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# Assessments for class xi

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### Assessments for class xi

1. 1. ASSIGNMENT OF TRIGONOMETRY <br /><ul><li>Special values in trigonometric functions
2. 2. There are some commonly used special values in trigonometric functions, as shown in the following table.
3. 3. Functionsin01cos10tan01cot10sec12csc21</li></ul>GENERAL SOLUTIONS OF TRIGONOMETRIC EQUATIONS<br /> 1. If sinx = 0 ⇨ x = nπ, ∀ n є Z<br /> 2. If cosx = 0 ⇨ x = (2n+1) π/2, ∀ n Є Z <br /> 3. If tanx = 0 ⇨ x = nπ, n Є Z<br /> 4. If sinx = six ⇨ x = nπ + (-1)n𝛂, ∀ n Є Z<br /> 5. If cosx = cos ⇨ x = 2nπ ±𝛂, ∀ n Є Z<br /> 6. If tanx = tan ⇨ x = nπ+𝛂, ∀ n Є Z. <br />EXAMPLE: Solve the equation: sin3θ + cos2θ = 0<br />SOLUTION: cos2θ = - sin3θ ⇨ cos2θ = cos (π2 +3θ)<br /> ⇨ 2θ = 2nπ± (π2 +3θ) , ∀ nЄZ<br /> -θ = nπ+ π2 and 5θ = nπ- π2 , ∀ nєz. <br />** Question 1 Solve: 2 sec∅ + tan∅ = 1[Solution 2 + sin∅ = cos∅ ⇨ cos∅ - sin∅ = 2 dividing by a2+b2 =2 ∵ a=1,b=1 (cos∅ - sin∅ )/ 2 = 1 ⇨ cos(п/4) cos∅ - sin(п/4) sin∅ =1 ⇨ cos(∅+п/4) = cos00 ∅+п/4 = 2nп±0, n∈Z ⇨ ∅ = 2nп – п/4.] ** Question2: If tan2A = 2tan2B + 1, prove that cos2A + sin2B =0.[Answer: L.H.S. 1-tan²A1+tan²B + sin2B = -2tan²B2(1+tan2B) + sin2B , by putting above result and simplify it.] ** Question3: Find the maximum and minimum values of sinx+cosx. [ Answer: max. value of (asinx+bcosx) = a2+b2 = √2 Mini. value of (asinx+bcosx) = - a2+b2 = - √2 Or √2[1√2 sinx +1√2 cosx] = √2 sin(x+п/4), as -1 ≤ sin(x+п/4) ≤1 ∀ x]<br /> **Question4: Find the minimum value of 3cosx+4sinx+8. [ answer is 3 as above result]<br />**Question5: ∀ x in (0,п/2), show that cos(sinx) > sin(cosx). [Answer: п/2> √2 [∵п/2 = 22/7=1.57 and √2 =1.4 We know that п/2 > √2 ≥ sinx+cosx ⇨ п/2 - sinx > cosx ⇨ cos(sinx) > sin(cosx).] ** Question6: If A = cos2x + sin4x ∀ x, prove that ¾ ≤ A ≤ 1 [ Answer: A = cos2x + sin2x. sin2x ≤ cos2x + sin2x ⇨ A ≤ 1 , A = (1 - sin2x) + sin4x = [sin2x – (½)]2 + (¾) ≥ ¾.] ** Question: (i) find the greatest value of sinx.cosx [ ½.(2sinx.cosx) ≤ ½] (ii) If sinx and cosx are the roots of ax2 – bx+c = 0, then a2 – b2 +2ac = 0. [ Hint: sum and product of roots ⇨ (sinx+cosx)2 = 1+2(c/a)] ** Question 7: If f(x) = cos2x+sec2x , then find which is correct f(x)<1, f(x)=1, 2< f(x)<1, f(x)≥2. [A.M. ≥ G.M.⇨ f(x)/2= (cos2x+sec2x)/2 ≥√ (cos2x.sec2x) , ANSWER IS f(x)≥2]<br />Question – 8 (i) Prove that cos 2π15 cos 4π15 cos 8π15 cos 14π15 = 116 . [Hint : L.H.S. ⇨ - cos 2π15 cos 4π15 cos 8π15 cos π15 [∵cos 14π15 = cos(π - π15 ) ] = - [ Sin 16A24 SinA ] , where A= π15 [∵ all angels are in G.P. , short-cut Method]<br />(ii) Prove that: tan300+ tan150+ tan300. tan150 =1. <br />(ii) [hint: take tan450 = tan(300+150)<br />Question – 9 Solve: 3cos2x - 2 3 sinx .cosx – 3sin2x = 0. [Hint: 3 cos2x - 3 3 sinx .cosx +3 sinx .cosx – 3sin2x = 0 ⇨ <br />3cosx ( cosx - 3 sinx) + 3 sinx (cosx - 3 sinx) = 0] <br />Question – 10 Prove that (1+cos π8 ) (1+ cos 3π8 ) (1+ cos 5π8 ) (1+ cos 7π8 ) = 18 .<br /> [Hint : cos 7π8 = cos( π - π8 ) = - cos π8 , cos 5π8 = cos ( π - 3π8 ) = - cos 3π8]<br />Question – 11 (i) Prove that sin200 sin400 sin600 sin800 = 3 16<br />[Hint : L.H.S. sin200 sin400 sin600 sin800 <br /> (3/2) <br />⇨ √3 2×2 (2sin200 sin400 sin800) ⇨ √3 4 [(cos200 – cos600) sin800 <br /> (ii) Prove that: cosπ7 cos2π7 cos4π7 = - 18 .<br />[Hint: let x = π7 , then 12sinx (2sinx cosx cos2x cos4x)]<br /> Sin2x <br />(iii) Prove that: tan200 tan400 tan800 = tan600.<br />[hint: L.H.S. sin200 sin400 sin800 cos200cos400cos800 solve as above method.]<br />Question – 12 If 𝛂, are the acute angles and cos2𝛂 = 3cos2β-13-cos2β , show that tan 𝛂 = √2 tan𝛃.<br /> [Hint: According to required result , we have to convert given part into tangent function By using cos2𝛂 = 1-tan²α1+tan²α ∴ we will get 1-tan²α1+tan²α = 3(1-tan²β1+tan²β)-13- 1-tan²β1+tan²β = 3-3tan²β-1-tan²β3+3tan²β-1+tan²β[<br />Question – 13 (i) Find the general solution of the following equation: 4sinxcosx+2sinx+2cosx+1 = 0 <br /> [ Hint: Above equation can be written as (4sinxcosx+2sinx) + (2cosx+1 ) = 0 ⇨ 2sinx (2cosx+1) + (2cosx+1) = 0]<br /> (ii) solve : tanx + tan2x + tan3x = tanx tan2x tan3x [Hint: take tan3x to the right side then take common]<br /> (iii) solve : tanx + tan(x+π/3)+ tan(x + 2π/3) = 3 [ use formula of tan(A+B)] Question – 14 (i) If cos(A+B) sin(C-D) = cos(A-B) sin(C+D) , then show that tanA tanB tanC + tanD = 0<br />(ii) If sinθ = n sin(θ+2𝛂), prove that tan(θ+𝛂) = 1+n1-n tan𝛂. [Hint: We can write above given result as cos(A+B)cos(A-B) = sin(C+D)sin(C-D) By C & D cosA+B+cos(A-B)cosA+B-cos(A-B) = sinC+D+sin(C-D)sinC+D-sin(C-D) <br />(ii) sin(θ+2α)sinθ = 1n , by C & D sinθ+2α+sinθsin⁡(θ+2α)-sinθ = 1+n1-n ] <br />Question – 15 Prove that tan1890 = cos36°-sin36°cos36°+sin36° [Hint: we can take both sides to prove above tan1890 = tan(1800 +90) ]Question16: If tanx+tany = a and cotx+coty = b, prove that 1/a - 1/b = cot(x+y). [Hint: L.H.S. = 1tanx+tany - 1coty+cotx = cosxcosy-sinxsinysinxcosy-cosxsiny {after simplification].