MATHEMATICAL TOOLS. Contents. Theory Exercise Exercise Advance Level Problems Answer Key

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1 MATHEMATICAL TOOLS Contents Topic Page No. Theoy - 6 Eecise Eecise Avance Level Poblems 8 - Answe Key - 9 Syllabus Diffeentiation, Integation & Vecto Name : Contact No. ARRIDE LEARNING ONLINE E-LEARNING ACADEMY A-79 ina Viha, Kota Rajasthan 5 Contact No. 8557

2 MATHEMATICAL TOOLS TRIGONOMETRY : q = Ac length Raius Angle Convesion fomulas Þ q = AB p egee = 8 (».) aian 8 aian» 57 egees Raians to egees : multiply by p RULES FOR FINDING TRIGONOMETRIC RATIO OF ANGLES GREATER THAN 9 : Step Ientify the quaant in which angle lies. Step (a) If angle = (np ± q) whee n is an intege. Then tigonometic function of (np ± q) = same tigonometic function of q an sign will be ecie by CAST Rule. é p ù (b) If angle = ê( n + ) ± qú whee n is an intege. Then ë û é p ù tigonometic function of ê( n + ) ± qú = complimentay tigonometic function of q ë û an sign will be ecie by CAST Rule. Values of sin q, cos q an tan q fo some stana angles. Degee Raians sinq cos q tan q p / 6 / / / 7 7p/8 /5 /5 / 5 p/ / / 5 5p/8 /5 / 5 / 6 p/ / / 9 p / p / / -/ - 5 p/ / -/ - 8 p - GENERAL TRIGONOMETRIC FORMULAS :. cos q + sin q = + tan q = sec q. + cot q= cosec q.. cos(a + B) = cos A cos B sin A sin B sin( A + B) = sin A cos B + cos A sin B tan A + tanb tan (A+B) = - tana tanb. sin q = sin q cos q ; cos q = cos q sin q = cos q = sin q + cosq cos q = ; sin q = cosq A-79 Ina Viha, Kota Rajasthan 5 Page No. #

3 . sine ule fo tiangles 5. cosine ule fo tiangles a b c a c g a q b b sin a sin b sin g = = a b c c = a + b ab cosq RULES FOR DIFFERENTIATION : c = ; n n - = n. (cu) = u c ; (u + u un ) = u u un (uv) = v u æ u ö u + v ; ç è v ø u v v - u v = (sin ) = cos ; (cos ) = -sin (tan ) = sec ; (sec ) = sec tan (cot ) = cosec ; (cosec ) = cosec cot ( log ) e = ; ( e ) e y = ; = f'[g()].g'() RULES FOR INTEGRATION : k f() = k () ; [f() ± g()] = f() ± g() Inefinite Integal Revese eivative fomula n+ æ n+ n. =,n ¹, n ational ö ç n + è n + ø = n = + = C (special case) () = cosk. sin k = - k æ cos ö ç - è kk ø = sin k sink. cos k = k æ sin k ö ç è k ø = cos k. sec = tan tan = sec A-79 Ina Viha, Kota Rajasthan 5 Page No. #

4 5. cosec = - cot 6. sec tan = sec ( cot ) = cosec sec = sec tan 7. cos ec cot = -cosec ( cosec ) = cosec cot VECTOR : DEFINITION OF VECTOR : A physical quantity is calle vecto if in aition to magnitue - (a) it has a specifie iection (b) it obeys the law of paallelogam of aition. (c) It obeys commutative law of aition A + B = B + A. If any of the above conitions is not satisfie the physical quantity cannot be a vecto. ADDITION OF VECTORS : Aition of vectos is one by paallelogam law o its coollay, the tiangle law : (a) Paallelogam law of aition of vectos : If two vectos A an B ae epesente by two ajacent sies of a paallelogam both pointing outwas (an thei tails coinciing) as shown. Then the iagonal awn though the intesection of the two vectos epesents the esultant (i.e., vecto sum of A an B ). R = A + B + ABcosq The iection of esultant vecto R fom A is given by MN tan f = = PN MN = PQ + QN - æ Bsin q ö f = tan ç è A + Bcosq ø Bsin q A + Bcosq (b) Tiangle law of aition of vectos : To a two vectos A an B shift any of the two vectos paallel to itself until the tail of B is at the hea of A. The sum A + B is a vecto R awn fom the tail of A to the hea of B, i.e., A + B = R. As the figue fome is a tiangle, this metho is calle tiangle metho of aition of vectos. If the tiangle metho is etene to a any numbe of vectos in one opeation as shown. Then the figue fome is a polygon an hence the name Polygon Law of aition of vectos is given to such type of aition. A-79 Ina Viha, Kota Rajasthan 5 Page No. #

