IIT JEE (2012) (Calculus)

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L.K. Gupta (Mathematic Classes) www.pioneermathematics.

1 L.K. Gupta (Mathematic Classes) MOBILE: , 4677 PAPER B IIT JEE (0) (Calculus) TOWARDS IIT JEE IS NOT A JOURNEY, IT S A BATTLE, ONLY THE TOUGHEST WILL SURVIVE TIME: 60 MINS MAX. MARKS: 80 MARKING SCHEME In Section I (Total Marks: 4), for each question you will be awarded marks if you darken ONLY the bubble corresponding to the correct answer and zero marks if no bubble is darkened. In all other cases, minus one () mark will be awarded. In Section II (Total Marks: 6), for each question you will be awarded 4 marks if you darken ALL the bubble(s) corresponding to the correct answer(s) ONLY and zero marks otherwise. There are no negative marks in this section. In Section III (Total Marks: 4), for each question you will be awarded 4 marks if you darken ONLY the bubble corresponding to the correct answer and zero marks otherwise. There are no negative marks in this section. In Section IV (Total Marks: ), for each question you will be awarded marks for each row in which you have darkened ALL the bubble(s) corresponding to the correct answer(s) ONLY and zero marks otherwise. Thus, each question in this section carries a maimum of 6 marks. There are no negative marks in this section. NAME OF THE CANDIDATE CONTACT NUMBER L.K. Gupta (Mathematics Classes) FOR SOLUTIONS KINDLY VISIT (In latest Updates) PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH

2 L.K. Gupta (Mathematic Classes) MOBILE: , 4677 Section I (Total Marks: 4) (Single Correct Answer Type) This section contains 8 multiple choice questions. Each question has four choices (a), (b), (c), (d) out of which ONLY ONE is correct.. If f() Lt... tonterms n ( )( ) ( )( ), then range of f() is (a) {0, } (b) {, 0} (c) {, } (d) [, ] Sol: (a) Sn = + + (n ) n = n if 0 But lt n if 0 n 0 if 0 f () = lt Sn = When 0 n 0 When 0 Range of f = {0, }. Let f be a continuous function on R such that f n 4 = (sin en n n ) e then f (0) = n (a) (b) 0 (c) (d) 4 Sol: (a) As f continuous on R, so f (/4 n ) = f (0) n n and lim (sine )e n n sine = lim n n e / n Since sin e n and lim n n n = n e 0 as n, thus f (0) = PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH

3 L.K. Gupta (Mathematic Classes) MOBILE: , 4677 π. If f() = cos [], < < and [] is the greatest integer less than or equal to, π then f is equal to (a) 0 (b) (c) (d) Sol: (a) As we know π π < < If = [] = π So, f () = cos = sin f () = cos. / π π π f =. Cos = 0 π f = 0 4. If the relation between subnormal SN and subtangent ST at any point S on the curve by = (+a) is p(sn) = q (ST), then p q is equal to a 8a 8b (a) (b) (c) 7b 7b 7b Sol: (d) Here by = ( + a), differentiating both sides, we get by dy d = ( + a). dy d = ( a) by Length of subnormal and length of subtangent p q (ST) (given) (SN) p q = (by ).b {( a) }.( a) SN = y dy d = ( a) b ST = y dy d = by ( a) {using (i)and (iii)}.() () (d) 8b 7 PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH

= L.K. Gupta (Mathematic Classes) www.pioneermathematics.com MOBILE: 985577, 4677 8b {( a) } 6 7 ( a) p q = 8b 7 {using, by = (+a) } = 8b 7 5.

δ c (abc)sc (c) = (d) none of these 4s 4s 4 s Sol: (d) Since = s(s a)(s b)(s c).

4 = L.K. Gupta (Mathematic Classes) MOBILE: , b {( a) } 6 7 ( a) p q = 8b 7 {using, by = (+a) } = 8b 7 5. With the usual meaning for a, b, c and s if be the area of a triangle, then the error in, (δ ) resulting from a small error in the measurement of c (δ c) is given by δc (a) δ = (b) δ = Δ. δ c (abc)sc (c) = (d) none of these 4s 4s 4 s Sol: (d) Since = s(s a)(s b)(s c). = {s (s a) (s b)(s c)} / Taking logarithm of both sides, we get in = {In s + In (s a) + In (s b) + In (s c) dδ ds d(s a) d(s b) d(s c)... () Δ dc s dc s a dc (s b) dc s c dc But s = (a + b + c) ds d(s a) ds da Now, 0 = dc dc dc dc, d(s b) ds db 0 dc dc dc d(s c) ds and dc dc Now from (), Δ. δδ.... δc s (s a) (s b) (s c) = 4 s (s a) (s b) (s c) Hence δ = Δ 4 S s a s b s c δc Let f() = In and g() =. If c (4, 5) then c In 6 equals to 5 (a) c In 5 8 (b) (c In 4 8) (c) (c In 5 8) (d) c In 4 8 Sol: (b) Let ϕ() = In (4) 6 In, which is continuous [4, 5] and differentiable on (4, 5), so by LMVT, PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH 4

