MATHEMATICS PAPER IB COORDINATE GEOMETRY(2D &3D) AND CALCULUS. Note: This question paper consists of three sections A,B and C.

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1 MATHEMATICS PAPER IB COORDINATE GEOMETRY(D &D) AND CALCULUS. TIME : hrs Ma. Marks.75 Not: This qustion papr consists of thr sctions A,B and C. SECTION A VERY SHORT ANSWER TYPE QUESTIONS. 0X =0.If th portion of a straight lin intrcptd btwn th as of co-ordinats is bisctd at (p, q), writ th quation of th straight lin.. Transform quation ( + 5k) ( + k) y + ( k) = 0 into form L L 0and find th point of concurrncy of th family of straight lins rprsntd by th quation.. P is a variabl point which movs such that PA = PB. If A = (,, ) and B =(,,)prov that P satisfis th quation + y + z + 8 y + 0z 47 = Show that th plan through (,, ), (,, ) and ( 7,, 5) is paralll to y- ais. 5. Chck th continuity of f givn by point. f 9 if 0 5 and.5 if At th 6.Show that f, givn by f 0 is continuous on R 0. n n 7. If y a b thn prov that y" n y

2 8. find th drivativ of sin tan 9. Find approimat valu of 8 0. It is givn that Roll s thorm holds for th function f() = + b + a on [, ] with c t. Find th valus of a and b. SECTION B SHORT ANSWER TYPE QUESTIONS. ANSWER ANY FIVE OF THE FOLLOWING 5 X 4 = 0. Find th quation of locus of P, if th lin sgmnt joining (, ) and (, 5) subtnds a right angl at P..Whn th as ar rotatd through an angl 4, find th transformd quation of 0y y 9. Find th points on th lin 4y = 0 which ar at a distanc of 5 units form th point (, ). 4. Lt A containr is in th shap of an invrtd con has hight 8 m and radius 6 m at th top. If it is filld with watr at th rat of m /minut, how fast is th hight of watr changing whn th lvl is 4 m? 6. Find th angl btwn th curv y and y-ais.

3 7.. If y = y dy log, thn show that. d ( log ) SECTION C LONG ANSWER TYPE QUESTIONS. ANSWER ANY FIVE OF THE FOLLOWING 5 X 7 =5 8. If (h, k) is th imag of (, y) w.r.t th lin a + by + c = 0 (a 0, b 0), h k y ( by c) a b a b thn prov that 9. If th quation a + hy + by + g + fy + c = 0 rprsnts a pair of intrscting lins, thn show that th squar of th distanc of thir point of intrsction from th origin is distanc is f g h b. c(a b) f g ab if th givn lins ar prpndicular. h. Also show that th squar of this 0. Show that th straight lins y 4y 0 and from a paralllogram and find th lngths of its sids. 4y 4y 5 0y 4 0. If a ray maks angl,, and with th four diagonals of a cub find cos cos cos cos. If y a a log a, show that a dy d. If th tangnt at any point on th curv / / / y a AB is constant. intrscts th coordinat as in A, B show that th lngth 4. A window is in th shap of a rctangl surmountd by a smi-circl. If th primtr of th window b 0 ft, find th maimum ara.

4 SOLUTIONS. If th portion of a straight lin intrcptd btwn th as of co-ordinats is bisctd at (p, q), writ th quation of th straight lin. Sol. Lt a, b b th intrcpts of th lin and AB b th lin sgmnt btwn th as. Thn points A =(a, 0) and B = (0, b) Equation of th lin in th intrcpt form is y a b --- () Mid -point of AB is M= a b, (p,q) a b p, q a 4p, b 4q y y 4 4p 4q p q Substituting in (), givn. Transform quation ( + 5k) ( + k) y + ( k) = 0 into form L L 0and find th point of concurrncy of th family of straight lins rprsntd by th quation. Sol. Givn quation is ( + 5k) ( + k) y + ( k) = 0 ( - y + ) + k (5 6y ) = 0 which is of th form L L 0whr L y 0 and L 5 6y 0 thrfor givn quation rprsnts a family of straight lins. Solving abov two lins, y y 5 5, y Th point of concurrncy is P(5, 4).

