IIT JEE (2011) PAPER-B

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1 L.K.Gupta (Mathematic Classes) MOBILE: , 4677 IIT JEE () (Integral calculus) TOWARDS IIT- JEE IS NOT A JOURNEY, IT S A BATTLE, ONLY THE TOUGHEST WILL SURVIVE TIME: 6 MINS MAX. MARKS: 75 MARKING SCHEME PAPER-B. For each question in Section I : you will be awarded 5 marks if you have darkened only the bubble corresponding to the correct answer and zero mark if no bubbles are darkened. In all other cases, minus two ( ) mark will be awarded.. For each question in Section II : you will be awarded marks if you darken the bubble corresponding to the correct answer and zero mark if no bubble is darkened. No negative marks will be awarded for incorrect answers in this Section.. For each question in Section III : you will be awarded marks if you darken only the bubble corresponding to the correct answer and zero mark if no bubbles are darkened. In all other cases, minus one ( ) mark will be awarded. 4. For each question in Section IV : you will be awarded marks for each row in which you have darkened the bubble (s) corresponding to the correct answer. Thus, each question in this section carries a maximum of 8 marks. There is no negative marks awarded for incorrect answer (s) in this Section. NAME OF THE CANDIDATE PHONE NUMBER L.K. Gupta (Mathematics Classes) Pioneer Education (The Best Way To Success) S.C.O., Sector 4- D, Chandigarh Ph: , PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O., SECTOR 4 D, CHANDIGARH

2 L.K.Gupta (Mathematic Classes) MOBILE: , 4677 Section I This Section contains 6 multiple choice questions. Each question has 4 choices A), B), C) and D) out of which ONLY ONE is correct.. Which of the following functions are differentiable in (-,)? (a) x (logx) dx (b) x x sinx dx (c) x x x t + t dt (d) none of these. + t + t Sol. x = f(x) (logx) x dx f'(x) = (logx) (logx) f'(x) does not exist for all x in (-, ) let x sinx sinx sin x g(x) =, dx g'(x) = x x x x g'(x) does not exist at x = and so is not differentiable in (-, ). Lat h(x) = dt Then, x t + t = + t + t x + x h'(x) =, which is defined for all x in (-,) + x + x as + x + x.. The slope of the tangent to the curve y = x dx at the point where x= is. + x (a) /4 (b) / (c) (d) none of these. Sol. dy dy =, = dx + x dx x=. Differential equation of the family of circles touching the line y = at (, ) is (a). x + (y ) + (y ) = (b). dx dy dx x + (y ) x y = dy PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O., SECTOR 4 D, CHANDIGARH

3 L.K.Gupta (Mathematic Classes) MOBILE: , 4677 (c). Sol. dx x + (y ) + + y (y ) = dy Equation of circle will be (d). None of these x + (y ) + λ(y ) = Differentiating, we get x + (y ) dx dy + λ dy = dx dx theequationisx + (y ) (y ) x + y 4 = dy 4. The solution of the equation dy x(logx + ) = dx siny + ycosy is (a). x ysiny = x logx + + c (b) ycosy = x (logx + ) + c (c). Sol. x ycosy = x logx + + c (d) (ycosy + siny)dy = (xlogx + x)dx ysiny sinydy + sinydx = x logx x dx + xdx + c x ysiny = x logx + c ysiny = x logx + c 5. The slope of the tangent at (x, y) to a curve passing through a point (, ) is then the equation of the curve is (a) (c) (x y ) = x (b) x(x y ) = 6 (d) (x y ) = 6y x(x + y ) = x + y xy, PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O., SECTOR 4 D, CHANDIGARH

4 L.K.Gupta (Mathematic Classes) MOBILE: , 4677 Sol: dy x + y =..() dx xy dy dv Puty = vx = v + x dx dx equation() transforms to dv x + v x + v v + x = = dx xvx v dv v v x = + v = dx v v vdv dx = v x logx + log( v ) = logc x( v ) = C y x = C x x y = Cx It passes through (, ) 4 = C C = x (x y ) x x y = = 6. The area bounded by y sec x,y cosec = = x and line x = is (a) (c) Sol. π log( + ) sq. units (b) π log( + ) sq. units π logesq. units (d) None of these. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O., SECTOR 4 D, CHANDIGARH 4

5 L.K.Gupta (Mathematic Classes) MOBILE: , 4677 integrating along x-axis, we get A (cosec = x sec x)dx Integrating along y-axis, we get π/4 A = (secy )dy π/4 = [logsecy + tany y] π π = log + = log( + ) sq.units. 4 SECTION II (Integer Type) This section contains 5 questions. The answer to each question is a single-digit integer, ranging from to 9. Correct digit below the question no. in the answer sheet is to be bubbled. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O., SECTOR 4 D, CHANDIGARH 5

