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1. DIFFERENTIALEQUATION &AREAUNDER CURVE There are 70 questions in this question bank. Q.12/AUC Area common to the curve y = 9 2 x & x² + y² = 6 x is : (A)  3 4 (B)  3 4 (C) 3           3 4 (D*) 3           3 3 4 [Hint: x2 + y2 = 9 ....(1) ; x2 + y2 – 6x = 0 ....(2) On solving x=3/2 ; y2 = 9 –9/4 = 27/4  y = 2 3 3  A = dx x 9 2 3 2 / 3 2   Q.22/DE Spherical rain drop evaporates at a rate proportional to its surface area. The differential equation corresponding to the rate of change of the radius of the rain drop if the constant of proportionality is K > 0, is (A*) dr dt  K = 0 (B) dr dt  K = 0 (C) dr dt  Kr (D) none [Hint: dt dV = – k4r2 ....(1) but V = 3 r 3 4   dt dV = 4r2 dt dV ....(2) hence dt dr = – K  (A) ] Q.33/AUC If y = 2 sin x + sin 2x for 0  x  2, then the area enclosed by the curve and the x-axis is (A) 9/2 (B*) 8 (C) 9 (D) 4 [Hint: A =      0 dx x 2 sin x sin 2 2 =      0 0 dx x 2 sin 2 dx x sin 4 = 8 + 0 = 8 Q.45/DEThegeneral solution of the differential equation, y + y (x)  (x). (x)= 0 where (x)isa known functionis: (A*) y = ce(x) + (x)  1 (B) y = ce+(x) + (x)  1 (C) y = ce(x) (x) + 1 (D) y = ce(x) + (x) + 1 where c is an arbitraryconstant . [Sol. dx dy + y '(x) =  (x).'(x) I.F. = ) x ( dx ) x ( e e     hence y.e(x) =  e(x).(x).'(x) dx =  et.t dt where  (x) = t = tet – et + C = (x).e(x) – e(x) + C  y = ce–(x) + (x) – 1  A ]
2. Q.824/DE The solution of the differential equation, xy dx dy = 2 2 x 1 y 1   (1 +x + x2) given that when x = 1, y= 0 is (A) ln 1 2  y = lnx + tan1 x   2 (B*) ln 1 2 2  y x = 2 tan1 x   2 (C) ln 1 2 2        y x =  4  2 tan1 x (D) none [Hint:   2 y 1 dy y = dx ) x 1 ( x x x 1 2     2 1 ln (1 + y2) = tan–1x + ln x + C ln 2 2 x y 1 = 2 tan–1x + C ] Q.96/AUC The area of the figure bounded by the curves y = ex , y= e-x & the straight line x = 1 is (A) e + 1 e (B) e  1 e (C*) e + 1 e  2 (D) none Q.1035/DE Acurve is such that the area of theregion bounded bythe co-ordinate axes, the curve& the ordinate of anypoint on it is equal to the cube of that ordinate .The curve represents (A) apairofstraight lines (B) a circle (C*) a parabola (D) an ellipse [Sol. dx ) x ( f x 0  = y3 Differentiating f (x) = 3y2. dx dy y = 3y2 dx dy  y = 0 (rejected) or 3y dy = dx 2 y 3 2 = x + c  parabola  C] Q.117/AUC The area bounded by the curve y = x²  1 & the straight line x + y = 3 is : (A) 9 2 (B) 4 (C) 2 17 7 (D*) 6 17 17 [Hint: 3 – x = x2 – 1  x2 + x – 4 = 0 x1 + x2 = –1 x1 x2 = – 4 ....(1) A =    dx 1 x ) x 3 ( 2 1 x x 2     =  dx x x 4 2 1 x x 2    use(1)]
3. Q.1823/AUC The area common to y  x & x >  y and the curve x² + y² = 2 is : (A)  4 (B) 3 2  (C)  (D*)  2 [Hint: dx x x 2 A 0 1 2 2            + dx x x 2 A 1 0 2           = 2  note that the area is equal to the sector AOB with central angle 900  1/4 (the area of the circle) ] Q.1952/DE Therealvalueofm forwhichthesubstitution, y= um willtransform thedifferential equation, 2x4y dy dx + y4 = 4x6 into a homogeneous equation is : (A) m = 0 (B) m = 1 (C*) m = 3/2 (D) no value of m [Hint: y = um  dx dy = m um  1 . Hence 2 x4 . um . m um  1 . dx du + u4m = 4 x6 . du dx = 4 2 6 4 4 2 1 x u m x u m m    4 m = 6  m = 3 2 and 2m – 1 = 2  m = 3 2  (C) ] Q.2024/AUC The area enclosed by x+ y = 1 & the axis of x is : (A*) 1 (B) 1 2 (C) 2 (D) none Q.2127/AUC The area bounded by x²+y²2x = 0 & y = sin x 2 in the upper half of the circle is : (A*)   2 4  (B)   4 2  (C)    8 (D)    2 2 Q.2254/DE The solution of the differential equation, x2 dy dx .cos 1 x  ysin 1 x = 1, where y1 as x  is (A*) y = sin 1 x – cos 1 x (B) y = x x x  1 1 sin (C) y = cos 1 x  sin 1 x (D) y = x x x  1 1 cos [Hint: dy dx  y x2 tan 1 x =  sec 1 x . 