Linear Inequalities: Each of the following carries five marks each: 1. Solve the system of equations graphically.

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Todd Mason
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1 Liner Inequlities: Ech of the following crries five mrks ech:. Solve the system of equtions grphiclly. x + 2y 8, 2x + y 8, x 0, y 0 Solution: Considerx + 2y 8.. () Drw the grph for x + 2y = 8 by line.it psses through (8,0) nd (0,4).Join these points, put x = 0 And y = 0 in(), we get which is true. Solution set of () is contining the origin. Consider2x + y 8.. (2) Drw the grph for 2x + y = 8 by line. it psses through (4,0) nd (0,8).Join these points, Put x = 0 nd y = 0 in we get which is true. Solution set of (2) is contining the Solution of x 0, y 0 is every point in the first qudrnt. Binomil Theorem: Ech of the following crries five mrks ech:. Stte nd prove the Binomil Theorem. Solution: 2. Solve the system of equtions grphiclly. 5x + 4y 40, x 2, y 3. Solution: Consider5x + 4y 40.. () Drw the grph for 5x + 4y = 40by line It psses through (8,0) nd (0,0) Join these points, put x = 0 nd y = 0 in () We get which is true. Solution set of () is contining the origin. Considery 3.. (2) Drw the grph for y = 3by line Solution set of (2) is not contining the origin. Considerx 2.. (3) Drw the grph for x = 2 by line Solution set of (3) is not contining the origin. Reltions nd Functions Ech of the following crries one mrk ech:. Let A = {, 2} nd B = {3, 4}. Find the number of reltions from A to B. Solution: We hve, A X B = {(, 3), (, 4), (2, 3), (2, 4)}. Since n (AXB ) = 4, the number of subsets of A X B is 2 4. the number of reltions from A into B will be 2 4.

2 2. Let A = {,b,c} nd B = {,5}. Find the number of reltions from A to B. Solution: We hve, A X B = {(,), (,5), (b,), (b,5), (c,), (c,5)}. Since n (AXB ) = 4, the number of subsets of A X B is 2 6. the number of reltions from A into B will be 2 6. Ech of the following crries five mrks ech:. Define Identity function, drw the grph nd write the domin nd rnge. Solution: Let R be the set of rel numbers. Define the rel vlued function f : R R by y = f(x) = x for ech x R. Such function is clled the identity function. Here the domin nd rnge of f re R. The grph is stright line. It psses through the origin. Solution: Let R be the set of rel numbers. Modulus function is rel vlued function f : R R given by, f(x) = x for ech x R. For ech non-negtive vlue of x, f(x) is equl to x. But for negtive vlues of x, the vlue of f(x) is the negtive of the vlue of x, i.e., Here, f x = x, if x 0 x, if x < 0 Domin of f is : R Rnge of f is : R + including {0}. 2. Define Constnt function, drw the grph of y=3 nd write the domin nd rnge. Solution: Let R be the set of rel numbers. Define the rel vlued function f : R R by y = f(x) = c ; where c is constnt nd x R. Here domin of f is R nd its rnge is {c}. Grph: y = 3, The grph is line prllel to x- xis. Here domin of f is R nd its rnge is {3}. 4. Define Signum function, drw the grph nd write the domin nd rnge. Solution: Let R be the set of rel numbers. Signum function is rel vlued function f : R R defined by,, if x > 0 f x = 0, if x = 0, if x < 0 The domin of the signum function is R nd the rnge is the set {, 0, }. The grph of the signum function is given by, 3. Define Modulus function, drw the grph nd write the domin nd rnge.

