Operating C 1 C 1 C 2 and C 2 C 2 C 3, we get = 0, as R 1 and R 3 are identical. Ans: 0

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Clarence Dean
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1 Q. Write the value of MATHEMATICS y y z z z y y y z z z y Operating R R + R, we get y z y z z y z y ( y z) z y Operating C C C and C C C, we get ( y z) z y y ( y z)( ) z y y = 0, as R and R are identical. Ans: 0 Q. Write the sum of the order and degree of the following differential d dy 0 Given differential equation is d dy 0 dy d y. 0 Order =, degree = Order + degree = Ans: Q. Write the integrating factor of the following differential equation: y y y dy ( ) ( cot ) 0 Given differential equation is y y y dy ( ) ( cot ) 0 y y y dy ( ) cot y cot y dy y 05 Wiley India Pvt. Ltd.

2 It is a linear differential equation of the form P Q; where P and Q are functions of y and its dy integrating factor is given by Ans: ( + y ) y dy y ln( y ) Pdy e e e ( y ) Q4. If aˆ, bˆ and c ˆ are mutually perpendicular unit vectors, then find the value of aˆb ˆ cˆ. Given that ab ˆ, ˆ and ĉ are mutually perpendicular unit vectors. aˆ bˆ cˆ and aˆ. bˆ bˆ. cˆ cˆ. aˆ 0 () Now aˆ bˆ cˆ ( aˆ bˆ cˆ ).( aˆ bˆ cˆ ) 4 aˆ aˆ. bˆ aˆ. cˆ bˆ. a bˆ bˆ. cˆ cˆ. aˆ cˆ. bˆ cˆ 4() () 6 aˆ bˆ cˆ 6, but modulus of a vector is non-negative quantity. Ans: 6 aˆ bˆ cˆ 6 Q5. Write a unit vector perpendicular to both the vectors a iˆ ˆj kˆ and b iˆ ˆ. j a iˆ ˆj kˆ, b iˆ ˆj A unit vector perpendicular to vectors a and b is given by ( a b) a b () Now, iˆ ˆj kˆ a b iˆ(0 ) ˆj (0 ) kˆ ( ) 0 iˆ ˆj 0kˆ and a b ( ) () (0) Required unit vector ( a b) ( iˆ ˆj ) ˆ i a b ˆj or iˆ ˆj Ans: ˆ ˆ i j or iˆ ˆj Q6. The equations of a line are 5 = 5y + 7 = 0z. Write the direction cosines of the line. Given line is 5 = 5y + 7 = 0z y 0 z y z Direction ratios of line are <6,, > or < 6,, > 05 Wiley India Pvt. Ltd.

3 Direction cosines are 6,, ; k k k or 6,, ; k (6) () ( ) 7 k k k Direction cosines of given line are 6,, or 6,, Ans: 6,, or 6,, Section B Q7. To promote the making of toilets for women, an organization tried to generate awareness through (i) house calls (ii) letters, and (iii) announcements. The cost for each mode per attempt is given below: (i) ` 50 (ii) ` 0 (iii) ` 40 The number of attempts made in three villages X, Y and Z are given below: (i) (ii) (iii) X Y Z Find the total cost incurred by the organization for the three villages separately, using matrices. Write one value generated by the organization in the society. Cost of per attempt through house calls = 50 ` Cost of per attempt through letters = 0 ` Cost of per attempt through announcements = 40 ` 50 We have a column matri B 0 defining per attempt cost through the given medium respectively. 40 Also the matri defining the number of attempts done in X, Y, Z Villages is A Ependitures incurred by the organization is given by E = A B , ,000 5, ,000 5, ,000 Ependiture on Village X = 0, 000 ` Ependiture on Village Y =,000 ` Ependiture on Village Z = 9,000 ` 05 Wiley India Pvt. Ltd.