1 2 x . IF = e x x dx  1 2 1 tan = sec 1 x  y . sec 1 x =         2 2 x 1 x 1 sec dx = tan 1 x + c if y  1 then x  c =  1  y = sin 1 x  cos 1 x ]
4. Q.2657/DE The latus rectum of the conicpassingthrough theorigin and having the propertythat normal at each point (x, y) intersects the xaxis at ((x +1), 0) is : (A) 1 (B*) 2 (C) 4 (D) none [Sol. Given y dx dy + x = x + 1 c x 2 y2   x = 0, y = 0  c = 0  y2 = 2x  latus rectum= 2  B ] Q.2735/AUC The line y = mx bisects the area enclosed by the curve y = 1 + 4x  x2 & the lines x = 0 , x = 3 2 & y = 0 . Then the value of m is : (A*) 13 6 (B) 6 13 (C) 3 2 (D) 4 Q.2859/DE The solution of the differential equation, 2x2y dy dx = tan (x2y2)  2xy2 given y(1) =  2 is (A*) sinx2y2 = ex–1 (B) sin(x2y2) = x (C) cosx2y2 + x = 0 (D) sin(x2y2) = e.ex [Sol. put x2y2 = z Given ) y x ( tan x 2 . y dx dy y 2 . x 2 2 2 2   ) y x tan( ) y x ( dx d 2 2 2 2  put x2 y2 = z nowgivenexpression transformsto dx dz = tan z  dx =  dz z cot x = ln (sin z) + c when x = 1 , y = 2   z = 2   c = 1  x = ln sin (x2y2) + 1  ln sin(x2y2) = x – 1 sin (x2 y2) = ex–1] Q.2936/AUC The area bounded by the curve y = f (x), the x-axis & the ordinates x =1 & x = b is (b1)sin(3b+ 4). Then f(x) is : (A) (x  1) cos (3x +4) (B) sin(3x+4) (C*) sin (3x + 4) + 3 (x  1) . cos (3x +4) (D) none [Sol: dx ) x ( f b 1  = (b – 1) sin (3b + 4)
5. Q.3344/AUC The area of the region of the plane bounded by max   x y ,  1 & xy  1 2 is : (A) 1 2 + ln 2 (B) 31 4 (C) 1 + 2 ln 2 (D*) 3 + ln 2 [Hint: shaded area in 1st quadrant = dx x 2 1 1 1 2 / 1         = 1 2 / 1 nx 2 1 x     l = 2 1 n 2 1 2 1 l  = 2 1 n 2 1 2 1 l   2 times the shaded area = 1 – ln2  Required area = 2 – (1–ln2) + 2 = 3 + ln2  D ] Q.3474/DE If y= | x c | n x l (where c is an arbitraryconstant) is the general solution of the differential equation dx dy = x y +          y x then the function          y x is : (A) 2 2 y x (B) – 2 2 y x (C) 2 2 x y (D*) – 2 2 x y [Hint: ln c + ln |x| = y x diff.w.r.t.x, x 1 = 2 1 y y x y  dx dy x y x y2   dx dy = 2 2 x y x y            y x = – 2 2 x y  D ] Q.3545/AUC The area of the region for which 0 < y < 3  2x  x2 & x > 0 is : (A)   3 2 2 1 3    x x dx (B)   3 2 2 0 3    x x dx (C*)   3 2 2 0 1    x x dx (D)   3 2 2 1 3    x x dx Q.3677/DE Ifthefunctiony=e4x +2e–x isasolutionofthedifferential equation, K y dx dy 13 dx y d 3 3   thenthe value of K is : (A) 4 (B) 6 (C) 9 (D*) 12
