SET-I SECTION A SECTION B. General Instructions. Time : 3 hours Max. Marks : 100

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1 General Instructions. All questions are compulsor.. This question paper contains 9 questions.. Questions - in Section A are ver short answer tpe questions carring mark each.. Questions 5- in Section B are short answer tpe questions carring marks each. 5. Questions - in Section C are long answer I tpe questions carring marks each.. Questions -9 in Section D are long answer II tpe questions carring marks each. Time : hours Ma. Marks : 00 SECTION A. If for an square matri A, A( adj A) , then write the value of A.. Determine the value of k for which the following function is continuous at : ( + ), f ( ) k,. Find : sin cos sin cos SET-I 7. The volume of a cube is increasing at the rate of 9 cm /s. How fast is its surface area increasing when the length of an edge is 0 cm? 8. Show that the function f ( ) + 00 is increasing on R. 9. The -coordinate of a point on the line joining the points P(,, ) and Q( 5,, ) is. Find its z-coordinate. 0. A die, whose faces are marked,, in red and, 5, in green, is tossed. Let A be the event number obtained is even and B be the event number obtained is red. Find if A and B are independent events.. Find the distance between the planes + z 5 and z 0. SECTION B 5. If A is skew-smmetric matri of order, then prove that det A 0.. Find the value of c in Rolle's theorem for the function f ( ) in [, 0 ].. Two tailors, A and B, earn ` 00 and ` 00 per da, respectivel. A can stitch shirts and pairs of trousers while B can stitch 0 shirts and pairs of trousers per da. To find how man das should each of them work and if it is desired to produce at least 0 shirts and pairs of trousers at a minimum labour cost, formulate this as an LPP.. Find : 5 8

2 SECTION C + π. If tan + tan, then find the + value of.. Using properties determinants, prove that a + a a + a + a + a ( ) Find matri A such that 8 0 A 9 b 5. If + a,then find d. If e ( + ), then show that d d.. Find : 7. Evaluate : cos θ d ( sin θ) ( 5 cos θ) θ + 0 π tan sec + tan Evaluate : { + + } 8. Solve the differential equation (tan ) ( + ) d. 9. Show that the points A, B, C with position vectors i j + k, i j 5k and i j k respectivel, are the vertices of a right-angled triangle. Hence find the area of the triangle. 0. Find the value of λ, if four points with position vectors i + j + 9 k, i + j + k, i + j + k and i + j + λk are coplanar.. There are cards numbered,, 5 and 7, one number on one card. Two cards are drawn at random without replacement. Let X denote the sum of the numbers on the two drawn cards. Find the mean and variance of X.. If the students in a school, it is known that 0% have 00% attendance and 70% students are irregular. Previous ears results report that 70% of all students who have 00% attendance attain A grade and 0% irregular students attain A grade in their annual eamination. At the end of the ear, one student is chosen at random from the school and he was found to have an A grade. What is the probabilit that the students has 00% attendance? Is regularit required onl in school? Justif our answer.. Maimise and minimise Z + subject to the constraints , 0 Solve the above LPP graphicall. SECTION D. Determine the product 7 are 5 use it to solve the sstem of equations + z, z 9 and + + z. 5. Consider f : R R given b + f ( ). Show that f is bijective. Find the + inverse of f and hence find f ( 0) and such that f ( ). Let A Q Q and let * be a binar operation on A defined b ( a, b) * ( c, d) ( ac, b + ad) for ( a, b) ( c, b) A. Determine, whether * is commutative and associative. Then, with respect to * on A (i) Find the identit element in A. (ii) Find the invertible elements of A.. Show that the surface area of a closed cuboid with square base and given volume is minimum, when it is a cube. 7. Using the method of integration, find the area of the ABC, coordinates of whose vertices are A (, ), B (, ) and C ( 8, ). Find the area enclosed between the parabola and the straight line + 0.