4 and total ependiture = 0,000 +, ,000 = 9,000 The value generated by the organization in the society is RESPECT AND CARE FOR WOMEN IN OUR SOCIETY. Q8. Solve for : tan ( ) tan ( ) tan 8 tan ( ) tan ( ) tan 8 We know that, y tan tan ytan y for y < y tan y y tan y for y >, > 0, y > 0 for y >, < 0, y < 0 if y =,, y, > 0; if y =,, y, < 0 Case (i) ( )( ) t(, ), then tan ( ) tan ( ) tan tan tan tan tan ( ) 8 6 = = 0 ( tan is an injective function) = 0 4 ( + 8) ( + 8) = 0 (4 ) ( + 8) = 0 = /4 or = 8 But (, ) 4 Case (ii) ( + ) ( ) > ; + > 0, > 0 > ; > ; > >, >, > >, > > i.e., (, ), then tan ( ) tan ( ) tan 8 tan 8 tan 8 = /4 or 8, both (, ) Case(iii) ( + ) ( ); + < 0, < 0 05 Wiley India Pvt. Ltd.

5 > ; < ; < ( 8, ), then tan ( ) tan ( ) tan 8 8 tan tan tan 8 tan which is impossible as L.H.S. 0,, Case(iv) ( + ) ( ) = ; + > 0, > 0, <, >, then tan ( ) tan ( ) tan 8 8 tan which is impossible Case (v) ( + ) ( ) = +; + < 0, < 0, then tan ( ) tan ( ) tan 8 where as R.H.S., 8 tan ; which is again impossible. Ans: Thus = /4 is the only solution. Q8. Prove the following: y yz z y y z z cot cot cot 0 y yz z cot cot cot y y z z (0 < y, y, z < ) (cot y cot ) (cot z cot y) cot cot z = 0 Q9. Using properties of determinants, prove the following: a bc ac c a ab b ac 4 a b c. ab b bc c a bc ac c a ab b ac ab b bc c 05 Wiley India Pvt. Ltd.

6 a c a c abc a b b a (Taking a, b and c common from C, C and C respectively) b b c c Operating R R + R + R, we get ( a b) ( c b) ( a c) abc a b b a b b c c a b c b a c abc a b b a b b c c Operating C C + C + C, we get ( a b c) c b a c a b c c b a c abc ( a b) b a 4abc a b b a ( b c) b c c b c b c c Operating C C C and C C C gives a b c a b a b c 4abc a b a b 4a b c a b b c 0 b b c 0 Operating R R R and R R R gives abc 4a b c c 0 0 4a b c a 0 Hence proved Q0. Find the adjoint of the matri A C C C C C C Ajoint(A) C C C C C C ; A and hence show that A.(adj A) = A I. C C C C C C Where C = ( ) ( 4) = ; C = ( ) ( + 4) = 6; C = ( ) 4 ( 4 ) = 6; C = ( ) ( = 6; C = ( ) 4 ( + 4) = ; C = ( ) 5 ( + 4) = 6; C = ( ) 4 (4 + ) = 6; C = ( ) 5 ( + 4) = 6; C = ( ) 6 ( + 4) = 6 6 Adjoint(A) 6 6 Ans. 6 6 Also A = ( )C C RC = ( ) ( ) ( 6) ( 6) = + + = 7 Now 05 Wiley India Pvt. Ltd.

7 A.(Adj. A) I A I 0 0 Hence A.(Adj. A) = A I Q. Show that the function f() = + +, for all R, is not differentiable at the points = and =. f() = + + ; ε R ; if f ( ) ( ) ( ) if ( ) ( ) if ; if ; if ; if f() is continuous at = and at = as Q. If Q. If also, lim f ( ) lim ( ) ; ( ) ( ) ; if f ( ) 0; if ; if lim f ( ) lim ( ) f () () () clearly f ( ) is discontinuous at = and = i.e., L.H.D. =, R.H.D. = 0 at = and L.H.D. = 0, R.H.D. = at = f() is non-differentiable at = and at =. msin y e, then show that ( ) d y dy m y 0. dy y e e msin msin dy my dy. m ( ) my d y dy m dy.. () d y dy dy ( ) m m( my) (using ()) ( ) d y dy m y 0 Hence proved f ( ) ; g( ) and h() =, then find f [ h { g ( )}]. 05 Wiley India Pvt. Ltd.