L.K. Gupta (Mathematic Classes) www.pioneermathematics.com MOBILE: 985577, 4677 ф(5) ф(4) = Φ (c), c (4, 5) 5 4 5 4 Now, Φ (5) Φ (4) = In 6 5 and Φ (c) = 5 4 c (c In 4 8 ) c In 6 = (c In 4 8) 5 7.

5 L.K. Gupta (Mathematic Classes) MOBILE: , 4677 ф(5) ф(4) = Φ (c), c (4, 5) Now, Φ (5) Φ (4) = In 6 5 and Φ (c) = 5 4 c (c In 4 8 ) c In 6 = (c In 4 8) 5 7. When the determinant cos sin cos 4 sin cos cos cos4 cos cos is epanded in power of sin, then the constant term in that epansion is (a) (b) 0 (c) (d) Sol: (c) f() = sin sin 8sin ( sin ) sin sin sin 8sin ( sin ) sin sin The required constant term is f(0) = = (0 ) = 0 8. The product of the matrices A = B = cos θ cosθsinθ and cosθsinθ sin θ cos ф cosфsinф is a null matri if ϕ = cosфsinф sin ф π π (a) (n+) (b) n π (c) n π (d) n PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH 5

6 L.K. Gupta (Mathematic Classes) MOBILE: , 4677 Sol: (a) cos θ cosθsinθ cos ф cosфsinф AB = cosθsinθ sin θ cosфsinф sin ф cos θcos ф cosθcosфsinθsinф = cos фcosθsinθ sin θsinфsinф cos θcosфsinф sin фsinθcosθ cosθcosфsinθsinф sin θsin ф cosθcos ф(cosθcosф sinθsin ф) sinθcos ф(cosθcosф+ sinфsin ф) cosθsin ф(cosθcosф sinθsin ф) sinθcos ф(cosθcosф sinθsin ф) = cosθcosфcos( θ ф) cosθsinфcos( θ ф) sinθcosфcos( θ ф) sinθsinфcos( θ ф) Clearly AB is the zero matri if cos ( Φ) = 0 i.e. Φ is an odd multiple of π. Section II (Total Marks: 6) (Multiple Correct Answer (s) Type) This section contains 4 multiple choice questions. Each question has four choices (a), (b), (c), and (d) out of which ONE or MORE may be correct. 9. Let f : D R be defined by f () = In (In (In (In ))) then (a) f () is into (b) f () is one one (c) f () is onto (d) D = (e e, ) Sol: (b, c, d) For f () to be real > 0, In > 0, In (In ) > 0 and In (In )) >0 > 0, >, > e and > e e D = (e e, ) Clearly Range of f () = R f () is onto e Also, f () = 0 if e In()In(In) f () in one one in its domain. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH 6

L.K. Gupta (Mathematic Classes) www.pioneermathematics.com MOBILE: 985577, 4677 0.

7 L.K. Gupta (Mathematic Classes) MOBILE: , The function f () =, 0 is ([] represent the greatest integer ) (a) continuous at = (b) discontinuous at = (c) Discontinuous at infinitely many points (d) Continuous everywhere Sol: (b, c) We have f () f( ) = Let > then 0 < < = 0 f() = 0 > Also, if <, then > 0 < < f () = 0 < Hence f () =, f ( ) = and f () = 0 If > f () cannot be continuous at = and = Again Let < < < < = < < = 0 f () = 0 if ϵ,, Net, Let < < < < = < < = 0 f () = 0 if ϵ,, At = ±, =. = f() = Similarly at =,, 5. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH 7

L.K. Gupta (Mathematic Classes) www.pioneermathematics.com MOBILE: 985577, 4677 f () is discontinuous at infinite nu

