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- 1. MATRIX:Definition: A matrix is defined as an ordered rectangulararray of numbers. Elementary Transformation 3. Ri → Ri + kRj means multiply each element of jthrow by k and add it to the corresponding elements of ithrow. 4. In applying one or more row operations while finding A-1 by elementary row operations, we obtain all zeros in one or more, then A-1 does not exist. NOTE If A is symm. As well as skew-symm., then A isa null matrix.( if A = AT then A is Symm. And if A = - ATthen A is skew- symm.) A= is symmetric and B isskew-symmetric. NOTE: (i) If A and B are symmetric matrices, thenBA-2AB is neither symm. nor skew-symm.
- 2. (ii) If A is symm. matrix then BTAB is symm. (iii) If Aand B are symmetric matrices of same order, then AB issymm. iff AB=BA (iv) Zero matrix is both symm. and skew-symm. (v) Sum of two skew-symm. matrices is always skew-symm (vi) If A is a symm., then A3 is a symm. and if A isskew-symm., then A2 is a symm.The Determinant of a MatrixDEFINITION: Determinants play an important role infinding the inverse of a matrix and also in solving systems oflinear equations. In the following we assume we have a squarematrix (m = n). The determinant of a matrix A will bedenoted by det(A) or |A|. Firstly the determinant of a 2×2 and3×3 matrix will be introduced, then the n×n case will beshown. Consistent and Inconsistent SolutionsConsistent system : A system of equation is said to beconsistent if its solution ( one or more ) exists.Inconsistent system : A system of equation is said to beinconsistent if its solution does not exist. Working rule to check consistency:Case I When A0 System is consistence and has uniquesolution. Case II When A=0 .Find Adj(A) and then find Adj(A).B. If Adj(A) .B 0 then system is inconsistence .Case III If Adj(A).B=0 Then it may have infinite solutions then itis consistence or have no solution then it is inconsistence.Properties of Determinants:Property 1. If each element of a row (column)ofdeterminant iszero , then value of determinant is zero.Property 2. Value of a determinant is not changed bychanging the rows into columns and columns into rows.
- 3. Property 3. If two adjacent rows (columns)of a determinant areinterchanged , then the sign of the determinant is changed butits numerical value is unchanged.Property 4. If two rows (columns) are identical, then the valueof the determinant is zero.Property 5. If every element of a row (column) is multiplied bysome constant k,the value of the determinant is multiplied byk.Property 6 .If each element in any row (column) consist of twoterms , then the determinant can be expressed as the sum ofthe determinants of same order.Property 7 . The value of a determinant remain unchanged if toeach element of a row (column) be add ( or subtracted)equimultiplies of the corresponding elements of one or morerows (columns) of the determinant.Property 8. The value of the determinant of a diagonal matrixis equal to the product of the diagonal elements.Property 9. The value of the determinant of a skew-symmetricmatrix of odd order is always zero.Property 10. The determinant of a symmetric matrix of evenorder is always a perfect square.ASSIGNMENT(matrices)Qoestion.1 Using matrices, solve the following system of equations (i) x+2y+z = 1 , 2x – y+z = 5 , 3x+y – z = 0.[Hint use AX = B ⇨ X = A-1 B, |A|=15≠0 means A is invertible. Adj(A) = ,A-1 = Ans. x =1, y=-1, z=2](ii) 2x+y – 3z = 13, x + y – z = 6 , 2x – y+4z = -12.[ Ans. |A| = 9, adj(A) = , x=1, y=2, z=-3.]
- 4. (iii) 2x+y+z = 1 , x – 2y – z = 3/2 , 3y – 5z = 9.[Hint |A| = 34, adj (A) = , x=1, y= 1/2., z=-3/2.]Question.2 Use the product to solvethe equations x – y+z = 4, x – 2y – 2z = 9, 2x+y+3z = 1.[Hint take product of above two matrices, we get identity matrix,then use AB=BA = I means B is the inverse of A Or A is the inverse of B. ⇨ = 8I3 ,according to above equation let A let B (1/8) B is the inverse of A. Ans. x=3, y=-2, z=-1.]Question.3 Solve the following system of homogenous equations:2x+3y – z = 0, x – y – 2z = 0, 3x+y+3z = 0.Solution: system of homogenous equations can be written as AX = O,A= ,|A| = -33So, the system has only the trivial solution given by x=y=z=0. If |A| =0 then system has non-trivial solution.]Question.4 Show that system of equations x+y – z = 0, x – 2y+z = 0,3x+6y – 5z = 0 has non-trivial solution. Find sol.Answer: |A| = 0, ithas infinitely many solutions ∴ let z = k a arbitrary x+y = z = k , x –2y = -z = -k , 3y = 2k i.e, y = 2k/3 ⇨ x = k/3 from first equation byputting the values of x, y & z in third equation, we get 0 which is true.The required solution is z = k, y = 2k/3, x = k/3 where k is arbitrary.