<br />Question 17 Prove that (a) sin3x + sin3(2π3+x) + sin3(4π3 +x) = - 34 sin3x. [Hint: Use sin3A = 3sinA – 4sin3A ⇨ 4sin3A = 3sinA - sin3A ⇨ sin3A = ¼[3sinA - sin3A]] Question-18 Find in degrees and radians the angle subtended b/w the hour hand and the minute hand Of a clock at half past three. [answer is 750 , 5π/12] Question-19: (i) Show that 1sin10° - √3cos10° =4 [Hint: 2 [ 1/2sin10° - 3/2cos10° ] = 2 [ sin30°sin10° - cos30°cos10° ] (ii) Prove that : 1+sinx-cosx1+sinx+cosx = tan(x/2) <br /> [ Hint: use 1 – cosx = 2sin2x/2 , 1+cosx = 2cos2x/2 and sinx = 2sinx/2.cosx/2]<br />Question-20 If 𝛂 , 𝛃 are the distinct roots of acosθ + bsinθ = c, prove that sin(𝛂+𝛃) = 2aba²+b² .[Hint: If 𝛂 , 𝛃 are the distinct roots of acosθ + bsinθ = c, then acos𝛂 + bsin𝛂 = c & acos𝛃 + bsin𝛃 = c <br /> By subtracting , we get a(cos𝛂 – cos𝛃) + b(sin𝛂 – sin𝛃) = 0⇨ a(cos𝛂 – cos𝛃) = b(sin𝛂 – sin𝛃) ⇨ 2a sin α+β2 sin α-β2 = 2b cosα+β2 sinα-β2 Use tanα+β2 = b/a for finding above result]<br />ASSIGNMENT(SETS)with valuable points.<br />1. Union of sets AB ={x:xA or xB } 2. Intersection of sets AB ={x:xA and xB } 3. Complement of a set A’ = {x: xU and xA}, A’ = U-A 4. Difference of sets A-B = {x: xA, xB} and B –A = {x: xB, xA} 5. Properties of the Operation of Union. a. Commutative Law: A B = B A b. Associative Law: (AB) C = A (BC) c. Law of Identity A = A d. Idempotent law A A = A e. Law of U U A = U 6. Properties of Operation of Intersection i) Commutative Law: A B = B A ii) Associative Law: (AB) C = A (BC) iii) Law of and U A =, U A = U iv) Idempotent law A A = A<br />v) Distributive law A (B C) = (A B) (A C)<br />7. Properties of complement of sets: a. Complement laws: i. A A’ = U ii. A A’ =  b. De-Morgan’s law: i. (A B)’ = A’ B’ ii. (A B)’ = A’ B’ c. Law of double complementation: (A’)’ = A d. Laws of empty set and universal set: ’ = U and U’ =  8. Counting Theorems a. If A and B are finite sets, and A B = then number of elements<br />in the union of two sets n(AUB) = n(A) + n(B) b. If A and B are finite sets, A B = then n(AU B ) = n(A) + n(B) - n(A ∩B)<br />c. n(A B) = n(A – B) + n(B – A) + n(A B) d. n(A B  C) = n(A) + n(B) + n(C) – n(B∩C) – n(A∩B) – n(A∩C) + n(A∩B∩C) 9. Number of elements in the power set of a set with n elements =2n. Number of Proper subsets in the power set = 2n-2<br />Question: 1 If U = {1,2,3......,10} , A = {1,2,3,5}, B = {2,4,6,7}, then find (A-B)’.[Answer is {2,4,6,7...10} Question: 2 In an examination, 80% students passed inMathematics,72% passed in science and 13% failed in both the subjects, if 312 students passed in both the subjects.Find the total number of students who appeared in the examination. [Answer number of students failed in both the subjects = n(M’∩S’)=13% of x=0.13x n(U) – n{(MUS)’} = 1.52x – 312 ⇨x=480.]<br /> Question: 3 If U = {x :x ≤ 10, x∈ N}, A = {x :x ∈ N, x is prime}, B = {x : x∈ N, x is even} Write A ∩B’ in roster form. [ Answer is {3,5,7}] Question: 4 In a survey of 5000 people in a town, 2250 were listed as reading English Newspaper, 1750 as reading Hindi Newspaper and 875 were listed as reading both Hindi as well as English. Find how many People do not read Hindi or English Newspaper. Find how many people read only English Newspaper. [Answer: People do not read Hindi or English Newspaper n[(EUH)’] = n(U) – n(EUH) = 1875, people read only English Newspaper n(E’∩H) = n(E) – n(E∩ H) = 1375. ] <br />Question 5 The Cartesian product AXA has 9 elements among which are found (-1,0) and (0,1).Find the set A and the remaining elements of AXA. [Answer (-1,0) and (0,1)∈AXA ⇨ A = {-1,0,1} and AXA = {(-1,-1) ,( -1,0),(1,1),(0,1),(0,0),(0,1),(1,1),(1,0),(1,1)}]<br />Question 6 A and B are two sets such that n(A-B) = 14 + x, n(B-A) = 3x and n(A ∩B) =x. Draw the venn diagram to illustrate information and if n(A) = n(B) then find the value of x. [Answer n(A-B) = 14 + x= n(A ∩B’) = n(A) - n(A ∩B)⇨n(A) 14+2x , n(B) = 4x ⇨ x=7]<br />Question:7Let A and B be two sets, prove that:(A – B)UB = A iff B⊂A [(A ∩B’) U B=A ⇨ (AUB) ∩U =A⇨ B⊂ A If B⊂ A ,(A ∩B’) U B = (AUB) ∩U=A.] Question:8 In a survey of 100 students , the number of students studying the various languages were found to be: English only 18,English but not Hindi 23,English and Sanskrit 8, English 26, Sanskrit 48, Sanskrit and Hindi 8, no language 24.Find: (i) How many students were studying Hind? (ii) How many students were studying English and Hindi? [Hint: answer (i) 18 (ii) 3 , use venn diagram] Question: 9 In a survey of 500 television viewers produced the following informations; 285 watch football, 195 watch hockey, 115 watch basketball, 45 watch football and basketball, 70 watch football and hockey, 50 watch hockey and basketball, 50 do not watch any of three games. How many watch all the three games? How many watch exactly one of the three games?[Hint: 20 ,325] Question: 10 (i) Write roster form of {x: nn²+1 and 1≤ n ≤3 , n∈ N} (ii) Write set-builder form of {-4,-3,-2,-1,0,1,2,3,4} [ answer { ½,2/5,3/10} , {x: x∈Z , x2 <20} Question:11 If set A = {x:x=1/y, where y∈N},then which of the following belongs to A: 0, 1, 2, 2/3. Question:12 If n(A) = 3, n(B) = 6 and the number of elements in AUB and in A∩B.[ Answer: A ⊆ B ⇨n(AUB)=n(B), n(A∩B)=n(A).] SOME Difficult questions **Question: Prove that for non-empty sets (AUBUC)∩(A∩B’∩C’)’∩C’ = B∩C’. [Answer: L.H.S.⇨ (AUBUC)∩(A’UBUC)’∩C’=(A∩A’)U(BUC)∩C’=(B∩C’)U(C∩C’)=R.H.S.] **Question: Let A = {(x,y):y=ex ,x∈R} and B = {(x,y):y=e-x ,x∈R}. Is A∩B empty? If not find the ordered pair belonging to A∩B.[Answer: ex = e-x ⇨ e2x =1⇨ x=0, for x=0,y=1⇨ A and B meet on (0,1) and A∩B=∅.]