IB ANS. -30 = -10 k k = 3

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1 IB ANS.. Find the value of k, if the straight lines 6 0y and k 5y are parallel. Sol. Given lines are 6 0y and k 5y a b lines are parallel a b k k 3. Find the condition for the points (a, 0), (h, k) and (0,b)where ab 0 to be collinear. Sol. A(a, 0), B(h, k), C(0, b) are collinear. k 0 b ak bh + ab Slope of AB Slope of AC h a a h k bh + ak ab + a b 3. Find the fourth verte of the parallelogram whose consecutive vertices are (, 4, ), (3, 6, ) are (4, 5, ). Sol. ABCD is a parallelogram where A (, 4, ), B (3, 6, ), C (4, 5, ) Suppose D(, y, z) is the fourth verte A B C D is a parallelogram Mid point of AC Mid point of BD y z, +, + +, +, + 7 B A C D

2 y 9 y 3 z 0 z Coordinates of the fourth verte are : D (3, 3, ) 4. Find the angle between the planes -y + z 6 and + y + z 7. Sol. Equation of the plane are -y + z 6 and + y + z 7. Let θ be the angle between the planes, then cos θ a a + bb + cc a b c a b c θ cos 3 B A B A C C D D 5. Lt Sol : Lt Lt ( rationalise Dr. ) ( 3 )( + + ) Lt Lt. Lt ( ) ( log 3)( ).log 3 6. Lt

3 Sol : as ( here is positive) Lt Lt 3 3 Lt 7. If y log ( sin ( log ) ), find dy d. y log ( sin ( log ) ) dy d d log ( sin ( log ) ) ( sin ( log ) ) d d ( sin ( log ) ) d d ( cos( log ) ) log ( sin ( log ) ) d ( cos( log ) ) ( sin ( log ) ) 8. Let be the side and A be the area of the Square. percentage error in is δ 00 4 Area A Applying logs on both sides Log A log Taking differentials on both sides δa δ δ A. δ A A 4 8. There fore percentage error in A is 8% 9. Let f() + 4. f is continuous on [ 3, 3] since f( 3) f(3) and

4 f is differentiable on [ 3, 3] By Rolle s theorem c (,) Such that f (c) 0 f () 0 f (c) 0 c 0 c 0 The point c 0 ( 3, 3) 0.If Sol. 3, find dy y (cot ) d. 3 3 u cot,u,y u du du, 3 dv + u d dy 3 u cot ( ) d + dy dy du dv d d dv d cot ( ) cot ( ) The ends of the hypotenuse of a right angled triangle are (0, 6) and (6, 0). Find the equation of locus of its third verte. ANS. Given points A(, 3), B(, 5). Let P(, y) be any point in the locus. P(,y) 90 A(0,6) B(6,0)

5 Given condition is : APB 90 (slope of AP ) (slope of BP ) y 6 y (y)(y 6) + ()( 6) 0 + y 6 6y 0 Locus of P is + y 6 6y 0. When the aes are rotated through an angle 45, the transformed equation of a curve is 7 6y+ 7y 5.Find the original equation of the curve. 0 Sol. Angle of rotation is θ 45. Let ( X,Y) be the new coordinates of (, y) + y X cosθ + ysinθ cos 45 + ysin 45 + y Y sinθ + ycosθ sin 45 + ycos 45 The original equation of 7X 6XY + 7Y 5 is + y + y + y + y ( + y + y) ( y ) ( + y y) y + 34y 6y y 34y y y 5 is the original equation

6 3. If the straight lines a + by + c 0, b + cy + a 0 and c + ay + b 0 are concurrent, then prove that a 3 + b 3 + c 3 3abc. Sol: The equations of the given lines are a + by + c 0 ---() b + cy + a 0 ---() c + ay + b 0 ---(3) Solving () and () points of intersection is got by y b c a b c a b c y ab c bc a ca b ab c bc a Point of intersection is, ca b ca b ab c bc a c + a + b 0 ca b ca b c ab c + a bc a + b ca b 0 ( ) ( ) ( ) abc c + abc a + abc b a + b + c 3abc. 4. f ( ) Lt h 0 f ( + h) f ( ) h Lt h 0 SIN ( + h) cos h + h + + h cos.sin Lt h 0 h cos Lt h 0 h h ( + ).sin( ) h h Lt cos( + h) Lt h 0 0 sin ( h) h cos. cos

7 5. Show that the length of sub-normal at any point on the curve y a varies as the cube of the ordinate of the point. Sol: Equation of the curve is y a. a y dy a m d Let P(,y) be a point on the curve. Length of the sub-normal. y a y 4 a a y y m ( a) 3 y y a 3 l.. st α y i.e. cube of the ordinate. 6. equation of the curve y Diff. w.r.t. t, dy d 4. dt dt Given d and 4. dt dy dt 4.4 ( ) 3 y co-ordinate is increasing at the rate of 3 units/sec. cos a cos b Lt f ( ) Lt Lt 0 ( a + b) ( b a) sin sin sin ( a + b) sin ( b a) Lt Lt 0 0