3 8. Find the particular solution of the differential equation ( ) ( + ), given that 0 when. 9. Find the coordinates of the point where the line through the points(,, 5) and(,, ), crosses the plane determined b the points (,, ), (,, ) and ( 0,, ). A variable plane which remains at a constant distance p from the origin cuts the coordinate aes at A,B,C. Show that the locus of the centroid of ABC is + + z p.. Given, A( adja) A( adja ) A adj( A ) 0 A A [ Q adja A n ] A A ± 8 ( + ). Given,, f( ) k, Since, f( ) is continuous at lim f( ) f( ) lim ( + ) k lim ( + ) k lim ( + ) ( + + ) k [Qa b ( a b) ( a + b)] lim ( ) ( + 9 ) k ( ) lim ( + 9) k + 9 k k. Let I sin cos sin cos sin cos sin cos sin cos sin cos cos sin (tan cot ) tan cot SOLUTIONS log sec [ log cosec ] log sec + log cosec log sec cosec. Given, + z 5 + z 5 0 and z 0 5[ z] 0 + z + z 8 + z 8 0 Clearl, planes (i) and (ii) are parallel. Distance between two parallel planes, d d d a + b 8 ( 5) ( ) + ( ) + [Qd 8, d 5, a, b and c ] Given, A is a skew-smmetric matri of order. A A A A A ( ) n A [ Q A A, ka k A ] A A A 0 A 0. Given, f( ) in [, 0] We know that, according to Rolle's theorem, f( ) is continuous at [, 0] and differentiable at (, 0 ). f ( c) 0, c [, 0] Now, f ( ) f ( c) c 0 [from Eq. (i)] [Q c] c ± But c [, 0 ], so neglecting positive value of c, we get c

4 7. Let volume of a cube be v cm. Then, we have dv dt 9 cm /sec and edge of cube, 0 cm Q v On differentiating both sides w.r.t. t, we get dv dt dt 9 [from Eq. (i)] dt (iii) dt Also, surface area of a cube, s On differentiating both sides w.r.t. t, we get ds dt dt ds dt [from Eq. (iii)] ds dt At 0, ds dt 0. /sec 8. Given, f( ) + 00 On differentiating both sides w.r.t., we get f ( ) ( + ) + ( ) + > 0 f ( ) > 0 This show that function f( ) is increasing on R. Hence proved. 9. The equation of line joining the points P(,, ) and Q(5,,) is Since, -coordinate is. z 5 z z z 0. When a die is thrown, the sample space is S {,,,, 5, } n( S) Also, A : number is even and B : number is red. A {,, } and B {,, } and A B { } n( A), n ( B) and n( A B) n( A) Now, P( A) n( S) n( B) P ( B) n( S) n ( A B) and P( A B) n( S) Now, P( A) P( B) P( A B) P ( A B) P( A) P( B) Thus, A and B are not independent events.. Suppose, tailor A stitch in das and tailor B stitch in das. The given data can be written in the tabular form as follows Tailor No. of Shirt No. of trousers Cost/da A ` 00 B 0 ` 00 Minimum requirement 0 Required linear programming problem is Min ( z) subject to constraints [dividing both sides b ] and [dividing both sides b ] 0, 0. Let I ( ) + ( ) 5 + [ + ( ) + ] ( + ) ( ) ( + ) log Q + + a + log a a a

5 . Given, tan tan π tan tan ( ) + + a + b [Qtan a + tan b tan ab and π tan ()] ( )( + ) + ( + ) ( ) ( ) ( + ) tan tan ( ) ( ) ( + ) ( ) ( + ) ( ) ( + ) ± 7. LHS a + a a + a + a + On appling R R R and R R R, we get a a 0 LHS a a 0 ( a )( a + ) a 0 ( a ) a 0 On taking (a ) common from R and R, we get a + 0 LHS ( a ) 0 On epanding along C, we get LHS ( a ) ( a + ) ( a ) ( a ) ( a ) Hence proved. Let order of A is m n. m, n Let A s t 8 8 Q 0 A 0 9 s t 9 s t 8 + s + t 9 On comparing both sides, we get s,, and t 8 At, s s s s and at, t 8, ( ) t 8 t 8 t On putting,, s and t in Eq. (i), we get A b 5. Given, + a Let v and u On putting these values in Eq. (i), we get v + u a b On differentiating both sides w.r.t., we get dv du + 0 (iii) Now, v [from Eq. (ii)] On taking log both sides, we get log logv log log v On differentiating both sides w.r.t., we get d dv + log v d dv + log v dv + log d (iv) [from Eq. (ii)] and u [from Eq. (ii)] On taking log both sides, we get log logu log log u On differentiating both sides w.r.t., we get d du + log u d du + log u