**Question: A and B are sets such that n(A-B)= 14+x, n(B-A)= 3x and n(A∩B)=x,draw venn diagram to illustrate the information and if n(A)=n(B), find x and n(AUB). [Answer: n(A)=n(B) ⇨ n(A-B)+n(A∩B)=n(B-A)+n(A∩B)⇨ x=7 n(AUB)= n(A-B)+n(A∩B)+n(B-A)=49.]**Question: If A ={1}, find number of elements in P[P{P(A)}].**Question: Two finite sets have m and n elements. The number of subsets of the first Set is 112 more than that of the second set. The values of m and n are, resp.(find) [Answer: 2m-2n =112⇨ 2n(2(m-n) – 1)= 24(23 – 1).] **Question: If X={8n – 7n – 1,n∈N} and Y = {49n – 49,n∈N}. Then find the relation b/w X,Y[X⊂ Y, Y⊂ X, X= Y, X ∩ Y=∅]Answer: X=Y.[X= (1+7)n - 7n – 1,by binomial expansion⇨ 49(C(n,2)+C(n,3)7+.....+C(n,n)7(n-2)) = 49(n-1)] <br />ASSESSMENT OF RELATIONS & FUNCTIONS (FOR CLASS—XI) Level ----1<br />Q.1 (x, y) Є R |x – y| ≤ 1 then R is <br /> (i) Reflexive & transitive (ii) symmetric & transitive (iii) reflexive & symmetric (iv) an equivalence <br /> Q.2 If binary * is defined on Z as a*b = 3a – b then (2 * 3) * 4 is<br /> (i) 2 (ii) 3 (iii) 4 (iv) 5<br /> Q.3 If f: R– {7/5} -> R – {3/5} is a function defined by f(x) = 3x+45x-7 , then find f -1 .<br />Q.4 If A={a, b, c, d} , B={p, q, r, s} the which of the following are relations from A to B? Give reason.<br />(i) R1 ={q ,b),(c ,s),(d ,r)} (ii) R2 ={(a ,p),(a ,q),(d ,p),(c ,r),(b ,r)}<br /> Q.5 Write the domain & range of f(x) = 1/ (5x – 7). <br />ANSWERS (Level--1) Ans.3 x= f -1 = 4+7y5y-3 , y ≠ 3/5 Ans. 4 (q, b) Є R1 but (q, b) ∉ AXB, so R1 ⊈ AXB, R1 is not relation and R2 ⊆ AXB , R2 is a relation. Ans.5 Domain = R – {7/5}, Range = R – {0}. Level.....2 Q.1 Let f(x) = x2 and g(x) = 2x+1 be two real functions. Find (f – g)(x) , (fg)(x) , (f/g)(x).<br />Q.2 Let f be the subset of Z×Z defined by f = {(a b, a+b): a, b ЄZ}.Is f a function from Z to Z? Justify your answer.<br />Q.3 Find the domain and range of f(x) = |x-4|x-4 . Q.4 If f(x) = x3, find f1.5- f(0)1.5-0 . Q.5 If f(x) = 3x-2 , if x<0x+1,if x≥0 , find f (-1) and f (0).<br />Q.6 Let R be the relation on the set N of natural numbers defined by R = {(a, b): a+3b = 12, a, bЄ N }.Find R, domain of R and range of R.<br /> ANSWERS: (Level—2)Ans. 1 f(x) – g(x) = x2 -2x – 1, f(x).g(x) = 2x3 + x2, f/g(x) = f(x)/g(x) = x²2x+1 , x ≠ - ½.Ans. 2 We observe that 1x6 = 6 and 2x3 =6 ⇨ (1x6, 1+6) Є f and (2x3, 2+3) Є f ⇨ (6,7) Є f and (6,5) Є f , so f is not a function from Z to Z. Ans. 3 Domain of f = R – {4}, Range of f = {-1, 1} Ans. 4 (1.5)2. Ans.5 f (-1) = -5, f(0) = 1.Ans.6 R={(9,1),(6,2),(3,3)} , Domain of R = {9,6,3} , Range of R = {1,2,3}.**Example Find the domain of the function given by :<br />f(x) = x(|x|-x) [Solution : The function is in rational form. The domain of the function in the numerator is "R". We are, now, required to find the value of “x” for which denominator is real and not equal to zero. ⇒ |x|−x>0⇒ |x|>x<br />If “x” is a non-negative number, then by definition, "|x| = x". This result, however, is contradictory to the inequality given above. ⇒ x<0Domain=(−∞,0)]<br /> ASSESSMENT OF COMPLEX NUMBERS & QUADRATIC EQUATIONS (XI) Level...1<br /> Q.1 Solve X2 – (3√2 – 2i) x - 6√2i = 0 (3√2, -2i)<br /> Q.2 1+ix-2i3+i + 2-3iy+i3-i = i, find real values of x and y. <br /> Q.3 If z = (1+i1-i ), then z4 is<br /> (i) 1 (ii) -1 (iii) 0 (iv) none of them.<br /> Q.4 If z = 11-I(2+3I) , then |z| is<br /> (i) 1 (ii) 1/√(26) (iii) 5/√(26) (iv) none <br /> Q.5 If x+iy = 3+5i7-6i , then y = <br /> (a) 9/85 (b) -9/85 (c) 53/85 (d) -53/85<br /> Q.6 If z = 11-cosφ-isinφ , then Re(z) is <br /> (a) 0 (b) ½ (c) cot(φ/2) (d) (½)cot(φ/2)<br /> Level......2<br /> Q.1 Find the real values of ϴ for which the complex number 1+icosϴ1-2icosϴ is purely real.<br /> Q.2 If (1 - i) (1 - 2i)(1 - 3i)...........(1 - ni) = (x - yi) , show that 2.5.10...........(1+n2) = x2+y2 .<br /> Q.3 Prove that arg(z) = 2π – arg(z) ,z ≠0<br /> Q.4 Express in polar form: -21+i√3 . <br /> Q.5 If iz3 +z2 – z+ i = 0, then show that |z| = 1. <br />Q.6 If a+ib = c+ic-i , then a2+b2 = 1 and b/a = 2cc²-1 <br />Q.7 If x = - 5 +2√(-4) , find the value of x4+9x3+35x2 – x+4. <br /> Q.8 Show that a real x will satisfy equation 1-ix1+ix = a – ib, if a2+b2 = 1 where a, b are real.<br /> Q.9 A variable complex z is such that arg ( z-1z+1 ) = π2 , show that x2+y2 – 1=0<br /> Q.10 Find the values of x and y if x2 – 7x +9yi and y2i+20i – 12 are equal. <br />Answers of Level—2 Ans.1. Rationalise it with 1+2icosϴ , then equate imaginary part to 0, value will be 2nπ±π2 , n ЄZ. Ans.2. By taking modulus or conjugate on the both sides. Ans.3. Let z = r( cosϴ + i sinϴ) , z = r (cosϴ - i sinϴ) = r (cos(2π-ϴ) + i sin(2π-ϴ). Ans.4. (cos2π/3 +isin2 π/3) Ans 5. take i outside, make factors as z2 (z – i)+ i (z – i) = 0 ⇨ z = i, z2= -i |z| = |i| =1, |z2| =|-i|=1 ⇨ |z|2 = |z2| =1⇨ |z| = 1. Ans.6. By taking conjugate on the both sides and use the given value. Ans.7. Divide given poly. By x2 +10x +41=0 as (x+5)2= (4i)2 Remainder is -160 -> answer. Ans. 8. By C &D then we will get x = 2b1+a2²+b² .Ans 9. Assume z = x+iy, arg ( z-1z+1 ) = π2 ⇨ arg( x+iy-1x+iy+1 ) = π2 ⇨ arg( x-1+iyx+1+iy x x+1-iyx+1-iy ) = π2 ⇨ tan-1π2 = 2xyx²+y²-1 .Ans.10. x =4, 3 and y =5, 4.<br /> ϴ, x>0, y>0<br /> π - ϴ, x<0, y>0<br /> arg (z) = ϴ - π , x<0, y<0 <br /><ul><li> 2π - ϴ, x>0, y<0</li></ul> it can be used in <br /><ul><li>finding principal argument of complex numbers.