3 5. Define Gretest integer function, drw the grph nd write the domin nd rnge. Solution: Let R be the set of rel numbers. Gretest integer function is rel vlued function f : R R defined by, f(x) = [x], x R. which ssumes the vlue of the gretest integer, less thn or equl to x. Such function is clled the gretest integer function. The grph of the function is, Here, Domin of f is : R Rnge of f is : Z Trigonometric Functions. Ech of the following crries one mrk ech:. Convert 40 20' into rdin mesure. Solution: We know tht 80 = πrdin. Hence, 40 20'= 40 degree = π 2 π 2 degree = rdin = 2π rdin ' = 2π rdin Convert 6 rdins into degree mesure. Solution: We know tht 80 = π rdin. Hence, 6 rdin= 80 π degree = degree = = ' + 2 degree = minute (becuse = 60') minute (becuse '= 60'') Solution: Angle subtended t the centre by n rc of length unit in unit circle (circle of rdius unit) is sid to hve mesure of rdin. Ech of the following crries three mrks ech:. If cos x = 3/5, x lies in the third qudrnt, find the vlues of other five trigonometric functions. Solution: Since cos x = -3/5, we hve sec x = -5/3 Now sin 2 x + cos 2 x =, i.e. sin 2 x = cos 2 x Or sin 2 x = 9 = Hence sin x = ± 4 5 Since x lies in third qudrnt, sin x is negtive. Therefore sin x = - 4/5 which lso gives cosec x = - 5/4 Further, we hve, sin x tn x = = 4. cos x 3 nd cot x = 3/4. 2. If cot x = 5/2 x lies in the second qudrnt, find the vlues of other five trigonometric functions. Solution: Since cot x = -5/2, we hve tn x = -2/5 Now, sec 2 x = + tn 2 x = + 44 = sec x = ± 3 5 Since x lies in second qudrnt, sec x is negtive. Therefore sec x = - 3/5 Which lso gives cos x = - 5/3 Further, we hve sin x = tn x cos x = 2/3 nd cosec x = /sin x = 3/2. Ech of the following crries six mrks ech:. Prove geometriclly tht cos (x + y) = cos x cos y sin x sin y, hence cos2x = cos 2 x sin 2 x Solution: Consider the unit circle with centre t the origin. Let x be the ngle P4OP nd y be the ngle POP2. Then (x + y) is the ngle P4OP2. Also let ( y) be the ngle P4OP3. P, P2, P3 nd P4 will hve the coordintes P(cos x, sin x), P2 [cos (x + y), sin (x + y)], P3 [cos ( y), sin ( y)] nd P4 (, 0) = ' + 0.9'' = ' '' pproximtely. Hence 6 rdin=343 38' '' pproximtely. 3. Define Rdin mesure.

4 2. Find the coordintes of the focus, xis, the eqution of the directrix nd ltus rectum of the prbol y 2 = 8x. Solution: The given eqution involves y 2, so the xis of symmetry is long the x-xis. The coefficient of x is positive so the prbol opens to the right. Compring with the given eqution y 2 = 4x, we find tht = 2. Thus, the focus of the prbol is (2, 0) nd the eqution of the directrix of the prbol is x = 2. Length of the ltus rectum is 4 = 4 2 = 8. Consider the tringles POP3 nd P2OP4. They re congruent (SAS). PP3 nd P2P4 re equl. By using distnce formul, we get (PP3) 2 = [cos x cos ( y)] 2 + [sin x sin( y)] 2 = (cos x cos y)2 + (sin x + sin y) 2 = [cos x cos y] 2 + [sin x + siny] 2 = cos 2 x + sin 2 x + cos 2 y + sin 2 y 2cosxcosy + 2sinxsiny = 2 2 (cos x cos y sin x sin y) Also (P2P4) 2 = [ cos (x + y)] 2 + [0 sin(x + y)] 2 = 2cos x + y + cos 2 x + y + sin 2 (x + y) = 2 2 cos (x + y) Since PP3 = P2P4, we hve (PP3) 2 = (P2P4) (cos x cos y sin x sin y) = 2 2 cos (x + y). Hence cos (x + y) = cos x cos y sin x sin y When x = y in the bove eqution, we get, cos2x = cos 2 x sin 2 x Conic Sections. Ech of the following crries three mrks ech:. Find the coordintes of the foci nd the vertices, the eccentricity, the length of the ltus rectum of the hyperbol 9 y2 6 = Solution: Compring the eqution with the stndrd eqution, 2 y2 Here, = 3, b = 4 nd c = 2 + b 2 = = 5 the coordintes of the foci re (±5, 0) nd tht of vertices re (±3, 0). The eccentricity : e = c = Find the coordintes of the foci, the vertices, the length of mjor xis, the minor xis, the eccentricity nd the ltus rectum of the ellipse 25 + y2 9 = Solution: Since denomintor of is lrger thn the denomintor of y 2, the mjor xis is long the x-xis. 9 Compring the eqution with the stndrd eqution, 2 y2 Here, = 5, b = 3 nd c = 2 b 2 = 25 9 = 4 the coordintes of the foci re ( 4,0) nd (4,0), vertices re ( 5, 0) nd (5, 0). Length of the mjor xis is 0 units length of the minor xis 2b is 6 units. The eccentricity : e = 4. 5 The ltus rectum : 2b 2 = 8 5 Ech of the following crries six mrks ech:. Define ellipse nd derive its eqution in stndrd form + y 2 =. 2 b 2 Solution: An ellipse is the set of ll points in plne, the sum of whose distnces from two fixed points in the plne is constnt. We will derive the eqution for the ellipse with foci on the x-xis. 25 The ltus rectum : 2b2 = 32 3 Let F nd F2 be the foci nd O be the midpoint of the line segment FF2. Let O be the origin nd the line from O through F2 be the positive x-xis nd tht through Fs the negtive x-xis. Let, the line through