5 IMPORTANT POINTS : To a vecto only a vecto of same type can be ae that epesents the same physical quantity an the esultant is a vecto of the same type. As R = [A + B + AB cosq] / so R will be maimum when, cos q = ma =, i.e., q = º, i.e. vectos ae like o paallel an R ma = A + B. The esultant will be minimum if, cos q = min = -, i.e., q = 8º, i.e. vectos ae antipaallel an R min = A B. If the vectos A an B ae othogonal, i.e., q = 9º, R = A + B As peviously mentione that the esultant of two vectos can have any value fom (A ~ B) to (A + B) epening on the angle between them an the magnitue of esultant eceases as q inceases º to 8º Minimum numbe of unequal coplana vectos whose sum can be zeo is thee. The esultant of thee non-coplana vectos can neve be zeo, o minimum numbe of non coplana vectos whose sum can be zeo is fou. Subtaction of a vecto fom a vecto is the aition of negative vecto, i.e., A - B = A + (- B ) (a) Fom figue it is clea that A - B is equal to aition of A with evese of B = (b) A - B = [(A) + (B ) + AB cos (8º - q)] / A - B = A + B - AB cosq Change in a vecto physical quantity means subtaction of initial vecto fom the final vecto. A + B ³ A + B (The tiangle inequality) MULTIPLICATION OF VECTORS : THE SCALAR PRODUCT : The scala pouct o ot pouct of any two vectos A an B, enote as A. B (ea A ot B ) is efine as the pouct of thei magnitue with cosine of angle between them. Thus, A. B = AB cos q {hee q is the angle between the vectos} PROPERTIES : It is always a scala which is positive if angle between the vectos is acute (i.e. < 9º) an negative if angle between them is obtuse (i.e. 9º < q 8º) It is commutative, i.e., A. B = B. A It is istibutive, i.e. A. ( B ) = A. B + A. C A-79 Ina Viha, Kota Rajasthan 5 Page No. #

6 As by efinition A. B éa. B ù = AB cos q. The angle between the vectos q = cos - ê ú ë AB û A. B = A (B cos q) = B (A cos q) Geometically, B cos q is the pojection of B onto A an A cos q is the pojection of A onto B as shown. So A. B is the pouct of the magnitue of A an the component of B along A an vice vesa. Component of B along A A.B = B cosq = = Â. B A Component of A along B A.B = A cosq = = A. Bˆ B Scala pouct of two vectos will be maimum when cos q = ma =, i.e., q = º, i.e., vectos ae paallel Þ ( A. B ) ma = AB If the scala pouct of two nonzeo vectos vanishes then the vectos ae pepenicula. The scala pouct of a vecto by itself is teme as self ot pouct an is given by ( A ) = A. A = AA cosq = AAcosº = A Þ A = A. A In case of unit vecto nˆ, nˆ.nˆ = cos º = Þ nˆ.nˆ = î. î = ĵ. ĵ = kˆ. kˆ = In case of othogonal unit vectos î, ĵ an kˆ ; î. ĵ = ĵ. kˆ = kˆ. î = A. B = ( î A + ĵ A y + kˆ A z ). ( î B + ĵ B y + kˆ B z ) = [A B + A y B y + A z B z ] VECTOR PRODUCT : The vecto pouct o coss pouct of any two vectos A an B, enote as A B (ea A coss B ) is efine as : A B = AB sin q nˆ Hee q is the angle between the vectos an the iection nˆ is given by the ight-han-thumb ule. Right-Han-Thumb Rule: To fin the iection of nˆ, aw the two vectos A an B with both the tails coinciing. Now place you stetche ight palm pepenicula to the plane of A an B in such a way that the finges ae along the vecto A an when the finges ae close they go towas B. The iection of the thumb gives the iection of nˆ. V = A B n q B A A-79 Ina Viha, Kota Rajasthan 5 Page No. # 5

7 PROPERTIES : Vecto pouct of two vectos is always a vecto pepenicula to the plane containing the two vectos i.e. othogonal to both the vectos A an B, though the vectos A an B may o may not be othogonal. Vecto pouct of two vectos is not commutative i.e. A B ¹ B A. But A B = B A = AB sin q The vecto pouct is istibutive when the oe of the vectos is stictly maintaine i.e. A (B ) = A B + A C. The magnitue of vecto pouct of two vectos will be maimum when sinq = ma =, i.e,, q = 9º A B ma = AB i.e., magnitue of vecto pouct is maimum if the vectos ae othogonal. The magnitue of vecto pouct of two non zeo vectos will be minimum when sinq = minimum =,i.e., q = º o 8º an A B min = i.e., if the vecto pouct of two nonzeo vectos vanishes, the vectos ae collinea. Note : When q = º then vectos may be calle as like vecto o paallel vectos an when q = 8º then vectos may be calle as unlike vectos o antipaallel vectos. The self coss pouct i.e. pouct of a vecto by itself vanishes i.e. is a null vecto. Note : Null vecto o zeo vecto : A vecto of zeo magnitue is calle zeo vecto. The iection of a zeo vecto is in eteminate (unspecifie). A A = AA sin º nˆ =. In case of unit vecto nˆ, nˆ nˆ = Þ î î = ĵ ĵ = kˆ kˆ = In case of othogonal unit vectos î, ĵ an kˆ in accoance with ight-han-thumb-ule, î ĵ = kˆ ĵ kˆ = î kˆ î = ĵ In tems of components, A B = î A B A B ĵ y y kˆ A B z z = Ay Az A Az î - ĵ + kˆ B B B B y z z A B A B y y A B = î (A B - A B ) + ĵ (A B - A B ) + kˆ (A y z z y z z y - B A y B ) A-79 Ina Viha, Kota Rajasthan 5 Page No. # 6

