CBSE Eamination Paper, Foreign-4 Time allowed: hours Maimum marks: General Instructions: As per given in CBSE Eamination Paper Delhi-4. SET I SECTION A Question numbers to carr mark each.. Let R = {(a, a ) : a is a prime number less than 5} be a relation. Find the range of R.. Write the value of cos + sin.. Use elementar column operation C C C in the matri equation 4 =. 4. If a 4 b 8 6 = a b 8 a 8b write the value of a b. 5. If A is a matri, A and A = k A, then write the value of k. 6. Evaluate: 7. Evaluate: / 4 sin cos tan 8. Write the projection of vector i j k along the vector j. 9. Find a vector in the direction of vector i j 6k which has magnitude units.. Find the angle between the lines r i 5 j k ( i j 6 k) and r 7 i 6 k (i j k).
564 Xam idea Mathematics XII SECTION B Question numbers to carr 4 marks each.. Let f : W W, be defined as f() =, if is odd and f() = +, if is even. Show that f is invertible. Find the inverse of f, where W is the set of all whole numbers.. Solve for : Prove that : cos (tan ) = sin cot 4 OR cos 7 + cot 8 + cot 8 = cot. Using properties of determinants, prove that a a z z a z a ( a z ) 4. If = a cos + b sin and = a sin b cos, show that d d. 5. If m n = ( + ) m + n, prove that d. 6. Find the approimate value of f(.), upto places of decimal, where f() = + 5 +. OR Find the intervals in which the function f() = 4 45 5 is (a) strictl increasing (b) strictl decreasing. 7. Evaluate: cos OR Evaluate: ( ) 4 8. Solve the differential equation ( ) d + ( + ) =, given that =, when =. 9. Solve the differential equation d + cot = cos, given that =, when =.. Show that the vectors a, b, c are coplanar if and onl if a b, and b c and c a are coplanar. OR Find a unit vector perpendicular to both of the vectors a b and a b where a i j k, b i j k.
Eamination Papers 4 565. Find the shortest distance between the lines whose vector equations are r ( i j ) ( i j k ) and r ( i j k ) ( i 5j k ).. Three cards are drawn at random (without replacement) from a well shuffled pack of 5 plaing cards. Find the probabilit distribution of number of red cards. Hence find the mean of the distribution. Question numbers to 9 carr 6 marks each. SECTION C. Two schools P and Q want to award their selected students on the values of Tolerance, Kindness and Leadership. The school P wants to award ` each, ` each and ` z each for the three respective values to, and students respectivel with a total award mone of `,. School Q wants to spend `, to award its 4, and students on the respective values (b giving the same award mone to the three values as school P). If the total amount of award for one prize on each value is `,, using matrices, find the award mone for each value. Apart from these three values, suggest one more value that should be considered for award. 4. Show that a clinder of a given volume which is open at the top has minimum total surface area, when its height is equal to the radius of its base. 5. Evaluate : tan sec tan 6. Find the area of the smaller region bounded b the ellipse and the line. 9 4 7. Find the equation of the plane that contains the point (,, ) and is perpendicular to both the planes + z = 5 and + z = 8. Hence find the distance of point P(, 5, 5) from the plane obtained above. OR Find the distance of the point P(, 5, ) from the point of intersection of the line joining the points A(,, ) and B(5,, 4) with the plane + z = 5. 8. A cottage industr manufactures pedestal lamps and wooden shades, each requiring the use of a grinding/cutting machine and a spraer. It takes hours on the grinding/cutting machine and hours on the spraer to manufacture a pedestal lamp. It takes hour on the grinding/cutting machine and hours on the spraer to manufacture a shade. On an da, the spraer is available for at the most hours and the grinding/cutting machine for at the most hours. The profit from the sale of a lamp is ` 5 and that from a shade is ` 5. Assuming that the manufacturer can sell all the lamps and shades that he produces, how should he schedule his dail production in order to maimise his profit. Formulate an LPP and solve it graphicall. 9. An insurance compan insured scooter drivers, 4 car drivers and 6 truck drivers. The probabilities of an accident for them are.,. and.5 respectivel. One of the insured persons meets with an accident. What is the probabilit that he is a scooter driver or a car driver?