4. 4. If |z1| = |z2| = 1 then |z1+z2| = |1/z1+1/z2| ⇨
5. 5. | ( z1z1)/ z1 + (z2 z2 )/ z2| z1z1 =|z1|2 =1 and 1/ z1 =(1/ z1 ) and also |Z1| =|z1| . </li></ul>ASSIGNMENT OF SEQUENCE & SERIES<br />S∞ = a1-r , a is first term , r is the common ratio. As n->∞ rn -> 0 for |r|<1.<br /> Geometric mean between two numbers a & b is ab i.e., G2 = ab or G = ab , Harmonic mean (H.M.)= 2aba+b<br /> If a , b, c are in G.P. then b/a = c/b ⇨ b2 = ac. A – G = a+b2 - ab ≥ 0 ⇨ A ≥ G≥H(Harmonic mean) R = (ba)1n+1 <br />: if a,b,c are in A.P. then b – a = c – b ⇨ 2b = a+c , d = b-an+1 and nth term = Tn = Sn – Sn-1 Arithmetic meam between two numbers a & b is (a+b)/2 is denoted by A.M.<br />ASSIGNMENT OF SEQUENCE & SERIES<br />Question:1 Find k so that 2/3, k, 5k/8 are in A.P.<br />Question:2 If the roots of (b-c)x2+(c-a)x+(a-b) = 0 are equal, then a,b,c are in A.P. [Hint: take Discriminant D=0]<br />Question:3 There are n A.M.’s between 7 and 85 such that (n-3)th mean : nth mean is 11 : 24.Find n. [Hint: 7, a2, a3, a4,…………..an+1, 85 ⇨ an+2 = 85 ⇨ d=78n+1 same as in your ncert book problem, find n=5] Question:4 Prove that the sum of n terms of the series <br />11+103+1005+ ……..is 109(10n - 1) + n2 [HINT: Sn = (10+1)+(100+3)+ (1000+5)+……….n terms<br /> = (10+102+103+……n term)+(1+3+5+……n terms)]<br />Question:5 How many terms of the series 2+2√2 +4+…..will amount to 30+14 √2 . [Hint: use Sn = a(1-rn)1-r = 30+14 √2 ,a=2 & r =√2, then n=7] Question:6 Find the sum of (1+122) + (12 + 124) + (122+126) +…….to ∞ [Hint: (1 + 12 + 122+…..) + (122+124+… ), answer is 7/3 ,use S∞ = a1-r , a is first term , r is the common ratio.]Question:7 Show that (x2+xy+y2) , (z2+zx+x2), (y2+zy+z2) are consecutive terms of A.P. if x, y, z are in A.P. [Hint: let a = (x2+xy+y2) , b = (z2+zx+x2) ,c = (y2+zy+z2) use b-a = c-b , then you will get if (x+z)2 – y2 = y(x+y+z) ⇨ x+z = 2y] <br />Question:8 If a, b, c, d are in G.P. prove that a2-b2, b2-c2, c2-d2 are also in G.P. [HINT: take a, b= ar, c=ar2, d = ar3 , show (b2-c2)/ (a2-b2)= (c2-d2)/ (b2-c2) by putting values of a, b, c, d] <br /> Question:9 If one geometric mean G and two arithmetic mean p and q be inserted between two quantities Show that G2 = (2p-q)(2q-p). [Hint: G2 = ab, a, p, q, b are in A.P. d(common difference)= (b-a)/3 .p = a+d= 2a+b3 , q = a+2d = a+2b3 put these values in R.H.S.] Question:10 If the value of 1+2+3+…….+n = 28, then find the value of 12+22+32+………..+n2 .<br />**Question:11 If g1, g2 be two G.M.’s between a and b and A is the A.M. between a & b, then prove that g1²g2 + g2²g1 = 2A. [Hint: 2A = a+b, b/g2 = g2/g1 = g1/a ⇨ a = g12 / g2 , b = g22 /g1.]<br />**Question:12 If a is the A.M. of b, c and two geometric means between b , c and G1,G2, then prove that G13 = g23 . [ Answer: 2a = b+c , c = ar3 ⇨ r = (cb)13 , put in G1= br & G2= br2 ].**Question:13 If x = 1+a+a2+a3+……….∞ and y = 1+b+b2+b3+………. ∞<br />Then prove that 1+ab+a2b2+….. ∞ = xyx+y-1 [Answer: x = 1+a+a2+a3+……….∞ = 11-a y = 1+b+b2+b3+………. ∞ = 11-b take R.H.S. xyx+y-1 = 11-ab = (1-ab)-1 = 1+ab+a2b2+….. ∞ [ by binomial expansion (1-x)-1 = (1+x+x2+x3+…..)] **Question:14 An A.P. consists of n(odd)terms and it’s middle term is m. Prove that Sn = mn [Answer: m = mid term = T(n+1)/2 = a+(n+12 - 1)d ⇨ 2m= 2a+(n-1)d , Sn = n/2[2a+(n-1)d]=mn.] **Question:15 Sum of infinity the series 12 + 12+4 + 12+4+6 +……….. [Hint: 12+4+6+……….n terms = 12(1+2+3+………n terms) = 12n(n+1)2 =1n(n+1) ⇨ S∞=1 (11.2 +12.3+......∞ =1 as given short-cut method on blog)]. **Question:16 Prove that 23rd term of sequence 17, 1615 , 1525 , 1435 ,………is the first negative term. [Hint: a=17, d=-4/5 and let nth term be first negative term ⇨ 17+(n-1)(-4/5) < 0 ⇨ n > 89/4 ⇨ n=23] **Question:18 Prove that 1+ 23 + 63² +1033 + 1434 + ………..∞ = 3. [Answer: we can write above series as 1 + 23 [1+ 33 + 532+…….],where a=1,d=2 and r= 1/3, then use formula of combined A.P.& G.P(Arithmetic –geometric series) [a1-r + dr1-r² ] or you can do by another method S - 13 S = {1+ 23 + 63² +1033 + 1434 + ………..∞} – { 13 + 232 + 633 +1034 + 1435 + ………..∞} ⇨ 2S3 = 1+ (23 - 13) + 432 {1+ 13+ 132+…..∞} =2 ⇨ S=3 <br />**Question: 19 If there are distinct real numbers a,b,c are in G.P. and a+b+c = bx , show that x ≤ -1 or x ≥ 3. [Hint: take D ≥ 0 , a+ar+ar2 = (ar)x ⇨ r2+(1-x)r+1=0] <br />**Question:20 (i) If first term of H.P. is 1/7 and 2nd term is 1/9, prove that 12th term is 1/29. [Hint: as 1/a,1/b,1/c are in H.P. ⇨ a,b,c are in A.P. therefore first and second terms are in A.P. will be 7,9 a=7, d= 2 , then find a12] (ii) In an increasing G.P., the sum of first and last term is 66, and product of the second and last but one term is 128. If the sum of the series is 126, find the number of terms in the series. [Hint: a+arn-1 = 66 , (ar)( arn-2) = 128 and Sn = 126 ⇨ r=2 and n = 6]<br />ASSIGNMENTof permutation & combination<br />Question 1 The principal wants to arrange 5 students on the platform such that the boy SALIM occupies the second position and such that the girl SITA is always adjacent to the girl RITA . How many such arrangements are possible?<br />Question 2 When a group- photograph is taken, all the seven teachers should be in the first row and all the twenty students should be in the second row. If the two corners of the second row are reserved for the two tallest students, interchangeable only b/w them, and if the middle seat of the front row is reserved for the principal, how many such arrangements are possible?<br />Question 3 If there are six periods in each working day of a school, in how many ways can one arrange 5 subjects such that each subject is allotted at least one period?<br />**Question 4 Three married couples are to be seated in a row having six seats in a cinema hall. If spouses are to be seated next to each other, in how many ways can they be seated? Find also the number of ways of their seating if all the ladies sit together. <br />**Question 5 In an exam. there are three multiple choice questions and each question has 4 choices. Find the no. Of ways in which a student fails to get all answer correct.<br />Question 6 Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?<br />Question 7 How many numbers greater than 4,00,000 can be formed by using the digits 0, 2, 2, 4, 4, 5?<br />Question 8 A boy has three library tickets and 8 books of his interest in the library. Of these 8, he does not want to borrow Chem. part II, unless Chem. Par I is also borrowed. In how many ways can be choose the three books to be borrowed?