CBSE 08 ANNUAL EXAMINATION DELHI (Series SGN Code No 65/ : Delhi Region) Ma Marks : 00 Time Allowed : Hours SECTION A Q0 Find the value of tan cot ( ) Sol 5 5 tan cot ( ) tan tan cot cot 6 6 6 0 a Q0 If the matri A 0 is skew symmetric, find the value of a and b b 0 T Sol As A is skew symmetric so, A A ie, aij a ji, if A a ij Therefore, a a a and, a a b ( ) Q0 Find the magnitude of each of the two vectors a and b, having same magnitude such that the o angle between them is 60 and their scalar product is 9 9 o Sol As ab a b cos a a cos 60 9 a a Hence a b Q0 If a * b denotes the larger of a and b and if a b (a *b), then write the value of (5) (0), where * and are binary operations Sol (5) (0) (5*0) 0 SECTION B Q05 Prove that : sin sin ( ),, Sol RHS: Let y sin [ ] Put sin sin (i) y sin [sin sin ] sin (sin ) As, sin sin 6 6 y sin LHS (By using (i) Q06 Given A 7, compute A and show that A 9 I A 7 Sol Here A 7 adja and A 7 adja 7 A A 7 0 9 0 7 Now A and, 9I A 9 0 7 0 9 7 Clearly, A 9 I A Q07 Differentiate cos tan with respect to sin Eclusively for Maths, click at : wwwtheopguptacom WhatsApp @ +996505080
Solved Papers Of CBSE Eams By OP Gupta (INDIRA Award Winner) cos cos Sol Let y tan tan tan cot tan tan sin sin cos dy d 0 d d Q08 The total cost C() associated with the production of units of an item is given by C() 0005 00 0 5000 Find the marginal cost when units are produced, where by marginal cost we mean the instantaneous rate of change of total cost at any level of output Sol Given C() 0005 00 0 5000 d [C()] 005 00 0 C () d C () 005 9 00 0 005 cos sin Q09 Evaluate d cos cos sin cos sin sin cos sin Sol Let I d d d cos cos cos I sec d tan c b 5 Q0 Find the differential equation representing the family of curves y a e, where a and b are arbitrary constants b 5 Sol Given that y a e (i) dy b5 ab e yb (By (i) d y yy y y b 0 yy (y ) 0 y y d y dy Hence the required differential equation is y 0 d d Q If is the angle between two vectors ˆ i j ˆ k ˆ and i ˆ j ˆ k ˆ, find sin Sol Let a ˆi j ˆ kˆ and b i ˆ j ˆ kˆ ˆi ˆj kˆ a b i ˆ 8j ˆ kˆ a b 6 6 6 6 6 Since sin sin a b 9 9 7 Q A black and a red die are rolled together Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than Sol Let E : sum of numbers appearing on the die is 8, F : the red die shows a number less than (,),(,),,(6,),(,), E {(,6),(6, ),(,5),(5,),(, )}, F (, ),,(6, ),(,),(,),,(6,) E F {(6, ),(5,)} Eclusively for Maths, click at : wwwtheopguptacom WhatsApp @ +996505080
Solved Papers Of CBSE Eams Annual Eam Delhi (08) P(E F) Therefore, P(E F) 6 P(E F) or P(F) 8 8 9 6 SECTION C Q Using properties of determinants, prove that y 9(yz y yz z) z Sol LHS : Let y By C C C, C C C z 0 y 0 Take common from both C and C z z 0 9 y 0 z z 9 0 y z yz 0 9 yz y yz z RHS Epanding along R, dy Q If ( y ) y, find d dy OR If a( sin ) and y a( cos ), find when d Sol Here ( y ) y ( y ) yy y y ( y ) yy y y ( y ) y( y )y y y dy y y y( y ) y y ( y ) d y y OR We have a( sin ) and y a( cos ) d dy a( cos ) a( cos ) and a(0 sin ) a sin d d dy dy d a sin sin cos cot d d d a( cos ) sin dy At, cot d d y dy Q5 If y sin(sin ), prove that tan ycos 0 d d dy Sol We have y sin(sin ) cos(sin ) cos (i) d d y And, cos(sin ) sin sin(sin ) cos d d y sin cos(sin ) cos y cos d cos Eclusively for Maths, click at : wwwtheopguptacom WhatsApp @ +996505080