6 u d du + d du + (v)[from Eq. (ii)] On substituting the values of dv du and from Eq. (iv) and Eq. (v) respectivel in Eq. (iii), we get d d + log + + log 0 d d + log + + log 0 d d log + log d log + log d log log + Given, e ( + ) On taking log both sides, we get log[ e ( + )] log loge + log( + ) log + log( + ) log On differentiating both sides w.r.t., we get d Again, differentiating both sides w.r.t., we get d 0 ( + ) d d 0 [Qfrom Eq. (i)] d d 0 d d cosθ. Let I dθ ( + sin θ) ( 5 cos θ)] cosθ dθ ( + sin θ) [ 5 ( sin θ)] cosθ d ( + sin θ) ( 5 + sin θ) θ cosθ d ( + sin θ) ( + sin θ) θ Let sinθ t cosθ dθ dt dt Then, I ( + t ) ( + t ) Hence proved. B Again, let + ( + t ) ( + t ) + At + t [b partial fraction] A B At t 0, + A + B (iii) At t, A B + 5A + 5B (iv) On solving Eqs. (iii) and (iv), we get A 5 and B 5 On putting A 5 and B in Eq. (ii), we get ( + t ) ( + t ) + t + t + ( + t )( + t ) 5( + t ) 5( + t ) On integrating both sides w.r.t. t, we get dt dt + dt ( + t ) ( + t ) t + t + + dt 5 5 t + t t tan / tan t c / Q tan + a a a sinθ tan + tan sinθ [put t sinθ] Let I π tan sec + tan 0 π ( π ) tan( π ) I 0 sec( π ) + tan( π ) I π ( π ) tan sec + tan 0 a [Q f( ) f( a ) ] 0 0 On adding Eqs. (i) and (ii), we get I π π tan 0 sec + tan I π π tan (sec tan ) + 0 (sec tan ) (sec tan ) π π 0 a [rationalising] (tan sec tan ) (sec tan ) ] [Q( a + b) ( a b) a b

7 π π tan sec sec + 0 [Qtan sec ] π π [sec tan + ] 0 π [sec π sec 0 tanπ + tan 0 + π 0] π π [ π 0] [ π ] Let I { + + } [( ) ( ) ( )] + [( ) + ( ) ( )] + [( ) + ( ) ( )] [ + + ] + ( + + ) + ( + + ) ( 5 ) + ( + ) + ( + ) Given, (tan ) ( + ) d d tan d ( + ) tan + + d tan The above differential equation is linear differential equation. On comparing Eq. (i) with d + P Q, we get P + and Q tan + tan Now, IF e P + e e Solution is given b IF Q IF + C tan tan tan e e + C + Put t tan dt + tan e t t e dt + C t e e dt + C tan e t t t e e + C tan tan tan t t [Integration b parts] e tan e e + C tan tan e (tan ) e + C [Qt e tan ] 9. Given, A i j + k, B i j 5k and C i j k We know that, two lines are perpendicular, if a a + bb c 0 Now, for lines with position vector A and B, we get ( ) + ( 5) Again, for lines with position vector A and C, we get ( ) + ( ) + 0 Again, for lines with position vector B and C, we get ( ) 5 ( ) This shows that A, B, C are the vertices of right angled ABC, right angled at B. Hence proved. Now, Area of ABC AB BC ( i j 5k i + j k ) ( i j k i + j + 5k ) ( i j k) ( i j + k) i j k ( i ) ( j + ) + k ( + ) + 8 i j 5 k sq units C A B