8 * Make Questions ae having moe than one coect option. Section - (A) : Function. If f() = + Fin f(f()) FUNCTION & DIFFERENTIATION. f() = log an g () = log Which of the following statement is / ae tue - (A) f() = g (B) f() = g() (C) f() = g() (D) f() = (g()). f() = cos + sin Fin f(p/) Section - (B) : Diffeentiation of Elementy Functions Diffentiate w..t. coesponing inepenent vaiable.. s = 5t t 5. y = tan + cot. y = 5 sin. y = y = + sin Fin the fist eivative & secon eivative of given functions w..t. coesponing inepenent vaiable. 6. y = ln + e 7. y = w = z 7 7z + z 9. y = sin + cos. = q q + q Section - (C) : Diffeentiation by Pouct ule Diffentiate w..t.. y = ( + ) ( ). y = sin cos. sin. y = ( ) ( + + ) Section - (D) : Diffeentiation by Quotient ule Fin eivative of given functions w..t. the inepenent vaiable.. Suppose u an v ae functions of that ae iffeentiable at = an that u () = 5, u () =, v( ) = v () = Fin the values of the following eivatives at =. (a) (uv) (b) æ u ö ç è v ø (c) æ v ö ç è u ø () (7v u) A-79 Ina Viha, Kota Rajasthan 5 Page No. # 7

9 . f(t) = t t - + t -, fin f (t) ; Whee : t ¹. z = + - z, Fin at =,. y = sin cos 5. y = Section - (E) : Diffeentiation by Chain ule y Fin as a function of æ ö æ ö. y = ( ) 9. y = ç -. y = ç - è 7 ø è ø y = sin 5. y = sin() + ln( ) + e 6. y = sin 5 7. y = sin (w + f) whee w an f constants 8. y = ( + ) 5 Section - (F) : Diffeentiation of Implicit functions y Fin. y + y = 6. ( + y) = Section - (G) : Diffeentiation as a ate measuement. Suppose that the aius an suface aea S = p of a sphee ae iffeentiable functions of t. Wite an s equation that elates to. t t. Suppose that the aius an aea A = p of a cicle ae iffeentiable functions of t.wite an equation that elates A / t to / t. Section - (H) : Maima & Minima. Fin the values of function at the points of maimum an minimum. Paticle's position as a function of time is given by = t + t + fin the maimum value of position cooinate of paticle. Section - (I): Miscellaneous y Given y = f(u) an u = g(), fin. y = cosu, u = -. y = sinu, u = +. y = 6u 9, u = (/). y = u, u = 8 A-79 Ina Viha, Kota Rajasthan 5 Page No. # 8

10 INTEGRATION Section - (A) : Integation of elementy functions Fin integals of given functions /. /. sin. sec 5. csc 6. sec tan /. ( - ) 5 Section - (B) : Integation by substitution metho. sec t tant t, (use, u = t). æ ç - cos è t ø ö sin t t, (use, u = cos t ). ( - ) 9, (use, u = ). -, (use, u = ) æ ö 5. cos ç, (use, u = ) 6. è ø sin, (use, u = ) 7. sin( ), (use, u = ) Integate by using a suitable substitution 8. ( - ) y 9. y. y + cos( z + ) z. sin( 8z - 5) z. cos( t + ) t t. + ) ( A-79 Ina Viha, Kota Rajasthan 5 Page No. # 9

11 Section - (C) : Definite integation. - æ ö ç + è ø 5. p. sinq q. e Section - (D) : Calculation of aea Use a efinite integal to fin the aea of the egion between the given cuve an the ais on the inteval [,b]. y =. y = + Use a efinite integal to fin the aea of the egion between the given cuve an the ais on the inteval [, p]. y = sin. y = sin p q VECTOR Section - (A) : Definition of vecto & angle between vectos. Rain is falling vetically ownwas with a spee 5 m/s. If unit vecto along upwa is efine as ĵ, epesent velocity of ain in vecto fom.. The vecto joining the points A (,, ) an B (,, ) & pointing fom A to B is - (A) î + ĵ 5 kˆ (B) î + ĵ + 5 kˆ (C) î ĵ + 5 kˆ (D) î ĵ 5 kˆ.. Vectos A, B an C ae shown in figue. Fin angle between (i) A an B, (ii) A an C, (iii) B an C.. The foces, each numeically equal to 5 N, ae acting as shown in the Figue. Fin the angle between foces? A-79 Ina Viha, Kota Rajasthan 5 Page No. #

