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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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. Find the condition for the points (a, 0), (h, k) and (0,b)where ab 0 to be collinear.. Fund the value of P, if the straight lines 3 + 7y = 0 and 7 p y + 3 = 0 are mutually perpendicular. 3. Find the distance between the midpoint of the line segment AB and the point (3,, ) where A = (6, 3, 4) and B = (,, ). 4. Find the constant k so that the planes y + kz = 0 and + 5y z = 0 are at right angles. 5. 0 sin a b sin a b 6. If f is given by f k k if if is a continuous function on R, then find the values of k.

7. y log7 log 0 8. If 4 4 y a y 0,then dy find 9. The time t of a complete oscillation of a simple pendulum of length l is given by the equation t = g where g is gravitational constant. Find the approimate percentage error in the calculated g, corresponding to an error of 0.0 percent is the value of t. 0. Verify the Rolle s theorem for the function ( )( ) on [, ]. Find the point in the interval where the derivate vanishes. SECTION B SHORT ANSWER TYPE QUESTIONS. ANSWER ANY FIVE OF THE FOLLOWING 5 X 4 = 0. A(5, 3) and B(3, ) are two fied points. Find the equation of locus of P, so that the area of triangle PAB is 9.. When the origin is shifted to the point (,3), the transformed equation of a curve is 3y y 7 7y 0. Find the original equation of the curve. 3. A straight line parallel to the line y 3 passes through Q,3 and cuts the line 4y 7 0 at P. Find the length of PQ. 4. Evaluate a sin a asin a

5. A balloon which always remains spherical on inflation is being inflated by pumping on 900 cubic centimeters of gas per second. Find the rate at which the radius of balloon increases when the radius in 5 cm. 6. Show that the curves +y = and 3 4 have a common tangent at the point (, ). 7. If f and g are differentiable functions at and g() 0 then the quotient function f g is differentiable at and f g( ). f ( ) f ( ) g ( ) ( ) g [ g ( )] SECTION C LONG ANSWER TYPE QUESTIONS. ANSWER ANY FIVE OF THE FOLLOWING 5 X 7 =35 8. If p and q are lengths of the perpendiculars from the origin to the straight lines sec ycosec aand cos ysin a cos, prove that 4p q a. 9. Write down the equation of the pair of straight lines joining the origin to the points of intersection of the lines 6 y + 8 = 0 with the pair of straight lines 3 4y 4y y 6 0. Show that the lines so obtained make equal angles with the coordinate aes. 0. Show that the lines joining the origin to the points of intersection of the curve y y 3 3y 0 and the straight line y 0 are mutually perpendicular.. Show that the lines whose direction cosines are given by l m n 0 m 3nl 5lm 0 are perpendicular to each other. If tan cos y sin, find dy

3. If the tangent at a point P on the curve m n meets the coordinate aes in A, B. show that AP : PB is constant. 4. Prove that the radius of the right circular cylinder of greatest curved surface area which can be inscribed in a given cone is half of that of the cone. solutions m n y a (mn 0) SECTION A. 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 h a a Slope of AB = Slope of AC h k bh ak ab a b. Fund the value of P, if the straight lines 3 + 7y = 0 and 7 p y + 3 = 0 are mutually perpendicular. Sol. Given lines are 3 + 7y = 0, 7 p y + 3 = 0 aa bb 0 lines are perpendicular 3.7 7 p 0 7p p 3 3. Find the distance between the midpoint of the line segment AB and the point (3,, ) where A = (6, 3, 4) and B = (,, ). Sol. Given points are : A= (6, 3, 4), B = (,, ) Coordinates of Q are : 6 3 4,, (,, ) Coordinates of P are : (3,, ) PQ (3 ) ( ) ( ) 4 9 4 units.

4. Find the constant k so that the planes y + kz = 0 and + 5y z = 0 are at right angles. Sol. Equations of the given planes are y + kz = 0 and + 5y z = 0 since these planes are perpendicular, therefore 5 k( ) 0 0 k k 8 5. 0 Sol : sin sin 0 a b sin a b a b sin a b cos a.sin b sin b cos a.. b 0 0 b cos a. b b. cos a. 6. If f is given by f k k if if is a continuous function on R, then Sol : find the values of k. f f k k k k Given f is continuous at 0 f f k k Given f is continuous on R, hence it is continuous at =. Therefore L.L =R.L k k 0 => k k 0 => k or

7. y log7 log 0 sol : y log7 log 0 dy.. log log 7 7 e log e 7 e log log log 8. If 4 4 y a y 0,then sol : Differentiate w. r. to d 4 y 4 a y 3 3 4 4. dy dy y a. y. 0 3 3 4 4. dy dy y a a y 0 dy 4y a a y 4 0 3 3 3 dy a y 4 3 4y a dy find 9. The time t of a complete oscillation of a simple pendulum of length l is given by the equation t = where g is gravitational constant. Find the approimate percentage error in the calculated g, corresponding to an error of 0.0 percent is the value of t. Sol: percentage error in t is =0.0 Given t= Taking logs on both sides log t = log ( ) + {(log (l) log g} g