8 L.K. Gupta (Mathematic Classes) MOBILE: , 4677 f () is discontinuous at infinite number of points given by ϵ, n N. n. If F () = f () g() and f () g () = c, then " " f g F" f g c (a) F = c f ' g' (b) (c) F f g fg Sol: (a, b, c) Given F() = f(). g().. () Differentiating both sides w.r.t., we get F () = f (). g() + g ().f () F''' f ''' g''' (d) F f g F''' f ''' g''' F" f " g" F () = f () g () f() g() f '() g'() f g F = c f ' g' (a) is correct Again differentiating both sides w.r.t., we get F () = f ().g()+g ().f()+f ().g () F () = f ().g()+g (). f() + c Dividing both sides by F() = f(). g() then { f ().g () = c} F"() f "() g"() c F() f() g() f()g() () or F" f " g" c (b) is correct. f f g fg Again given f () g ()= c Differentiating both sides w.r.t., we get f ()g () + g () f () = 0 From (), F ()= f ().g() + g ().f()+c Differentiating both sides w.r.t., we get F () = f (). g () + f ().g()+g ().f ()+f().g ()+0 =f ().g()+g ().f()+(0) [from ()] PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH 8

9 L.K. Gupta (Mathematic Classes) MOBILE: , 4677 Now dividing both sides by F() =f() g() Then F'''() f '''() g"'() F() f() () or F"' f "' g"' F f g n n n. Let f (n) = n n P P n p n n n n n n Cn cn Cn where the symbols have their usual meanings. The f(n) is divisible by (a) n + n + (b) (n + )! (c) n! (d) none of these Sol: (a, c) n n n f(n) = n! (n )! (n )! n n! nn! (n )(n )! using 0 0 C C C = C C C = (n+)(n+)! nn! = n! = n![(n+) n] = n! (n + n + ) Thus, f(n) is divisible by n! and n + n+. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH 9

L.K. Gupta (Mathematic Classes) www.pioneermathematics.com MOBILE: 985577, 4677 Section III (Total Marks: 4) (Integer Answer Type) This section contains 6 questions.

10 L.K. Gupta (Mathematic Classes) MOBILE: , 4677 Section III (Total Marks: 4) (Integer Answer Type) This section contains 6 questions. The answer to each of the questions is a single digit integer, ranging from 0 to 9.The bubble corresponding to the correct answer is to be darkened in the Answer sheet. y f() f (y). Let f, y R If f (0) eists and equals and f(0) =, then the value of f ( ) is equal to Sol: () Put y = 0 in the given relation, then f() f(0) f f = f () + f () = f() () Now, f( h) f() ( h) f () lim lim f f() h0 h h0 h f() f(h) f() f(h) lim lim h0 h h0 h [from ()] f(h) f(0) lim = f (0) = h0 h f() = + c. Put = 0 = 0 + c c = 0 f() = + f( ) = 4. If y y = then y d y dy 7 d d Sol: (9) Given y y = Differentiate both sides with respect to, we get is equal to (y ) dy d = dy () d (y ) Again differentiating both sides with respect to, we get dy.6y d y d Using () we get d (y ) PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH 0

11 L.K. Gupta (Mathematic Classes) MOBILE: , 4677 d y 4y () d (y ) d y dy Now, 7 d d 4 = 7 (y ) (y ) = y (y ) 4y y(y ) 4 7 (y ) (y ) ( y y = ) {7y (y ) 4} ( 4y) y(y ) = 08 (y ) (y ) [From () and ()] y 54y (y ) 8 9(y ) = 9 (y ) (y ) y ( α)( α ) 8 ( α ) y = ( = y ) 9 α α 9 5. If the acute angles between the curves y = and y = at their points of intersection be such that tan = m 7 then m 0 is equal to Sol: () Given curves are y = () y = (), or y =,. (), or and y =, (4) Equating the two values of y from () and () we get = or = ± ( ) = ± From (), when = ±, y = Let A (, ) and B (, ) Here A and B are the points of intersection of curves () and() Angle of intersection between curves () and () at A (, ): dy From (), ( 0) ml d (say) at(,) PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH

L.K. Gupta (Mathematic Classes) www.pioneermathematics.com MOBILE: 985577, 4677 dy From (4), ( ) m d (say) at(,) Let be the acute angle between curves () and () at A, then m m 4 tan = mm 8 7 m = 4 6.

12 L.K. Gupta (Mathematic Classes) MOBILE: , 4677 dy From (4), ( ) m d (say) at(,) Let be the acute angle between curves () and () at A, then m m 4 tan = mm 8 7 m = 4 6. If f () = cos ( cos sin ) + sin (cos + sin ) w.r.t. then find df() 7 d Sol: () f () = cos ( cos sin ) + sin (cos + sin ) = cos cos tan + sin sin tan = cos cos tan sin sin tan sin tan f () = + cos tan cos tan sin tan cos tan sin tan sin tan cos tan f (/4) = + = Now let g() = g () = g (/4) = /5 f (/4)/g (/4) = 0/. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH

L.K. Gupta (Mathematic Classes) www.pioneermathematics.com MOBILE: 985577, 4677 7.

(a) k (b) (c) k (d) None of these Sol: () V = and the percent error in measuring = d 00 = k The percent error in measuring volume = dv V 00 Now, dv = dv = d dv V = d d dv V d 00 = 00 = k a b c 8.