- 5. Question.5 Show that system of equations3x+2y +7 z = 0, 4x – 3y - 2z= 0, 5x+9y +23z = 0 has non-trivial solution. Find the solution. [Hintx = -k, y = -2k, z = k]Question.6 The system of equations 2x+3y = 7 , 14x+21y = 49 has(a) only one solution (b) finitely many solution (c) no solution (d)infinitely many solution . [give reason]Question.6 Find the inverse (using elementary transformations) offollowing matrices: (i) A =[Hint: A-1 = ,R1↔R2,R3→R3 –3R1,R3→R3+5R2,R1→R1 – 2R2,R2→R2 – R3, R1→R1+(1/2)R3,R3→(1/2)R3](ii) A = [Hint: A-1 =,R1↔R3,R3→R3 – 3R1,R1→R1+R2,R2→1/2R2,R3→R3 – 10R2,R3→-R3,R1→R1 – 3R3,R2→R2+3/2R3](iii) A = [Hint: A-1 = ,R2→R2+R1+R3,R3→R3 –2R1,R1→R1+3R3,R3→8R3,R3→R3+R2 R3→1/25R3,R1→R1 – 10R3,R2→R2+7R3,R2→1/8R2]Question. If A is singular matrix then underwhatcondition set ofequations AX = B may beconsistent. [answer if (adjA)B = O ,theneqns. Will have infinitly many sols. Hence consistent.]
- 6. Question. If A is a square matrix of order 3 such that |adjA| = 289,find |A|. [ |A| = ±17 ∵ |adjA| = |A|n-1.] ASSIGNMENT ( WITH HINTS)(determinant)Question: (i) Let = and = , then - =0[Hint = ](ii) If f(x) = , then which is correctf(a)=0 ,f(b)=0, f(0)=0 and f(1)=0 [ Hint f(0)=0 det.(skewsymm.matrix)=0].**(iii) Let f(t) = , then is equal to 0,1,2,3.[Hint 0, = → as t→ ].(iv) There are two values of a which makes determinant = = 86, then sum of these numbers is 4,5,-4,9. [Hinta=-4, operate R2 – 2R1]Question.1 Prove that the points P (a, b+c), Q(b, c+a), R(c,a+b) are collinear.Answer : If P,Q and R are collinear then =0
- 7. By applying C2 → C2+C1 = (a+b+c) =0 ( ∵ C2, C3 are identical)Question.2 Find the value of k if the area of the triangle withvertices (-2,0),(0,4) and (0,k) is 4 square units.Answer: Area of =½ =4 ⇨ the absolute value of ½ (-2)(4 – k) = 4 ⇨ the absolutevalue of (k – 4) = 4 ⇨ |k – 4| = 4 ⇨ k – 4 = 4, -4 ⇨ k = 8, 0.Question.3 Without expanding, show that(i) =0 Operating C1 → C1+C2+C3, we get = 0.(ii) = 0 Taking out (-1) from C1,C2 and C3, weget, = (-1)(-1)(-1) = -1 =- (by interchanging rows and columns) 2 =0 ⇨ =0(iii) =0 ⇨ = = abc
- 8. ( Operating C3 → C3+C1) abc(ab+bc+ac) x 0 = 0 ( two cols. Are identical)(iv) =0 ⇨ - = - - = 0 (PassC3overthefirsttwocolumns.)(v) =R.H.S. = ( applying C1 ↔C2)= - (apply C3 → C3 – (ab+bc+ca))C1) = = (apply C2 ↔ C3 and C1 ↔ C2)Q. If a,b,c are +ve and are the pth,qth and rth terms resp. Of aG.P.,show without expanding that
- 9. ** (vi) =0 (put a=xyp-1 ,b=xyq-1 ,c=xyr-1 ,apply C1 →C1-logx.C3 ,C1→C1+C3)(vii) = (same method as given below)(viii) = ( Multiply by abc as R1 with a,R2 by b and R3 by c then divide withabc ) Find the values of: (ix) (Operate C2 → C 2.and value is 0(x)(Operating C3 → C3 – cos .C2+ sin .C1 and value is 0)Q. Prove that : (a) = (1+a2+b2)3 ( Apply C1 → (C1 - bC3) and C2 → (C2+aC3)(b) = 1 (Apply R3 → sin R3 +cos R1)
- 10. (C) = xyz(x+y+z)3 (Apply R1→ x R1, R2 → y R2, R3 → z R3 and take x,y,z common fromC1,C2,C3 resp.)(d =(a-b)(b-c)(c-a)(a+b+c)(a2+b2+c2 )( Apply C1 → (C1+C2 – 2C3)(e) = 2(a+b)(b+c)(c+a)(Apply C1 →(C1+C3 )and C2 →(C2+C3)) (f) = (a+b+c)(a2+b2+c2)(g) = 2 (apply C1→C1-C2-C3,C2→C2-C1,C3→C3-C1, C2↔C3)(h) = abc ( + + +1(ab+bc+ca+abc).(Hint taking a,b,c common from each row , applyR1→R1+R2+R3 then expand along first row).(i) = )3 Apply C1→C1-b C3, C2→C2+a C3, we get
- 11. =( expand along C1, Weget ( .**(h) Evaluate where =C(x,1) ( binomialcoefficient) Solution: = ( taking x,y,z common fromR1,R2,R3 resp. and ½!,1/3!From C2,C3 resp.) ( by formula of C(n,r) = )Apply C3→C3 + C2 and put a= x-1, b=y-1, c=z-1 = (a-b)(b-c)(c-a) = (x-y)(y-z)(z-x).Question: If x,y,z are all different and if =0, prove that xyz = -1Solution: = + = + xyz =0
- 12. (1+xyz) = 0 ⇨ (x-y)(y-z)(z-x)(1+xyz) = 0 ⇨xyz=-1 ∵ x ≠y≠ z.Question: By using properties of determinant,show that = 1+a2+b2+c2[Hint: multiply and divide by a,b,c with R1,R2,R3respectively,taking a,b,c common from C1,C2,C3 respectivelyR1→R1+R2+R3]Question: show that =(a+b+c)3.[Hint:R1→R1+R2+R3+Question(i) Using matrix method, solve the following system ofequations: + + = 4, - + = 1, + - = 2; x, y, z ≠ 0. [X=2,Y=3,Z=5,|A|=1200,adjA = ] (ii) - + =4, + - = 0, + + =2 [ x=1/2,y=-1,z=1 adjA= |A| = 10] (iii) - + = 10, + + = 10, - + =13; X, Y, Z ≠ 0.