8 ( b + a) ( b a) b a 8. A(,3 ), B(, ), C( 4,0) are the vertices of Δ ABC. Let S be the circumcentre of the ABC. Let D be the midpoint of BC A (-, 3) S E B(, -) D C(4, 0) D , 3, Slope of BC 0 4 SD is perpendicular to BC Slope of SD m Equation of SD is y+ ( 3) y+ 4( 3) y 0 ---() Let E be the midpoint of AC

9 Co-ordinates of E are +, +, Slope of AC 4 6 SE is perpendicular to AC Slope of SE m 3 Equation of SE is y ( ) y 3 4( ) y 0 ---() 4 + y 0 ---() Adding (), () Substitute this in (), 3 y y 5 Co-ordinates of S are 3 5, Show that the lines joining the origin to the points of intersection of the 9. curve y+ y y 0 and the straight line y 0 are mutually perpendicular.

10 Sol. Le t A,B the the points of intersection of the line and the curve. Equation of the curve is y+ y y 0.() Equation of the line AB is y 0 y y.() Homogenising, () with the help of () combined equation of OA, OB is y + y y.. 0 ( ) ( ) y y y+ y y 0 3 y+ y + y y+ y 0 ( ) ( ) 3 3 y+ y + y + y y y y 0 coefficient of +coefficient of y 3 3 a+ b 0

11 OA, OB are perpendicular. 0. Let (). a + hy + by 0 represent the lines l + my 0 -- () and l + my 0 -- Then ll a, lm + l m h, m m b. l The equations of bisectors of angles between () and () are + my l + m y ± l + m l + m l+ my l + m y 0 l + m l + m and l+ my l + m y + 0 l + m l + m The combined equation of the bisectors is l my l my my my + + l + l l + m l + m + m + m l l l + my l+ my 0 l m l m + + ( ) ( ) l + m ( l + m y) l + m ( l + m y) 0 ( ) ( ) ( ) ( ) y lm( l m ) lm( l m) l l + m l l + m y m l + m m l + m ( ) ( ) ll + lm ll lm y lm + mm ml mm y lml + lmm lml lmm 0 ( ) ( ) l m l m y l m l m y[ ll ( lm lm) mm( lm lm) ]

12 ( ) ( )( ) ( y) lm lm yll mm lm lm ( y )( l m + l m ) y( l l m m ) h ( y) ya ( b) h ( y) ( a by ) OR y y a b h. Given 3l + m+ 5n 0 6mn nl + 5lm 0 From (), m ( 3l + 5n) Substituting in () n( l n) nl l( l n) ln 30n nl 5l 5ln 0 5l 45ln 30n 0 l + 3ln + n 0 ( l n)( l n) l + n 0or l + n 0 Case (i) : l n l+ n 0 n l; n l; But ( ) ( ) m 3l + 5n 3n + 5n n m n + l m n

13 D.rs of the first line l are(,, ) Case (ii) : l + n 0 l n l n ( 3 5 ) ( 6 5 ) m l + n n + n n m n l m n D.rs of the second line l are(,,) Suppose ' θ ' is the angle between the lines l and l cosθ aa+ bb + cc a + b + c a + b + c ( ) + + ( ) θ 6 ( ) cos / 6 log sin. Let u ( sin ), v so that y u + v, u ( sin ) log applying logs, log log u log{ ( sin ) } log.log( sin )

14 Differentiating w. r. to d d log u log.log( sin ) d d dy. log.cos + log ( sin ) u d sin dy log ( sin ) u cot.log + d ( ) log log sin sin cot.log. + v sin logv ( log ) sin sin.log dv. sin. + ( log.cos) v d dv d sin v.cos.log sin sin + cos.log dy du dv sin ce y u+v + d d d ( ) ( ) log log sin sin sin sin cot.log. + + ( + cos.log ) 3. Show that the curves,. Sol: Equation of the first curve is y 0 and y y 3 touch each other at

15 y 5 6 m 5. dy d atp, dy dy 5. 5 d d 5 6 Equation of the second curve is dy 8 + 6y. 0 d dy 6y. 8 d 4 + 8y 3 dy 8 d 6y y m dy d atp, m m The given curves touch each other at P, 4. let r be the radius and h be the height of the cylinder. O H A r R B From Δ OAB, OA + AB OB h h r + R ; r R 4 4 Curved surface area π rh

16 h π R.h 4 πh 4R h Let f ( h) πh 4R h f ( h) π h. ( h) + 4R h. 4R h ( ) h + 4R h π R h π. 4R h 4R h For ma or min f ( h) 0 ( ) π R h 0 4R h R h 0 h R h R 4R h ( h) + ( R h ) d 4R h f'when dh ( h R) π 4R h 4π h + 0 < 0 4r h f( h ) is greatest when h R i.e., Height of the cylinder R P.S.RAVI KUMAR

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

MATHEMATICS PAPER IB COORDINATE GEOMETRY(D &3D) AND CALCULUS. TIME : 3hrs Ma. Marks.75 Note: This question paper consists of three sections A,B and C. SECTION A VERY SHORT ANSWER TYPE QUESTIONS. 0X =0.