6 L.K.Gupta (Mathematic Classes) MOBILE: , If Sol. 6 tanx ( tanx cot x)dx atan + = + c, then the value of btanx LetI = ( tanx + cotx)dx = (sinx + cosx) dx (sinxcosx Put sinx cosx = t sinx = t (cosx + sinx)dx = dt dt t t t Then,I = = sin t = tan + c sinx cosx tan = + c sinx tan x tan = + c tanx Wegeta =,b = 4 5 Then,a + b = 4 + = a + b = 6 = a + b must be 8. If tan x b tan x sin4x.e dx = acos x.e + c, then the value of Sol. tan x LetI = sin4x.e dx tan x = sinxcosx.e dx b a 55must be PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O., SECTOR 4 D, CHANDIGARH 6

7 L.K.Gupta (Mathematic Classes) MOBILE: , 4677 = + tan x tan x tan x 4 sinxcosx..e dx = 6 tan x 4 tanxsec xcos x( tan x).e dx Put tan x = t tanxsec xdx = dt { + } t t ( t)e dt (t ) e dt Then,I = = ( + t) (+ t) = ( + t) ( + t) t e dt t e = + c ( + t) 4 tan x = cos x.e + c b 8 a =,b = 4,thena = ( ) = If x (+ x) Sol. f(t)dt = x, then the value of 5.f() must be Differentiating both sides w.r.t. x, then f(x ( + x)) (x + x ) = Atx = f() = 5 5f() = 5 = 5. If I = x[x]dx, where [.] denotes the greatest integer function, then the value of PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O., SECTOR 4 D, CHANDIGARH 7

8 L.K.Gupta (Mathematic Classes) MOBILE: , 4677 I 7 Must be Sol. I = x[x]dx / / =.dx + xdx + xdx + x dx / / / x x = + + [x ] + / / 9 9 = = = = I = 8 4 = 7 = = 7 4. If Sol. sinα π π dx < α < and I = then the value of 4cos α x sinα ( ) x sin α I = sin = sin cosα cosα = sin (sinα) I + α π must be PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O., SECTOR 4 D, CHANDIGARH 8

9 L.K.Gupta (Mathematic Classes) MOBILE: , 4677 π = π α π < α < I + α π α + α = = π π SECTION III (Paragraph Type) This section contains paragraphs. Based upon each of the paragraphs multiple choice questions have to be answered. Each of these question has four choices A), B), C) and D) out of which Only One is correct. Paragraph for questions 4 Two curves / / C [f(y)] + [f(x)] = and / / C [f(y)] [f(x)] + =, Satisfying the relation: f(x y)f(x + y) (x + y)f(x y) = 4xy(x y ).. The area bounded by C and C is (a). π sq. units (b). π+ sq. units (c). π+ 6 sq. units (d). π sq. units. The area bounded by the curve C and x + y = is (a). π sq. units (b). 6 sq. units (c). 6 sq. units (d). None of these 4. The area bounded by C and x + y + = is (a). 5/ sq. units (b) 7/ sq. units (c). 9/ sq. units (d). None of these PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O., SECTOR 4 D, CHANDIGARH 9

10 L.K.Gupta (Mathematic Classes) MOBILE: , 4677 Sol.. (b) Given (x y)f(x + y) (x + y)f(x y) = 4xy(x y ) = (x y )[(x + y) (x y) ] = (x y)(x + y) (x + y)(x y) f(x + y) = (x + y) f(x) = x,f(y) = y Now equations of given curves are y + x =.() x + y = () Solving equations () and (), we get x =,y = ± The area bounded by curves A = x dx + x dx π/ I = x dx = cos θdθ π/ = ( + cos θ)dθ π/ π/ PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O., SECTOR 4 D, CHANDIGARH

11 L.K.Gupta (Mathematic Classes) MOBILE: , 4677 π/ sinθ π π = θ + = + 4 π/ π = = π. 6 4 [( x) ] 4 / / I = x dx = = [ ] / = 4. A = π + 4 = π + sq. units.. (a). The required area is =area of circle area of square = π 4 sq units 4. (c). The Required area PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O., SECTOR 4 D, CHANDIGARH

12 L.K.Gupta (Mathematic Classes) MOBILE: , 4677 ( y ( )) = y dy y y = + y 4 8 = = squnits Paragraph for questions 5 7 A certain radioactive material is known to decay at a rate proportional to the amount present. Initially there is 5 kg of the material present and after two hours it is observed that the material has lost percent of its original mass. Based on these data answer the following questions. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O., SECTOR 4 D, CHANDIGARH

13 L.K.Gupta (Mathematic Classes) MOBILE: , The expression for the mass of the material remaining at any time t (a). (c). N = 5e (/)(ln.9)t (b). 5e (/4)(ln9)t N = 5e (ln.9)t (d). None of these 6. The mass of the material after four hours (a)..5ln 9 5 (b). 5e ln9 (c). ln.9 5e (d). None of these 7. The time at which the material has decayed to one half of its initial mass. (a). (ln/)/(/ln9)hr (b).(ln)/( /ln.9)hr (c).(ln/)/( /ln.9)hr (d). None of these Sol. 5. (a). Let N denote the amount of material present at time t. Then, dn kn dt = This differential equation is separable and linear, its solution is N = ce kt. () At t =, we are given that N = 5. Therefore, from equation (), 5 = ce k (), or c = 5. or c = 5. Thus, N = 5 e kt.. () At t =, percent of the original mass of 5 mg or 5 mg, has decayed. Hence, at t =, N = 5 5 = 45. Substituting this value into equation () and solving for k, we have k = 5e ork = log 5 Substituting this value into (), we obtain the amount of mass present at any time t as N = 5 e (/)(ln.9)t () where t is measured in hours. 6. (c). We require N at t = 4. Substituting t = 4 into () and then solving for N, we find N = 5e ln.9 PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O., SECTOR 4 D, CHANDIGARH