12 Section - (B) : Aition of Vectos. Two foce of F = 5 N ue east an F = 5 N ue noth, Fin F F?. Two vectos a an b incline at an angle q w..t. each othe have a esultant c which makes an angle b with a. If the iections of a an b ae intechange, then the esultant will have the same (A) magnitue (C) magnitue as well as iection (B) iection (D) neithe magnitue no iection.. Two vectos A an B lie in a plane. Anothe vecto C lies outsie this plane. The esultant A + B of these thee vectos (A) can be zeo (B) cannot be zeo (C) lies in the plane of A & B (D) lies in the plane of A & A + B. The vecto sum of the foces of N an 6 N can be (A) N (B) 8 N (C) 8 N (D) N. 5. A set of vectos taken in a given oe gives a close polygon. Then the esultant of these vectos is a (A) scala quantity (B) pseuo vecto (C) unit vecto (D) null vecto. 6. The vecto sum of two foce P an Q is minimum when the angle q between thei positive iections, is (A) p (B) p (C) p (D) p. 7. The vecto sum of two vectos A an B is maimum, then the angle q between two vectos is - (A) º (B) (C) 5 (D) 6 8. Given : C = A + B. Also, the magnitue of A, B an C ae, 5 an units espectively. The angle between A an B is (A) º (B) p (C) p (D) p. 9. If P + Q = P Q an q is the angle between P an Q, then (A) q = º (B) q = 9º (C) P = (D) Q =. The sum an iffeence of two pepenicula vectos of equal lengths ae (A) of equal lengths an have an acute angle between them (B) of equal length an have an obtuse angle between them (C) also pepenicula to each othe an ae of iffeent lengths (D) also pepenicula to each othe an ae of equal lengths.. A man walks m Noth, then m East an then m South. Fin the isplacement fom the stating point?. Two foce F an F ae acting at ight angles to each othe, fin thei esultant?. A vecto of magnitue an iection eastwas is ae with anothe vecto of magnitue an iection Nothwas. Fin the magnitue an iection of esultant with the east. A-79 Ina Viha, Kota Rajasthan 5 Page No. #

13 Section - (C) : Resolution of Vectos. One of the ectangula components of a velocity of 6 km h is km h. Fin othe ectangula component?. If.5 î +.8 ĵ kˆ is a unit vecto. Fin the value of C. The ectangula components of a vecto ae (, ). The coesponing ectangula components of anothe vecto ae (, ). Fin the angle between the two vectos. The an y components of a foce ae N an N. The foce is (A) î ĵ (B) î + ĵ (C) î ĵ (D) î + ĵ 5. Fin the magnitue of î + ĵ + kˆ? 6. If A = î + ĵ then fin  7. What ae the an the y components of a 5 m isplacement at an angle of º with the -ais (anti clockwise)? Section - (D) : Poucts of Vectos. Thee non zeo vectos A, B& C satisfy the elation A. B= & A. C =. Then A can be paallel to : (A) B (B) C (C) B. C (D) BC.* The magnitue of scala pouct of two vectos is 8 an that of vecto pouct is 8. The angle between them is : (A) º (B) 6º (C) º (D) 5º. If A = î + ĵ + kˆ an B = î + ĵ fin (a) A.B (b) A B. If A =, B = an q = 6 in the figue, Fin (a) A.B (b) A B B q A A-79 Ina Viha, Kota Rajasthan 5 Page No. #

14 PART - I : OBJECTIVE QUESTIONS Single Coect Answe Type. If the angle between two foces inceases, the magnitue of thei esultant (A) eceases (B) inceases (C) emains unchange (D) fist eceases an then inceases. A ca is moving on a staight oa ue noth with a unifom spee of 5 km h when it tuns left though 9º. If the spee emains unchange afte tuning, the change in the velocity of the ca in the tuning pocess is (A) zeo (B) 5 km h S-W iection 5 km h S-W (C) 5 km h N-W iection 5 km h N-W (D) 5 km h ue west. 5 km h. Which of the following sets of isplacements might be capable of binging a ca to its etuning point? (A) 5,, an 5 km (B) 5, 9, 9 an 6 km (C),, 9 an km (D),, an 9 km. When two vecto a an b ae ae, the magnitue of the esultant vecto is always (A) geate than (a + b) (B) less than o equal to (a + b) (C) less than (a + b) (D) equal to (a + b) 5. Given : A = î + ĵ an B = 5 î 6 ĵ. The magnitue of A + B is (A) units (B) units (C) 58 units (D) 6 units. 6. Given : A = î ĵ + kˆ an B = î ĵ + kˆ. The unit vecto of A B is (A) î + kˆ (B) î (C) kˆ (D) - î - kˆ 7. Vecto A is of length cm an is 6º above the -ais in the fist quaant. Vecto B is of length cm an 6º below the -ais in the fouth quaant. The sum A + B is a vecto of magnitue - (A) along + y-ais (B) along + -ais (C) along ais (D) along ais 8. Si foces, 9.8 N each, acting at a point ae coplana. If the angles between neighboing foces ae equal, then the esultant is (A) N (B) 9.8 N (C) 9.8 N (D) 9.8 N. 9. A vecto A points vetically ownwa & B points towas east, then the vecto pouct A B is (A) along west (B) along east (C) zeo (D) along south A-79 Ina Viha, Kota Rajasthan 5 Page No. #