8 0. Let (,, z ) (,, 9), (,, z ) (,, ) (,, z ) (,, ) and (,, z) (,, λ) We know that, four points are coplanar if and onl if z z z z z z 9 9 λ λ λ 9 ( ) λ ( λ + 7 0) + ( λ ) ( 0 + ) 0 λ 5 λ λ 0 λ. Here, S {(, ), (, 5), (, 7), (, )(, 5), (, 7), ( 5, ), ( 5, ), ( 5, 7) ( 7, ), ( 7, ), ( 7, 5 )} n ( S) Let random variable X denotes the sum of the numbers on two cards drawn. So, the random variables X ma have values,, 8, 0 and. At X, P( X) At X, P( X) At X 8, P( X) At X 0, P( X) At X, P ( X) Therefore, the required probabilit distribution is as follows X 8 0 P( X ) Mean, E( X) Σ X P( X) Also, Σ X P( X) Var ( X) ΣX P( X) ΣX P ( X)] [ ] 9 0. Let E : the event that student have 00% attendance. E : the event that student are irregular Let E : a student attains A grade Then, E and E are mutuall eclusive and ehausitve events P( E ) 0%, P ( E) 70% P ( E / E) 70%, P( E / E) 0% B using Bae's theorem, we get P( E / E) P( E) P ( E / E) P( E/ E ) P ( E ) + P( E / E ) P( E ) 7 P ( E / E ) 0 0 / / / / 00 8/ 00 Regularit is required for good result in an tpe of work. 0. Our problem is to minimise and maimise Z + Subject to constraints, (iii) + 00 (iv) 0, 0 (v) Table for line + 00 is So, the line + 00 is passing through the points ( 0, 50) and ( 00, 0 ). On putting (0, 0) in the inequalit + 00, we get , which is not true. So, the half plane is awa from the origin. Table for line 0 is

9 So, the line 0 is passing through the points (0, 0) and (0, 0). On putting (5, 0) in the inequalit 0, we get , (which is not true) So, the half plane is towards Y-ais. Table for line + 00 is So, the line + 00 is passing through the points (0, 00) and (00, 0). On putting (0, 0) in the inequalit + 00, we get , which is true. So, the half plane is towards the origin. Also,, 0. So, the region lies in the I quadrant A(0, 50) 0 0 X (0, 0) O 0 Y 00 D(0, 00) Y B(0, 0) (0, 0) C(50, 00) On solving equations 0 and + 00, we get B(0, 0). Again, solving the equations 0 and + 00, we get C(50, 00). Feasible region is ABCDA. The corner points of the feasible region are A(0, 50), B(0, 0), C(50, 00) and D(0, 00). The values of Z at corner points are given below: Corner points Z + (00, 0) A(0, 50) Z B(0, 0) Z C(50, 00) Z D(0, 00) Z The maimum value of Z is 00 at D(0, 00) and the minimum value of Z is 00 at all the points on the line segment joining A(0, 50) and B(0, 0). X Consider the product I I 8 5 Now, AX B X A B Here, A and B 9 X z , and z 5. Given, f : R R + defined b f( ) + Let, R such that f ( ) f( ) ( + )( + ) ( + )( + ) f is one-one. Let R, then

10 The function f is onto if there eist R, such that f ( ) Now, f( ) ( + ) + + ( ) R Thus, for an R, there eist R f is onto. f R ( ) f ( ) f 0 0 ( ) 0 and also f ( ) ( ) Given, binar operation on A is defined as ( a, b) ( c, d) ( ac, b+ ad). We have to show that is commutative and associative. Commutative Here, ( a, b) ( c, d) ( ac, b+ ad) Also, ( c, d) ( a, b) ( ca, d+ cb) Since, ac ca but b+ ad d+ cb ( a, b) ( c, d) ( c, d) ( a, b) Therefore, is not commutative. Associative Here, ( a, b) [( c, d) ( e, f)] ( a, b) ( ce, d+ cf ) ( ace, b+ ad+ acf) Also, [( a, b) ( c, d)] ( e, f) ( ac, b+ ad) ( e, f ) ( ace, b+ ad+ acf) ( a, b) [( c, d) ( e, f)] [( a, b) ( c, d)] ( e, f) Therefore, is associative. (i) Let ( e, f) be the identit element. Then, ( a, b) ( e, f) ( a, b) ( ae, b+ af) ( a, b) On comparing, we get ae a and b+ af b From Eqs. (i) and (ii), we get e and f 0 (, 0) is the identit element. (ii) Let ( c, d) be the inverse element. Then, ( a, b) ( c, d) (, 0) ( ac, b+ ad) (, 0) On comparing, we get ac and b+ ad 0 c b and d a a b, is the inverse element. a a. Let V be the fied volume of a closed cuboid with length, breadth and height. Let S be its surface area then, V V Now, S ( + + ) S ( + ) V + [using Eq. (i)] V S + On differentiating, we get ds V and d S 8V + Now, put ds 0 V 0 V V V 7. Now, when, we have V d S 8 + > 0 So, S is minimum when length, breadth and height, i.e. when it is cube. X Y Y A (, ) B (, ) 8 C (8, ) X

