(i) b : Curves y = cos x and y = x + 1 meet at point C(O, 1).

(ii) C : curve y=cosx meet the x axis at \(A^{\prime}\left(\frac{-\pi}{2}, 0\right)\) and \(A\left(\frac{\pi}{2}, 0\right)\) .
(iii) (a) : \(\int_{-1}^{0}(x+1) d x=\left[\frac{x^{2}}{2}+x\right]_{-1}^{0}=0-\left(\frac{1}{2}-1\right)=\frac{1}{2}\) 
(iv) (d) : \(\int_{0}^{\pi / 2} \cos x d x=[\sin x]_{0}^{\pi / 2}=\sin \frac{\pi}{2}-\sin 0=1\) 
(v) (b) : Required area \(\int_{-1}^{0}(x+1) d x+\int_{0}^{\pi / 2} \cos x d x\) 
\(=\frac{1}{2}+1=\frac{3}{2} \text { sq. units }\)

(i) (a) : Points (0, 2) and (3, 0) pass through both line and ellipse.
(ii) (b) :

(iii) (c) : \(\frac{2}{3} \int_{0}^{3} \sqrt{9-x^{2}} d x=\frac{2}{3} \int_{0}^{3} \sqrt{(3)^{2}-x^{2}} d x\) 
\(=\frac{2}{3}\left[\frac{1}{2} x \sqrt{9-x^{2}}+\frac{9}{2} \sin ^{-1}\left(\frac{x}{3}\right)\right]_{0}^{3}\) 
\(=\frac{2}{3}\left[\frac{3}{2} \sqrt{0}+\frac{9}{2} \sin ^{-1}(1)-\frac{1}{2}(0)-\frac{9}{2} \sin ^{-1}(0)\right]\) 
\(=\frac{2}{3}\left[\frac{9}{2} \cdot \frac{\pi}{2}\right]=\frac{3 \pi}{2}\) 
(iv) d : \(2 \int_{0}^{3}\left(1-\frac{x}{3}\right) d x=2\left[x-\frac{x^{2}}{6}\right]_{0}^{3}\) 
\(=2\left(3-\frac{9}{6}-0-0\right)=2 \times \frac{3}{2}=3\) 
(v) (d) : Area of smaller region bounded by the mirrorand scratch
\(=\frac{2}{3} \cdot \int_{0}^{3} \sqrt{9-x^{2}} d x-2 \int_{0}^{3}\left(1-\frac{x}{3}\right) d x\) 
\(=\frac{3 \pi}{2}-3=3\left(\frac{\pi}{2}-1\right) \text { sq. units }\)

(i) (b) : We have, (x - 1)²+y² = 1
\(\Rightarrow y=\sqrt{1-(x-1)^{2}}\)   ...(i) 
Also, \(x^{2}+y^{2}=1 \Rightarrow y=\sqrt{1-x^{2}}\) ...(ii)
From (i) and (ii), we get
\(\sqrt{1-(x-1)^{2}}=\sqrt{1-x^{2}}\) 
\(\Rightarrow(x-1)^{2}=x^{2} \Rightarrow 2 x=1 \Rightarrow x=\frac{1}{2}\) 
(ii) (c):
 