<br />Question 9 A polygon has 44 diagonals. Find the number of its sides. Question10 In how many ways can the letters of the word PERMUTATIONS be arranged if the <br />(i) Words start with P and end with S? (ii) Vowels are all together ? (iii) T’s are together? (iv) There is no restriction? <br /> (v) P comes before S? (vi) Order of vowels remain unchanged? (vii) There are always 4 letters b/w P & S? <br />Question 11 Find the no. of per. Of n different things taken r at a time such that two specific things occur together. <br />**Question 12 There are 10 points in a plane , no three of which are in the same st. Line excepting 4 points, which are collinear. Find the (i)<br /> No. Of st. Lines obtained from the pairs of these points , (ii) no. Of ∆ that can be formed with the vertices as these points.<br /> ANSWERS WITH HINTS<br />Answer 1 . We have to arrange the remaining 4 students, according to condition two seats III, IV OR IV , V May be occupied by SITA & RITA in 4 ways, answer is 8.Answer 2 6 teachers can be arranged in the front row . The remaining 18 students can be arranged in second row ∴ total no. Of ways = 6!.(18)!.2! Answer 3 Five subjects can be allotted 5 periods out of the six periods in 6P5 ways. Now one period is left and it can be allotted to any one of the 5 subjects in 5 ways. So, total no. Of ways = 6P5 . 5 = 3600. Answer 4 (i) Let A, B, C be three married couples can be seated = 3! & can sit in 2 ways =2! ∴ req. Ways of seating = 3!.2!.2!.2!=48 .<br /> (ii) three ladies can be seated in 4 ways which are at seat number (1,2,3), (2,3,4), (3,4,5), (4,5,6). They can interchange their seats =3! Ways. Men can be seated at three remaining seats in 3! , so req. No. Of ways = 4.3!.3!= 144 . Answer 5 Req. No. Of ways= 4.4.4 – 1=63. Answer 6 Required number of ways = (210 x 120) = 25200.<br />Answer 7 ∴ total no. Of ways 90. Answer 8 ∴ total no. Of ways = 6C1 + 7C3=41. Answer 9 No. Of diagonals = nC2 - n = 44 ( nC2 = no.of st. Lines of polygon n sides). Answer 10 (i) 10!/2! (ii) consider 5 vowels as one letter (8!/2!) .(5!) (iii) consider 2 T’s as one letter = (11)! (iv) 12!/2! (v) 12 letters in 12 places as P****S****** leaving 4 places in between , Thus P & S may be filled up in 7 ways, same for S & P , remaining 10 letters are arranged in 10!/2! Ways ∴ req. No. of ways = 151200. <br />Answer 11 Req. Per. = 2! (r-1) . n-2Pr-2 . <br /> Answer 15 (i) No. Of st. Lines formed by joining 10 points , taking 2 at a time = 10C2 = 45. No. Of st. Lines joining 4 points = 4C2 = 6 , but 4 are collinear , when join pairwise give only one line. Req. No. of lines = 45 – 6 + 1=40<br /> (ii) No. of ∆s formed by joining the points , taken 3 at a time = 10C3 = 120, no. of ∆s formed by joining 4 points , taken 3 at a time = 4C3 = 4<br /> Req. No. of ∆s = 120 – 4 = 116.<br /> <br /> ASSIGNMENT OF BINOMIAL EXPANSION <br />Use the binomial theorem formula to determine the fourth term in the expansion <br />Answer<br />Results: (i) nCr + nCr-1 = n+1Cr<br />(ii)     nCx = nCy x = y or x + y = n<br />(iii)    nCr = n/ r.  n-1Cr-1<br />ASSIGNMENT <br />Question.1    The 3rd, 4th and 5th terms in the expansion of (x+a)n are respectively ‘84’, ‘280’ and ‘560’, find the value of ‘x’, ‘a’ and ‘n’.<br />Question.2  If the coefficients of (2r + 1)th term and (r + 2)th term in the expansion of (1 + x)43 are equal, find ‘r’.<br />Question3 .: Find the 4th term from the end in the expansion of  [ x32  -  2x4 ] 7<br />**Question4  Find the coefficient of x-7 in the expansion of  ( ax-1bx² )11 <br /> Question.5 Let ‘n’ be a positive integer. If the coefficients of second, third and fourth terms in (1+x)2 are in arithmetic progression, then find the value of n.<br />**Question:6      If the coefficient of ‘4’th and ‘13’th terms in the expansion of  [x2 +  (1/x)] n be equal, then find the term which independent of ‘x’.                                     <br />Question.7 How many terms are there in the expansion of <br /> [(3x + y2)9]4 ?<br /> Question.8 Find the middle term in the expansion of <br /> (2x23+ 32x2 )10<br /> Question.9 If the co=efficient of x7 and x8 in the expansion of<br /> (2+x3)n , n∊N , are equal, then find n.                                                                                               <br />Question 10 If the co-efficient of x in (x2 + kx )5 is 270 , then find k.<br />Question11 Show that 2(4n+4) – 15n – 16 is divisible by 225<br /> ∀ n∊N.    <br /> **Question12 Find the co-eff. Of x50 in the expansion of<br /><ul><li> (1+x)1000+ x(1+x)999+ x2(1+x)998+........+x1000. </li></ul> ANSWERS<br />[Hint :1    Tr+1  = nCr  xn-r. ar Putting r  =  2, 3 and 4 respectively<br />            T3 = nC2 xn-2. a2          = 84 ------------(i)<br />            T4 = nC3 xn-3 a3           = 280----------(ii)<br />and      T5 = nC4 xn-4 a4           = 560 --------(iii)<br />eqn (i) × eqn(iii) : [nC2 xn-2 a2] [nC4 xn-4 a4] =  84 × 560 or divide as you done in example of ncert. x=1 a=2 & n=7] [Hint: 2 According to the questions: 43C2r  =  43Cr+1<br />  2r + r + 1 = 43]<br />[Hint 3:    4th term from end = (7-4+2)th or 5th term from beginning. T5 = T4+1 = 7C4  (x3/2)7-4 . (-2/x4) 4 , answer is 70x]<br />[Hint:4GeneralTerm,Tr+1= 11Cr(ax)11-r  - (1/bx2)r            <br /> Tr+1 =(-1)r11Cr.(a11-r/br) x113r                                            the coefficient of x-7 in  ax- (1 / b x2) 11 is 11C6a5b-6 ]  <br /> [Hint: 5 General Term : Tr+1 = nCr  xr 2nd Term : T2 = nC1 x<br /> Coefficient  =  nC1 3rd Term :  T3 = nC2 x2 Coefficient  =  nC2        Similarly coefficient of 4th term  = nC3 These are in A. P., so. 2 nC2 = nC1   +  nC3 , answer is n= 7]<br />[Hint: 6.: T4  = T3+1   =  nC3 (x2)n-3.     1/ x3 Coefficient =  nC3<br />T13  =  T12+1  =  nC12 (x2)n-12   1 / x12   Coefficient =nC12     ,According to the question: nC3  =  nC12   n  =  12 + 3 , n  =  15 ,      r     =  10 from general term<br />T11  =  15C10 = 15!/(10! . 5!)  = (15 . 14 . 13 . 12 . 11) / (5 . 4 . 3 . 2 . 1) =  3003.( independent of x)]<br />[Hint:8 use general term , answer is 252]<br />[Hint: 9 nC7 . 2(n-7)/ 37 = nC8 . 2(n-8)/ 38 ⇨ n=55]<br />[Hint: 10 calculate 5Cr . kr . x (10 – 3r) ⇨ r = 3 ⇨ k =3 ]<br /> [ Hint: 11 write 2(4n+4) – 15n – 16 as (1+15)n+1 – 15n – 16 <br /> = n+1C0(15)0 + n+1C1(15)1+ n+1C2(15)2+.......+n+1Cn+1(15)n+1 – 15n -16]<br /><ul><li>SOL.12 It is in G.P. with ratio = x/(1+x) , n= 1001 , a= (1+x)1000 Sum = (1+x)1000[1-(x1+x1001 ]1- x1+x = (1+x)1001 – x1001 ,