- 13. [X=1/2,Y=1/3,Z=1/5,|A|=-9, adjA = ]Inverse Trigonometric FunctionsTable of domain and range of inverse trigonometric function Inverse-forward identities are
- 14. Forward-inverse identities areRelation between inverse functions:
- 15. For suitable values of x and y sin-1 x + sin-1 y= sin-1 (x√1-y2+ y√1-x2) sin-1 x - sin-1 y=sin-1 (x√1-y2- y√1-x2) cos-1 x + cos-1 y= cos-1 (xy- √1-x2√1-y2)
- 16. cos-1 x - cos-1 y= cos-1 (xy+√1-x2√1-y2) tan-1 x + tan-1y = ; xy<1 tan-1 x – tan-1 y= ; xy>-1 2tan-1 x= = =Trigonometry examplesExample 1:Solve the following equation:Suggested answer:
- 17. Problems – Solve Inverse Trigonometric FunctionsProblems:Problem 1:
- 18. Prove that + = , x< Solution: Let x = tan θ , then θ = tan-1 x. we have You will take R.H.S to prove the given expression R.H.S = = tan-1 (tan 3θ) = 3 θ = 3 tan-1 x = tan-1 x + 2 tan-1 x = + = L.H.S Hence, the given expression will be proved. ORWecantake L.H.S. = + = By usingtan-1 x + tan-1y = =R.H.S. Example problem 2: Solve tan-1 2x + tan-1 3x = π/4 Solution: Given: tan-1 2x + tan-1 3x = π/4 Or
- 19. tan-1 ((2x + 3x)/(1 - 2x . 3x)) = π/4 tan-1 ((5x)/(1 - 6x2)) = π/4 ∴ (5x)/(1 - 6x2) = tan π/4 = 1 or 6x2 + 5x – 1 = 0 That means, (6x – 1)(x + 1) = 0 Which gives x = 1/6 or x = -1 since x = -1 does not satisfy the equation ,the equation of the L.H.S is negative, so x = 1/6 is the only solution of the given equation. ASSIGNMENT: Question.1 Evaluate: (i) sin-1(sin10) (ii) cos-1 (cos10)(iii) tan-1(tan(-6)) Answer: (i) if –π/2 ≤x ≤π/2, then sin-1(sinx)=x but x= 10radians does not lie between –π/2 and π/2 3π – 10 lies between –π/2 and π/2 ∴ sin-1(sin(3π-10)) = 3π-10. Similarly for (ii) cos-1 (cos10) = cos-1 (cos(4π-10)) = 4π-10. [10 radians does not lie between 0 and π. ∴ 0≤4π-10≤π] For (iii) tan-1(tan(-6)) = tan-1(tan(2π-6)) = 2π-6 . { -6radians does not lie in [ –π/2 , π/2]} Question.2 If x = cos-1(cos4) and y = sin-1(sin3), thenwhich holds? (give reason) (i) x=y=1 (ii) x+y+1=0 (iii) x+2y=2 (iv) tan(x+y) = -tan7.