11 The given points are A(, ), B(, ) and C( 8, ). The equation of line AB is 5 ( ) ( ) ( ) Equation of line BC is ( ) ( ) 8 ( ) Equation of line AC is ( ) 8 ( ) + Required area (Area under line segment AB) + (Area under line segment BC) (Area under line segment AC) ( ) {( 5 5) ( 0 )} + {( + 9) ( 8 + 7)} {( ) ( 8)} ( 9 + ) + ( 5) ( 8 + ) sq units OR Given parabola is represents an upward parabola with verte (0, 0) and equation of line is + From Eqs. (i) and (ii), we get ( + ) ( )( + ) 0, 8 8 X when, then + when, then ( ) + Thus, intersection points are (, ) and (, ). + Required area sq units 8. Given differential equation is ( ) d ( + ) d + The above differential equation is homogeneous. Putting v d v + dv, we get v dv v + + v dv v + + v v dv v + v v dv v v v + ( ) v dv (, ) v v v + + v dv v v + + v v dv v + v + v dv v + v + Now, v A d ( dv v + v + ) + B v A( v + ) + B v Av + A + B Y (, 0) (, 0) O (0, ) Y + (, ) (, 0) X

12 Equating coefficient of v and constant term from both sides, we get A A and A + B + B B v v + ( ) Then, from Eq (i), we get + ( v ) dv v + v + + v dv dv v + v + v + v + log v + v + dv + + v log log v + v + / tan v + / / log v + log v + v + tan log + log + + tan log + log + + tan log Put and 0 in Eq. (ii), we get log tan log π π c tan Then, from Eq. (ii), we get + π log + + tan log + which is the required particular solutions. 9. Equation of an plane through (,, ) is a( ) + b( ) ( z ) 0 Since, it passes through the points (,, ) and (0,, ), So, a( ) + b( ) ( ) 0 a + 0b c 0 and a( 0 ) + b( ) ( ) 0 a + b + 0 c 0 (iii) Solving Eqs. (ii) and (iii) b cross multiplication, we get a b c a b c a b c λ (sa) a λ, b λ, c λ Substituting the value of a, b, c in Eq. (i), we get λ( ) + λ( ) + λ( z ) z z 7 0 Now, equation of line joining (,, 5) and (,, ) is z z λ (sa) λ +, λ, z λ 5 Therefore, an point on the line is of the form ( λ +, λ, λ 5 ). This point lies on the plane + + z 7 0 So, ( λ + ) + ( λ ) + ( λ 5) 7 0 5λ 0 0 λ Hence, the coordinates of the required point are (,, 5) (,, 7) Let the equation of the variable plane is z + + a b c Since, above plane (i) meets the X-ais, Y-ais and Z-ais at the point A( a, 0, 0 ), B ( 0, b, 0) and C( 0, 0, c), respectivel and let ( α, β, γ) be the coordinates of the centroid of ABC Then, α + +, β + b + and γ + a α, β b and γ c a α, b β and c γ p length of the perpendicular from (0, 0, 0) to the plane (i) p a b c p a b c a b c + + a b c 9p 9α 9β 9γ 9 [using Eq. (ii)] + + α β γ p Hence, the locus of the centroid is + + z p Hence proved.