15 . If A + B = A = B, then the angle between A an B is (A) º (B) 6º (C) 9º (D) º.. Given : a + b + c =. Out of the thee vectos a,b an c two ae equal in magnitue. The magnitue of the thi vecto is times that of eithe of the two having equal magnitue. The angles between the vectos ae: (A) 9º, 5º,. 5º (B) º, 6º, 9º (C) 5º, 5º, 9º (D) 5º, 6º, 9º. A hall has the imensions m m m. A fly stating at one cone ens up at a iametically opposite cone. The magnitue of its isplacement is nealy (A) 6 m (B) 7 m (C) 8 m (D) m.. A vecto is not change if (A) it is isplace paallel to itself (C) it is coss-multiplie by a unit vecto (B) it is otate though an abitay angle (D) it is multiplie by an abitay scala. Moe than one choice type. Which of the following is a tue statement? (A) A vecto cannot be ivie by anothe vecto (B) Angula isplacement can eithe be a scala o a vecto. (C) Since aition of vectos is commutative theefoe vecto subtaction is also commutative. (D) The esultant of two equal foces of magnitue F acting at a point is F if the angle between the two foces is º. 5. Which of the aangement of aes in Fig. can be labelle ight-hane cooinate system? As usual, each ais label inicates the positive sie of the ais. (i) (ii) (iii) (iv) (v) (vi) (A) (i), (ii) (B) (iii), (iv) (C) (vi) (D) (v) A-79 Ina Viha, Kota Rajasthan 5 Page No. #

16 PART - II : SUBJECTIVE QUESTIONS FUNCTION & DIFFERENTIATION. y = f() = Fin f(y) - -. If f() = í ì +, < î, ³ Evaluate f(), f() an f(). If f() = - then fin f {f()} + Fin the fist eivative an secon eivative of given functions w..t. the inepenent vaiable.. y = 7 + tan 5. y = ln + sin Fin eivative of given functions w..t. the coesponing inepenent vaiable. æ ö 6. y = ç + ( + ) 7. y = sin + cos sin è ø 8. y = cos sin cos 9. = ( + sec q) sin q. y = e tan. y = sin + cos - Fin eivative of given functions w..t. the espective inepenent vaiable.. p = tanq + tanq. y = sin + cos cos. y = cot + cot y Fin as a function of æ 5. y = ç è 8 ö sin ( + ) ø q 7. y = ( +) - / 8. q = -, fin 9. y = sin + sin. y = sin sin sin A-79 Ina Viha, Kota Rajasthan 5 Page No. # 5

17 . y = sin e y Fin. + y = 8 y. The aius an height h of a cicula cyline ae elate to the cyline s volume V by the fomula V = p h. (a) If height is inceasing at a ate of 5 m/s while aius is constant, Fin ate of incease of volume of cyline. (b) If aius is inceasing at a ate of 5 m/s while height is constant, Fin ate of incease of volume of cyline. (c) If height is inceasing at a ate of 5 m/s an aius is inceasing at a ate of 5 m/s, Fin ate of incease of volume of cyline.. A sheet of aea m in use to make an open tank with a squae base, then fin the imensions of the base such that volume of this tank is maimum. 5. Fin two positive numbes & y such that + y = 6 an y is maimum - 6. y = cosu, u = sin 7. y = sinu, u = cos 8. Fin y if (a) y = cosec, (b) y = sec. Fin integals of given functions. INTEGRATION - +. ( +. ( ) t ). t + t t t +. t t t 5. cot 6. ( - cot ) 7. cos q (tanq + sec q) q 8. ( ) æ ö 9. ç + è - 5 ø Integate by using the substitution suggeste in backet /. sin ( - ), (use, u = / ). csc q cot q q (a) Using u = cot q (b) Using u = csc q (a) Using u = (b) Using u = Integate by using suitable substitution.. - s s. q ( - q ) q 5. 8q q - q 6. (+ ) ( + 7. ) 8. sec ( + ) A-79 Ina Viha, Kota Rajasthan 5 Page No. # 6

18 9. tan sec æ p ö æ p ö sin(t + ). sec ç u + tanç u + u. è ø è ø t cos (t + ) 6cost. ( + sint) -5 t. 8(7 - ), (use, u = 7 ). (y + y + ) (y + y) y, (use, u = y + y + ) æ u - p ö æ u - p ö 5. csc ç cotç u 6. è ø è ø X p p 8. cos 9. sin. p Use a efinite integal to fin the aea of the egion between the given cuve an the ais on the inteval [,b], p q q. y = b -. y = VECTOR. The esultant of two vectos of magnitues A an A acting at an angle q is A. Fin the value of q?. Vecto A points N E an its magnitue is kg ms it is multiplie by the scala l such that l = secon. Fin the iection an magnitue of the new vecto quantity. Does it epesent the same physical quantity o not? 5. A foce of N is incline at an angle q to the hoizontal. If its vetical component is 8 N, fin the hoizontal component & the value of q. 6. Two vectos acting in the opposite iections have a esultant of units. If they act at ight angles to each othe, then the esultant is 5 units. Calculate the magnitue of two vectos. 7. A vecto B which has magnitue 8. is ae to a vecto A which lies along the -ais. The sum of these two vectos is a vecto which lies along the y-ais & has magnitue that is twice the magnitue of A. Fin the magnitue of A. 8. Fin the esultant of the thee vectos OA, OB an OC each of magnitue as shown in figue? 9. If A = î + ĵ an B = î + ĵ + kˆ then fin out unit vecto along A + B. The an y components of vecto A ae m an 6m espectively. The,y components of vecto A+ B ae m an 9m espectively. Fin the length of B an angle that B makes with the ais.. Ù Ù Ù Ù. If a= i+ y j & b = i + y j. Fin the conition that woul make a & b paallel to each othe.. The angle q between iections of foces A an B is 9º whee A = 8 yne an B = 6 yne. If the esultant R makes an angle a with A then fin the value of a? A-79 Ina Viha, Kota Rajasthan 5 Page No. # 7