Taking differentials on both sides, = 0 + Multiplying with 00, Percentage error in g = -0.0 0. Verify the Rolle s theorem for the function ( )( ) on [, ]. Find the point in the interval where the derivate vanishes. Sol. Let f() = ( )( ) = 3 + f is continuous on [, ] since f( ) = f() = 0 and f is differentiable on [, ] By Rolle s theorem c (,) Let f (c) = 0 f () = 3 4 3c 4c = 0 c 4 6 4 8 6 6 c 7 3 SECTION B. A(5, 3) and B(3, ) are two fied points. Find the equation of locus of P, so that the area of triangle PAB is 9.

Sol. Given points are A(5, 3), B(3, ) Let P(, y) be a point in the locus. Given, area of APB = 9. 5 y 3 3 5 3 5 y 3 5 8 9 5 5 y 6 8 5 y 9 8 5 y 9 8 5 y 9 8 or 5 y 9 8 5 y 0 or 5 y 37 0 Locus of P is (5 y )(5 y 37) = 0.. When the origin is shifted to the point (,3), the transformed equation of a curve is 3y y 7 7y 0. Find the original equation of the curve. Sol. New origin =(,3) = (h,k) Equations of transformation are X h, y Y k X h, Y y k y 3 Transformed equation is 3y y 7 7y 0 ( here, y can be treated as upper case letters) Original equation is 3 y 3 y 3 7 7 y 3 0 4 4 3y 9 6y 8 y y 8 7 34 7y 0 Therefore, original equation is 3y y 4 y 0 0 3. A straight line parallel to the line y 3 passes through Q,3 and cuts the line 4y 7 0 at P. Find the length of PQ. Sol: PQ is parallel to the straight line y tan 3 tan 60º 3

60º Q,3 is a given point Q(, 3) P + 4y - 7 =0 Co-ordinates of any point P are r cos, y r sin r cos60º, 3 r sin 60º r 3 P, 3 r y 3 P is a point on the line 4y 7 0 r 3 4 3 r 7 0 4 r 3r 7 0 r 3 7 6 3. 3 3 3 4. Sol : a sin a asin a sin a asin a a a (Adding and subtactin a sina in Nr.) a a sin a a sin sin a a a a cos.sin sin a a. a a sin a acos a sin a acos a sin a asin a asin asin a a a a sin a sin sin a a a a a a sin cos a sin a a. a a a

5. A balloon which always remains spherical on inflation is being inflated by pumping on 900 cubic centimeters of gas per second. Find the rate at which the radius of balloon increases when the radius in 5 cm. dv Sol. 900c.c./ sec dt r = 5 cm 4 3 Volume of the sphere v r 3 dv 4 dr dr 3r 900 4 (5) dt 3 dt dt 900 dr 900 dr 4 5 dt 900 dt dr dr cm / s dt dt 6. Show that the curves +y = and 3 4 have a common tangent at the point (, ). Sol: Equation of the first curve is Differentiating w, r, to dy y 0 dy y At p (, ) slope of the tangent = - Equation of the second curve is Differentiating w. r. to, dy 4 6 y 3 y y y At p(, ) slope of the tangent = = - dy y y The slope of the tangents to both the curves at (, ) are same and pass through the same point (, ) The given curves have a common tangent p (, ) 3 y 4y. dy 6 y. 4 3 dy y. 4 6

7. If f and g are differentiable functions at and g() 0 then the quotient function f g is differentiable at and f g( ). f ( ) f ( ) g ( ) ( ) g [ g ( )] Proof: Since f and g are differentiable at, therefore f ( h) f ( ) f '() = h 0 h g( h) g( ) and = g '() h 0 h = = = = = f ( f / g)( h) ( f / g)( ) g h 0 h f ( h) f ( ) ( ) ( ) h 0 h g h g f ( h) g( ) g( h) f ( ) ( ) ( ) h 0 h g g h f ( h) g( ) f ( ) g( ) f ( ) g( ) f ( ) g( h) h 0 g( ). g( h) h f ( h) f ( ) g( h) g( ) g( ) f ( ) [ g ( )] h 0 h h 0 h f ( h) f ( ) g( h) g( ) g( ) f ( ) [ g ( )] h 0 h h 0 h = [ g ( )] g( ) f ( ) f ( ) g ( ) f g( ) f ( ) f ( ) g ( ) ( ) g [ g ( )] SECTION C 8. If p and q are lengths of the perpendiculars from the origin to the straight lines sec ycosec aand cos ysin a cos, prove that Sol: Equation of AB is sec ycosec a y a cos sin sin ycos a sin cos sin ycos a sin cos 0 4p q a.