566 Xam idea Mathematics XII OR Five cards are drawn one b one, with replacement, from a well shuffled deck of 5 cards, Find the probabilit that (i) all the five cards are diamonds. (ii) onl cards are diamonds. (iii) none is a diamond. SET II Onl those questions, not included in Set I, are given 9. Evaluate : / 4 sin. Write a unit vector in the direction of vector PQ, where P and Q are the points (,, ) and (4, 5, 6) respectivel. 9. Using properties of determinants, prove that: ( 5 )( ). If e e e, prove that d e.. Find a particular solution of the differential equation d tan = sin, given that =, when =.. Find the shortest distance between the following lines : = z ; 5 z 7 7 6 8. A window is of the form of a semi-circle with a rectangle on its diameter. The total perimeter of the window is m. Find the dimensions of the window to admit maimum light through the whole opening. 9. Evaluate : SET III a cos b sin Onl those questions, not included in Set I and Set II are given. 9. Write the value of the following : i ( j k ) j ( k i ) k ( i j ). Evaluate : e
Eamination Papers 4 567 9. Find the distance between the lines l and l given b l : r i j 4k ( i j 6k ) ; l: r i j 5k ( 4i 6j k ).. Solve the differential equation log d + = log.. If cos = cos (a + ), where cos a, prove that d. Prove the following, using properties of determinants: a a ab ab b bc b bc ac c ac c 4a b c cos ( a ). sin a 8. The sum of the perimeters of a circle and a square is k, where k is some constant. Prove that the sum of their areas is least when the side of the square is equal to the diameter of the circle. 9. Evaluate: / 4 sin cos sin 9 6
568 Xam idea Mathematics XII SET I SECTION A. Here R = {(a, a ) : a is a prime number less than 5} R = {(, 8), (, 7)} Hence Range of R = {8, 7}. We have, cos + sin cos cos cos Also sin = = sin sin 6 [ cos ( ) = cos ] [, ] = 6, 6 cos sin 6 = [Note: Principal value branches of sin and cos are, and [, ] respectivel.]. Given 4 = Appling C C C, we get 4 6 = 4 4. Given a 4 b 8 6 = a b 8 a 8b On equating, we get a + 4 = a +, b = b +, a 8b = 6 a =, b = Now the value of a b ( ) 5. Here, A = k A A = k A [ ka = k n A where n is order of A] Solutions k = 7 7 A = k A
Eamination Papers 4 569 6. Let I = sin. cos = cosec. sec = ( cot ). sec = sec cot. sec = tan + sec tan = tan + z dz = tan + z + c = tan c z = tan / 4 7. Let I = tan = [Let tan = z sec = dz] c [Putting z = tan ] tan sin cos Let cos = z sin = dz sin = dz For limit, if =, z = ; = 4 z = I dz z = dz z [log ] log log z log log 8. Required projection = ( i j k ). j = 9. Required vector = j i j 6k = = ( ) 6 = i j 6k 49 = i j 6 k = ( i j 6k ) 7 = 6i 9j 8k
57 Xam idea Mathematics XII. Given two lines are r i 5 j k ( i j 6 k) r 7 i 6 k (i j k). Parallel vectors of both lines are k i j 6k, k i j k Required angle = angle between k and k. If be required angle. then k. k cos = k. k cos = 4 49 9 cos = 9 SECTION B cos = 9 7 = cos 9. In order to prove that f is invertible, we have to prove f is one-one onto function. For one-one Case I Let, both be odd numbers Now f( ) = f( ) =, W i.e. Case II f is one-one. = Let, both be even number Now f( ) = f( ) + = + i.e. Case III i.e. f is one-one. = Let be even and be odd. If f( ) = f( ) + = = = Which is not possible as the difference of even and odd is alwas odd. f( ) f( ) when is even and is odd. i.e. f( ) f( ) Hence f is one-one function....(a) For Onto f() = if is odd f() = + if is even even number Wc odd number ( + ) Wd as f pre image and odd number Wc we have even number ( ) Wd as f pre image. Hence f is onto function...(b) A and B implies that f is one-one and onto function i.e. f is invertible function.