- 20. Question.3 if + = , then prove that -cos + = sin2 [Hint: + = - ]= ⇨ cos )2 = )2 Simplify it] Question.4 *(i) sin-1x + sin-1y + sin-1z = π, then prove that X4+y4+z4+4x2y2z2 = 2(x2y2+y2z2+z2x2) (ii) If + + = π/2 ; prove thatxy+yz+xz = 1. (iii) If + + = π , prove that x+y+z =xyz. [Hint: for (i) sin-1x + sin-1y = π - sin-1z ⇨ cos(sin-1x + sin-1 y) =cos( π - sin-1z) Use cos(A-B) = cosAcosB – sinAsinB and cos(π – )=-cos It becomes - xy = - and simply it. [Hint: for (ii) tan-1 x + tan-1y = ] Question.5 Write the following functions in the simplestform: (i) ) (ii) ) (iii) ,-a<x<a
- 21. [Hint: for (i) write cosx = cos2x/2 – sin2x/2 and 1+sinx=(cosx/2 +sinx/2)2 , then use tan(A-B), answer is π/4 – x/2 ] [ Hint: for (ii) write cosx = sin(π/2 – x) and sinx = cos(π/2– x), then use formula of 1-cos(π/2 – x)= 2sin2(π/4 – x/2) andsin(π/2 – x) = 2 sin(π/4 – x/2) cos(π/4 – x/2) Same method can be applied for (i) part also. Answer isπ/4 + x/2] [ for (iii) put x=a cos , then answer will be ½ ] Question.6 If y = )- , prove thatsiny = tan2(x/2). [Hint: y = - 2 , use formula2 = )]Question.7 (i) Prove that + + = π.(ii) Prove that )+ )+ ) = 0. [Hint: for (i) = - = , then useformula of tan-1 x + tan-1y = ] (ii) [Hint: write = ]Question.8 Solve the following equations:(i) + = .(ii) + = .(i) [Hint: write = - , put = y](ii) [Hint: use ]
- 22. Question.9 Using principal values, evaluate )+ ). [answer is π] Question.10 Show that tan( ) = and justifywhy the other value is ignored? [ Hint: put =∅ ⇨ ¾ = sin2∅ = 2tan∅/(1+tan2∅),find tan∅- ** SOME HOT QUESTIONS: 1. Which is greater tan 1 or tan-11? 2. Find the value of sin(2 ) + cos( 3. Find the value of x which satisfies the equation+ = . 4. Solve the equation: + ) = -π/2. 5. Show that tan ) = ). 6. If = - ), then find thegeneral value of . ANSWERS WITH HINTS: 1. Since 1> π/4 ⇨ tan1> 1> tan-11.
- 23. 2. sin(2 ) + cos( = sin2x + cosy ⇨ + = + = . 3. + = sin( ) by usingsin(A+B)=sinA cosB + cosA sinB ⇨ x + (1-x) = ∵sin( )= ⇨ 2x – x2 = 1 ⇨ x = 0 or ½. 4. = - - ) ⇨ 6x = sin[- - )] = -cos[ ] =-cos[ ]=- etc. 5. 1/2(2 tan )) , use formula 2 =and tan2x/2 = . 6. Put tan = t and use sin2 = and cos2 =then put t/3 = T,answer is = nπ, nπ+π/4. - = ½ ⇨ = ½ =½ = ½ (2T), then tan = 0 ,1.ASSESSMENT OF Relations & functions forclass—XII Level—1
- 24. Q.1 Let f(x) = Find f(-1) ,f(4) and f(5). Q.2 If f(x) = x2 - , then find the value of f(x) + f ( . Q.3 Let Q be the set all rational numbers and relation on Q defined byR = {(X, Y): 1+XY > 0}. Prove then R is reflexive and symmetric but nottransitive.Q.4 Write the identity element for the binary operation *defined on set R by a*b = 3ab/8 ∀ a, b ЄR.Q.5 Show that the function f: R → R defined by f(x) = sin x is neither 1-1 nor onto. Answers (Level—1) Ans.1 f (-1) = 2, f (4) = 14, f(5)= 30. Ans.2 0. Ans.3 Consider any x, y Є Q,since 1+x.x =1+x2 ≥ 1⇨ (x,x)Є R ⇨ reflexiveLet (x,y) Є R ⇨ 1+xy > 0 ⇨ 1+yx > 0 ⇨ (y,x) ЄR ⇨ symmetric.But not transitive . Since (-1, 0) and (0, 2) ЄR, because 1 > 0 by puttingvalues. But (-1, 2) ∉ R because -1<0. Ans.4 Let e be the identityelement in R. Then a *e =a =e*a ∀ aЄR ⇨ a*e =a ∀ a ЄR ⇨ e = 8/3 inR. Ans.5 f is not 1-1 because sin 0 = 0 =sin π,so the different elementso, π have same images. f is not onto because -1 ≤ sin x ≤ 1 for all x ЄR ∴the range of f =[-1,1], which is a proper subset of R. LevelQ.1 If f: R→ R is given by f(x) = (3 – x3)1/3 show that fof =Ig where Ig is theidentity map on R.