(iii) (a) : \(\int_{0}^{1 / 2} \sqrt{1-(x-1)^{2}} d x\) 
\(=\left[\frac{x-1}{2} \sqrt{1-(x-1)^{2}}+\frac{1}{2} \sin ^{-1}\left(\frac{x-1}{1}\right)\right]_{0}^{1 / 2}\) 
\(\begin{array}{r} =\frac{1}{2}\left(\frac{1}{2}-1\right) \sqrt{1-\frac{1}{4}}+\frac{1}{2} \sin ^{-1}\left(-\frac{1}{2}\right)-\left(-\frac{1}{2}\right)(0) -\frac{1}{2} \sin ^{-1}(-1) \end{array}\) 
\(=\left[\frac{-1}{4} \cdot \frac{\sqrt{3}}{2}-\frac{1}{2} \cdot \frac{\pi}{6}+0+\frac{1}{2} \cdot \frac{\pi}{2}\right]=\frac{-\sqrt{3}}{8}-\frac{\pi}{12}+\frac{\pi}{4}\) 
\(=\frac{\pi}{6}-\frac{\sqrt{3}}{8}\) 
(iv) (c : \(\int_{1 / 2}^{1} \sqrt{1-x^{2}} d x=\left[\frac{x}{2} \sqrt{1-x^{2}}+\frac{1}{2} \sin ^{-1} x\right]_{1 / 2}^{1}\) 
\(=0+\frac{1}{2} \sin ^{-1}(1)-\frac{1}{4} \sqrt{1-\frac{1}{4}}-\frac{1}{2} \sin ^{-1}\left(\frac{1}{2}\right)\) 
\(=\frac{\pi}{4}-\frac{\sqrt{3}}{8}-\frac{\pi}{12}=\frac{\pi}{6}-\frac{\sqrt{3}}{8}\) 
(v) (d) : Required area
\(=2\left[\int_{0}^{1 / 2} \sqrt{1-(x-1)^{2}} d x+\int_{1 / 2}^{1} \sqrt{1-x^{2}} d x\right]\) 
\(=2\left[\frac{\pi}{6}-\frac{\sqrt{3}}{8}+\frac{\pi}{6}-\frac{\sqrt{3}}{8}\right]\) 
\(=2\left[\frac{\pi}{3}-\frac{\sqrt{3}}{4}\right]=\left(\frac{2 \pi}{3}-\frac{\sqrt{3}}{2}\right) \text { sq. units }\) 

(i) (a) : Equation of line AB is
\(y-0=\frac{3-0}{1+1}(x+1) \Rightarrow y=\frac{3}{2}(x+1)\) 
(ii) (c) : Equation of line BC is \(y-3=\frac{2-3}{3-1}(x-1)\) 
\(\Rightarrow y=-\frac{1}{2} x+\frac{1}{2}+3 \Rightarrow y=\frac{-1}{2} x+\frac{7}{2}\) 
(iii) (d) : Area of region ABCD
= Area of \(\triangle A B E\) + Area of region BCDE
\(=\int_{-1}^{1} \frac{3}{2}(x+1) d x+\int_{1}^{3}\left(\frac{-1}{2} x+\frac{7}{2}\right) d x\) 
\(=\frac{3}{2}\left[\frac{x^{2}}{2}+x\right]_{-1}^{1}+\left[\frac{-x^{2}}{4}+\frac{7}{2} x\right]_{1}^{3}\) 
\(=\frac{3}{2}\left[\frac{1}{2}+1-\frac{1}{2}+1\right]+\left[\frac{-9}{4}+\frac{21}{2}+\frac{1}{4}-\frac{7}{2}\right]\) 
= 3 + 5 = 8 sq. units
(iv) (a) : Equation of line AC is \(y-0=\frac{2-0}{3+1}(x+1)\) 
\(\Rightarrow y=\frac{1}{2}(x+1)\) 
\(\therefore \text { Area of } \Delta A D C=\int_{-1}^{3} \frac{1}{2}(x+1) d x=\left[\frac{x^{2}}{4}+\frac{1}{2} x\right]_{-1}^{3}\) 
\(=\frac{9}{4}+\frac{3}{2}-\frac{1}{4}+\frac{1}{2}=4 \text { sq. units }\) 
(v) (b) : Area of \(\Delta A B C\)= Area of region ABCD - Area of \(\Delta A C D=8-4=4 \mathrm{sq} . \text { units }\)

(i) (c) : For point of intersection, we have sin x = cos x
\(\Rightarrow \frac{\sin x}{\cos x}=1 \Rightarrow \tan x=1 \Rightarrow x=\frac{\pi}{4}\) 
(ii) (a) : \(\int_{0}^{\pi / 4} \sin x d x=[-\cos x]_{0}^{\pi / 4}=-\cos \frac{\pi}{4}+\cos 0\) \(=1-\frac{1}{\sqrt{2}}\) 
(iii) (b) : \(\int_{\pi / 4}^{\pi / 2} \cos x d x=[\sin x]_{\pi / 4}^{\pi / 2}=\sin \frac{\pi}{2}-\sin \frac{\pi}{4}\) \(=1-\frac{1}{\sqrt{2}}\) 
(iv) (c) : \(\int_{0}^{\pi} \sin x d x=[-\cos x]_{0}^{\pi}=[-\cos \pi+\cos 0]=2\) 
(v) (b) : \(\int_{0}^{\pi / 2} \sin x d x=[-\cos x]_{0}^{\pi / 2}=\left[-\cos \frac{\pi}{2}+\cos 0\right]\) = 0+1=1