19 DIFFERENTIATION Fin fist eivative an secon eivative of epenent vaiable.. y = -. y = +. y = + + Fin the eivative of functions using quotient ule.. Suppose u an v ae iffeentiable functions of an that u () =, u () = v () = 5 v () =. Fin the values of the following eivatives at =. (a) (uv) (b) æ u ö ç è v ø (c) æ v ö ç è u ø () (7v u). ] 5. g() = y = + 5 sin 7. y = cosec y = + cos s Fin t 9. s = + cosect -cosect. s = sin t - cost. s = tan t t. s = t sec t + t p Fin q. p = ( + cosec q) cos q. p = 5 + y Fin as a function of. cot q 5. y = 5 cos y = sin Fin the eivatives of the functions æ pt ö 7. s = sin ç è ø æ pt ö + cos ç è ø 8. = (cosecq + cotq) - 9. = (secq + tanq) -. s = sin t + cos 5t p 5p A-79 Ina Viha, Kota Rajasthan 5 Page No. # 8

20 INTEGRATION Fin an antieivative fo each function. Do as many as you can mentally.check you answe by iffeentiation.. (a) (b) (c). (a) -/ (b) -/ (c) / p p. (a) p cos p (b) cos p (c) cos + p cos p p. (a) csc cot (b) csc 5 cot 5 (c) p csc cot 5. ( + cos ) 6. (a) 6 (b) 7 (c) (a) - (b) + (c) + Evaluating Integals Check you answes by iffeentiation. æ ö 8. ( 5-6) 9. ç è t t + t. æ ø ö ç t + t è ø t. -/. æ ö ç + è ø æ. ö ç8 y - / y è y ø -. (- ) 5. - cost) ( t 6. (- 5sint) t q 7. 7 sin q 8. cos5q q 9. (- csc ) æ. ö ç sec - csc è. ø q cot q q. sec q tanq 5 q. (sec tan - sec ). (csc - csc cot ) 5. (sin - csc ) 6. ( cos - sin) 7. sin y cos y y 8. 7 y 9. + cost t. - cos6t t. ( + tan q) q csc q. csc q - sin q q. ( + ) A-79 Ina Viha, Kota Rajasthan 5 Page No. # 9

21 Checking Integation Fomulas We will see whee fomulas like these come fom. - ( - 5). ( - 5) = - 5. sec (5 - ) = 5 tan (5 ) æ -ö 6. csc æ -ö ç = cot ç è ø è ø 7. ( + ) = + 8. Right, o wong? Say which fo each fomula an give a bief eason fo each answe. (a) ( + ) = + (b) ( + ) = + (c) ( + ) = ( + ) 9. ( 7 - ) = (7 - ) 8 Evaluate Integals by substitution metho.. y 7 - y y. sin 5 7 cos.. tan sec. 5 æ ö ç -. è8 ø æ 5 ö ç7 / / - 5. sin( - 8) è ø. æ u - p ö æ u - p ö 6. csc ç cotç u 7. è ø è cot y csc y ø sec ztan z y 8. sec z z æ ö 9. cosç - t 5. t è t ø 5s + s Fin the efinite integals of following Functions 5. - / 5. / ( +) b 5. Evaluate efinite integals of following Functions p / 5. q q VECTOR. A sail boat sails km ue East, 5km 7 South of East an finally has an unknown isplacement. If the final isplacement of the boat fom the stating point is 6km ue East, the thi isplacement is.. The esultant of two vectos u an v is pepenicula to the vecto u an its magnitue is equal to half of the magnitue of vecto v. Fin the angle between u an v.. Let the esultant of thee foces of magnitue 5N, N & N acting on a boy be zeo. If sin = (5/), fin the angle between the 5N foce & N foce.. Two vectos A & B have the same magnitue. Une what cicumstances oes the vecto A + B have the same magnitue as ½ A ½ o ½ B ½. When oes the vecto iffeence A - B have this magnitue. 5. The esultant of P an Q is R. If Q is ouble, esultant is ouble in magnitue, when Q is evese, R is again ouble, fin P : Q : R. A-79 Ina Viha, Kota Rajasthan 5 Page No. #