14 L.K.Gupta (Mathematic Classes) MOBILE: , (c). We require t when N = 5/ = 5. Substituting N = 5 into equation () and solving for t, we find 5 = 5 e (/)(ln.9)t or t = (ln /) / ( / ln.9) hr. SECTION IV (Matrix type) This section contains questions. Each question has four statements (A, B, C and D) given in Column I and five statements (p, q, r, s and t) in Column II. Any given statement in Column I can have correct matching with one or more statements(s) given in Column II. For example, if for a given question, statement B matches with the statements given in q and r, then for that particular question, against statement B, darken the bubbles corresponding to q and r in the ORS. 8. If at every point x of an interval [ a, b],the inequalities g(x) f(x) h(x) are fulfilled, b b b then ( ) ( ) ( ) g xdx f xdx h x dx,a < b a a a Match the entries from the following two columns: (a) Column I If µ < 7 x dx ( + ) 8 x < λ,then Column II (p) [λ +µ ] =, Where [.] denotes the greatest integer function (b) If µ < dx ( + x ) 6 < λ,then (q) [λ +µ ] = 4, Where [.] denotes the greatest integer function (c) If µ < dx ( 4 x x ) < λ,then (r) [λ-µ] =, Where [.] denotes the greatest integer function (s) [λ -µ ] =, Where [.] denotes the greatest integer function (t) [λ +µ ] =, Where [.] denotes the greatest integer function Sol: PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O., SECTOR 4 D, CHANDIGARH 4

15 L.K.Gupta (Mathematic Classes) MOBILE: , x 7 x 7 Since < < x < x < Then dx dx x dx 8 < < 8 + x + x ( ) ( ) 7 x dx Hence, < < λ =,μ = [λ + μ] =,[λ μ] = o(r,t) ( + x ) Since ( x ) < ( + x 6 ) < ( + x ) x (,) > > x (,) ( x ) ( + x 6 ) ( + x ) dx dx dx < < 6 ( ) ( + x ) ( + x ) ( x ) 6 ( + x ) dx ln{x + + x } < {sin x} < 6 ( + x ) dx π π ln < < λ.57μ ln.69[λ μ [λ μ] (p,r) = = + = = Since, 4 x > 4 x x > 4 x x (,) 4 x > 4 x x > 4 x x (,) < < x (, ( ) ( ) ( ) dx dx dx < < ( 4 x ) ( 4 x x ) ( 4 x ) sin ( 4 x ) ( 4 x x ) ( 4 x ) x dx x π dx π < < < < sin ( 4 x x ) ( 4 x x ) π π μ = π.5 λ = 4.4and [λ + μ] = 4,[λ μ] = (q,s) 9. Match the statements in Column I with the Column II. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O., SECTOR 4 D, CHANDIGARH 5

16 L.K.Gupta (Mathematic Classes) MOBILE: , 4677 (a) Column I Column II ( ) ( ) ( t + ) 4 / t t + t tanx dx = Aln + Btan + c (p) A = 4 Where t = tan / x (b) sinx + sin x cosx + (q) dx = Acos x + Bln + c A = cosx cosx (c) dx x = Atan x + Btan + c + + ( x )( x 4) (r) A = (s) B = (t) B= 4 / Ans. (a) Let I = (tanx) dx put tanx = t sec xdx = t dt dx = ( ) ( ) ( ) sinx + sin x dx + sin x sinxdx (b) let I = = cosx cos x ( cosx) t 6 ( + t ) ( cos x)sinx dt = dx put cosx = t sinxdx = then, t dt (t 4) (t ) (t ) I = = t = dt t In c t = + t + dt PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O., SECTOR 4 D, CHANDIGARH 6

17 L.K.Gupta (Mathematic Classes) MOBILE: , 4677 ( ) cosx = cosx In + c 4 cosx + cosx = cosx In + c A = (r),b = (t) 4 4 cosx + t dt t.dt ( + t ) + (t ) I = = 6 Put zdz zdz t = z tdt = dzthan I = = + z ( + z)( z + z ) ( + z) + dz = In( + z) + dz In( z) In( z z ) = ( z + z ) 4 4 z + z dz = In( + z) + In( z + z ) z + 4 z + z z t t + t = In. tan c In tan c + 4 ( z) 4 + = (t + ) A = (P):B = (S) (c) Let dx x (x + )(x + 4) x + x + 4 x = tan x tan c A (Q),B 6 + = = 6 I = = dx = tan x tan + c PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O., SECTOR 4 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