13 SET-II (Onl Uncommon Questions from Set I) SECTION B. The length of a rectangle is decreasing at the rate of 5 cm/min and the width is increasing at the rate of cm/min. If 8 cm and cm, then find the rate of change of area of the rectangle. SECTION C. Solve the following linear programming problem graphicall. Maimise Z + 5 Subject to constraints + 00, + 70,, 0. Given that length of a rectangle is decreasing at the rate of 5 cm/min. 5 cm / min dt Also, width of a rectangle is increasing at the rate of cm/min. d cm / min dt We know that, area of rectangle A On differentiating both sides w.r.t. t, we get da d + (iii) dt dt dt Now, we have 8cm and cm On putting Eqs. (i) and (ii) in Eq. (iii), we get da dt 8( ) + ( 5) 0 cm / min SOLUTIONS. We have the following LPP Maimise Z + 5 Subject to the constraints + 00, + 70,, 0 Now, considering the inequations as equations, we get + 00 and + 70 Table for line + 00 is Find the value of such that the points A(,, ), B (,, 5 ), C (,, ) and D(, 5, ) are coplanar.. Find the general solution of the differential equation ( + ) d 0. SECTION D 8. AB is the diameter of a circle and C is an point on the circle. Show that the area of ABC is maimum, when it is an isosceles triangle If A, then find A. Hence using A, solve the sstem of equations + 5z, + z 5 and + z. So, the line passes through the points ( 0, 00) and ( 00, 0 ). On putting ( 0, 0) in the inequalit + 00, we get , which is true. So, the half plane is towards the origin. Table for line + 70 is / So, the line passes through the points (5, 0) and (0, 70/). On putting (0, 0) in the inequalit + 70, we get , which is true. So, the half plane is towards the origin. Also,, 0, so the region lies in the st quadrant. The graphical representation of the sstem of inequations is as given below. Y (0,00) + 00 (0, 70/) B (5, 0)A (00,0) X X (0, 0) O + 70 Y Clearl, the feasible region is OAB, where the corner points are O( 0, 0 ), A( 5, 0) and B( 0, 70/ ).

14 Now, the values of Z at corner points are as follow : Corner points Z + 5 O( 0, 0) Z A( 5, 0) Z B( 0, 70/ ) Z Hence, the maimum value of Z is 90.. Given, A (,,, ) B (,, 5 ), C (,, ) and D (, 5, ) AB i + ( ) j + k, AC i k and AD i + j k Since, AB, AC and AD are coplanar. [ AB AC AD ] 0 ( ) 0 0 ( 0 + 9) ( )( + 9) + ( 0) 0 9 ( )( 7) We have, ( + ) d 0 + d d which is a linear differential equation. IF d e e log Hence, required solution of the differential equation is IF Q IFd + C d + C + C + C 8. Let the side of ABC be, and r be the radius of circle. Also, C 90 [Qangle made in semi-circle is 90 ] In ABC, we have ( AB) ( AC) + ( BC) ( r) ( ) + ( ) r + We know that, A C B Area of ABC,( A) On squaring both sides, we get A Let A S Then, S S ( r ) [from Eq. (i)] S ( r ) On differentiating both sides w.r.t., we get ds r 8 ( ) For maima or minima, put ds 0. ( 8 r ) 0 8r 8r r r From Eq. (i), we get r r r r Here,, so triangle is an isosceles triangle. d S d Also, r 8 ( ) 8 ( r ) r d S At r, r ( r ) r < 0 Hence, area is maimum when triangle is an isosceles triangle. Hence proved Given, A Now, A ( + ) + ( + ) + 5( ) Hence, A eists. Now, the cofactors of elements of A are A ( ) 0 + A ( ) ( + ) A ( ) 5 A ( ) 5 ( )

15 A ( ) ( ) ( + ) 5 5 ( ) 0 5 ( ) 8 5 ( ) ( ) + 9 A 5 A A 5 A A A A 0 adj( A) A A A 9 5 A A A T T Now, A adj( A) A Given sstem of equations can be written in matri form as 5 5 z 0 i.e. AX B X A B X z 5 + 9, and z SET-III (Onl Uncommon Questions from Set I & II) SECTION A. The volume of a sphere is increasing at the rate of 8 cm /s. Find the rate at which its surface area is increasing when the radius of the sphere is cm. SECTION C 0. Solve the following linear programming problem graphicall: Maimise Z subject to the constraints , 0 e. Find: ( e ) ( e + ). If a i j k and b 7 i + j k, then epress b in the form of b b + b,where b is parallel to a and b is perpendicular to a.. Find the general solution of the differential equation d sin SECTION D 9. A window is in the form of a rectangle surmounted b a semicircular opening. The total perimeter of the window is 0 m. Find the dimensions of the window to admit maimum light through the whole opening.