Solved Papers Of CBSE Eams By OP Gupta (INDIRA Award Winner) d y dy tan ycos (By (i) d d d y dy Hence, tan ycos 0 d d Q6 Find the equations of the tangent and the normal to the curve 6 9y 5 at the point (, y ), where and y 0 OR Find the intervals in which the function f () 5 is (a) strictly increasing, (b) strictly decreasing Sol Let P (, y ) where and y 0 Given curve is 6 9y 5 (i) Clearly P shall satisfy (i) so, 6 9y 5 y (as y 0 Therefore, point of contact is P(, ) dy dy 6 Now 8y 0 d d dy 6 m 9y d T at P 9 7 Eq of tangent : y ( ) 7y 5 7 7 Eq of normal : y ( ) 7 y 0 OR Here f () 5 f () 0 ( )( )( ) For f () 0, ( )( )( ) 0,, Interval Sign of f () Nature of f () (, ) ve Strictly decreasing (, ) +ve Strictly increasing (, ) ve Strictly decreasing (, ) +ve Strictly increasing Q7 An open tank with a square base and vertical sides is to be constructed from a metal sheet so as to hold a given quantity of water Show that the cost of material will be least when depth of the tank is half of its width If the cost is to be borne by nearby settled lower income families, for whom water will be provided, what kind of value is hidden in this question? Sol Let length and width of the base of the open bo be and its depth be y Also let volume of bo V be V So, V y y (i) Let k (in `) be the cost per sq units V So, the cost of material used, C k y k (By (i) dc V V d C V k k, k d d dc V / For 0, k 0 V ie, (V) d d C V / Clearly k 6k 0 d so, C is least when (V) / V at (V) Eclusively for Maths, click at : wwwtheopguptacom WhatsApp @ +996505080
Solved Papers Of CBSE Eams Annual Eam Delhi (08) width of bo Now V y y y ie, depth of bo Any appropriate value cos Q8 Find d ( sin )( sin ) cos Sol Let I d [Put sin t cos d dt ( sin )( sin ) dt I (i) ( t)( t ) Consider A B(t) C A( t ) B(t)( t) C( t) ( t)( t ) t t t On comparing the coefficients of like terms, we get : A, B, C By (i), dt t I dt dt t t log t log t tan t c t I log sin log sin tan (sin ) c Q9 Find the particular solution of the differential equation e tan yd ( e ) sec ydy 0, given that y when 0 Sol OR Find the particular solution of the differential equation dy y tan sin d, given that y 0 when e d sec ydy We have e tan yd ( e ) sec ydy 0 0 e tan y Put e t e d dt in st integral, and, tan y u sec ydy du in nd integral dt du tan y 0 t log u log t log c log u e tan Given that y when 0 so, c c 0 e Hence required solution is, tan y e or, y tan ( e ) log c tan y e OR We have dy y tan sin d dy ( tan )y sin d This differential equation is of the form dy P()y Q(), where P() tan, Q() sin d tan d logsec So, integrating factor, IF e e sec So, the solution is : y sec sec sin d c ysec sec tan d c sec c Given that y 0 when so, 0sec sec c c c Eclusively for Maths, click at : wwwtheopguptacom WhatsApp @ +996505080 5
Solved Papers Of CBSE Eams By OP Gupta (INDIRA Award Winner) Hence the required solution is ysec sec or, y cos cos Q0 Let ˆ ˆ ˆ ˆ ˆ ˆ a i 5j k, b i j 5k and c i ˆ ˆj kˆ Find a vector d which is perpendicular to both c and b and da Sol As d which is perpendicular to both c and b d c b ie, d (c b) ˆi ˆj kˆ d (i ˆ 6j ˆ k) ˆ 5 As da so, (i ˆ 6ˆj k)(i ˆ ˆ 5ˆj k) ˆ ( 80 ) 6ˆj kˆ ˆi Hence d (i ˆ 6ˆj k) ˆ Q Find the shortest distance between the lines r (i ˆ ˆj) (i ˆ j ˆ k) ˆ and r (i ˆ ˆj k) ˆ (i ˆ j ˆ 5k) ˆ Sol Given lines are r (i ˆ ˆj) (i ˆ j ˆ k) ˆ and r (i ˆ ˆj k) ˆ (i ˆ j ˆ 5k) ˆ Here ˆ ˆ a ˆ ˆ ˆ i j, b i j k and a ˆ ˆ ˆ i j k, b ˆ ˆ ˆ i j 5k ˆi ˆj kˆ a a iˆ k, ˆ b b i ˆ ˆj So, 5 (a a )(b b ) ( i ˆ k)(i ˆ ˆ ˆj) 6 6 Therefore, SD units b b i