- 25. Q.2 Show that the function f: [-1, 1+ →R defined by f(x) = is 1-1 .Find the range of f. Also find the inverse of the function f: [-1, 1+ →range of f.Q.3 Show that the function f: R → R defined by f(x) = cos (5x+2) isneither 1-1 nor onto? Q.4 If f: R → R be given by f(x) = sin2x +sin2(x+π/3) +cosx .cos(x+π/3) ∀x Є R, and g: R → R be a function such that g(5/4) =1 , then prove that(gof) : R → R is a constant function. Q.5 Let R1=R – {-1} and an operation * is defined on R1 by a*b = a + b +ab ∀ a, b Є R1 . Find the identity element and inverse of an element.ANSWERS OF Level—2Ans.1 As f: R → R, fof exists and fof : R → R is given by (fof) (x) = f(f(x)) =f(3 – x3)1/3 = (3 – ((3 – x3)1/3 )3 )1/3 = (3 – (3 – x3))1/3 =x ∀ x ЄR Ans.2 f is1-1, as consider any x1, x2 Є *-1, 1] such that f(x1) = f(x2) ⇨ = ⇨x1x2+2x1 = x1x2 +2x2 ⇨ x1 = x2 For the range of f Let y = f(x) ⇨ y = ⇨ xy +2y =x ⇨ (y – 1) x= -2y ⇨ x = As x Є *-1, 1], so -1 ≤ ≤1 , but (y – 1)2 >0 , y ≠ 1⇨ -(y – 1)2 ≤ (y- 1)2≤ (y – 1)2 ⇨ -(y2 – 2y +1) ≤ -2y2+2y ≤ y2 – 2y +1 ,y≠ 1⇨ Y2 – 1 ≤ 0 and 0 ≤ 3y2 – 4y +1 ⇨ y ε *-1,1] and (y – 1/3) (y – 1) ≥ 0 ,y≠ 1 ⇨y Є *-1,1+ and y ε (-∞ ,1/3+ U *1,∞) , y ≠ 1⇨ y Є *-1,1+ and y ε (-∞ ,1/3+ U (1,∞)⇨ y Є*-1,1/3].∴ its inverse exists as f is 1-1 and onto, to find f-1 = y ⇨ xy +2y =x ⇨ 2y = x (1 – y) ⇨ x= f-1(y) = x = . Ans. 3 For f is not 1-1, 5x+2 = π/2 ⇨ x = (π – 4)/10
- 26. ∴ 5x+2 =π/2, again 5x+2 = 3π/2 ⇨ x = (3π – 4)/10, Now f ((π – 4)/10)) = cos*5((π – 4)/10) +2+ = cosπ/2 =0f ((3π – 4)/10) = cos*5((3π – 4)/10) +2+ = cos3π/2 = 0.For f is not onto, as -1 ≤ cos (5x+2) ≤1, then -1≤ y ≤1, range of f = *-1, 1]= {y : -1≤ y ≤1 - ≠ co-domain R.Ans. 4 ½[ 2sin2x +2sin2(x+π/3) +2cosx cos(x+π/3)+f (x)= ½[ 1 – cos2x +1 – cos (2x+2π/3)+ cos (2x+π/3)+cosπ/3+( As we know that 2sin2x= 1 – cos2x and 2cosA cosB= Cos(A+B) + cos(A-B).)½[5/2 – ,cos2x + cos(2x+2π/3)- + cos(2x+π/3) ⇨ ½[5/2 – 2cos(2x+π/3)cos π/3 + cos(2x+π/3)+ = 5/4 ∀ x ЄR∴ for any x Є R , we have (gof)(x) = g(f(x)) = g(5/4)=1 ,so it is constantfunction.Ans. 5 * can be shown to be a binary operation on R1 as let a ≠ -1, b ≠ -1 . a*b = a+b+ab Є R – {-1} ⇨ a+b+ab ≠ -1⇨ a(1+b)+(b+1) ≠0 ⇨ (a+1) (1+b) ≠0 ⇨ a ≠ -1 and b ≠ -1 which istrue.Now if e is the identity element, then a*e =a ⇨ a+e+ae =a ⇨ e(1+a) = 0 ⇨ e =0 or a = -1 ⇨e =0 , 0 is the identity w.r.t. *Let a’ be inverse of a, then a*a’ =0 ⇨ a+a’+aa’ = 0 ⇨ a’(1+a) = - a∴ a’ = - a/(1+a) , is the inverse of a w.r.t. *.ASSIGNMENT(continuity & differentiability) (XII)**Question 1 Determine a and b so that the function f given by f(x) = , x<п/2
- 27. =a, x=п/2 = , x>п/2 Is continuous at x=п/2. Answer [a = 1/3 , b = 8/3] **Question 2 Find k such that followingfunctions are continuous at indicated point (i) f(x) = at x=0 (ii) f(x) = (2x+2 - 16)/(4x – 16) , x≠2 = k, x = 0 at x=2. Answer [ (i) k=1,(ii) k=1/2] **Question 3 The function f is defined asIf f(x) is continuous on [0,8], find the values of a and b. Answer[a=3,b=-2]** Question 4 If f(x) = is continuous inthe [-1,1], find p. Answer [p=-1]**Question 5 Find the value of a and b such that the f(x) defined asf(x) = is continuous for allvalues of x in [0,п]. ANSWER [a=п/6 , b=-п/12]** Question 6 Prove that = -4 [ Hint: Nr. Can be written as tanx(tanx-1)(tanx+1) =- [tanx(cosx-sinx)(tanx+1)]/cosx
- 28. Cosx-sinx = cos )]**Question 7 Prove that (i) = [ Hint:put x= sinѲ] (ii) ) = -3/2. [Hint: = & useformula of ]Question 8 f(x) = , =1 & =1, then p.t.f(-2)=f(2)=1. [ Hint: =0]Question 9 [Dr. = 2|sinx/2| & =1 |sinx/2| =+ve & -ve as x→0+ & x→0- , ⇨ limit does not exist]Question 10 Show that the function f(x) is continuous at x=0. [Hint: use =1 , =1]Question11 Show that f(x) = |x-3|,x∊R is cts. But not diff. at x=3.[Hint:show L.H.lt=R.H.lt by |x-3| = x-3, if x ≥3 and –x+3, if x<3, L.hd=-1≠1(R.h.d)Question 12 Discuss the continuity of the fn. f(x) = |x+1|+|x+2|, at x =-1 & -2 [Hint:f(x) =yes cts. At x=-1,-2Question 13 Find the values of p and q so that f(x)= is diff. at x = 1. [ answer is p=3 , q=5]