8 f g ( ) ; ( ) ; To find f [ h[ g ( )]]. h() = Here, f ( ) ; g( ) Also h( ) h( g ( )) R f [ h( g( )] f () 5 () ( ) ( )( ) ( ) ( ) which is defined R Ans: / 5 Q4. Evaluate: ( ). ( ). ( ( ). ( )( ) / / ( ) ( ).( ) 9 / ( ).( ) 4 4 / ( ).( ) / ( ) sin C / ( ) 9 / sin ( ) C 4 OR Q4. Evaluate: ( )( ) ( )( )? Let A B C ( )( ) + + = (A + B) ( + ) + C( + ) For = ; = 5C C = /5 For = 0; = B + C B = C = /5 = /5 B = /5 Comparing the coefficients of, we get = A + C A = C = /5 = /5 05 Wiley India Pvt. Ltd. ()

9 A = /5 Substituting, the values of A, B and C in equation (), we get 5 5 /5 ( )( ) ( ) ( ) ( )( ) 5 ( ) 5 ( ) 5 5 ( ) C ln( ) tan ln or tan ln ( )( ) C 5 Ans: 5 Q5. Find: [tan ln ( )( ) ] /4 0 cos sin C /4 /4 cos sin 0 0 (sin ) (cos ) / 7/ /4 /4 sec (sin ) (cos ) / / 0 0 /4 ( tan )sec tan Put 0 tan 4 ( ) 0 sec / (sin ).cos / (cos ) t.sec dt tan I t dt as at = 0, t = 0 and at, t = 4 5 t 6 t Ans: 6/5 Q6. Find: log ( ) log (log ). ( ) ( ) Integrating by parts, we get 05 Wiley India Pvt. Ltd. I (log ) (log ). II d ( ) ( )

10 (log ) ( ) log log ( ) ( ) ( ) ( ) log log log( ) C Ans. ( ) Q7. If a iˆ ˆj kˆ, b iˆ ˆj and c iˆ 4 ˆj 5 kˆ, then find a unit vector perpendicular to both of the vectors ( a b) and ( c b). a iˆ ˆj kˆ, b iˆ ˆj, c iˆ 4 ˆj 5 kˆ, To find a unit vector perpendicular to both vectors ( a b) and ( c b) Unit vector perpendicular to ( a b) and ( c b) ( a b) ( c b) ( a b) ( c b) () Now iˆ ˆj kˆ ( a b) ( c b) iˆ(0) ˆj (4) R(4) 5 5 4ˆj 4kˆ ( 4 ˆj 4 kˆ) ( 4 ˆj 4 kˆ) Unit vector = Unit Vector ˆ ˆ j k or ˆ j k ˆ Ans Q8. Find the equation of a line passing through the point (,, 4) and perpendicular to two lines r (8iˆ 9 ˆj 0 kˆ) (iˆ 6 ˆj 7 kˆ) and r (5i ˆ 9 ˆj 5 kˆ) (iˆ 8ˆj 5 kˆ). The required line is perpendicular to lines r 8iˆ 9 ˆj 0 kˆ (iˆ 6 ˆj 7 kˆ) and r (5i ˆ aj ˆ 5 kˆ) (iˆ 8 ˆj 5 kˆ) line is along the vector (iˆ 6 ˆj 7 kˆ) (iˆ 8 ˆj 5 kˆ) b (say) iˆ ˆj kˆ b 6 7 iˆ(80 56) ˆj ( 5 ) kˆ (4 48) 8 5 4iˆ 6 ˆj 7kˆ Thus the line is passing through the point (,, 4) and along vector b is r ( iˆ ˆj 4 kˆ) (4iˆ 6 ˆj 7 kˆ) 4 In vector form y z or y z 4 Ans. 6 OR 05 Wiley India Pvt. Ltd.