13 L.K. Gupta (Mathematic Classes) MOBILE: , If there is an error of k% in measuring the edge of a cube, then the percent error in estimating its volume nk then what is value of n? (a) k (b) (c) k (d) None of these Sol: () V = and the percent error in measuring = d 00 = k The percent error in measuring volume = dv V 00 Now, dv = dv = d dv V = d d dv V d 00 = 00 = k a b c 8. Given a matri A = b c a where a, b, c are real positive numbers, abc = and A T A = I, c a b then find the value of a + b + c. Sol: A T A = I a b c a b c 0 0 b c a b c a = 0 0 c a b c a b 0 0 a b c ab bc ca ab bc ca 0 0 ab bc ca a b c ab bc ca = 0 0 ab bc ca ab bc ca a b c 0 0 a + b + c = () and PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH

14 L.K. Gupta (Mathematic Classes) MOBILE: , 4677 ab + bc + ca = 0 () Now, (a + b + c ) = (a + b + c)(a + b + c ab bc ca) + abc = (a + b + c) + () Now, (a + b + c) = a + b + c + (ab + bc + ca) = + 0 = a + b + c = (Since a, b, c are real positive numbers) Now from Eq. (), a + b + c = + = 4 Alternative solution: A T A = I A T A = I AI = (a + b + c abc) = a + b + c abc = (since a, b, c are positive real number) a + b + c abc ( A.M. G.M.) a + b + c = 4 Section IV (Total Marks: ) (Matri Match Type) This section contains questions. Each question has three statements (a, b and c) given in Column I and five statements (p, q, r, s and t) in Column II. Any given statements in Column I can have correct matching with ONE or MORE statements(s) given in Column II. For eample, if for a given question, statement B matches with the statements given in q and r, then for the particular question, against statement B, darken the bubbles corresponding to q and r in the ANSWER SHEET. 9. Observe the following columns : Column I tan (a) lim 0, where [.] denotes the greatest integer function, is equal to sin a sin (b) If the lim 0 be a finite number then a can be equal to / / e e (c) If f () = g (), / / e e 0, where g() = n h (), h () being a continuous function, then n p. q. 0 r. Column II PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH 4

15 L.K. Gupta (Mathematic Classes) MOBILE: , 4677 can be equal to (d) Let f () = +, s. g () = +, then go f () is continuous if is equal to Sol: A q; B s; C r,s,t,;d p,q,r,s,t) t. (A) Comparing the graphs of y = and y = tan, we get that if > 0, > tan > 0 0 < tan < Also, if < 0 then < tan < 0 0 tan <. So, tan lim 0 = 0 sin asin cos acos (B) lim lim 0 0 In order that the limit eits, a = 0 a =. then limit = 4sin asin 8cos acos lim lim which is a finite number (= ) if a =. (c) e e e lim lim e e e / / / / / 0 0 = and / / / e e e lim lim e e e = / / / 0 0 Clearly e limg() 0 e e / / / e / can eist if and only if lim 0 g() = 0 n PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH 5

16 L.K. Gupta (Mathematic Classes) MOBILE: , 4677 if, if (D) f () =, g () = if, if g(f()) = f() if f() f() if f() ( ) if and ( ) if and if and if and g (f()) = if if Clearly g(f()) is continuous everywhere. 0. Observe the following columns : Column I (a) The normal line to y = be /a where it crosses y ais, has slope equal to p. (b) Subnormal length to y = a b at any point (, y) is p then p y is equal a b q. a b Column II to (c) The length of sub tangent at any y point (, y) on the ellipse = is a b p p then is equal to y (d) If m be slope of tangent at any point y (, y) on the curve = then my a b is equal to r. a b s. b a PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH 6

17 L.K. Gupta (Mathematic Classes) MOBILE: , 4677 Sol: A q; B r; C p; D s (A) Point of intersection (0, b), dy d = be /a ; a dy b m = d a (0,b) Slope of normal = a b (B) d y, Subnormal = dy (c) m = d dy. y y y y y d y. a b a b y b ; ya Length of subtangent = y y y a dy b b d ya y y dy (D) = = 0 a b a b d dy b d a y PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH 7

IIT JEE (2013) (Trigonometry and Algebra)

IIT JEE (2013) (Trigonometry and Algebra) L.K. Gupta (Mathematic Classes) www.pioneermathematics.com MOBILE: 985577, 4677 PAPER B IIT JEE () (Trigonometry and Algebra) TOWARDS IIT JEE IS NOT A JOURNEY, IT S A BATTLE, ONLY THE TOUGHEST WILL SURVIVE