(i) (a) : Here, teacher explained about cosine curve.
(ii) (c) : Required area \(\int_{0}^{\pi / 2} \cos x d x\) 
\(=[\sin x]_{0}^{\pi / 2}=\sin \frac{\pi}{2}-\sin 0=1-0=1 \text { sq. unit }\) 
(iii) (b) : Required area = \(\left|\int_{\pi / 2}^{3 \pi / 2} \cos x d x\right|=\left|[\sin x]_{\pi / 2}^{3 \pi / 2}\right|\) 
\(=\left|\sin \frac{3 \pi}{2}-\sin \frac{\pi}{2}\right|=|-1-1|=|-2|\) 
= 2 sq. units [Since, area can't be negative]
(iv) (a) : Required area = \(\int_{3 \pi / 2}^{2 \pi} \cos x d x=[\sin x]_{3 \pi / 2}^{2 \pi}\) 
\(=\sin 2 \pi-\sin \frac{3 \pi}{2}=0-(-1)=1 \text { sq. unit }\) 
(v) (d) : Required area
\(=\int_{0}^{\pi / 2} \cos x d x+\left|\int_{\pi / 2}^{3 \pi / 2} \cos x d x\right|+\int_{3 \pi / 2}^{2 \pi} \cos x d x\) 
= 1 + 2 + 1 = 4 sq. units.

(i) (d) : 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]
(ii) (c) : We have, y² = 4x and x + y = 3
From (i) and (ii), we have y² =  4(3 - y)
\(\Rightarrow y^{2}+4 y-12=0 \Rightarrow y^{2}+6 y-2 y-12=0\) 
\(\Rightarrow y(y+6)-2(y+6)=0\) 
\(\Rightarrow (y+6)(y-2)=0 \Rightarrow y=2, y=-6\) 
From (ii), x = 3 - 2 = 1 or x = 3 + 6 = 9
\(\therefore\) Required points of intersection are (1, 2), (9, - 6)
(iii) (b) :

(iv) (d): \(\int_{-6}^{2}(3-y) d y=\left[3 y-\frac{y^{2}}{2}\right]_{-6}^{2}\) 
\(=\left[6-\frac{4}{2}-\left[3(-6)-\frac{(-6)^{2}}{2}\right]\right]=4+36=40\) 
(v) (c): Required area = \(\int_{-6}^{2}(3-y) d y-\int_{-6}^{2} \frac{y^{2}}{4} d y\) 
\(=40-\frac{1}{4}\left[\frac{y^{3}}{3}\right]_{-6}^{2}=40-\frac{1}{4}\left[\frac{8}{3}-\frac{(-6)^{3}}{3}\right]\) 
\(=40-\frac{2}{3}-\frac{216}{12}=\frac{480-8-216}{12}=\frac{256}{12}=\frac{64}{3} \mathrm{sq} . \text { units }\)

(i) (b) : We have, y² + 1 = 4  ...(i)
and \(x=\sqrt{3} y\)   ...(ii)
From (i) and (ii), we get
\(3 y^{2}+y^{2}=4 \Rightarrow 4 y^{2}=4 \Rightarrow y^{2}=1 \Rightarrow y=\pm 1\) 
From (ii), \(x=\sqrt{3},-\sqrt[4]{3}\) 
\(\therefore\) Points of intersection of pizza and edge of knife are \((\sqrt{3}, 1),(-\sqrt{3},-1)\) .
(ii) (a) :