5 . P is a variabl point which movs such that PA = PB. If A = (,, ) and B = (,, ) prov that P satisfis th quation + y + z + 8 y + 0z 47 = 0. Sol. Givn points ar : A(,, ) and B = (,, ) Lt P(, y, z) b any point on th locus. Givn condition is : PA = PB 9 PA = 4 PB 9[( + ) + (y ) + (z ) ] = 4[( ) + (y + ) + (z ) ] 9( y 4y z 6z + 9) = 4( y + 6y z 6z + 69) 9 + 9y + 9z + 6 6y 54z + 5 = 4 + 4y + 4z y 04z y +5z y +50z 5 = 0 Dividing with 5 locus of P is : + y + z + 8 y + 0z 47 = Show that th plan through (,, ), (,, ) and ( 7,, 5) is paralll to y- ais. Sol. Equation of th plan through (,, ), (,, ) and ( 7,, 5) is y y z z y y z z y y z z 0 y- z z 0 D.rs of normal to th plan ar,0,-4

6 d.rs of y ais ar 0,,0 aa b b cc Normal to th plan is prpndicular to th y-ais. hncplna is paralll to Y-ais. 5. Chck th continuity of f givn by point. Sol : Givn f() =.5. f 9 if 0 5 and.5 if At th Lt f Lt 9 Lt f f is continuous at. 6. Show that f, givn by f 0 is continuous on R 0., 0 Sol :. f, 0 0, 0 => f, 0 Lft limit at = 0 is Right limit at =0 is Lt f Lt 0 0 Lt f Lt Lt f Lt f 0 0 Lt f dos not ist. 0 Hnc th function is not continuous at =0.

7 Whn <0, f() =, a constant. And it is continuous for all <0. Whn >0, f() = 0, which is continuos for all >0. Hnc th function is continuous on R-{0}. n n 7. If y a b thn prov that y" n n Sol : y a b n y n y na nb n n. y n a n b n n y" n a b. n n y 8. find th drivativ of sin tan Diff. w.r.t., dy d sin tan d d cos tan. tan cos tan.cos tan 9. Find approimat valu of Sol: lt f() =, = 8, = Now f ( ) f f., put = 8, = It is givn that Roll s thorm holds for th function f() = + b + a on [, ] with c t. Find th valus of a and b. Sol. Givn f() = + b + a 8

8 f () b a f '() 0 c bc a 0 c c b 4b a 6 b b a b b a b b a and b 6 and b a 6 a a a Hnc a =, b = 6.. Find th quation of locus of P, if th lin sgmnt joining (, ) and (, 5) subtnds a right angl at P. Sol. Givn points A(, ), B(, 5). Lt P(, y) b any point in th locus. P(,y) 90 A(,) B(,5) Givn condition is : APB = 90 (slop of AP ) (slop of BP ) = y y 5 (y )(y 5) ( )( ) 0 y 8y 0 Locus of P is + y 8y + = 0

9 .Whn th as ar rotatd through an angl 4, find th transformd quation of 0y y 9 Sol. Givn quation is 0y y () Angl of rotation of as is.lt (X,Y) b th nw co-ordinats of. 4 y X Y X cos Y sin = X cos y sin 4 4 X Y y X sin Y cos X sin Y cos 4 4 Transformd quation of () is X Y X Y X Y X Y X XY Y X Y X XY Y X 6XY Y 0X 0Y 6X 4Y 8 0 8X Y 9 X 6XY Y 8 0. Find th points on th lin 4y = 0 which ar at a distanc of 5 units form th point (, ). Sol. Equation of th lin is 4y = 0 slop of th lin is tanθ = ¾ sinθ = /5 and cosθ =4/5 Givn Point is (,) = (, y ) and r = 5. Co-ordinats of any point on th givn lin at a distanc r ar 4. ( r cos, y r sin ) Co-ordinats of th points ar 4 5., 5. (, ) 5 5 Lt , 5. (7,5) And 5 5

10 Lt 0 Sol : Lt Lt Lt A containr is in th shap of an invrtd con has hight 8 m and radius 6 m at th top. If it is filld with watr at th rat of m /minut, how fast is th hight of watr changing whn th lvl is 4 m? Sol. h = 8 m = OC r = 6 m = AB dv dt m /minut OAB and OCD ar similar angl thn A r C D h O B CD OC AB OA r h r h Volum of con v r h

11 9 v h h 6 v h 6 dv dh h h 6 dt 6 dt dh dh (6) 6 dt 9 dt 6. Find th angl btwn th curv y and y-ais. Sol: Equation of y-ais is = 0 Th point of intrsction of th curv Th angl mad by th tangnt to th curv y at P with ais is givn by Furthr, if is th angl btwn th y ais and y, thn w hav Th angl btwn th curv and th y-ais is tan If y = y dy log, thn show that. d ( log ) Sol. y = y y log log ylog y y tan 0, tan tan cot 4 dy d 4 4 0, y and 0 is P 0,

12 y log ( log ) dy d ( log ) log log ( log ) ( log ) 8. If (h, k) is th imag of (, y) w.r.t th lin a + by + c = 0 (a 0, b 0), h k y ( by c) a b a b thn. Proof: Lt A(, y), B(h, k) Mid pointof isp = h, y k Sinc B is th imag of A,thrformid pointp lis on a + by + c = 0. h y k a b +c =0 a + by + ah + bk + c = 0 ah + bk = a + by c. k y Slop of AB is h And Slop of givn lin is AB is prpndicular to th givn lin k y a h b k y h b a By th law of multiplirs in ratio and proportion a b