22 Single choice type OBJECTIVES 5. In Figue, E + D C equals (A) A (B) A (C) B (D) B 6. Two foces P an Q acting at a point ae such that if P is evese, the iection of the esultant is tune though 9º. Then (A) P = Q (B) P = Q (C) P = Q (D) No elation between P an Q. 7. The sum of the magnitues of two foces acting at a point is 6 N. The esultant of these foce is pepenicula to the smalle foce an has a magnitue of 8 N. If the smalle foce is of magnitue, then the value of is (A) N (B) N (C) 6 N (D) 7 N 8. Foces popotional to AB, BC an CA act along the sies of tiangle ABC in oe. Thei esultant epesente in magnitue an iection as (A) CA (B) AC (C) BC (D) CB 9. A given foce is esolve into components P & Q equally incline to it. Then : (A) P = Q (B) P = Q (C) P = Q (D) none of these. The esultant of two foces P & P is R, if fist foce is ouble, the esultant is also ouble. Then the angle between the foces is : (A) º (B) 6º (C) º (D) 5º. The esultant of two foces acting at an angle of 5º is kg wt an is pepenicula to one of the foces. The othe foce is : (A) kg wt (B) kg wt (C) kg wt (D) kg wt. The esultant of two equal foces is ouble of eithe of the foces. The angle between them is : (A) º (B) 9º (C) 6º (D) º. A foce of 6 kg wt. an anothe of 8 kg wt. can be applie togethe to pouce the effect of a single foce of: (A) kg wt. (B) kg wt. (C) 5 kg wt. (D) kg wt.. Let u be a constant vecto an v be a vecto of constant magnitue such that v = u an u ¹. Then the maimum possible angle between u an u + v is : (A) (B) 6 (C) (D) 5 5. In Figue, E equals (A) A (B) B (C) A + B (D) ( A + B ) A-79 Ina Viha, Kota Rajasthan 5 Page No. #

23 6. In figue, D C equals (A) A (B) A (C) B Moe than one choice type (D) B 7.* Which of the following is coect? (A) The minimum numbe of vectos of unequal magnitue equie to pouce zeo esultant is. (B) When A is multiplie by, the iection of A is evese but magnitue becomes thee times. (C) The angle between A + B an A B can vay between º an 8º. (D) None of these. 8. In the Figue which of the ways inicate fo combining the an y components of vecto a ae pope to etemine that vecto? (i) (ii) (iii) (iv) (v) (vi) (A) (iii) (B) (iv) (C) (vi) (D) (i), (ii) an (v). 9*. Let a an b be two non-null vectos such that a + b a = a - b. Then the value of b may be : (A) (B) 8 (C) (D).* Given : A + B = C. If B is evese, then the esultant becomes D. Which of the following is incoect? (A) C + D = (A + B) (B) C + D = (A + B ) (C) C + D = A + B (D) C D = (A B ). A-79 Ina Viha, Kota Rajasthan 5 Page No. #

24 Section - (A) Eecise # FUNCTION & DIFFERENTIATION. 7. (C). Section - (B). s = 5 t 5 t t. sec cosec. y = 5 cos. y y y y = + 5. = + cos 6. = + e, = + e 7. y y = + -, = - 8. w w = z 6 z z + z Þ = 6z 5 z + 9. z y y = cos sin, = sin cos. = q + q q q 5 Þ q = q 8q 5 + q 6 Section - (C). y = + + Section - (D). cos sin. sin + cos. y =. (a) (b) 7 (c) 5 7 t - t + (). f (t) = (t + t - ). z = ( -) z at = = sec 5. y = ( - ) Section - (E). y = 7( ) æ ö ç -. è 7 ø y æ ö = 5 ç -. è ø - cos sin y 5. = cos() + () + e. = cos() + + e Ans cos 5 7. w cos(w+f) 8. ( + ) Section - (F). - y - y + y. y = A-79 Ina Viha, Kota Rajasthan 5 Page No. #

25 Section - (G). Section - (H) s = 8p t t. A = p. t t. 9, 8. 8 Section -(I). y = sin Section - (A). cos ( + )... INTEGRATION y = 8 (8 ). + c. + c. - + c. - + c 5. + c c 7. + c 8. + c 9. / + c. / + c. / + c. / + c. cos + c. tan + c 5. cot +c 6. sec + c 7. l n + c c c. + + c. + +c. Section - (B). / sec t. + / 6 + c. / + c. æ t ö ç - cos. è ø ( ). 6 ( ) / 5. sin 6. cos 7. cos ( ) y + cos( 8z - 5) sin (z + ).. sin ( t + ) 8 ( + ). 8 Section - (C).... e Section - (D) : Calculation of aea b b. Aea = b ( + b) = b units. + b = units. units. p / units Section - (A). V R = -5 ĵ p VECTOR. (C). (i) 5º, (ii) 5º, (iii) 5º.. º A-79 Ina Viha, Kota Rajasthan 5 Page No. #

