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Questions:

Case Study Questions

5 x 4 = 20

Consider the following equations of curves y = cos x, y = x + 1 and y = 0. On the basis of above information, answer the following questions.
(i) The curves y = cos x and y = x + 1 meet at

(a) (1, 0)

(b) (0, 1)

(c) (1, 1)

(d) (0,0)

(ii) y = cos x meet the x-axis at

(a) \(\left(\frac{-\pi}{2}, 0\right)\)

(b) \(\left(\frac{\pi}{2}, 0\right)\)

(c) both (a) and (b)

(d) None of these

(iii) Value of the integral \(\int_{-1}^{0}(x+1) d x\) is

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

(b) \(\frac{2}{3}\)

(c) \(\frac{3}{4}\)

(d) \(\frac{1}{3}\)

(iv) Value of the integral \(\int_{0}^{\pi / 2} \cos x d x\) is

(a) 0

(b) -1

(c) 2

(d) 1

(v) Area bounded by the given curves is

(a) \(\frac{1}{2} \mathrm{sq} . \text { unit }\)

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

(c) \(\frac{3}{4} \text { sq. unit }\)

(d) \(\frac{1}{4} \text { sq. unit }\)

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+1 = 1, while other circle represents the equation x² +1 = 1.

Based on the above information, answer the following questions.
(i) Both the circular pieces of cardboard meet each other at

(a) x = 1

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

(c) \(x=\frac{1}{3}\)

(d) \(x=\frac{1}{4}\)

(ii) Graph of given two curves can be drawn as

(iii) Value of \(\int_{0}^{1 / 2} \sqrt{1-(x-1)^{2}} d x\) is

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

(b) \(\frac{\pi}{6}+\frac{\sqrt{3}}{8}\)

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

(d) \(\frac{\pi}{2}-\frac{\sqrt{3}}{4}\)

(iv) Value of \(\int_{1 / 2}^{1} \sqrt{1-x^{2}} d x\) is

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

(b) \(\frac{\pi}{6}+\frac{\sqrt{3}}{8}\)

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

(d) \(\frac{\pi}{2}-\frac{\sqrt{3}}{4}\)

(v) Area of hidden portion of lower circle is

(a) \(\left(\frac{2 \pi}{3}+\frac{\sqrt{3}}{2}\right) \text { sq. units }\)

(b) \(\left(\frac{\pi}{3}-\frac{\sqrt{3}}{8}\right) \text { sq. units }\)

(c) \(\left(\frac{\pi}{3}+\frac{\sqrt{3}}{8}\right) \text { sq. units }\)

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

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

(i) Equation of line AB is

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

(b) \(y=\frac{3}{2}(x-1)\)

(c) \(y=\frac{1}{2}(x+1)\)

(d) \(y=\frac{1}{2}(x-1)\)

(ii) Equation of line BC is

(a) \(y=\frac{1}{2} x-\frac{7}{2}\)

(b) \(y=\frac{3}{2} x-\frac{7}{2}\)

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

(d) \(y=\frac{3}{2} x+\frac{7}{2}\)

(iii) Area of region ABCD is

(a) 2 sq. units

(b) 4 sq. units

(c) 6 sq. units

(d) 8 sq. units

(iv) Area of \(\Delta A D C\) is

(a) 4 sq. units

(b) 8 sq. units

(c) 16 sq. units

(d) 32 sq. units

(iv) Area of \(\Delta A B C\) is

(a) 3 sq. units

(b) 4 sq. units

(c) 5 sq. units

(d) 6 sq. units

Consider the following equation of curve I' = 4x and straight line x + y = 3.
Based on the above information, answer the following questions.
(i) The line x + y = 3 cuts the x-axis and y-axis respectively at

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

(b) (3, 3), (0, 0)

(c) (0, 3), (3, 0)

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

(ii) Point(s) of intersection of two given curves is (are)

(a) (1, -2), (-9, 6)

(b) (2, 1), (-6, 9)

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

(d) None of these

(iii) Which of the following shaded portion represent the area bounded by given curves?

(iv) Value of the integral \(\int_{-6}^{2}(3-y) d y\) is

(a) 10

(b) 20

(c) 30

(d) 40

(v) Value of area bounded by given curves is

(a) 56 sq. units

(b) \(\frac{63}{5} \text { sq; units }\)

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

(d) 31 sq. units

Consider the curve x² +y²= 16 and line y = x in the first quadrant. Based on the above information, answer the following questions.
(i) Point of intersection of both the given curves is

(a) (0, 4)

(b) \((0,2 \sqrt{2})\)

(c) \((2 \sqrt{2}, 2 \sqrt{2})\)

(d) \((2 \sqrt{2}, 4)\)

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

(iii) The value of the integral \(\int_{0}^{2 \sqrt{2}} x d x\) is

(a) 0

(b) 1

(c) 2

(d).4

(iv) The value of the integral \(\int_{2 \sqrt{2}}^{4} \sqrt{16-x^{2}} d x\) is

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

(b) \(2(\pi-8)\)

(d) \(4(\pi+2)\)

(v) Area bounded by the two given curves is

(a) \(3 \pi \text { sq. units }\)

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

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

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Answers Key:

Case Study Questions

5 x 4 = 20

(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) (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) (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) (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)
\(=\int_{0}^{2 \sqrt{2}} x d x+\int_{2 \sqrt{2}}^{4} \sqrt{16-x^{2}} d x\)
\(=4+2(\pi-2)=2 \pi \text { sq. units. }\)