(iii) (b) : Required area = \(\int_{0}^{\sqrt{3}} \frac{x}{\sqrt{3}} d x+\int_{\sqrt{3}}^{2} \sqrt{4-x^{2}} d x\) 
\(=\frac{1}{\sqrt{3}}\left[\frac{x^{2}}{2}\right]_{0}^{\sqrt{3}}+\left[\frac{x}{2} \sqrt{4-x^{2}}+\frac{4}{2} \sin ^{-1}\left(\frac{x}{2}\right)\right]_{\sqrt{3}}^{2}\) 
 \(=\frac{1}{\sqrt{3}}\left[\frac{3}{2}-0\right]+\left[2 \sin ^{-1}(1)-\left(\frac{\sqrt{3}}{2}+2 \sin ^{-1} \frac{\sqrt{3}}{2}\right)\right]\) 
\(=\frac{\sqrt{3}}{2}+\frac{2 \pi}{2}-\frac{\sqrt{3}}{2}-\frac{2 \pi}{3}=\frac{\pi}{3} \text { sq. units }\) 
(iv) (a) : We have,x²+y²=4
\(\Rightarrow(x-0)^{2}+(y-0)^{2}=(2)^{2}\) 
\(\therefore\) Radius = 2
Area of \(\frac{1}{4} \text { th slice of pizza }=\frac{1}{4} \pi(2)^{2}=\pi \mathrm{sq} . \text { units }\) 
(v) (d) : Area of whole pizza = \(\pi(2)^{2}=4 \pi \mathrm{sq} . \text { units }\) 

(i) (c) : We have, x² +y²= 16  ..(i)
and y = x ...(ii)
From (i) and (ii), \(2 x^{2}=16 \Rightarrow x^{2}=8 \Rightarrow x=2 \sqrt{2}\)  (\(\therefore\) x lies in first quadrant)
\(\therefore\) Point of intersection of (i) and (ii) in first quadrant is \((2 \sqrt{2}, 2 \sqrt{2})\) .
(ii) (b) : The shaded region which represent the areabounded by two given curves in first quadrant is shown below.
 
(iii)( d) : \(\int_{0}^{2 \sqrt{2}} x d x=\left[\frac{x^{2}}{2}\right]_{0}^{2 \sqrt{2}}=\frac{(2 \sqrt{2})^{2}}{2}=\frac{8}{2}=4\) 
(iv) (a) : \(\int_{2 \sqrt{2}}^{4} \sqrt{16-x^{2}} d x=\left[\frac{x}{2} \sqrt{16-x^{2}}+\frac{16}{2} \cdot \sin ^{-1}\left(\frac{x}{4}\right)\right]_{2 \sqrt{2}}^{4}\) 
 \(=8 \sin ^{-1}(1)-4-8 \sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)\) 
\(=8\left(\frac{\pi}{2}\right)-4-8\left(\frac{\pi}{4}\right)=4 \pi-4-2 \pi=2 \pi-4=2(\pi-2)\) 
(v) (d) : Required area = Area (OLA) + Area (BAL)
\(=\int_{0}^{2 \sqrt{2}} x d x+\int_{2 \sqrt{2}}^{4} \sqrt{16-x^{2}} d x\) 
\(=4+2(\pi-2)=2 \pi \text { sq. units. }\)

(i) (b) : We have, x² = y ...(i) and x = y. , ...(ii)
From (i) and (ii), x²= x \(\Rightarrow\)x² - x = 0
\(\Rightarrow\) x(x - 1) = 0 ~ x = 0, 1
From (ii), y = 0, 1
Required points of intersection are (0, 0), (1, 0)
(ii) (a) :
 
(iii) (c) : \(\int_{0}^{1} x d x=\left[\frac{x^{2}}{2}\right]_{0}^{1}=\frac{1}{2}-0=\frac{1}{2}\) 
(iv) (b) : \(\int_{0}^{1} x^{2} d x=\left[\frac{x^{3}}{3}\right]_{0}^{1}=\frac{1}{3}-0=\frac{1}{3}\) 
(v) (a) : Required area \(=\int_{0}^{1} x d x-\int_{0}^{1} x^{2} d x\) 
\(=\frac{1}{2}-\frac{1}{3}=\frac{1}{6} \text { sq. units. }\)