Eamination Papers 4 57 For Inverse Function Let f () be inverse of f() fof = I fof () = I f (f ()) = [ I() = ] f () = if f () is odd and f () + = if f () is even f () = + if is even f () = i.e. f = f.. Given cos(tan ) sin cot 4 if is odd cos(tan ) cos cot 4 tan cot 4 cot cot cot 4 cot 4 OR Note: sin cos tan cot We have, L.H.S. = cot 7 cot 8 cot 8 = tan tan tan 7 8 = tan 7 8 tan 8 7 8 = tan tan = tan 8 8 8 65 = tan 98 tan 65 tan 95 95 98 = cot = RHS 4 8 7 8 8
57 Xam idea Mathematics XII. L.H.S = a a z z a z Appling C C C C, we get a z = a z a z = (a + + + z) a a z z a z z z a z Appl R R R, we get = (a + + + z) Epanding along R, we get 4. Given a a z a z = (a + + + z) { + a (a + z z)} = a (a + + + z) = a cos + b sin = a sin + b cos d Also = a sin b cos d = a cos + b sin d d d = d = a cos b sin a sin b cos d d = a cos b sin b cos a sin d = Differentiating again w.r.t., we get d. d = = d = (i) (ii) [From (i) and (ii)]... (iii)... (iv)
Eamination Papers 4 57 Now d d + = + 5. Given m n = ( + ) m + n = + Taking logarithm of both sides, we get log m n = log ( + ) m + n = + [From (iii) and (iv)] = log m + log n = (m + n) log ( + ) [B law of logarithm] m log + n log = (m + n) log ( + ) Differentiating both sides w.r.t. we get m n d. = m n d m m m n m n n n d =. ( ) ( ). m n m n d = ( ) ( ) 6. Here f() = + 5 + Let = and =. + =. m m n = m n n d. d = B definition, we have approimatel f () = f ( ) f ( ) f () = f(. ) f( ) [Putting = and =.]. f () = f(. ) f( ). Now f() = + 5 +... () f () = 6 + 5 f () = Also f() = + 5 + = 7 + 5 + = 45 Putting in () we get = f (. ) 45. f(.) =. + 45 = 45.46 OR Here, f() = 4 4 45 + 5 f () = 6 9 f () = 6( 5) = 6( + ) ( 5)
574 Xam idea Mathematics XII Now for critical point f () = 6( + )( 5) = =,, 5 i.e.,, 5 are critical points which divides domain R of given function into four disjoint sub intervals (, ), (, ), (, 5), (5, ) For (, ) For (, ) For (, 5) For (5, ) f () = +ve ve ve ve = ve i.e. f() is decreasing in (, ) f () = +ve ve +ve ve = +ve i.e. f() is increasing in (, ) f () = +ve +ve +ve ve = ve i.e. f() is decreasing in (, 5) f () = +ve +ve +ve +ve = +ve i.e. f() is increasing in (5, ) Hence f() is (a) strictl increasing in (, ) (5, ) (b) strictl decreasing in (, ) (, 5) 7. Let cos I = Let cos = z = dz I = cos z. z dz = (z. sin z sin z dz c) ( z.sin z cos z c) z sin z cosz c dz I = cos. + c [ = cos z sin z = ] I = cos + c OR Let I = ( ) Let = A d ( + + ) + B = A ( + ) + B = A + (A + B) -ve +ve -ve +ve 5
Eamination Papers 4 575 Equating we get A = and A + B = A = and B = = 7 Now, I = 7 ( ) = ( ) 7 I = I 7 I... (i) Where, I = ( ) and I = Now, I = ( ) Let + + = z ( + ) = dz I = z dz = z / + c z c I = ( + +) + c... (ii) Again I = =.. 4 = I = + Putting value of I and I from (ii), (iii) in (i), we get. log c... (iii) I = ( ) 7 4 6 log + c 8. The given differential equation is ( ) d + ( + ) = ( ) d + ( + ) = ( ) ( ) d = ( + ) d = ( ) [where c c c ]