p = length of the perpendicular from O on AB = sin a sin.cos a. p a sin ---() Equation of CD is cos ysin a cos cos ysin a cos 0 q = Length of the perpendicular from O on CD Squaring and adding () and () 4p q a sin a cos a sin cos a. a 0 0 a sin cos sin 0 0 a cos cos sin cos a cos ---() 9. Write down the equation of the pair of straight lines joining the origin to the points of intersection of the lines 6 y + 8 = 0 with the pair of straight lines 3 4y 4y y 6 0. Show that the lines so obtained make equal angles with the coordinate aes. Sol. Given pair of line is 3 4y 4y y 6 0 () Given line is 6 y 6 y 8 0 8 Homogenising () w.r.t () y 6 -----() 8 y 6 3 4y 4y y 8 y 6 6 0 8 64 3 4y 4y 8 y 66 y y 6 y 36 y 0 936 56y 56y 34y 0 468 7y 0 4 y 0 ----(3) is eq. of pair of lines joining the origin to the point of intersection of () and ().

The eq. pair of angle bisectors of (3) is h y a b y 0 0 y 4 y 0 y 0 = 0 or y = 0 which are the eqs. is of co-ordinates aes The pair of lines are equally inclined to the co-ordinate aes 0. Show that the lines joining the origin to the points of intersection of the curve y y 3 3y 0 and the straight line y 0 are mutually perpendicular. Sol. Le t A,B the the points of intersection of the line and the curve. Equation of the curve is y y 3 3y 0.() Equation of the line AB is y 0 y y.() Homogenising, () with the help of () combined equation of OA, OB is y y 3. 3y.. 0 y y y y 3 y 0 3 y y y y y 0 3 3 y y y y y 0 3 3 y y 0

coefficient of +coefficient of y = OA, OB are perpendicular. 3 3 a b 0. Show that the lines whose direction cosines are given by l m n 0 m 3nl 5lm 0 are perpendicular to each other Sol: Given equations are l m n 0 mn 3nl 5lm 0 From (), l m n Substituting in () mn 3n m n 5m m n 0 mn 3mn 3n 5m 5mn 0 5m 4mn 3n 0 m m 5 4 3 0 n n m m 3 m m n n nn 5 3 5 From (), n = l m Substituting in (), m l m 3l l m 5lm 0 lm m 3l 3lm 5lm 0 3l 0lm m 0 l l 3 0 0 m m l l l l m m mm 3 3 Form (3) and (4) ll mm nn k say 3 5 l l k, m m 3 k, n n 5k ll mm nn k 3k 5k 0 The two lines are perpendicular 4 3

tan cos. If y sin, find dy Sol : Let tan u and v sin tan log u log tan log dy. tan log sec. u du tan u log.sec tan tan log.sec cos log v log sin cos ] cos.logsin dv. cos. cos log sin sin v sin cos sin sin log sin cos dv cos v sin log sin sin cos cos sin sin log sin sin dy du dv tan tan log sec + cos cos sin sin.log sin sin 3. If the tangent at a point P on the curve m n constant. m n y a (mn 0) meets the coordinate aes in A, B. show that AP : PB is

Y B P, y Sol: Equation of the curve is let P(, y ) be a point on the curve. Then. Differente given curve w. r. to, m n dy m n n.y m..y Slope of the tangent at Equation of the tangent at P is co-ordinates of A are the ratio in which P divides AB is m n AP X X m n n PB X 0 m m O A m n m n.y a P, y X y a and B are m n m n m n dy n m.ny. y.m 0 m dy m..y my m n n. y n ny ny my my my n my ny my ny m n y my n..y m n y m n y y m n m n..y m n OA m n. m n.ob m n m n.,0 m n my y y n m n 0,.y n 4. Prove that the radius of the right circular cylinder of greatest curved surface area which can be inscribed in a given cone is half of that of the cone. Sol. Let O be the center of the circular base of the cone and its height be h. Let r be the radius of the circular base of the cone. Then AO = h, OC = r

Let a cylinder with radius (OE) be inscribed in the given cone. Let its height be u. A P R Q i.e. RO = QE = PD = u Now the triangles AOC and QEC are similar. Therefore, QE EC OA OC i.e., u r h r h(r ) u r Let S denote the curved surface area of the chosen cylinder. Then S = u As the cone is fied one, the values of r and h are constants. Thus S is function of only. Now, ds d S 4 h h(r ) / r and r The stationary point of S is a root of ds 0 i.e., (r ) / r 0 i.e., ds 0 r for all, therefore ds B r / D O 0 Hence, the radius of the cylinder of greatest curved surface area which can be inscribed in a given cone is r/. E C