6. 6. co-eff. Of x50 = 1001C50 .</li></ul>ASSIGNMENT OF STRAIGHT LINES<br />Question 1 In what ratio is the line joining the points A(4,4) and B(7,7) divide by P(-1,-1)? [Hint: use section formula after assuming ratio k:1 , k= -5/8] Question 2 Determine the ratio in which the line 3x+y – 9 = 0 divides the segment joining the points (1,3) and (2,7). [ Hint: use section formula , k=3/4] Question 3 Find the equation of the straight lines which pass through the origin and trisect the intercept of line 3x+4y=12 b/w the axes. [ Hint: Let the line AB be trisected at P and Q, then AP : PB = 1:2 , A(4,0) , B(0,3) BY using section formula we get P(8/3 , 1) AQ : QB = 2 : 1 ⇨ Q=(4/3 ,2) then equation of line OP and OQ passing through (0,0) is 3x – 8y =0 and 3x – 2y =0 resp.] Question 4 The line segment joining A(2,3), B(-3,5) is extended through each end by a length equal to its original length. Find the co-ordinates of the new ends. [ Hint: answer is x= 7, y= 1 and α = -8 , β = 7<br /> (α ,β) (-3,5) (2,3) (x,y) <br /> C B A D ]<br /> Question 5 Two opposite vertices of a square are (3,4) and (1,-1). Find the co-ordinates of other vertices. [ Hint: Let A(3,4) , C(1,-1) then slope of AC = 5/2 , M mid point of AC & BD =(2, 3/2) Let m is the slope of a line making an angle of 450 with AC i.e., m is slope of lines AD or CD. USE formula of angle b/w two lines ⇨ m= -7/3, 3/7 then equations of AD , CD are 7x+3y – 33=0 , 3x – 7y – 10 =0 and D(9/2 ,1/2) ,M is mid point of BD ⇨ B(-1/2 , 5/2)]<br />Question 6 (i) Find the co-ordinates of the orthocenter of the ∆ whose angular points are (1,2), (2,3), (4,3) (ii) Find the co-ordinates of the circumcenter of the ∆ whose angular points are (1,2), (3,-4), (5,-6). [ answer (i) (1,6) (ii) (11,2)] Question 7 Find the equation of the line through the intersection of the lines x -3y+1=0 and 2x+5y -9=0 and whose distant from the origin is √5 . [ Hint: (x -3y+1) +k(2x+5y -9)=0 -------(1) , then find distant from (0,0) on the line (1) is √5 ⇨ k=7/8 ,put in (1) Answer is 2x+y – 5=0]<br /> <br />Question 8 The points (1,3) and (5,1) are the opposite vertices of a rectangle.The other two vertices lie on the line y = 2x+c. Find c and the remaining vertices.<br /> [Hint: D(α,β) C(5,1) <br />y=2x+c <br /> M (3,2) <br /> A(1,3) B(X,2X-4) <br />M(3,2) (by mid point formula) , it lies on BD ∴ c = -4 , use<br />(AB)2 + (BC)2 = (AC)2 ⇨ x=4 or 2 ∴ B(4,4) then D (2,0), if B(2,0) then D(4,4)] <br />Question 9 The consecutive sides of a parallelogram are 4x+5y=0 and 7x+2y=0. If the equation of one diagonal be 11x+7y=9, find the equation of other diagonal. <br /> 11x+7y=9<br /> [ Hint: D C <br /> 7x+2y=0 P <br /> <br /> O 4x+5y=0 B <br /> B(5/3,-4/3) , D(-2/3,7/3) by solving equations of OB & BD and OD & BD resp. Then find point P (1/2,1/2) & equation of OC i.e, OP is y=x.] <br />Question 10 One side of a rectangle lies along the line 4x+7y+5=0. Two of vertices are (-3,1) & (1,1). Find the equation of other three sides.<br />[ Hint: <br /> D slope=-4/7 C(1,1)<br /> <br /> Slope=7/4 slope=7/4 <br /> A(-3,1) 4x+7y+5=0 B<br /> <br />Equation of BC is 7x – 4y -3=0 , equation of AD & CD are 7x – 4y +25=0 & 4x+7y=11=0 resp.] <br />Question 11 The extremities of the base of an isosceles ∆ are the points (2a,0) & (0,a). The equation of the one of the sides is x=2a. Find the equation of the other two sides and the area of the ∆. <br />[ Hint: y<br /> C<br /> <br /> B(0,a) <br /> x+2y-2a=0 <br /> o A(2a,0) X <br /> <br />by solving CA2 = CB2 ⇨ Y=(5a)/2 i.e, C is (2a,5a/2), equation of BC is 3x – 4y+4a=0 & area of ∆ACB is 5a2/2 sq.units.] <br />Question 12 One side of a square is inclined to x-axis at an angle α and one of its extremities is at origin. If the sides of the square is 4, find the equations of the diagonals of the square.<br />[ Hint Y <br /> B(h,k) <br /> (-4sinα, 4cosα) <br /> C 4 4 A(4cosα, 4sinα) <br /> <br /> M O L <br />Angle COM=900-α , angle AOL=α ] <br />Take ∆OLA , find A as OL/4=cosα & AL/4=sinα and in ∆OMC ,find point C, then find equation of OB & AC ( by using mid pointof OB & AC) equations of OB & AC are x(cosα+sinα) – y(cosα – sinα)=0 , x(cosα - sinα) + y(cosα + sinα)=4 resp.]<br />Question 13 Prove that the diagonals of the //gm. Formed by the four lines.<br /> x/a + y/b = 1 ……(i) , x/b + y/a = 1 …….(ii) , x/a + y/b = -1 ……(iii) , x/b + y/a = -1 ……..(iv) are perp. to each other.<br /> [Hint: Find all co-ordinates & for perpendicularity of diagonals show <br /> ( -aba+b , -aba+b ) D line (iii) C (aba-b , -aba-b )<br /> Slope=1 <br /> Line(iv)<br /> Slope=-1 line(ii) (-aba-b, aba-b ) A line(i) B (aba+b , aba+b ) <br /> <br />product of slopes of AC & BD = -1×1=-1] <br />Question 14 On the portion of the line x+3y – 3 =0 which is intercepted b/w the co-ordinates axes, a square is constructed on the side of the line away from the origin. Find the co-ordinates of the intersection of its diagonals. Also find the equations of its sides.<br /> [Hint: <br /> Y C <br /> <br /> D(4,3) <br /> (0,1)B 45 P(2,2) <br /> X+3y-3=0 <br /> A(3,0) <br /> <br />P is the mid point of BD angle ABD=450 , use formula tan 450 =| 3m+13-m | ⇨ m =1/2 or -2 , equations of BD ,AC, CD AD & BC are x-2y+2=0, 2x+y-6=0, x+3y-13=0, 3x-y-9=0 & 3x-y+1=0 resp.]<br />Question 15 If one diagonal of a square is along the line 8x-15y=0 and one of its vertex is at (1,2), then find the equation of sides of the square passing through this vertex.<br /> [ Hint: use formula tan 450 = m1-8151+m1815 ⇨ m1 = 23/7 , <br /> (1,2) A B <br /> m1 8x-15y=0<br /> 450 m2 =8/15<br /> D C<br /> Equations of AD & AB ( Perp. to each other) are 23x-7y-9=0 , 7x+23y-53=0]<br /> ASSIGNMENT OF CONIC SECTION<br />Question.1 Find the eqn. of the circle which passes through the origin and cuts off intercepts 3 & 4 from the positive parts of the axes resp.<br />Question. 2 Find the eqn. of circle whose radius is 5 and which touches the circle x2+y2 – 2x – 4y – 20 = 0 externally at the point (5,5).<br />Question.3 Find the eqn. of circle of radius 5 which lies within the circle x2+y2+14x+10y – 26 = 0 and which touches the given circle at the point (-1,3).<br />Question.4 A circle of radius 2 lies in the first quad. And touches both the axes. Find the eqn. of the circle with centre at (6,5) and touching the above circle externally.<br />Question.5 Find the eqn. of circle circumscribing the ∆ formed by the lines x + y = 6, 2x+y = 4 & x+2y = 5.<br />Question.6 Find the eqn. of circle drawn on the diagonals of the rectangle as the diameter whose sides are x=6, x=-3, y=3 & y=-1.<br />** Question.7 Find the eqn. of circle touching the y-axis at a distant -3 from the origin and intercepting a length 8 on the x-axis.