576 Xam idea Mathematics XII d = Integrating both sides, we get d = d d = log = + c It is general solution Now putting = and = in general solution, we get log = + c c = we have particular solution as log + = + + 9. Given differential equation is d d + cot = cos + cot. = cos It is in the form d + P = Q. where P = cot, Q = cos cot I.F. = e = e log sin = sin. Therefore, general solution is. sin = cos. sin c sin = sin c sin = cos + c sin = cos + c Now put and in the above equation, we get sin cos c ( ) c ( cos ) c cos The particular solution is sin or sin ( cos ). If part: Let a, b, c are coplanar scalar triple product of a, b and c is zero [a b c] = a.( b c) b.( c a) c.( a b) =
Eamination Papers 4 577 Now, [ a b b c c a] = ( a b). {( b c) ( c a)} = ( a b).{ b c b a c c c a} = ( a b).{ b c b a c a} [ c c ] = a.( b c) a.( b a) + a.( c a) b.( b c) b.( b a) b.( c a) = [ a b c] [ b c a] [B propert of scalar triple product] = [ a b c] [ a b c] = [ a b c] = = [[ a b c] = ] Hence, a b, b c, c a are coplanar Onl if part: Let a b, b c, c a are coplanar. [ a b b c c a] ( a b).{( b c) ( c a)} ( a b).{ b c b a c c c a} ( a b).{ b c b a c a} a.( b c) a.( b a) a.( c a) b.( b c) b.( b a) b.( c a) [ a b c] [ b c a] [ a b c] [ [ a b c] [ b c a ] [ a b c] Hence, a, b, c are coplanar. OR a b = ( i j k ) ( i j k ) = i j 4k a b = ( i j k ) ( i j k ) = j k Perpendicular vector of ( a b) and ( a b) = ( a b) ( a b) = ( i j 4k ) ( j k ) = i j k 4 = ( 6 + 4) i ( 4 ) j + ( ) k = i 4j k
578 Xam idea Mathematics XII Required unit vector perpendicular to ( a b) and ( a b) = i 4j k ( ) 4 ( ) = i 4j k 4 6 4 = i 4j k = 4j i k 4 6 6 6 = i j k 6 6 6. Comparing the given equations with equations r a b and r a b, we get a i j, b i j k and a i j k, b i 5j k Therefore, a a ( i k ) and i j k b b ( i j k) ( i 5 j k ) b b 9 49 59 5 i j 7 k Hence, the shortest distance between the given lines is given b ( b b ) ( a a ) d 7 b b 59 59 units.. Let the number of red card in a sample of cards drawn be random variable X. Obviousl X ma have values,,,. Now P(X = )= Probabilit of getting no red card = Hence, 6 C 5 C = 6 = 7 P(X = )= Probabilit of getting one red card and two non-red cards = 6 C 5 C 6 C = 845 = 4 P(X = )= Probabilit of getting two red card and one non-red card = 6 C 5 C 6 C = 845 = 4 P(X = )= Probabilit of getting red cards = the required probabilit distribution in table as X P(X) 7 4 4 6 C 5 C = 6 7 = 7
Eamination Papers 4 579 Required mean = E(X) p i i = 7 = 6 4 4. According to question, + + z = 4 + + z = + + z = 4 4 6 6 = 7 4 The above sstem of equation ma be written in matri form as i.e., AX B X = A B where A 4, X, B z 7 = 5 4 = A = 4 ( ) ( 4 ) ( 4 ) 6 5 A eist Now, A = ( ) =, A = (4 ) =, A = (4 ) =, A = ( ) =, A = ( ) =, A = ( ) = A = (6 ) = 5, A = (9 4) = 5, A = ( 8) = 5 Adj (A) = 5 5 5 A = A (Adj A) = 5 X = A B. z z 5 5 T 5 5 5 5 5 5 5 5 5 5 5, 4, z 5 5 5 5 5 z 5 4 z 5 44 6 6 6 66 6 i.e., ` for tolerance, ` 4 for kindness and ` 5 for leadership are awarded. One more value like punctualit, honest etc ma be awarded.
