2\bigg[\frac{\pi}{6}-\frac{\sqrt{3}}{8}+\frac{\pi}{6}-\frac{\sqrt{3}}{8}\bigg]$

$=2\bigg[\frac{\pi}{3}-\frac{\sqrt{3}}{4}\bigg]=\bigg(\frac{2\pi}{3}-\frac{\sqrt{3}}{2}\bigg)\text{ sq.units}$

1. (d) (3, 0), (0, 3)

Solution:

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]

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

Solution:

We have y² = 4x and x + y = 3

From (i) and (ii), we have y² = 4(3 - y)

⇒ y² + 4y - 12 = 0

⇒ y² + 6y - 2y - 12 = 0

⇒ y(y + 6) - 2 (y + 6) = 0

⇒ (y + 6)(y - 2) = 0

⇒ 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)

3. (b) 

4. (d) 40

Solution:

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

$=\bigg[6-\frac{4}{2}-\bigg[{(-6)}-\frac{{(-6)}^2}{2}\bigg]\bigg]=4+36=40$

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

Required area $=\int\limits_{-6}^{2}(3-\text{y})\text{dy}-\int\limits_{-6}^{2}\frac{\text{y}^2}{4}\text{dy}$

$=40-\frac{1}{4}\Big[\frac{\text{y}^3}{3}\Big]^2_{-6}$

$=40-\frac{1}{4}\Big[\frac{8}{3}-\frac{(-6)}{3}^3\Big]$

$=40-\frac{2}{3}-\frac{216}{12}$

$=\frac{480-8-216}{12}$

$=\frac{256}{12}=\frac{64}{3}\text{ sq.units}$

1. (a) (0, 2), (3, 0)

Solution:

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

2. (b) 

3. (c) $\frac{3\pi}{2}$

Solution:

$\frac{2}{3}\int\limits_{0}^{3}\sqrt{9-\text{x}^2}\text{dx}=\frac{2}{3}\int\limits_{0}^{3}\sqrt{(3)^2-\text{x}^2}\text{dx}$

$=\frac{2}{3}\bigg[\frac{1}{2}\text{x}\sqrt{9-\text{x}^2}+\frac{9}{2}\text{sin}^{-1} \frac{\text{x}}{3}\bigg]^3_0$

$=\frac{2}{3}\bigg[\frac{3}{2}\sqrt{0}+\frac{9}{2}\text{sin}^{-1}(1)-\frac{1}{2}(0)-\frac{9}{2}\text{sin}^{-1}(0)\bigg]$

$=\frac{2}{3}\bigg[\frac{9}{2}.\frac{\pi}{2}\bigg]=\frac{3\pi}{2}$

4. (d) 3

Solution:

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

$=2\Big(3-\frac{9}{6}-0-0\Big)=2\times\frac{3}{2}=3$

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

Solution:

Area of smaller region bounded by the mirror and scratch

$=\frac{2}{3}\int\limits_{0}^{3}\sqrt{9-\text{x}^2}\text{dx}-2\int\limits_{0}^{3}\Big(1-\frac{\text{x}}{3}\Big)\text{dx}$

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

1. (c) $\frac{\pi}{4}$

Solution:

for point of intersection, we have

$\text{sin}\text{ x}=\text{cos}\text{ x}$

$\Rightarrow\frac{\text{sin}\text{ x}}{\text{cos}\text{ x}}=1$

$\Rightarrow\text{tan}\text{ x}=1$

$\Rightarrow\text{x}=\frac{\pi}{4}$

2. (a) $1-\frac{1}{\sqrt{2}}$

Solution:

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

$=1-\frac{1}{\sqrt{2}}$

3. (b) $1-\frac{1}{\sqrt{2}}$

Solution:

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

$=1-\frac{1}{\sqrt{2}}$

4. (c) 2

Solution:

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

5. (b) 1

Solution:

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

$=0+1+1$

1. (a) Cosine

Solution:

Here, teacher explained about cosine curve.