5 O perpendiculr to the x-xis be the y-xis. Let the coordintes of F be ( c, 0) nd F2 be (c,0) s shown in the figure. Let P(x, y) be ny point on the ellipse such tht the sum of the distnces from P to the two foci be 2 so given PF + PF2 = () Using the distnce formul, we hve (x + c) 2 + y 2 + (x c) 2 + y 2 = 2 i.e. (x + c) 2 + y 2 = 2 (x c) 2 + y 2 Squring both sides, we get (x + c) 2 + y 22 = {2 x c 2 + y 2 } 2 (x + c) 2 + y 2 = x c 2 + y 2 + x c 2 + y 2 which on simplifiction gives x c 2 + y 2 = c x Squring gin nd simplifying, we get 2xc + c 2 + y 2 = 2 2xc + c2 2 x2 c2 2 x2 + y 2 = 2 c 2 { c2 2} + y2 = 2 c 2 { 2 c 2 + y 2 = 2 c 2 y c 2 = Since c 2 = 2 b 2, we hve b 2 = 2 c 2 i.e. 2 + y2 Hence ny point on the ellipse stisfies 2 + y2 Conversely, let P (x, y) stisfy the bove eqution with 0 < c <. Then y 2 = b 2 { x2 2} PF= x + c 2 +y 2 = x + c 2 +b 2 { 2 = x + c 2 + ( 2 c 2 ){ 2 = 2 + c2 + 2xc 2 = 2 + c2 2 = { + xc }2 xc + 2 PF = + c x Similrly PF2 = c x Hence PF + PF2 = 2. So, ny point tht stisfies + y 2 2 b2 =, stisfies the geometric condition nd so P(x, y) lies on the ellipse. Hence, we proved tht the eqution of n ellipse with centre of the origin nd mjor xis long the x-xis is 2 + y2 2. Define hyperbol nd derive its eqution in stndrd form y 2 =. 2 b 2 Solution: A hyperbol is the set of ll points in plne, the difference of whose distnces from two fixed points in the plne is constnt. Let F nd F2 be the foci nd O be the mid-point of the line segment FF2. Let O be the origin nd the line through O through F2 be the positive x-xis nd tht through F s the negtive x-xis. The line through O perpendiculr to the x-xis be the y-xis. Let the coordintes of F be ( c,0) nd F2 be (c,0). Let P(x, y) be ny point on the hyperbol such tht the difference of the distnces from P to the frther point minus the closer point be 2. So given, PF PF2 = 2 Using the distnce formul, we hve (x + c) 2 + y 2 (x c) 2 + y 2 = 2 i.e. (x + c) 2 + y 2 = 2 + (x c) 2 + y 2 Squring both sides, we get (x + c) 2 + y 22 = {2 + x c 2 + y 2 } 2 (x + c) 2 + y 2 = x c 2 + y 2 + x c 2 + y 2 which on simplifiction gives x c 2 + y 2 = c x Squring gin nd simplifying, we get 2xc + c 2 + y 2 = 2 2xc + c2 2 x2 c2 2 x2 + y 2 = 2 c 2

{ c2 2} + y2 = 2 c 2 { 2 c 2 + y 2 = 2 c 2 2 + y2 2 c 2 = ; x2 2 y2 c 2 2 = Since, b 2 = c 2 2 i.e. 2 y2 Hence ny point on the ellipse stisfies 2 y2 Conversely, let P (x, y) stisfy the bove eqution with 0 < < c.

the right of the line x =, x >, c x >, c x becomes negtive. Thus, PF2 = c x. Therefore PF PF2 = + c x c x + = 2 Also, if P is to the left of the line x =, then PF = { + c x} nd PF2 = c x.

6 { c2 2} + y2 = 2 c 2 { 2 c 2 + y 2 = 2 c y2 2 c 2 = ; x2 2 y2 c 2 2 = Since, b 2 = c 2 2 i.e. 2 y2 Hence ny point on the ellipse stisfies 2 y2 Conversely, let P (x, y) stisfy the bove eqution with 0 < < c. Then y 2 = b 2 { x2 2 PF= x + c 2 +y 2 = x + c 2 +b 2 { x2 2 PF = + c x = x + c 2 + (c 2 2 ){ x2 2 = 2 + c2 + 2xc 2 = 2 + c2 xc = { + xc }2 Similrly,PF2 = x c In hyperbol c > ; nd since P is to the right of the line x =, x >, c x >, c x becomes negtive. Thus, PF2 = c x. Therefore PF PF2 = + c x c x + = 2 Also, if P is to the left of the line x =, then PF = { + c x} nd PF2 = c x. In tht cse P F2 PF = 2. So, ny point tht stisfying y 2 2 b2 = lies on the hyperbol. Thus, we proved tht the eqution of hyperbol with origin (0,0) nd trnsverse xis long x-xis is 2 y2 Sttistics. Ech of the following crries five mrks ech:. Find the men devition bout the men for the following dt. Solution We mke the following tble from the given dt : Here, i= f i = 40, i= f i x i = 800 x = N nd i= i= M. D. x = N f i x i x = 400 f i x i = = 45 i= f i x i x = Clculte the men devition bout medin for the following dt: Solution Form the following Tble 5.6 from the given dt The clss intervl contining Nth or 2 25th item is is the medin clss. We know tht, N Medin = l + 2 C h f