- 29. Question 14 For what choice of a, b, c if any , does the function F(x)= becomes diff at x=1,2 & show that a=b=c=0. Question15For what values a,b f(x)= is diff.at x=0 [Hint: L.H.d= 2 =1& R.H.d=a, since f‘(x)=0exists, a=2,b∊R] ASSIGMENT OF DIFFERENTITION Question 1 Show that y = aex and y = be –x cut at right angles aab=1 [ by equating , we get ex = ⇨ x= ½ log ( b/a) , find slopes(dy/dx) at pt. of intersection is (½ log ( b/a , ). Question 2 (i) If y +x = 1, prove that dy/dx= (-1)[Hint: put y=sinѲ & x= sin , use formula of sin((ii) If cos-1 ) = tan-1a , find dy/dx.[let cos(tan-1a )= k(constant), then assume c= 1-k/1+k , dy/dx= y/x](iii) If = , prove that dy/dx =(iv) If xm.yn = (x+y)m+n, then find dy/dx. [ y/x]Question 3 Differentiate w.r.t. x :**(i) Using logarithmic differentiation, differentiate: Solution:
- 30. x logx(ii) + (iii) (iogx) + xQuestion 4 (i) If = , prove that dy/dx =(ii) If f(1)= 4,f’(1)=2,find d/dx{logf(ex)} at the point x =0.[1/2](iii)If y = ,show that (2y – 1)dy/dx =1. a (t+1/t)(iv) If x = (t+1/t) , y= a where a>0,a≠1,t≠0, find dy/dx.[Hint: take dy/dt & dx/dt , then find dy/dx = ylogy/ax. ]Question5(i)differentiate: Sec-1(1/(2x2 – 1)),w.r.t.sin-1(3x –4x3).[Hint: let u=1st fn. & v= 2nd fn. , find du/dv = 1](ii)differentiate: tan-1 ( ),w.r.t. sin-1 ( ) if 1<x<1;x≠0[ du/dv= ¼, put x=tanѲ⇨ u=Ѳ/2, v=2Ѳ , u&v as assumed above- (msin-1x)(iii) If y = e , show that (1-x2)y2 – xy1 – m2y= 0.Question 6 Water is driping out from a conical funnel, at the uniform rateof 2cm3/sec. through a tiny hole at the vertex at the bottom. When theslant height of the water is 4cm.,find the rate of decrease of the slantheight of the water given that the vertical angle of the funnel is 1200 .[Hint: Let l is slant height ,V = 1/3. .l( /2)2.l/2= l3/8(vertical angle willbe 600 (half cone), take dv/dt=-2cm3/sec. ⇨l=-1/3 cm/s.]**Question 7(i) Let f be differentiable for all x. If f(1)=-2 and if f `(x) ≥2 ∀x∊[1, 6], then prove f(6) ≥8.[ use L.M.V.Thm.,f`(c)≥2,c∊[1, 6]](ii) If the function f(x)= x3 – 6x2+ax+b defined on [1, 3] satisfies therolle’s theorem for c = (2 +i)/ , then p.t. a = 11 & b∊R.[Hint: Take f(1)=f(3) , use rolle’s thm. f`(c)=0⇨ a=11]Question 8 (i) Show that f(x)= x/sinx is increasing in (0, п/2)
- 31. *HINT: f’(x)>0 , tanx >x+(ii) Find the intervals of increase and decrease for f(x) = x3 + 2x2 – 1.[Answer is increasing in (-∞, -4/3)U(0, ∞) & decreasing in (-4/3, 0)](iii) Find the interval of increase&decrease for f(x) =log(1+x)-(x/1+x) ORProve that x/1+x < log(1+x) < x for x > 0.[ Hint: f(x)strictly ↑ in ,0, ∞) , x>0 ⇨f(x)>f(0), let g(x)=x-log(1+x)g(x)>0 ↑ in ,0,∞) & f(x) ↓ in (-∞, 0-.-(iv) For which value of a , f(x)=a(x+sinx)+a is increasing.,Hint: f’(x) a(1+cosx) ≥0 ⇨ a>0 ∵ -1≤cosx≤1-**Question 9 Problem: Using differentials, approximate the expressionSolution: We letHence, x = 0.05 and y = /4.Differentiating, we obtainSubstituting, we getQuestion 10 For the curve y = 4x3 − 2x5, find all the points at which the tangentspasses through the origin. [Hint: eqn. Of tangent at (x0,y0) , put x,y=0,(x0,y0)lies on given curve]
- 32. Question 11 Find the stationary points of the function f(x) = 3x4 –8x3+6x2 and distinguish b/w them. Also find the local max. And localmini. Values, if they exist.* f’(x)=0⇨ x=0,1 f has local mini. At x=0∵f’’>0 & f’’(1)=0, f has pointof inflexion at x=1,f(1)=1]Question 12 Show that the semi – vertical angle of right circular coneof given total surface area and max. Volume is sin-1 1/3.[Hint: take S=Пr(l+r) ⇨ l= S/пr – r , take derivative of V² OR can usetrigonometric functions for l & h]Question 13 A window has the shape of a rectangle surmounted by anequilateral ∆. If the perimeter of the window is 12 m., find thedimensions of the rectangle so that it may produce the largest area ofthe window.[Hint: let x=length, y=breadth, then y=6 – 3y/2, A= XY+ X2 /4, takederivative of A & it is max. ,x=4(6+ )/11 ,y=6(5 )/11]ASSIGNMENT OF INTEGRATIONQuestion 1 Evaluate: (i)** Integrate .[ Use the power substitution Put ] ** (iii) Integrate . [ Use the power substitution Put ] (iii) [answer is (2 - √2)/3 ](iv) ∫ dx [multiply÷ by sin(a-b)] (v) dx[multiply & divide by ] (Vi)∫ dx [by partial fraction](v) dx [ use ∫ex(f(x)+f’(x))dx+ (vi) dx [put sinx= , cosx = , then put t=tanx/2. Answer is – ]