<br />** Question.8 For what value of c will the line y=2x+c be tangent/touch to the circle x2+y2=5? <br />**Question.9(i) Find the eqns. Of tangents to the circle<br /> x2+y2 – 4x – 6y – 3 = 0 which perp. to the line 4x + 3y + 5 = 0. <br /> (ii) Find the eqns. Of tangents to the circle x2+y2 – 25 = 0 which parallel to the line 2x – y + 4 = 0. <br />**Question.10 (i) Prove that the tangents to the circle x2+y2=169 at (5,12) & (12,-5) are perp. to each other. <br /> (ii) If the lines 5x + 12y – 10 =0 & 5x – 12y – 40 =0 touch a circle C1 of diameter 6 and if the centre of C1 lies in the first quad. , find the eqn. of a circle C2 which is concentric with C1 and cuts intercept of length 8 on these lines.<br />Question.11 A circle has radius 3 & its centre lies on the y = x-1. Find the eqn. of the circle if it passes through (7,3).<br />Question.12 Find the eqn. of circles whose centre is (3,-1) and which cuts off an intercept of length 6 from the line 2x-5y+18=0. <br />ASIGNMENT ( PARABOLA, ELLIPSE & HYPERBOLA)<br />Question.1 Find the eqn. of parabola whose focus is at (-1,-2) & the directrix is x – 2y + 3=0.<br />**Qestion.2 Find the eqn. of parabola with vertex (2,-3) & focus(0,5).<br />**Question.3 Find the eqn. of parabola whose vertex is at (2,1) and directrix is x = y – 1.<br />**Question.4 Find the eqn. of parabola whose focus is at (1,1) & the tangent at the vertex is x +y -1=0.<br />**Question.5 Find the eqn. of ellipse whose axes are parallel to the coordinate axes having it’s centre at the point (2,-3) and one vertex at (4,-3) & one focus at (3,-3).<br />Question.6 Find the eqn. of ellipse whose centre is at origin , foci are (1,0) & (-1,0) and e=1/2.<br />Question.7 Find the eqn. of ellipse whose foci are ( 2,3), (-2,3) and whose semi-minor axis is √5 .<br />Question.8 Find the eqn. of hyperbola, the length of whose latus-rectum is 8 and e = 3√5 .<br />Question.9 Find the eqn. of hyp. Whose conjugate axis is 5 and the distance b/w the foci is 13.<br />** Question.10 Find the eqn. of hyp. Whose foci are (8,3), (0,3) & e=4/3.<br />Question .11 Find eqn. of hyp. Whose focus is (1,2), directrix 2x+y=1 & e=√3.<br />Question.12 If the minor axis of an ellipse is equal to the distance b/w it’s foci, prove that its eccentricity is 1/√2.<br /> ANSWERS WITH HINTS CIRCLES<br /> Answer.1 OA=3, OB=4. ∴ OL=3/2 & CL=2. BY Pythagoras Thm. (∆OLC) OC=5/2, Eqn. is (x-3/2)2 + (y-2)2=(5/2)2 Answer.2 centreC1(1,2) and radius =5 of given circle , it touches externally another circle at P(5,5) of r=5 , then by mid pt. formula centre of anther circle is (9,8), so eqn. will be (x-9)2 + (y-8)2 =52 .Answer.3 Find centreC2 of small circle , radius by using mid point formula, P(-1,3) is pt. of intersection of circles &C1(-7,-5) of big circle, eqn. is (x+4)2 + (y+1)2 = 52. Answer.4 use C1C2 = C1P+C2P ⇨ C2P=3, Eqn. is (x-6)2 + (y-5)2=32.Answer.5 Let AB , BC & CA of given eqn. , find A , B & C (7,-1), (-2,8) & (1,2) resp. (by solving given eqns. ) f=-19/2, g=-17/2, c=50 .Answer.6 eqn. of circle is x2+y2 -3x-2y -21=0.Answer.7 The length of intercepts made by the circle x2+y2+2gx+2fy+c = 0 with X & Y axes are 2g2-c & 2f2-c resp.<br />Since it touches y-axis at (0,-3) & (0,-3) lies on circle ∴ c=f2 …..(i)<br /> 9-6f+c=0 ….(ii) by solving (i) & (ii) ⇨f=3 ,c=9 ∴ 2g2-c = 8 (given)⇨ g= ±5, eqn. of circle is x2+y2±10x+6y+9=0 .Answer.8 Let the eqn. of circle x2+y2=a2 & eqn. of line be y = mx+c R (radius of circle)= | c1+m2 | =length of per. From the centre (0,0) on the line( tangent on circle)., then c=±5.Answer.9 (i) eqn. of line perp. to 4x + 3y + 5 = 0 is 3x-4y+k=0…..(i) Per. Distant from centre(2,3) on line 3x-4y+k=0 = radius =4⇨k=26 or -14 put in (i) .(ii) eqn. of line parallel to 2x – y + k = 0 , find k same as above steps K= ±5√5 , put this value in above eqn. of line.Answer.10 (i) The eqn. of tangent at point P(x1,y1) to a circle x2+y2+2gx+2fy+c=0 & x2+y2=a2 are xx1+yy1+g(x+x1)+f(y+y1)+c=0 & xx1+yy1 = a2 resp. by using above result prove that product of slopes = -1. (ii) per. Distance from centre A(h,k) of small circle(C1) on the given lines are equal to 3 ⇨ |5h+12k=10|=39 , |5h-12k-40|=39…..(i) And they are opposite in sign by solving 5h+12k-10 & 5h-12k-40=-39 And 5h+12k-10=-39 & 5h-12k-40 =39 ⇨ h=5 , k=2 & h=5 , k=-9/2 ,A lies in 1ST ∴ A=(5,2), RS(chord of C2) = 8 ⇨ RL = 4 ⇨ AR(radius of C2)=5(Pythagoras thm.), eqn. is (x-5)2 + (y-2)2 = 52. (L is pt. of intersection of RS & radius of C1). Answer.11 Let centre is (h,k) lies on given line , then we get k=h-1 ..(i)Eqn. of circle (x-h)2 +(Y-K)2 = 9 passes through (7,3) & put the value of (i) ⇨ (7-h)2 + (4-h)2 = 9 ⇨ h=4,7 ∴ k = 3,6 then find eqns. Of circleAnswer.12 C(3,-1) , eqn. of AB is 2x-5y+18=0 , AB=6∴ AL=3, then CL = √(29) (per. Distance from centre on line), AC= √(38) , find the eqn. of circle.<br />PARABOLA, ELLIPSE & HYPERBOLA <br />Answer.1 Let P(x,y) be any point on the parabola whose focus is F(-1,-2) & the directrix x-2y+3 =0. Draw PM is per. From P on directrix , by defn. FP2=PM2 ⇨ (X+1)2 + (Y+2)2 =( X-2Y+31+4 )2 Eqn. will be 4x2+y2+4xy+4x+32y+16=0.Answer.2 Let (x1,y1) be the pt. of intersection of axis and directrix. By mid point formula x1=4, y1=-11, A be the vertex & F is the focus , slope of AF is -4 , then slope of directrix is ¼ Eqn. of directrix is x-4y-48=0 , . FP2=PM2 ⇨ 16x2+y2+8xy+96x-554y-1879=0.Answer.3 same as Q.2 , eqn. of line per. To x-y+1=0 is x+y+k=0 Required eqn. is x2+y2-14x+2y+2xy+17=0. Answer.4 same as Q.3 & use mid point formula ,find S(x1,y1)=(0,0) is point of intersection of directrix & axis of parabola .Answer.5 Let 2a , 2b be the major & minor axes , it’s eqn. is (x-2)2/a2 + (y+3)2/b2 = 1, C is the centre , F1, A are the one focus & vertex resp. , CF1 = ae =1 , CA=a =2 ⇨ e=1/2 , we know that b2 = a2(1-e2) ⇨ b2 = 3, find eqn. of ellipse. Answer.6 Let P(x,y) be any point on ellipse , F & F’ are (1,0), (-1,0) resp. ⇨ FF’ = 2 ⇨ 2ae = 2 ⇨ a=2.By defn. PF+PF’ =2a , we get 3x2+4y2-12=0.Answer.7 same as Q.6 , USE b2 = a2(1-e2) , put 2ae=4,b=√5 ⇨ a=3 , Eqn. is 5x2+9y2+72x-54y+36=0 Answer.8 use b2 = a2(1-e2), eqn. is x2/25 – y2/20 = 1 Answer.9 same as Q.8 , 2ae= 13 & 2b=5, eqn. is 25x2 – 144y2 = 900.Answer.10 centre of hyper. Is mid point of line joining foci will be (4,3) , dist. b/w foci is 2ae = 8 ⇨ a=3, use b2 = a2(1-e2)⇨ b2 =7, eqn. is 7x2 – 9y2 – 56x + 54y – 32=0.Answer.11 P(x,y) be any point on hyper. & F is focus , PM is per. Dist. From P on directrix use PF = e PM , eqn. is 7x2+12xy-2y2-2x+14y-22=0.Answer.12 2b=2ae , use b2 = a2(1-e2).<br />Figures of some answers of assignment:<br /> Q.1 B <br /> C C <br /> o lL A<br /> L<br />Q.4 <br /> P C C2 (6,5) <br /> O <br /> C1(2,2) <br />Q.12 <br /> C(3,-1)<br /> A B<br /> L<br />Q.5(Ellipse) <br /> F1(3,-3) <br /> A’ A(4,-3)<br /> C(2,-3)<br /> ASSIGNMENT OF LIMITS<br /><ul><li>INTERDETERMINATE FORMS