7 Here l = 20, C = 3, f = 5, h = 0 nd N = Medin = = = 28 5 Thus, Men devition bout medin is given by, 6 M. D. M = f N i x i M = i= = 0.6 Stright lines Ech of the following crries five mrks ech.. Derive formul to find the Distnce of Point From Line. Solution The distnce of point from line is the length of the perpendiculr drwn from the point to the line. Let L : Ax + By + C = 0 be line, whose distnce from the point P (x, y) is d. Substituting the vlues of re ( PQR) nd QR in (), we get PM = Ax + By + C A 2 + B 2 Or, d = Ax + By + C A 2 + B 2 Thus, the perpendiculr distnce (d) of line Ax + By+ C = 0 from point (x, y) is given by the bove eqution. 2. Derive the Norml Form of line. Solution Suppose non-verticl line is known to us with following dt: (i) Length of the perpendiculr (norml) from origin to the line. (ii) Angle which norml mkes with the positive direction of x-xis. Let L be the line, whose perpendiculr distnce from origin O be OA = p nd the ngle between the positive x-xis nd OA be XOA = ω. The possible positions of line L in the Crtesin plne re shown in the Fig. Now, our purpose is to find slope of L nd point on it. Drw perpendiculr AM on the x-xis in ech cse. Drw perpendiculr PM from the point P to the line L. If the line meets the x-nd y-xes t the points Q nd R, respectively. Then, coordintes of the points re Q C, 0 nd Q 0, C. Thus, A B the re of the tringle PQR is given by re PQR = PM. QR, i. e. PM 2 2 re PQR =. () QR Also, re PQR = 2 x 0 + C B + C A C B y + 0(y 0) = 2 x C B + y C A + C2 AB or 2re PQR = C. Ax + By + C nd QR = 0 + C A C AB A 2 + B 2 AB 2 + C B 0 2 = In ech cse, we hve OM = p cos ω nd MA = p sin ω, so tht the coordintes of the point A re (p cos ω, p sin ω). Further, line L is perpendiculr to OA. Therefore The slope of the line L = = = cosω Slope A tnω sinω Thus, the line L hs slope cosω nd point A pcosω, sinω psinω on it. by point-slope form, the eqution of the line L is y psinω = cosω x pcosω xcosω sinω + ysinω = p(sin 2 ω + cos 2 ω) or xcosω + ysinω = p Hence, the eqution of the line hving norml distnce p from the origin nd ngle ω which the norml mkes with the positive direction of x-xis is given by the bove eqution.

8 Limits nd derivtives Ech of the following crries five mrks ech. sin.prove tht lim 0 = nd hence deduce tn tht lim 0 =. ( is in rdins) Proof: Consider circle with centre O nd rdius r units. Let AOB = c. Join AB. From A drw AD OA, meeting OB produced t D. Let OB = OA = r units (rdius). 2 r2 sin Dividing by 2 r2 2 r2 sin tn sin cos sin cos cos sin Applying limit 0, 2 r2 tn (Dividing by sin ) lim 0 cos lim 0 sin lim 0 lim 0 sin By sndwich theorem, lim 0 sin = Now lim 0 tn = lim 0 sin = lim 0 lim 0 cos sin. cos =. = Hence proved. From the figure, we hve Are of ΔOAB Are of sector OAB Are of ΔOAD () Are of ΔOAB = OA.BC= OA.OB sin = r 2 sin [From le BOC, sin = BC OB BC = OB sin = rsin ] Are of sector OAB = 2 r2 Are of ΔOAD= OA.DA = OA.OA tn 2 2 = 2 r2 tn [From le DOA, tn = DA OA DA = OA tn = r tn ] Substituting in ()

I. Equations of a Circle a. At the origin center= r= b. Standard from: center= r=

I. Equations of a Circle a. At the origin center= r= b. Standard from: center= r= 11.: Circle & Ellipse I cn Write the eqution of circle given specific informtion Grph circle in coordinte plne. Grph n ellipse nd determine ll criticl informtion. Write the eqution of n ellipse from rel