13 h k y a h b k y a b a b ah bk a by a b a by c a by a b ( a by c) a b h k y ( a by c) Hnc a b a b 9. If th quation a + hy + by + g + fy + c = 0 rprsnts a pair of intrscting lins, thn show that th squar of th distanc of thir point of intrsction from th origin is distanc is f g h b Sol. Lt th quation c(a b) f g ab if th givn lins ar prpndicular. a + hy + by + g + fy + c = 0 rprsnt th lins h. Also show that th squar of this l + m y + n = 0 l + m y + n = 0 () () (l + m y + n )( l + m y + n ) = a + hy + by + g + fy + c l l = a, m m = b, n n = c l m l m h,l n l n g, m n m n f Solving () and () y m n m n l n l n l m l m Th point of intrsction,

14 mn mn ln ln P, l m l m l m l m OP (mn mn ) (ln ln ) (lm lm ) (m n m n ) 4m m n n (ln ln ) 4ll nn (l m l m ) 4l l m m 4f 4abc 4g 4ac 4h 4ab c(a b) f g ab h If th givn pair of lins ar prpndicular, thn a + b = 0 a = b OP 0 f g f g ( b)b h h b. 0. Show that th straight lins y 4y 0 and from a paralllogram and find th lngths of its sids. Sol. Equation of th first pair of lins is y 4y 0, y y 0 y 0or y 0 4y 4y 5 0y 4 0 Equations of th lins ar y = 0..()and y = 0..() Equations of () and () ar paralll. Equation of th scond pair of lins is y 5 y 4 0 y 4 y y 4 0 y y 4 y 4 0 y y 4 0 4y 4y 5 0y 4 0

15 y 0, y 4 0 Equations of th lins ar y 0..()and y 4 0 (4) Equations of () and (4) ar paralll. Solving (), () + + = 0, = - Co-ordinats of A ar (-, ) Solving (), () = 0, = - 7 Co-ordinats of D ar (-7,) Solving (), (4) = 0, = - 6 Co-ordinats of B ar (-6, ) AB AD Lngth of th sids of th paralllogram ar, 5. If a ray maks angl,, and with th four diagonals of a cub find cos cos cos cos

16 Sol: Lt OABC;PQRS b th cub. Lt a b th sid of th cub. Lt on of th vrtics of th cub b th origin O and th co-ordinat as b along th thr dgs OA, OB and OC passing through th origin. Th co-ordinat of th vrtics of th cub with rspct to th fram of rfrnc OABC ar as shown in figur ar A (a,o,o), B(o,a,o), C(0,o,a) P(a,a,a) Q(a,a,o) R(o,a,a) and S(a,o,a) Th diagonals of th cub ar OP, CQ, ARandBS. and thir d.rs ar rspctivly (a, a, a), (a, a, -a), (-a, a, a) and (a, -a, a). Lt th dirction cosins of th givn ray b l, m, n. Thn l m n If this ray is making th angls,, and with th four diagonals of th cub, thn cos Similarly, cos a l a m a n l m n a a a cos. l m n l m n l m n and cos cos cos cos cos = { l m n l m n l m n l m n } [ l m n l m n l m n l m n ]

17 4 [4 ] l m n (sincl m n ). If y a a log a, show that a sol : y a a log a diff. w.r.t, d d y a a log a d d dy d. a. a a a a dy d a a a a. a a a a a a a a a a a a. If th tangnt at any point on th curv / / / y a AB is constant, intrscts th coordinat as in A, B show that th lngth Y B P O A X Sol: Equation of th curv is Lt a cos, y a sin / / / y a

18 b th paramtric quations of th curv. Thn any point P on th curv is dy d a.sin. cos a.cos. sin sin cos Equation of th tangnt at P is sin y a sin a cos cos y a sin a cos sin cos y a sin cos a cos sin y a cos a sin int rcpt OA a cos y Now int rcpt OB= asin AB OA OB a cos a sin a sin cos a AB a,acons tan t a cos, a sin 4. A window is in th shap of a rctangl surmountd by a smi-circl. If th primtr of th window b 0 ft, find th maimum ara. Sol: Lt th lngth of th rctangl b and bradth b y so that radius of th smi-circl is. Primtr y. 0 y 0 y 0. Ara y.

19 0 0 Lt f() 0 f ' 0 4 and f" =-4- + = -4- for ma or min for ma or min f ' 4 0 f has a maimum whn Maimum ara f y y sq.ft

MATHEMATICS PAPER IIB COORDINATE GEOMETRY AND CALCULUS. Note: This question paper consists of three sections A, B and C.

MATHEMATICS PAPER IIB COORDINATE GEOMETRY AND CALCULUS. Tim: 3hrs Ma. Marks.75 Not: This qustion papr consists of thr sctions A, B and C. SECTION -A Vry Short Answr Typ Qustions. 0 X = 0. Find th condition