- 33. (vii) dx [ + = ∫+ve dx+∫ -ve dx ,answer is 5/2п- 1/п2] (viii) [ write sin2x = 1-cos2x answer is п/6](ix) + dx * answer is √2 ] (x) dx [ put x=atan2Ѳ, answer is a/2(п-2) ] (xi) dx [ use property dx = dx , dx = dx ∵f(2a-x) = f(x) , then put t=tanx, answer isп²/2√2 - (xii) dx , where f(x) =|x|+|x+2|+|x+5|. [ dx + dx , answer is 31.5 ] (xiii) Evaluate dx [use (f(x)+f’(x))dxQuestion 2 Using integration, find the area of the regions: (i) { (x,y): |x-1| ≤y≤ }(ii) *(x,y):0≤y≤x2+3; 0≤y≤2x+3; 0≤x≤3+[(i) A= dx- dx - dx = 5/2 [+ ] – ½ ] [(ii) dx + dx , answer is 50/3](iii) Find the area bounded by the curve x 2 = 4y & the line x = 4y – 2.[A = dx - dx = 9/8 sq. Unit.]**(iv) Sketch the graph of f(x) = ,evaluate dx[hint: dx = dx + dx = 62/3.]**Question 3 evaluate dx [ mult. & divide by , put 1+x=A.(d/dx)(x2+x)+B ,find A=B=1/2, integrate]Definite integral as the limit of a sum , use formula : dx , where nh=b-a & n→∞ Question 4 Evaluate ) dx (ii) dx[ use = 1 for part (i) , use formulas of special sequences, answer is 6]
- 34. Some special case :(1) Evaluate: [ put x+1=t²] (2) [ put x+1 = t² ](3) Evaluate: (4) Evaluate: [ put x=1/t for both](5) Evaluate: [ divide Nr. & Dr. By x2 , then write x²+1/x²=(x-1/x)² +2according to Nr. , let x-1/x=t](6) Evaluate dx [ let x=A(d/dx) ( 1+x-x²) +B](7) Integrating by parts evaluate =(8) Evaluate dx = dx [ put sinx=Ad/dx(sinx+cosx)+B(sinx+cosx)+CIf Nr. Is constant term then use formulas of sinx,cosx as Ques. No. 1 (vi) part]
- 35. ASSESSMENT OF DIFFERENTIAL EQUATIONS FOR CLASS—XI Level--1 Q.1 Find the order and degree of the following differential equations.State also whether they are linear or non-linear. (i) X2( )3 + y ( )4 y4 =0. (ii) = .Q.2 Form the differential equation corresponding to y2 = a (b – x)(b+ x)by eliminating parameters a and b.Q.3 Solve the differential equation (1+e2x) dy + (1+y2) ex dx = 0, whenx= 0, y =1.Q.4 Solve the differential equation: = .Q.5 Verify that y = A cosx – Bsinx is a solution of the differentialequation + y = 0.Answers of Level—11. (i) order =2 , degree=3 , non linear( because degree is more than 1 ), (ii) order 2 , degree 3 , non-linear .2. y2 = a(b – x)(b+ x) = a (b2 – x2), 2y =-2ax ⇨ y = -ax, againdifferentiate Y + )2 = -a , by using the value of a from abovestep , we will get , x{ Y + )2 } = y .3. = - , Integrating both sides, we get =-∫ , put ex = t⇨ =- +c Using x=0, y=1, we have y = 1/ex. 4. y = 2 tan(x/2) – x +c , put tan(x/2) =5. = - A sinx – B cosx , = - A cosx + B sinx = -y.Level---2
- 36. Q.1 Solve: y dx + x log ( ) dy – 2x dy = 0 .Q. 2 Which of following transformations reduce the differentialequation + = into the form + P(x) u = Q(x) ? (i) u = (ii) u = ex (iii) u = -1 (iv) u = 2Q.3 Solve: +xy = xy3Q.4 Solve: = +Q.5 Solve: +x = x3 cos2yAnswers of Level ---2 1. put x = vy , answer = 1+ log ( ) = ky .2. (iii) differentiate w.r.t. x = - . , put the value of inthe given differential equation. 3. put = t, answer is =1+c . 4.put x+y = v, answer is = x + c.5. put tan y = v, I.F. = , also use = 2siny cosy.ASSESSMENT OF PROBABILITY FOR CLASS –XIILevel—1 Q. 1 If the mean and variance of a binomial distribution are 4 and 4/3respectively, find P(X 1). Q. 2 If P(A) = 3/8 , P(B) = ½ and P(A∩B) = ¼ , find P( / ). Q.3 A bag contains 4 white and 2 black balls. Another bag contains 3white and 5 black balls. If one ball is drawn from each bag, find the probability that (i) Both are white balls. (ii) One is white and one is black.Q.4 If A and B are independent events and P(A∩B) = 1/8, P( A’ ∩ B’)=3/8 , find P(A) and P(B).