16 . Let r be the radius, v be the volume of sphere and s be the surface area of sphere. Then, we have dv dt 8 cm / s To find ds, when r cm dt Since, v πr dv dr r dr π 8 π r dt dt dt dr cm/s dt πr ds d Now, consider ( π r ) π r dr dt dt dt 8πr [using Eq. (i)] πr r ds At r, dt cm /sec r 0. Given, linear progarmming problem is Maimise Z subject to constraints and 0, 0 Now, table for of + 0 is Putting (0, 0) in the inequalit + 0, we get (which is true) So, half plane is towards the origin. X (0, 80) (0, 0) Y (0, 0) (0, 0) Y D A C (0, 0) B (0, 0) SOLUTIONS X Table for + 80 is Putting (0, 0) in the inequalit + 80, we get (which is true) So, half plane is towards the origin. 0 is a line parallel to Y-ais. On putting (0, 0) in the inequalit 0, we get 0 0 (which is not true) So, the half plane is awa from the origin. Now, Intersecting point of 0 and line + 0 is 0, 00. Intersecting point of line + 0 and + 80 is (0, 0). The feasible region is ABCDA. The corner points of the feasible regions are A( 0, 0 ), B ( 0, 0 ), C ( 0, 0) and D 0, 00. The values of Z at these points are as follows Corner points Z A( 0, 0) B ( 0, 0) C( 0, 0) (maimum) D 0, So, maimum value of Z is 0 at C( 0, 0 ). e. Let I ( e ) ( e + ) Put e t e dt dt I t ) ( t + ) A B C Now, let + + ( t ) ( t + ) ( t ) ( t ) ( t + ) A( t ) ( t + ) + B( t + ) + C ( t ) A( t + t ) + B( t + ) + C( t t + ) t ( A + C) + t( A + B C) A + B + C On comparing the coefficient of t, t and the constant term from the both sides, we get A + C 0, A + B C 0 and A + B + C A C, A + B C

17 and A + B + C (iii) On substituting A C in Eqs. (ii) and (iii), we get C + B C B C (iv) and C + B + C C + ( C) + C [from Eq. (iv)] 9C C 9 Now, from Eqs. (i) and (iv), we get A 9, B Now, I ( t ) ( t ) 9 ( t + ) dt + dt dt + 9 t ( t ) 9 t + + ( t ) log t + 9 ( + ) log 9 t t e log 9 e + ( t ) dt + log t [putting t ( e ). It is given that b is parallel to a therefore, e ] b λ a for some scalar λ It is also given that b b + b b b b b λ a also b is perpendicular to a 0 b a ( b λ a ) a 0 b a λ( a a ) 0 b λ a a a Now, a i j k and b 7 i + j k b a ( 7 i + j k) ( i j k) + 8 (iii) and a a ( i j k) ( i j k) Substituting these values in Eq. (iii), we get 8 λ 9 b λ a ( i j k) i j k and b b λ a ( 7 i + j k) ( i j k) 7 i + j k i + j + k i + j + k Hence, b i j + k and b i + j + k d. We have, sin, which is a linear differential equation of the form d + P Q, where P & Q are functions of or constant. Here, P and Q sin P ( ) IF e e e Now, the solution of given differential is given b (IF) (IF) Q + C e e sin + C Let I e sin II B using the method of integration b parts, we get I I sin e e cos ( ) ( ) sin e + e cos Again, b using integration b parts, we get I sin e os e e ( ( ) sin ) ( ) sin e cos e e sin II sin e cos e I I e (sin os ) e I (sin os ) Then, from Eq. (i), we get e e (sin os ) + C (sin os ) + Ce I 9. Let be the length and be the width of the window. Then, radius of semicircular opening m Since, perimeter of the window is 0 m.

18 π π 0 ( π + ) ( π + ) Note that, to admit maimum light area of window should be maimum. Here, area of window A area of rectangle + area of semicircular region + π 0 ( π + ) A + π A 0 ( π + ) + π On differentiating both sides w.r.t., we get da 0 ( π + ) + π 0 π + π 0 π For maimum, put da 0 0 π + 0 π + Again, on differentiating both sides of Eq. (ii), we get d A π 0 d A At ( π + ) < 0 π + 0 π + Thus, area is maimum when 0 π + Now, on substituting the value of in Eq. (i), we get 0 0 ( π + ) π + 0 π + π π + 0 π + 0 π + 0 Hence, length of window m and width of π + 0 window m, to admit maimum light through π + the whole opening.