ˆ ˆj 5 Hence required shortest distance is 6 5 units 5 Q Suppose a girl throws a die If she gets or, she tosses a coin three times and notes the number of tails If she gets,, 5 or 6, she tosses a coin once and notes whether a head or tail is obtained If she obtained eactly one tail, what is the probability that she threw,, 5 or 6 with the die? Sol Let A : getting or, B : getting,, 5 or 6, E : getting eactly one tail So, P(A), P(B) 6 6 PE A C (using binomial distribution), P(E B) 8 By Bayes theorem, P E B P(B) PB E = P E A P(A) P E B P(B) P B E = 6 8 6 6 8 8 P B E 8 6 Eclusively for Maths, click at : wwwtheopguptacom WhatsApp @ +996505080
Solved Papers Of CBSE Eams Annual Eam Delhi (08) Q Two numbers are selected at random (without replacement) from the first five positive integers Let X denote the larger of the two numbers obtained Find the mean and variance of X Sol Since X denotes the larger of the two numbers obtained from,,,, and, 5 So values of X :,,, 5 X 5 P(X) 5 0 5 0 6 6 5 0 8 8 5 0 6 8 0 Now, mean X P(X) 5 0 0 0 0 0 6 8 And, variance X P(X) Mean 5 0 0 0 0 8 6 96 00 6 7 6 0 SECTION D Q Let A { Z : 0 } Show that R {(a, b) : a,b A, a b is divisible by } is an equivalence relation Find the set of all elements related to Also write the equivalence class [] OR Show that the function f : R R defined by f (), R is neither one-one nor onto Also, if g : R R is defined as g(), find fog () Sol We have R (a, b) : a, b A, a b is divisible by and A Z : 0 ie, A {0,,,,} Refleivity : For any a A, we have a a 0, which is divisible by That is, (a,a) R so, R is refleive Symmetry : Let (a, b) R a,b A We have a b is divisible by As a b b a So b a is also divisible by That is, (b, a) R so, R is symmetric Transitivity : Let a, b,c A such that (a, b) R and (b,c) R then, a b is divisible by and b c is divisible by This implies that a b m and b c p where m,p Z As (a b) (b c) (m p) a c (m p) a c (m p) That is a c is also divisible by so, R is transitive Since the relation R is refleive, symmetric and transitive so, it is an equivalence relation Also the set of all elements related to in relation R is {, 5, 9} Now let (, ) R so that is divisible by A That is,,6,0 Hence [] = {, 6, 0} OR We have f : R R defined by f (), R Let m,n R so that f (m) f (n) Eclusively for Maths, click at : wwwtheopguptacom WhatsApp @ +996505080 7
Solved Papers Of CBSE Eams By OP Gupta (INDIRA Award Winner) m n That implies, mn m nm n m n mn(m n) m n m n mn(m n) (mn )(m n) 0 Either m n 0 or, mn 0 ie, m n or, m n So, f (m) f (n) does not necessarily imply m n for all m, n R Hence f () is not one-one y Let y f () y y 0 y Now for to be real, we must have y 0 ie, ( y)( y) 0 ie, y, That is for all R (domain), we do not have y R (codomain) ie, range codomain Hence f () is not onto Now fog() f (g()) f ( ) ( ) ( ) 5 Q5 If A, find A Use it to solve the system of equations y 5z, y z 5, y z OR Using elementary row transformations, find the inverse of the matri A 5 7 5 5 5 Sol Here A A 0 ( ) 5() 0 A eists Consider C ij be the cofactor of corresponding element a ij of matri A C 0, C, C, C, C 9, C 5, C, C, C 0 adja 9 5 0 adja A 9 A 5 Now y 5z, y z 5, y z 5 Let A, X y and B 5 z 0 As AX B X A B 9 5 5 By equality of matrices, we get :, y, z OR Given that A 5 7 5 y z 8 Eclusively for Maths, click at : wwwtheopguptacom WhatsApp @ +996505080