- 37. Q.5 The probabilities of P, Q and R solving a problem are ½, 1/3 and ¼respectively. If the problem is attempted by all simultaneously, find the probability of exactly one of themsolving it.Answers of Level—1 Ans.1 np = 4, npq = 4/3 ⇨ q=1/3 ⇨ p = 1 - 1/3= 2/3 ⇨ n=6 ⇨ P(X 1) =1 – C(6,0) (2/3)0 (1/3)6 = 1 - = . Ans.2 P( / ) = P( ∩ ) /P(B) = = = ¾. Ans.3 (i) P(A ∩ B) = P(A).P(B) = (2/3).(3/8) [A,B are independentevents] (ii) P(A’ ∩ B) + P(A ∩ B’) = P(A’).P(B)+P(A).P(B’) =( 1/3).(3/8)+(2/3).(5/8)=13/24. *A’, B are indep. Events, B’ A are indep. events+,where A = drawing a white ball from first bag. B= drawing a same ballfrom second bag.A’ = drawing a black ball from first bag and B’=drawing from second bag. Ans.4 P(A∩B) = P(A).P(B) = 1/8 let x=P(A),y= P(B), P( A’ ∩ B’) =3/8 = P( A’) .P( B’) =(1- X)(1 – Y) ⇨ X+Y – XY = 5/8 ⇨X=1/2 , Y= ¼.Ans. 5 P(A’)=1/2 ,P(B’) = 1-1/3=2/3 , P(C’)=3/4 ∴ Req. Prob. =P(A)P(B’)P(C’) = P(A’)P(B)P(C’)+ P(A’)P(B’)P(C) *A,B,C are indep. events+= (1/2)(2/3)(3/4)+(1/2)(1/3)(3/4)+(1/2)(2/3)(1/4)=11/24. Level---2Q.1 If A ∩ B = ф, show that P(A/B) =0, where A and B are possibleevents.Q.2 A pair of dice is thrown if the sum is even, find the probability thatat least one of the dice Shows three.Q.3 Let X denotes the number of hours you study during a randomlyselected school day.The probability that X can take the value x, has thefollowing form, where k is some unknown constant P(X=0)=0.1 andP(X=x) =
- 38. (i) Find k. (ii) What is the probability that you study at least two hour?Exactly two hour? At most two hours?Q.4 Six dice are thrown 729 times. How many do you expect at leastthree dice to show a 5 or 6? Q.5 In a class; 5% of the boys and 10% of the girls have an I.Q. of morethan 150. In this class 60% of the students are boys. If a student isselected at random and is formed and is found To have an I.Q. of morethan 150, find the probability that the student is a boy.Answers of Level—2Ans.1 A and B are possible events ⇨A ≠ ф⇨ P(A)≠ 0 , P(B) ≠0 ButA∩B = ф ⇨ P(A∩B) = P(ф) = P(A/B) = =0. Ans. 2n(S)=36, n(A)=18 Out of these 18, the cases which at least one dieshows up 3 are (1, 3),(3,1),(3,3),(3,5),(5,3) Requiredprobability=5/18. Ans.3 X 0 1 2 3 4 P(X) 0.1 K 2K 2K K(i) k=0.15 (ii) 0.75, 0.3, o.55.Ans.4 P(success)= 2/6=1/3 ∴ q=2/3 P(x success i.e., getting a 5 or 6)= C(6, x) Px q6-x P(at least threesuccesses in six trials) = P(x≥3)=1 – [p(0)+p(1)+p(2)]By using above result we get 1 – (16/81)(31/9) = 233/729 ∴required answer is 233/729x729=233. Ans.5 Let E1: The studentchosen is a boy. P(E1)=60/100 ∴ E2: ........................................girl. P(E2)= 40/100 E1, E2 are mutually exclusive. A: a student has an I.Q. ofmore than 150. P(A/ E1)= 5/100, P(A/ E2)= 10/100 By Baye’stheorem P(E1/A) = 3/7.

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