2. (c) ${1}\text{ sq.unit}$

Solution:

Required area  $=\int\limits_{0}^{\frac{\pi}{2}}\text{cos x dx}$

$=\big[\text{sin}\text{x}\big]^\frac{\pi}{2}_0=\text{sin}\frac{\pi}{2}-\text{sin}0=1-0=1\text{ sq.unit}$

3. (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]

4. (a) 1 sq. unit

Solution:

Required area  $=\int\limits_{\frac{3\pi}{2}}^{2\pi}\text{cos x dx}=\big[\text{sin}\text{ x}\big]^{2\pi}_\frac{3\pi}{2}$

$=\text{sin } 2\pi-\text{sin}\frac{3\pi}{2}$

$=0(-1)=1\text{ sq.unit}$

5. (d) 4 sq. units

Solution:

Required area $=\int\limits_{0}^{\frac{\pi}{2}}\text{cos x dx}+​​​​=\begin{vmatrix}\int\limits_\frac{\pi}{2}^{\frac{3\pi}{2}}&\cos\text{x dx}\end{vmatrix}+\int\limits_{\frac{3\pi}{2}}^{2\pi}\text{cos x dx}$

$=1+2+1=4\text{ sq.units}$

1. (b) (0, 1)

Solution:

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

2. (c) Both (a) and (b).

Solution:

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

3. (a) $\frac{1}{2}$

Solution:

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

4. (d) 1

Solution:

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

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

Solution:

Required area $=\int\limits_{-1}^{0}(\text{x}+1)\text{dx}+\int\limits_{0}^{\frac{\pi}{2}}\cos\text{x dx}$

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

1. (c) $(2\sqrt{ 2},2\sqrt{2})$

Solution:

We have x² + y² = 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 $(2\sqrt{2},(2\sqrt{2})$

2. (b)

Solution:

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

3. (d) 4

Solution:

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

4. (a) $2(\pi-2)$

Solution:

$\int\limits_{2\sqrt{2}}^{4}\sqrt{16-\text{x}^2}$

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

$=8\text{sin}^{-1}(1)-4-8\text{sin}^{-1}\Big(\frac{1}{\sqrt{2}}\Big)$

$=8\Big(\frac{\pi}{2}\Big)-4-8\Big(\frac{\pi}{4}\Big)$

$=4\pi-4-2\pi=2\pi-4=2(\pi-2)$

5. (d) $2\pi\text{ sq.units}$

Solution:

Required area = Area (OLA) + Area (BAL)

$\int\limits_{0}^{2\sqrt{2}}\text{x}\text{dx}+\int\limits_{2\sqrt{2}}^{4}\sqrt{16-\text{x}^2}\text{dx}$

$4+2(\pi-2)=2\pi\text{ sq.units}$

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

Solution:

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

2. (c) $\text{y}=\frac{-1}{2}\text{x}+\frac{7}{2}$

Solution:

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

$\Rightarrow\text{y}=-\frac{1}{2}\text{x}+\frac{1}{2}+3\Rightarrow\text{y}=\frac{-1}{2}\text{x}+\frac{7}{2}$

3. (d) 8 sq. units

Solution:

Area of region ABCD = Area of $\triangle\text{ABE }+$ Area of region BCDE

$=\int\limits_{-1}^{1}\frac{3}{2}(\text{x}+1)\text{dx}+\int\limits_{1}^{3}\bigg(\frac{-1}{2}\text{x}+\frac{7}{2}\bigg)\text{dx}$

$=\frac{3}{2}\bigg[\frac{\text{x}^2}{2}+\text{x}\bigg]^1_{-1}+\bigg[\frac{\text{-x}^2}{4}+\frac{7}{2}\text{x}\bigg]^3_{1}$

$\frac{3}{2}\bigg[\frac{1}{2}+1-\frac{1}{2}+1\bigg]+\bigg[\frac{-9}{4}+\frac{21}{2}+\frac{1}{4}-\frac{7}{2}\bigg]$

$=3+5=8\text{ sq.units}$

4. (a) 4 sq. units

Solution:

Equation of line AC is $\text{y}-0=\frac{2-0}{3+1}(\text{x}+1)$

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

$\therefore\text{Area of }\triangle\text{ADC}=\int\limits_{-1}^{3}\frac{1}{2}(\text{x}+1)\text{dx}=\bigg[\frac{\text{x}^2}{4}+\frac{1}{2}\text{x}\bigg]^3_{-1}$

$=\frac{9}{4}+\frac{3}{2}-\frac{1}{4}+\frac{1}{2}=4\text{ sq.units}$

5. (b) 4 sq. units

Solution:

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