Solved Papers Of CBSE Eams Annual Eam Delhi (08) 0 0 Using elementary row operations, we have : A I A ie, 5 7 0 0 A 5 0 0 0 0 By R R R, R R R, 0 0 A 0 0 0 0 5 0 By R R R, R R R, 0 0 A 0 0 0 0 0 By R R R, 0 0 A 0 0 0 Using I A A, we get A 0 Q6 Using integration, find the area of the region in the first quadrant enclosed by the -ais, the line y and the circle y Sol We have y (i), y (ii) Solving (i) & (ii), we get : points of intersections :, Required area y d y d i ii 0 d d 0 sin 0 6 0 6 6 8 8 8 squnits Q7 Evaluate : Sol OR / sin cos d 6 9sin 0 Evaluate / ( e )d, as the limit of the sum sin cos Let I d 6 9sin 0 Put sin cos t (cos sin )d dt and, When 0 t, when t 0 5 So, 0 0 5 t I dt dt 5 9t log 5 (t) 5 5 t (sin cos ) t sin t 0 Eclusively for Maths, click at : wwwtheopguptacom WhatsApp @ +996505080 9
Solved Papers Of CBSE Eams I log log log log 0 8 0 5 OR Let I ( e )d b By OP Gupta (INDIRA Award Winner) We know f d lim h f a f a h f a h f a (n )h a n As n, h 0 nh b a b n a f d lim h f a rh (i) Here n r0 f () e, a, b a rh f (a rh) (a rh) (a rh) e rh rh f rh rh rh e 5rh r h ee n rh e d lim h 5rh r h ee n r0 By using (i), n n n n rh I lim h h r 5h r e e n r0 r0 r0 r0 n(n )(n ) n(n ) I lim h h 5h n e e h e h e (n )h n 6 nh(nh h)(nh h) nh(nh h) nh h I lim 5 nh e (e ) n h 6 e Since n, h 0 nh ( 0)( 0) 5 6 So, I ( 0) e (e ) (e e) 6 6 Hence I ( e )d e e Q8 Find the distance of the point (, 5, 0) from the point of intersection of the line r i ˆ ˆj k ˆ (i ˆ ˆj k) ˆ and the plane r(i ˆ ˆj k) ˆ 5 Sol Since the line r i ˆ ˆj k ˆ (i ˆ ˆj k) ˆ and plane r(i ˆ ˆj k) ˆ 5 intersects each other so, i ˆ ˆj k ˆ (i ˆ j ˆ k) ˆ (i ˆ ˆj k) ˆ 5 That is, ( ) 5 0 the point of intersection is r i ˆ ˆj kˆ 0(i ˆ ˆj k) ˆ i ˆ ˆj kˆ ie, Q(,, ) Also let P(, 5, 0) So, the required distance is, PQ ( ) ( 5) ( 0) 9 6 units Q9 A factory manufactures two types of screws A and B, each type requiring the use of two machines, an automatic and a hand-operated It takes minutes on the automatic and 6 minutes on the hand-operated machines to manufacture a packet of screws A while it takes 6 minutes on the automatic and minutes on the hand-operated machine to manufacture a packet of screws B Each machine is available for at most hours on any day The manufacturer can sell a packet of screws A at a profit of 70 paise and screws B at a profit of ` Assuming that he can sell all the screws he manufactures, how many packets of each type should the factory owner produce in a day in order to maimize his profit? Formulate the above LPP and solve it graphically and find the maimum profit 0 Eclusively for Maths, click at : wwwtheopguptacom WhatsApp @ +996505080,
Solved Papers Of CBSE Eams Annual Eam Delhi (08) Sol Let and y be the number of packets of screws of type A and type B respectively To maimize : Z (070 y) in ` Subject to constraints :, y 0, 6y 0 y 0, 6 y 0 y 80 Corner Points Value of Z A(0, 0) 8 B(0, 0) C(0, 0) 0 O(0, 0) 0 Hence maimum profit of ` is obtained at B(0, 0) Therefore, number of packets of screw of type A is 0 and that of type B is 0 # Note that the papers of Delhi, All India and Foreign regions were same All Rights Reserved with OP Gupta Disclaimer : All care has been taken while preparing this solution draft Still if any error is found, please bring it to our notice Kindly forward your concerns/feedbacks through message or WhatsApp @ +996505080 or mail at theopgupta@gmailcom Let s learn Maths with smile:-) Eclusively for Maths, click at : wwwtheopguptacom WhatsApp @ +996505080