6. [Sol. y = c1cos(x + c2) – (c3 4 c x e   ) + (c5sin x)  y = c1 (cos x cos c2 – sin x sin c2) – (c3 4 c e e–x)+ (c5 sin x)  y = (c1 cos c2 ) cosx – (c1 sin c2 – c5) sinx – (c3 4 c e ) e–x  y = l cosx + m sinx – n e–x ....(i) where l, m, n are arbitraryconstant  dx dy = – l sinx + m cosx + n e–x ....(ii)  2 2 dx y d = – l cosx – m sinx – n e–x ....(iii)  3 3 dx y d = l sinx – m cosx + n e–x ....(iv) (i)+(iii)gives 2 2 dx y d + y = – 2n e–x ....(v) (ii) + (iv) gives dx dy dx y d 3 3  = 2n e–x ....(vi) From (v) and (vi) we get             y dx y d dx dy dx y d 2 2 3 3 or 0 y dx dy dx y d dx y d 2 2 3 3     is therequired differential equation ] Q.4263/AUC Let f(x) = x  x2 & g(x) = ax . If the region above the graph of g and below the graph of f has an area equal to 9/2 then 'a' is equal to : (A) 2 (B*) 4 (C*)  2 (D) 3 Q.4380/DE The curve, with the propertythat the projection of the ordinate on the normal is constant 'a', is (A*) c y a y n a x 2 2           l (B) c y a x 2 2    (C) (y – a)2 = cx (D) ay = tan–1 (x + c) [Sol. Ordinate = PM. Let P  (x, y) Projection of ordinateon normal = PN  PN = PM cos = a (given)  a tan 1 y 2     y = 2 1 ) y ( 1 a   a a y dx dy 2 2        dx a y dy a 2 2  c x | a y y | n a 2 2     l ] Q.4464/AUC The area bounded by the curves y =  x and x =  y where x, y  0 (A) cannot be determined (B*) is 1/3 (C) is 2/3 (D) is same as that of the figure bounded by the curves y = x ; x  0 and x = y ; y  0
7. Q.4868/AUC Consider the graph of continuous function y = f(x) for x  [a  b, a + b] a, b  R+ and b > a. If the origin is shifted to (a + b, 0) such that new axes are parallel to the old axes, then the area bounded bythe given curve, the X-axis and the new ordinates X = a, X = b can be written as : (A) b a a b    2 2 f(x) dx (B)    a b f(x)dx (C*) a b  f(a + b  x)dx (D) 1 2 a b a b    f(x)dx [Hint: Required area =Area of the regionABB'PA' =    a b dX ) X ( f =  b a dx ) x ( f =    b a dx ) x b a ( f  (C) ] x – (a + b) = X if X = – a then x = b X = – b then x = a ] Q.4971/AUC The curvilineartrapezoid is bounded bythe curvey= x2 +1 and thestraight lines x=1 andx=2.The co-ordinates ofthepoint (on thegivencurve)with abscissax [1,2] wheretangent drawn cutofffrom the curvilineartrapezoidan ordinarytrapezium ofthegreatest area,is (A) (1,2) (B) (2,5) (C*)       4 13 , 2 3 (D) none [Hint: 1 1y x dx dy    = 2x1 T : y – y1 = 2x1 (x–x1) ....(1) For co-ordinates of the point P put x = 1 in (1) y = y1 + 2x1 (1–x1) = 1 + 2 1 x +2x1 + 2 2 1 x = 1 + 2x1 – 2 1 x Hence P(1, 1 + 2x1 – 2 1 x ) For Q put x= 2 in (1) y = y1 + 2x1 (2 – x1) = 1 + 2 1 x + 4x1 – 2 2 1 x = 1 + 4x1 – 2 1 x  A = (1 + 2x1 – 2 1 x ) + (1+ 4x1 – 2 1 x ) = 1 + 3x1 – 2 1 x 1 dx dA = 3 – 2x1  x1 = 3/2 ] Q.5089/DE If  x a dt ) t ( y t = x2 + y (x) then yas a function of x is (A*) y = 2 – (2 + a2) 2 a x 2 2 e  (B) y = 1 – (2 + a2) 2 a x 2 2 e  (C) y = 2 – (1 + a2) 2 a x 2 2 e  (D) none
8. [Sol. (a, 0) lies on the given curve  0 = sin2a – 3 sina  sina = 0 or cosa = 3 /2  a = 6  (as a >0 and thefirst point of intersection with positive X-axis) and A =    6 / 0 dx ) x sin 3 x 2 (sin = 6 / 0 x cos 3 2 x 2 cos          = 3 4 7 3 2 1 2 3 4 1                     4A + 8 cosa = 7 ] Q.5390/DE General solution of the differential equation 2y' ln x + x y = y–1cos x is (A*) y2 ln x = c + sin x (B) y2 ln x = c – sin x (C) y2 ln x = c + cos x (D) y2 ln x = c – cos x [Hint: put y2 = t ] Q.5473/AUC Area of the region enclosed between the curves x = y2 – 1 and x = |y| 2 y 1 is (A) 1 (B) 4/3 (C) 2/3 (D*) 2 [Hint: dy ) 1 y ( y 1 y 2 A 1 0 2 2            = 2 ] Q.5591/DE Solution of the equation  1 0 dt ) t x ( y = n y(x) is (A) y = c x1/n (B*) y = c n n 1 x  (C) y = c n 1 1 x  (D) y = c x– 1/n [Hint: start x t = y ] Q.5677/AUC Let y= g (x) be the inverse of a bijective mapping f : R  R f (x) = 3x3 + 2x. The area bounded by the graph of g (x), the x-axis and the ordinate at x = 5 is : (A) 4 5 (B) 4 7 (C) 4 9 (D*) 4 13 [Hint: noteforinversefunctionyaxis willbethe x axis andx axis will bethe yaxis requiredarea =Area of rectangle –  1 0 dx ) x ( f = 5 –   1 0 3 dx ) x 2 x 3 ( = 5 – ( 4 3 + 1) = 3 4 1 = 4 13 Ans ]
9. Q.609/DE Equationofacurvepassingthrough theorigin iftheslopeofthe tangent drawnat anyofitspoint(x, y) is cos(x + y) + sin(x + y), is (A) y = 2 tan–1(ex – 1) + x (B*) y = 2 tan–1(ex – 1) – x (C) y = 2 tan–1(ex) – x (D) y = 2 tan–1(ex) + x [13th test (24-03-2005)] [Sol. dx dy = cos(x + y) + sin(x + y) ; put x + y = u ; 1 + dx du dx dy  dx du – 1 = cos u + sin u  dx du = (1 + cos u) + sin u = 2cos2 2 u + 2sin 2 u cos 2 u = 2cos2 2 u (1 + tan 2 u ) du 2 u tan 1 2 2 u sec2         = dx ;  tan 2 u = t ;     dx t 1 dt x = ln (1 + t) + C t = 0 ; C = 0 1 + t = ex  t = ex – 1 ;        2 y x tan = ex – 1  2 y x  = tan–1(ex – 1) y = 2 tan–1(ex – 1) – x ] Q.6165/auc The area bounded bythe curves y= x (1  lnx) ; x = e1 and positive X-axis between x = e1 and x = e is : (A) e e 2 2 4 5         (B*) e e 2 2 5 4         (C) 4 5 2 2 e e         (D) 5 4 2 2 e e         [Hint: y = x (1 – lnx) = 0  x = e (as x > 0) dx dy = – lnx  in (0,1) and  in (1, ) also 0 x Lim  x (1 – lnx) = 0 A =   e e / 1 dx ) nx 1 ( x l ] Q.6279/auc Which one of the following DOES NOT represent the area enclosed bythe curves y = sec–1x , y = cosec–1x and the line x – 1 = 0 ? (A*) dx ) 1 x ec (cos 2 / 0    (B) 2 dx ) 1 x (sec 4 / 0    (C) dx ) 1 x ec (cos 2 2 / 4 /     (D) dx ) x sec x ec (cos 2 1 1 1    
10. Q.6586/auc Area bounded by the curve y = min {sin2x, cos2x} and x-axis between the ordinates x = 0 and x = 4 5 is (A) 2 5 square units (B) 4 ) 2 ( 5   square units (C*) 8 ) 2 ( 5   square units (D)         2 1 8 square units [Hint: A =           4 5 4 3 2 4 3 4 2 4 0 2 dx x sin dx x cos dx x sin ] Q.6615/DE Waterisdrainedfrom averticalcylindrical tank byopeningavalveat thebase ofthetank.It isknown that therateat whichthewaterlevel drops isproportional to the squareroot of waterdepth y, where the constant of proportionalityk > 0 depends on the acceleration due to gravity and the geometryof the hole. If t is measured in minutes andk = 15 1 then the time to drain thetank if the wateris 4 meter deep to start with is (A)30min (B)45min (C*) 60 min (D)80 min [Hint: dt dy = – k y  0 4 y dy = –  t 0 dt k 0 4 y 2 = – kt = – 15 t 0 – 4 = – 15 t  t = 60 minutes  (C) ] Q.6787/auc If the area bounded between x-axis and the graph of y = 6x – 3x2 between the ordinates x = 1 and x = a is 19 square units then 'a' can take the value (A) 4 or – 2 (B) two values are in (2, 3) and one in (–1, 0) (C*) two values one in (3,4) and one in (–2,–1) (D) none of these [Hint: I =   dx ) x 3 x 6 ( 2 = 3 x 3 2 x 6 3 2  = 3x2 – x3 = x2(3 – x) A1 = I(2) – I(1) = 4 – 2 = 2 units A2 = I(2) – I(3) = 4 – 0 = 4 units A3 = I(3) – I(4) = 0 – (–16) = 16 units  onevalue of awill lie in (3, 4) usingsymmetric otherwill lie in (–2, –1) ]
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