58 Xam idea Mathematics XII 4. Let r and h be radius and height of given clinder having volume V. If S is surface area then V r h S = rh r V h r V S = r. r S = V r r r ds V r dr r For etremum value of S, ds dr = V V r = r r r r = V r = V Again d S 4V dr r Now, d S dr V Ve r Hence, for r = V S is minimum. Therefore, for minimum value of surface area r = V r = r h [V r h] h r r r h r = h i.e. radius = height tan 5. Let I = sec tan a a As f( ) f( a ) I ( ) tan ( ) sec ( ) tan ( ) (i) ( ) tan (ii) sec tan B adding equations (i) and (ii), we get I tan sec tan Multipling and dividing b (sec tan ), we get
Eamination Papers 4 58 I tan (sec tan ) sec tan (sec tan tan ) sec tan sec sec tan ( ) ( ) ( ) I ( ) I ( ) 6. Given curves are and 9 4 We have 9 and ( ) So, area of required region ( ) [ 9 ( )] [ ( ) ( )] 9 9 sin 9 9 9 9 9 sq units. 4 7. Equation of plane containing the point (,, ) is given b a( ) + b ( + ) + c (z ) =... (i) (i) is perpendicular to plane + z = 5 a + b c =... (ii) Also (i) is perpendicular to plane + z = 8 a + b c =... (iii) From (ii) and (iii) a 9 4 = b 6 = c 4 a b c (sa) 5 4 a = 5, b = 4, c = Putting these values in (i) we get 5 ( ) + 4( + ) + (z ) = 5 ( ) + 4( + ) + (z ) = 5 + 5 + 4 + 4 + z = 5 + 4 + z + 7 = X + = 9 4 Y Q(, ) O Y P(, ) + = X
58 Xam idea Mathematics XII 5 4 z 7 =... (iv) It is required equation of plane. Again, if d be the distance of point p (, 5, 5) to plane (iv) Then d = 5 ( 4 ) 5 ( ) 5 7 5 ( 4) ( ) 5 7 = = 4 = 4 unit 5 6 4 OR The vector form of line and plane can be written as r ( i j k ) ( i 4 j k ) (i) r. ( i j k ) 5 (ii) For intersection point, we solve equations (i) and (ii) b putting the value of r from (i) in (ii). [( i j k ) ( i 4j k )].( i j k ) 5 ( ) ( 4 ) 5 5 5 Hence, position vector of intersecting point is i j k. i.e., coordinates of intersection of line and plane is (,, ). Hence, Required distance ( ) ( 5) ( ) = 9 6 44 69 units 8. Let the manufacturer produces padestal lamps and wooden shades; then time taken b pedestal lamps and wooden shades on grinding/cutting machines ( ) hours and time taken on the spraer ( ) hours. Since grinding/cutting machine is available Y for at the most hours. and spraer is available for at most hours. Thus, we have 9 C(,) 8 7 Now profit on the sale of lamps and 6 shades is, 5 Z 5 5. 4 B (4, 4) So, our problem is to find and so as to Maimise Z 5 5 (i) Subject to the constraints: (ii) (iii) (iv) + = X O Y + = A (6, ) 4 5 6 7 8 X
Eamination Papers 4 58 (v) The feasible region (shaded) OABC determined b the linear inequalities (ii) to (v) is shown in the figure. The feasible region is bounded. Let us evaluate the objective function at each corner point as shown below: Corner Points Z 5 5 O (, ) A ( 6, ) 5 B ( 4, 4 ) 6 Maimum C (, ) 5 We find that maimum value of Z is ` 6 at B( 4, 4 ). Hence, manufacturer should produce 4 lamps and 4 shades to get maimum profit of `6. 