26 Section - (B). 5 5 N, tan () W of N. (A). (B). (B) 5. (D) 6. (D) 7. (A) 8. (C) 9. (D). (D). m East. F +. 5, 5º with East Section - (C) F. km h.. ±. 5º. (A) î + ĵ 5 Section - (D) 7. 5 cos º an 5 sin º. (D).* (BC). (a) (b) î + ĵ kˆ. (a) 6 (b) 6 Single choice type Eecise # PART-I. (A). (B). (B). (B) 5. (C) 6. (A) 7. (B) 8. (A) 9. (D). (D). (A). (D). (A) Moe than one choice type. (ABD). (ABC) PART-II FUNCTION & DIFFERENTIATION.. f() =, f() =, f() = 5. Section- (B). 6 - y 7 = 7 y + sec, = tan sec 5. y y = + cos, = - sin Section - (C) 6. y = cos 8. y = sin 9. = cos q + sec q q. e (tan + sec ). sin + sin cos + cos - + cos - sin Section - (D) sec q. (+ tanq). y - csc = sec. (+ cot ) A-79 Ina Viha, Kota Rajasthan 5 Page No. # 5

27 Section - (E) æ 5. With u = ö ç æ ö y y è 8 + ç, y = u : = ø è ø u u æ = u. ç è + + ö æ ö = ç æ ö + - ç + + ø 8 è ø è ø 6. sin ( + ) cos ( + ) 7. / ( + ) 9. sin cos + cos. cos [sin {sin }] cos {sin } cos. Section - (F) y. = Section - (G) 8y - y sin cos e V. (a) = p t h t = 5p V V h (b) = ph = ph (c) = p + ph = 5p + ph t t t t t Section - (H). m 5. = & y = Section - (I) 6. sin (sin ) cos. 7. y = cos ( cos ) ( + sin ) 8. (a) cosec cosec (b) sec sec Section (A): Integation of elementy functions INTEGRATION. / + /.. t t. t t / 5. cot +C 6. + cot 7. cos q + q 8. Section - (B) ( / ) sin ( / ) 6. (a) (cot q) (b) (csc q) é ù. ê ú. ( s) /. ( q ) 5/ ë û 5 5. (q ) / 6. (-) (+ ) 7. (+ ) A-79 Ina Viha, Kota Rajasthan 5 Page No. # 6

28 8. tan æ p ö tan ( + ) 9.. sec çu + è ø. cos(t + ) -. ( + sint). (7 ) æ u - p ö. (y + y +) 5. csc ç è ø Section - (C) 6. 5 l n = æ 5 ö ln ç 7. è ø Section - (D) p. pb. Using n subintevals of length D = n b an ight en point values : b Aea = = b VECTOR. 5º. B = la = N-E = S-W No it oes not epesent the same physical quantity.. N ; 7 appo = 576 = N. P = ; Q = ( + ) 7. î + 5ĵ + kˆ , tan - 9. y = y. 7º ADVANCE LEVEL PROBLEMS PART-I DIFFERENTIATION. y = + y = 5 A-79 Ina Viha, Kota Rajasthan 5 Page No. # 7

29 . y y =, =. y y = + + ; = (a) (b) (c) () 7 5. g () = 5 ( +.5) 6. y = + 5 cos 7. csc cot - cosec t cot t 8. sin 9. (- cosec t). s = t cost -. sec t. s p = t sec t tan t +. = sing q cosec q t q. sec q 5. y y 6. With u = sin, y = u : = u u = u cos = sin (cos ) y = sin cos 5 p é æ pt ö ù 7. ê ö æ pt cos ç - sin ç ú ë è ø è ø û 8. cosecq cot q + cosec q 9. sec q sec q + tanq. INTEGRATION (cos t sin 5t) p (a) (b) (c) (a) / (b) / (c) - / p p. (a) sin p (b) sin (c) sin p + p sin. (a) csc (b) csc (5) 5 æ p ö (c) csc ç è ø 5. + sin + sin 6. (a) (b) 8 8 (c) (a) - - (b) 6 + (c) - + t t +. t + t 6. /. + /. y 8 y / sin t 6. 5 cos t q 7. cos 8. sin 5q 9. cot 5 A-79 Ina Viha, Kota Rajasthan 5 Page No. # 8

30 . - tan. csc q. sec q 5. sec tan. cot + csc 5. cos + cot 6. sin + cos 7. y sin y +C t + sin t 8.. tan q. t sin 6t y sin y + 8. tan q +. Right 5. Right 6. Right 7. Right 8. (a) Wong : (b) Wong : (c) Wong : + + = + æ C ö ç + + è ø + = ( + ) = + + æ ç è ö + ø ) =. ( + ) /. = ( + ) 9. Right. (7 y ) /.. æ ö 6 tan 8. ç - 8. è ø æ ö sin 6 ç è ø æ 5 ö ç7 - è ø æ u - p ö 5. cos ( / 8) 6. csc ç 7. (cot y) è / ø æ ö 8. sec z 9. sin ç - 5. è t ø (5s + ) / 5 5. Aea =.5 squae units 5. Aea = squae units 5. 9b 5. VECTOR. BC = ĵ = km in noth. 5.. when the angle between A & B is ; when it is 6 5. Hence P : Q : R = : : Single choice type PART-II OBJECTIVES 5. (D) 6. (A) 7. (C) 8. (A) 9. (C). (C). (C). (D). (B). (A) 5. (D) 6. (A) Moe than one choice type 7.* (ABC) 8. (ABC) 9*. (CD).* (ACD) p A-79 Ina Viha, Kota Rajasthan 5 Page No. # 9