11 Q8. Find the equation of the plane passing through the points (,, 0), (,, ) and parallel to the line y z. Let the equation of plane passing through the points A(,, 0) and B(,, ) be a( + ) + b(y ) + c(z 0) = 0 () It passes through B(,, ), a + 0b c = 0 () Also the required line is parallel to line y z (a) + (b) + ( )(c) = 0 a + b c = 0 () Solving () and (), we get, a = b, c = b c = b = a direction ratios of normal to the plane are <,, > The equation of required plane will be ( + ) + (y ) + z = 0 or + y + z = 0 Ans Q9. Three cards are drawn successively with replacement from a well shuffled pack of 5 cards. Find the probability distribution of the number of spades. Hence find the mean of the distribution. X = Number of spades drawn in three successive draws with replacement. Clearly X can take values 0,, and Now P( = 0) = P(All three cards non spades) P( = ) = P(eactly one card spade and other two non-spades) 9 9 () (4) 64 P( = ) = P(eactly one non-spade and other two spades) 9 () () (4) P( = ) = P(All three spades) = Probability distribution table is as shown below, X i 0 P(X i ) Mean X ip( i) (0) Ans : 4 OR Q9 For 6 trials of an eperiment, let X be a binomial variate which satisfies the relation 9P(X = 4) = P(X = ). Find the probability of success. Here, n = 6, Let P = Probability of success According to question, qp( = 4) = p( = ) p C ( p) ( p) C ( p) ( p) Wiley India Pvt. Ltd.

12 p 9 p ( p) p p but p, p > 0 p p = p p 4 05 Wiley India Pvt. Ltd. SECTION C Q0. Consider f : R + [ 9, ] given by f() = Prove that f is invertible with 54 5y f ( y). 5 f: R + [ 9, ) and f() = f ( s) for > /5, 0, 5 f() is injective function for f() is also injective on R +, f() being a polynomial function is continuous and also increasing Range of f ( ) [ f (0), f ( )) [ 9, ) co domain f() is also subjective f() is invertible (as bijective) Let y = y = 0 (5) 6 (6) 4(5)( 9 y) y 6 0y y 54 5y But 5y 54 R 0 5 5y 54 f ( y) ; 5 Hence proved Q0. A binary operation * is defined on the set = R { } by * y = + y + y,, y X. OR Check whether * is commutative and associate. Find its identity element and also find the inverse of each element of X. *:X X X; where X = R < > and * y = + y + y,, y X. Commutativity: * y = + y + y = y + + y = y * * is commutative in X. Associativity:-, y, z X, then *(y * z) = * (y + z + yz) = + (y + z + yz) + (y + z + yz) = ( + y + y) + z + yz + (z + yz) = [( * y) + z] + (y + + y)z R

13 =[( * y) + z] + ( * y).z = ( * y) * z Thus * is associate in X Clearly 0 X R and for X, * (0) = (0) = 0 is the identity element in X. Let y X and be the inverse of X ( * y) = 0 (y * ) y X R Thus for any X, inverse of it is Q. Find the value of p for when the curves = 9P (9 y) and = p(y + ) cut each other at right angles. Given curves are = 9p(9 y) and = p(y + ) At the point of intersection 9p(9 y) = P(y + ) 8 9y = y + y = 8 p Point of intersection is ( p,8) Q (say) For curve, = 9p(9 y), dy dy 9p 9p m (at Q) ( p) 9 p For curve, = p(y + ), dy dy p p ( p) m (at Q) p For curves to cut at light angle, m. m = ( ) 6 p = 4 9 p Ans: p = 4 Q. Using integration, prove that the curves y = 4 and = 4y divide the area of the square bounded by = 0, = 4, y = 4, and y = 0 into three equal parts. Given curve are y = 4 and = 4y We are to prove A = A = A. 05 Wiley India Pvt. Ltd.

14 A [64 0] sq.units / 4 ( ) 4 / (4) sq.units A A / A A sq.units Also A + A + A = Area of square OABC = 6 sq. units Thus A A A 6 sq.units Q. Show that the differential equation dy y y is homogeneous and also solve it. Given differential equation is dy y y / y y/ dy ( y/ ) f ( y/ ) ( y/ ) dy n y Which is a homogeneous equation as it is of the form f ; Now put dy dv y v V, dv v dv v v v dv v v v where n is a whole number. v dv ln v ln v c v = ln + ln v = v + c y = ln (v) = v + c ln y c = ln y = y + c Q. Find the particular solution of the differential equation (tan y )dy = ( + y ), given that = when y = 0. Given differential equation is (tan y )dy = ( + y ), and At =, y = 0 tan y dy y y or OR tan y dy y y Which is a linear differential equation of the type P Q; dy where P tan y, Q y y Integrating factor dy y Pdy e e e tan y Solution is given by, ( I. F.) Q( I. F) dy C tan y tan y tan y. e. e dy c y 05 Wiley India Pvt. Ltd.