9. Let E, E, E and A be events such that E = Selection of scooter drivers E = Selection of car drivers. E = Selection of truck drivers. A = meeting with an accident. 4 6 P( E ), P( E), P( E) 6 P(A/E ) =. = P(A/E ) =. = P(A/E ) =.5 = 5 P( E). P( A / E) P(E /A) = P( E ). P( A / E ) P( E ). P( A / E ) P( E ). P( A / ) = 6 5 5 5 5 6 5 5 6 45 5 Therefore, required probabilit = P E A = 45 7 5 5 OR Let number of diamond cards be taken as random variable X. X ma have values,,,, 4, 5. Here, p = probabilit of drawing diamond card in one draw E
584 Xam idea Mathematics XII SET II = 5 4 q = probabilit of drawing non diamond card in one draw = 4 4 Here, drawing a card is "Bernoullian trails" therefore we can appl P(X = r) = n C r p r q n r where n = 5. (i) P (getting all the five cards diamond) = P(X = 5) = 5 C5p 5. q = 5 5 5 C 5 4 4 4 4 (ii) P (getting onl cards diamond) = P(X = ) = 5 C p. q = 5 C 4 4 (iii) P (getting no card diamond) = P(X = ) = 5 5 C. p q = 5 C 4 4 / 4 9. I = sin cos = 4 = [ ] = = cos cos. PQ = (4 ) i ( 5 ) j ( 6 ) k = i j 6k 9. L.H.S. Required unit vector = i j 6k 6 5 = i j 6 k 49 45 5 4 4 6 i j k 7 7 7 5 5 5 [Appling C C C C ] ( 5 ) [Taking out ( 5 ) common from C ]
Eamination Papers 4 585 ( 5 ) Epanding along C, we get. Given e e e ( 5 ) ( ) R.H.S. Differentiating both sides we get d d e e. e [Appling R R R and R R R] e + e. d = e+ + e +. d d ( e e ) e d ( e e e ) e e e [ e e e (given)] d e. e d e. Given differential equation is d tan. sin Comparing it with d P Q, we get P tan, Q sin tan I. F. e log sec e Hence general solution is. sec sin. sec C d e e d e e log sec sec [ log e z z ].sec sec. tan C. sec sec C cos C cos Putting and, we get cos C. cos C C 4 Required solution is cos cos. Let 5 z 7 z and k 7 6 Now, let s take a point on first line as e
586 Xam idea Mathematics XII A (, 5, 7 ) and let B ( 7k, 6k, k ) be point on the second line The direction ratio of the line AB 7k 4, 6k 6, k 8 Now as AB is the shortest distance between line and line so, ( 7k 4) ( 6k 6) ( ) ( k 8)...(i) and ( 7k 4) 7 ( 6k 6) ( 6) ( k 8)...(ii) Solving equation (i) and (ii), we get and k A (, 5, 7 ) and B (,, ) AB ( ) ( 5 ) ( 7 ) 6 6 64 6 units = 9 units 8. Let ABCED be required window having length and width. If A is the area of window. Given, Perimeter Then A ( ) Obviousl, window will admit maimum light and air if its area A is maimum. Now, da E For maima or minima of A da ( 4 ) d A Also, ( 4 ) < 4 For maimum value of A, 4 and thus 4 Therefore, for maimum area, i.e., for admitting maimum light and air, Length of rectangular part of window = 4 and Width = 4 D A C B
Eamination Papers 4 587 9. Let I I I a cos b sin a cos ( ) b sin ( ) a cos b sin a (i) [using f( ) f( a ) ] a (ii) Adding (i) and (ii) I a cos b sin I Divide numerator and denominator b cos sec I a b tan sec I a b tan Let b tan t b sec dt a When, t and, t [using f( ) f( ) ] a cos b sin a I b dt. tan a t b a t a I (tan tan ). ab ab I ab 9. i ( j k ) j ( k i ) k ( i j ) SET III = i j i k j k j i k i k j =. I = e k j i k j i = Let = z = dz = dz Also = z =, = z = I = z e dz = [ e z ] = ( e e ) = (e )