7. 7. EXAMPLES:
8. 8. 00 ( limX->1X²-1X-1 )
9. 9. ∞∞ ( limX->∞X²+1X³+2 )
10. 10. 0.∞ ( limX->π/2(π2-X))
11. 11. ∞-∞ ( limX->11X²-1 - 2X⁴-1 )
12. 12. 1∞ ( limX->π/2(sinxtanx) )
13. 13. 00 ( limX->0(sinxtanx) )
14. 14. ∞0 (limX->0(cotxsinx) </li></ul>PROBLEM : Determine if the following function is continuous at x=0 . <br />SOLUTION : Function f is defined at x=0 since i.) f(0) = 2 . The left-hand limit <br /> = 2 . The right-hand limit = 2 . Thus, exists with ii.) . Since <br />iii.) , all three conditions are satisfied, and f is continuous at x=0 . <br />**Question limx->0ex-11-cosx [Dr. = √2|sinx/2| &limx->0ex-1x =1<br /><ul><li> |sinx/2| =+ve & -ve as x->0+ & x->0- , ⇨ limit does not exist] </li></ul>Question.1 limx->0tanx-sinxx³ [hint: put tanx=sinx/cosx, answer is 1/2]<br />Question.2 limx->0sinx-2sin3x+sin5xx [Hint: use sinx-siny =2cos(x+y)/2.sin(x-y)/2 , answer is 0]<br />Question.3 limx->π/2cotx-cosxcos³x [ answer is ½]<br />Question.4 limx->01-cosxsin²2x [ answer is 1/8]<br />Question.5 Let f(x) = 3-x² ,x≤-2ax+b , -2<x<2x22,x≥2<br />Find a,b so that limx->2f(x) and limx->-2f(x) exist.<br />[Hint: limx->2f(x) exists limx->2⁻fx = limx->2 (ax+b)= limx->2+f(x) =limx->2 x2/2 2a+b=2……(1)<br />Similarly -2a+b = -1……(2)<br />a=3/4,b=1/2]<br />Question. 6 limx->3+x[x] and limx->3-x[x] where [x] denotes the integral part of x. Are they equal?<br />[HINT: limx->3+x[x] = 3/3=1,x>3<br />limx->3-x[x] = 3/2 ,x<3]<br />Question.7 limx->5-0[-x] and limx->5+0[-x], are they equal? <br /> -5 as –x -> -5 -6 <br />Question.8 Test the continuity of function at x=0,f(x) = e1x1+e1x , x ≠0<br /> 0 , x=0 [Hint e-∞ =0,e∞ =∞]<br />**Question Determine a and b so that the function f given by <br /> f(x) = 1-sin²x3cos²x , x<п/2<br /> =a, x=п/2<br /> = b(1-sinx)п-2x², x>п/2Is continuous at x=п/2. Answer [a = 1/3 , b = 8/3] <br /> **Question Find k such that following functions are continuous at indicated point (i) f(x) =1-cos4x 8x², x≠0k , x=0 at x=0<br /> (ii) f(x) = (2x+2 - 16)/(4x – 16) , x≠2<br /> = k, x = 0 at x=2. Answer [ (i) k=1,(ii) k=1/2] <br />**Question The function f is defined as x²+ax+b , 0≤x<23x+2 , 2≤x≤42ax+5b , 4<x≤8 <br /> If f(x) is continuous on [0,8], find the values of a and b. [Answer [a=3,b=-2] <br /> ** Question If f(x) = 1+px- 1-pxx , -1≤x<02x+1x-1 , 0≤x≤1 is continuous in the [-1,1], find p. [ Answer [p=-1]<br />**Question Find the value of a and b such that the f(x) defined as f(x) = x+a2sinx , 0≤x<п/42xcotx+b , п/4≤x≤п/2acos2x-bsinx , п 2<x≤п is continuous for all values of x in [0,п]. [ ANSWER [a=п/6 , b=-п/12]<br />ASSIGNMENT OF PROBABILITY(with valuable points)<br />SOME USEFUL RESULTS:<br />1. P(A∩B’) = P(A-B) = P(A) – P(A∩ B) = P( Only A) <br /> 2. P(A’∩ B’) = P(AUB)’ = 1 – P(AUB) and P(A’UB’) = 1 – P(A∩B)<br /> 3. P(at least one) = 1 – P(None) = 1 – P(0)<br />4. For any two events A and B, P(A∩B) ≤ P(A)≤P(AUB)≤P(A)+P(B)<br />EXTRA QUESTIONS:<br />Question: Three squares of chess board are selected at random. Find the porb. Of getting 2 squares of one Color and other of a different color .<br />Answer Total 64 squares in chess board , 32 are of white and 32 of black colour , selecting as 1w & 2b or 1b &2w<br />Req. prob. = c32,2×c(32,1)×2c(64,3) = 16/21.<br />Question: A box contains 100 bolts and 50 nuts, It is given that 50% bolt and 50% nuts are rusted. Two Objects are selected from the box at random. Find the probability that both are bolts or both are rusted.<br /> Answer: A= event of getting a bolt , B= event of getting a rusted item and A ∩B= event of getting a rusted bolt. <br /> Required probability = P(A)+P(B)-P(A ∩B) = C(100,2)C(150,2) + C(75,2)C(150,2) - C(50,2)C(150,2) = 260/447.<br />Ques.1. A pair of dice is rolled. Consider the following events: A : the sum is greater than 8, B : 2 occurs on either die , C : the sum is atleast 7 and a multiple of 3. Which pair of events are mutually exclusive?<br />Ques.2. A card is drawn at random from a deck of 52 playing cards. Find the probability that it is: (i) an ace (ii) a jack of hearts (iii) a three of clubs or a six of diamonds (iv) a heart (v) any suit except heart (vi) a ten or a spade (vii) neither a four nor a club (viii) an honours card (ix) a face card (x) a spade or a face card<br />Ques.3. Four cards are drawn at random from a pack of 52 cards. Find the probability of getting: (i) all the four cards of the same suit (ii) all the four cards of the same number (iii) one card from each suit (iv) all four face cards (v) two red cards and two black cards (vi) all cards of the same color (vii) getting four aces<br />Ques.4. An urn contains 9 red, 7 white and 4 black balls. If two balls are drawn at random, find probability that: (i) both the balls are red (ii) one ball is white (iii) the balls are of the same color (iv) one is white and other is red <br />Ques.5. A five digit number is formed by the digits 1, 2, 3, 4, 5 without repetition. Find the probability that the number is divisible by 4.<br />Ques.6. One number is chosen from numbers 1 to 200. Find the probability that it is divisible by 4 or 6.<br />Ques.7. The letters of word “SOCIETY” are placed at random in a row. What is the probability that three vowels come together?<br />Ques.8. Fine the probability that in a random arrangement of the letters of the word “UNIVERSITY” the two I’s come together.<br />Ques.9. The letters of the word “MUMMY” are placed at random in a row. What is the chance that letters at the extreme are both M ?<br />**Ques. 10. A bag contains 50 tickets numbered 1, 2, 3… ,50 of which five are drawn at random and arranged in ascending order of magnitude x1 < x2 < x3< x4 <x5 . Find the probability that x3= 30 .ANSWERS: 1. A, B; B, C are mutually exclusive; but A, C are not.<br />2.(i ) 1/13 (ii ) 1/52 (iii ) 1/26 (iv ) 1/4 (v ) 3/4 (vi ) 4/13 (vii ) 9/13 (viii ) 4/13 (ix ) 3/13 (x ) 11/26<br />3. (i ) 44/4165 (ii ) 1/20825 (iii ) 2197/20825 (iv ) 99/54145 (v ) 325/833 (vi ) 92/833 (vii ) 1/270725<br />4.(i ) 18/95 ( ii) 91/190 (iii ) 63/190 (iv ) 63/190<br /> 5. 1/5 , 6. P(A)=50/200, P(B)= 33/200 & P(A∩B)=16/200, P(AUB)= 67/200 , 7. (5!.3!)/7! = 1/7 , 8. 1/5 , 9. 3/10 , 10. X1,x2 < 30 , they come from tickets 1 to 29 & x4,x5>30 come from 20 (31to 50)<br /> REQ. PROB. =( 29C2.20C2)/50C5 = 551/15134. <br />