15 t t. e dt c; where t = tan y t t te. e dt c t t tan y tan y te e c (tan y) e e c But f() = 0 = + c c = tan y tan y. e e (tan y ) is the particular solution. or tan y (tan y ) e Q4. Find the distance of the point P(, 4, 4) from the point, where the line joining the points A(, 4, 5) and B(,, ) intersects the plane + y + z = 7. Equation of line AB is, y 4 z 5 r 4 5 Any point on line is ( + 4, 4 r, 5 6r) = Q (say) ( + r) + ( 4 r) + ( 5 6r) = 7 r = Q (,,7) PQ ( ) (4 ) (4 7) 7 Ans: 7 Q5. A company manufactures three kinds of calculators : A, B and C in the two factories I and II. The company has got an order for manufacturing at least 6400 calculators of kind A, 4000 of kind B and 4800 of kind C. the daily output of factory I is of 50 calculators of kind A, 50 calculators of kind B, and 0 calculators of kind C. The daily output of factory II is of 40 calculators of kind A, 0 of kind B and 40 of kind C. the cost per day to run factory I is `,000 and of factory II is ` 5,000. How many days do the two factories have to be in operation to produce the order with the minimum cost? Formulate his problem as an LPP and solve it graphically. Let the factory I is operated for days and factory II is operated for y days. Production of calculators of type A = y 6400 (given) Production of calculators of type B = y Wiley India Pvt. Ltd.

16 Production of calculators of type C =0 + 40y 4800 Cost = y. Thus the L.P.P. is To minimize cost c = y; Subject to constraints: y 6400 or 5 + 4y 640 () y 4000 or 5 + y 400 () and y 4800 or + 4y 480 () 0, y 0 (4) Draw the graphs of 5 + 4y = 640, 5 + y = 400 and + 4y = 480 on same frame of reference and then find the feasible region satisfying the inequalities () to (4) as shown below. The feasible region is unbounded. At A(60, 0) Cost = 9,0,000 ` At B (80, 60) Cost = 8,60,000 ` At C (, 0) Cost =, 84,000 ` At D (0, 00) Cost = 0,00,000 ` We observe that the minimum cost appears to be at B(80, 60). But the feasible region is unbounded so we are to analyze further. Draw the region represented by inequality y < 8,60,000 or + 5y < 80 or 4 + 5y < 60 The corresponding region is represented by shaded region below line EF (ecluding the line itself). Thus we observe, that there is no common region between the feasible region and the region represented by inequality 4 + 5y < 60 or y < 8,60,000 Cost below 8,60,000 is impossible 05 Wiley India Pvt. Ltd.

17 Minimum cost = 8,60,000 ` at = 80, y = 60 i.e., Factory A should work for 80 days and Factory B should work for 60 days. Q6. In a factory which manufactures bolts, machines A, B and C manufacture respectively 0%, 50% and 0% of the bolts. Of their outputs, 4 and percent respectively are defective bolts. A bolts is drawn at random from the product and is found to be defective. Find the probability that this is not manufactured by machine B. Let E = Event that the bolt is manufactured by Machine A; P(E ) = 0/00 E = It is produced by machine B; P(E ) = 50/00 E = It is produced by machine C; P(E ) = 0/00 A Event that the drawn bolt is defective E We are to find P A Conditional Probability that the defective drawn bolt is manufactured by machine B. Also A P E Probability of defective bolt produced by machine A. 00 A 4 Similarly, P E and 00 By Baye s theorem, A P E 00 A P. P( E ) E E P A A A A P P E P P E P P E E E E. ( ). ( ) ( ) Ans. 05 Wiley India Pvt. Ltd.

Mathematics. Class 12th. CBSE Examination Paper 2015 (All India Set) (Detailed Solutions)

Mathematics. Class 12th. CBSE Examination Paper 2015 (All India Set) (Detailed Solutions) CBSE Eamination Paer (All India Set) (Detailed Solutions) Mathematics Class th z z. We have, z On aling R R R, we get z z z z (/) Taking common ( z) from R common from R, we get ( z)( ) z ( z)( ) [ R R