588 Xam idea Mathematics XII 9. Given lines are l : r i j 4k ( i j 6k ) ; l: r i j 5k ( 4i 6j k ) After observation, we get l l Therefore, it is sufficient to find the perpendicular distance of a point of line l to line l. The co-ordinate of a point of l is P(,, 4) Also the cartesian form of line l is = = z 5 4 6... (i) Let Q(,, ) be foot of perpendicular drawn from P to line l Q(,, ) lie on line l 4 = 6 = 5 = (sa) 4, 6, 5 Again, PQ is perpendicular to line l. P(,,-4) Q(,,) l l PQ. b =, where b is parallel vector of l ( ). 4 ( ). 6 ( 4). 4 4 6 48 4 6 4( 4 ) 6( 6 ) ( 5) 6 6 8 44 6 96 96 98 Co-ordinate of Q 4 6,, 5 98 98 98 6,, 5 45 44,, 49 49 49 49 49 Therefore required perpendicular distance is 45 49 44 5 4 49 49 96 46 55 49 49 49 96 6 5 457 49 49 96 46 55 49 7 9 49 9 7 units 5 49
Eamination Papers 4 589. Given differential equation is log d + = log I.F. = e d. (Divide each term b log ).log It is in the form d P Q where P =. log, Q P e log z dz logz put log z e e z log General solution is. log = log. c log log = c Let log = z = dz, Also log = z = ez z log = dz c z e z log = z. e dz c z z log = z. e e dz c z z log = [ ze e dz] c log = ze z e z + c log = log e log e log + c log = log. log c e e log = ( + log ) + c. Given, cos cos ( a ) cos cos ( a ) Differentiating w.r.t. to on both sides, we have cos ( a ) ( sin ) cos [ sin ( a )] d cos ( a ) d cos sin ( a ) sin cos ( a ) cos ( a ) log
59 Xam idea Mathematics XII d d d sin ( a ) cos ( a ) sin a cos ( a ) cos ( a ) sin a. L.H.S. = a bc ac c a ab b ac ab b bc c Taking out a, b, c from C, C and C abc a c a c a b b a b b c c Appling C C C C c a c abc b b a b b c c Taking b from C ab c c a c b a b c c Appling R R R ab c c a c c a c b c c Epanding b I column, we get c a c ab c c a c ab c( ac c ac c ) ab c( ac) 4a b c = R.H.S. 8. Let side of square be a units and radius of a circle be r units. It is given, k 4a 4a r k where k is a constant r Sum of areas, A a r
Eamination Papers 4 59 A a k 4a a k 4a 4 ( ) Differentiating w.r.t., we get da ( k 4a) a ( k 4a). ( 4) a da 4 For minimum area, da da ( k 4a) a ( k 4a) a r a ( ) [As k 4a r given] a r Now, again differentiating equation (i) w.r.t. d A 8 ( 4) da d A 8 at a =, da For a r, sum of areas is least. Hence, sum of areas is least when side of the square is double the radius of the circle. 9. Let I / 4 sin cos 9 6 sin Here, we epress denominator in terms of sin cos which is integral of the numerator. We have, (sin cos ) sin cos sin cos sin sin (sin cos ) / 4 sin cos I 9 6 { (sin cos ) } I / sin cos 4 5 6 (sin cos ) Let sin cos t. Then, d (sin cos ) dt (cos sin ) dt Also, t sin cos and 4 t sin cos 4 4 I dt 5 6t (i)
59 Xam idea Mathematics XII I I dt I 6 5 t 6 6 5 log 4 dt 6 5 t 4 5 t 4 5 t 4 / 4 log log 4 9 / 4 I 4 9 log log 9 4