MATHEMATICS (SET -3) Labour cost Z 300x 400y (to be minimized) The constraints are: SECTION - A 1. f (x) is continuous at x 3 f (3) lim f (x)

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1 8 Class th (SET -) BD PPER -7 M T H E M T I C S () SECTION -. f () is continuous at f () lim f () ( ) 6 k lim ( )( 6) k lim ( ) k. adj I 8 I 8 I 8I 8. P : z 5 5 P : 5 5z z 8 Distance between P & P sin cos d tan cot d sin cos ln sec ln sin C. SECTION - B d d 5. Let I d.. 5 d ( ) Put t d dt dt t t log C t ( ) log C ( ) MTHEMTICS (SET -) d d a b c 8 5 a d log C a a a 6. Let work for das and B works for das. Labour cost Z (to be minimized) The constraints are: Here sample space S {,,,, 5, 6} : Number obtained is even B : Number obtained is red. Clearl {,, 6} B,, B {} Here P(), P(B), P( B) 6 6 Since P( B) P()P(B), therefore, and B are not independent events. P,, 8. Given Q 5,, 9. Equation of line passing through P and Q is z Substituting z z z z-coordinate is f() 6 / f () 6 6 ( ) / f () R Therefore, the function f is increasing on R.. Given f (),, Since f () is a polnomial function hence it is continuous in,,. f f () Rolle's Theorem is applicable. f (c) c c, and differentiable in CTJ EE, Raman Niwas, Near kashvani, Meh moorganj, Varanasi, Ph: (5) 655. www. catjee.in

2 8 Class th (SET -) BD PPER -7 M T H E M T I C S () Clearl c, Hence, c. is a skew - smmetric matri T T det() ( ) det() det() n ( det(k) k det()) det(). Let r be radius of sphere, V be volume and S be surface area. V r and S r where r is a function of time. dv dr r dt dt dv 8 cm / sec dt dr dr 8 r dt dt r r ds dr 8 r dt 8r dt r 6 6 (as, r cm) r cm / sec SECTION C. X = Sum of the numbers on the two drawn cords. Clearl X, 5, 7, 5, 7, 5 7, 6, 8,8,, Probabilit distribution of X is X 6 8 P(X) / 6 / 6 / 6 / 6 / 6 Mean of X: X Xipi Variance of X, Var(X) X i pi X Here C ˆi j ˆ 5kˆ CB i ˆ ˆj kˆ B ˆ i j ˆ 5k ˆ i ˆ ˆ j k ˆ ˆ ˆ ˆ ˆ ˆ ˆ C i ˆ j ˆ k ˆ C CB i j 5k i j k 5 C CB Hence angle C 9, B, C are vertices of a right angled triangle. lso, C 5 5 CB 6 rea of triangle BC C CB 5 6 sq. units 5. Let E Selected student have % attendance. E Selected student is irregular = student obtained '' grade. Clearl P(E ). and P(E ).7 P/ E Probabilit that selected student obtained '' grade when he/she has % attendance. P / E Probabilit that selected student obtained '' grade when he/she is irregular Given P/ E.7, P/ E. P(E PE / ) P() 6. P(E )P / E P(E )P / E P(E )P / E (Baes' Theorem) Yes regularit in a school is important to understand the core concepts of the subjects properl which ensures good grade point in the eams. tan tan CTJ EE, Raman Niwas, Near kashvani, Meh moorganj, Varanasi, Ph: (5) 655. www. catjee.in

3 8 Class th (SET -) BD PPER -7 M T H E M T I C S () 7. Let tan tan tan tan tan tan a a a D a a B R R R and R R R a a D a a a (a ) Epanding along C, we get D (a ) (a ) (a ) (a ) (a ) a a a a a (a ) 8 9 a Hence, is a matri. Let c b d 8 a b c d 9 a c b d 8 a b a c b d 9 a c, b d 8, a, b () c, ( ) d 8, a, b c, d, a, b b, b d 8 d a b c d b 8. Given a Let u, v b u v a Differentiating both sides with respect to : du dv (i) d d But u ln u ln du d ln u d d du d ln d d lso, v ln v ln dv d ln v d d (ii) dv d ln (iii) d d From (i), (ii) and (iii) we get d d ln ln d d d ln d ln e ( ) e ln( ) d d ( ) CTJ EE, Raman Niwas, Near kashvani, Meh moorganj, Varanasi, Ph: (5) 655. www. catjee.in

4 8 Class th (SET -) BD PPER -7 M T H E M T I C S () lso, d d d ( ) d 9. tan Let I d sec tan (i) a a Using f ()d f (a )d we get ( ) tan( ) I d sec( ) tan( ) ( ) tan I.(ii) sec tan From (i) and (ii) tan I d sec tan / tan d sec tan I / tan d sec tan / tan d I sec tan /sin d I sin /sin ( sin ) I d sin / sin sin d cos / (tan sec sec )d sec tan / / sin cos / cos cos sin / cos () sin ( ) Let I d d ( )d d ( )d (5 )d ( )d 5 6 ( ) Given constraints are, 6 6 (, ) (, 8) B (, ) O (, ) (, ) 6+ = B solving 6 and 6 (6, ) + 6 = We get B (, ) The shaded region in the figure is feasible region which is bounded.. Let Corner point Corresponding value of Z 7 (, ) (, ) 8 (, ) (, ) Hence, maimum value of Z is attained at the point (, ). I e d (e ) (e ) Put e t e d dt I Let dt (t ) (t ) B C t (t ) (t ) (t ) t (t )(t ) B(t ) C(t ) CTJ EE, Raman Niwas, Near kashvani, Meh moorganj, Varanasi, Ph: (5) 655. www. catjee.in

5 8 Class th (SET -) BD PPER -7 M T H E M T I C S (5) Put t, we get B B t, we get 9C C 9 lso, C 9 / 9 / / 9 t (t ) (t ) (t ) t / 9 / / 9 dt dt dt dt t (t ) (t ) (t ) (t ) log t log t C 9 (t ) 9 t log C 9 t (t ) e log C 9 e ( e ). Given, a i ˆ ˆj kˆ b 7i ˆ j ˆ kˆ a and b are parallel vectors. b a (i ˆ ˆj k) ˆ where is a scalar. lso, b b b b b b 7i ˆ j ˆ k ˆ (i ˆ ˆj k) ˆ (7 )i ˆ ( ) ˆj ( )kˆ b is perpendicular to a b a (7 )i ˆ ( ) ˆj ( )k ˆ (i ˆ ˆj k) ˆ (7 ) ( ) ( ) b ˆ ˆ ˆ (i j k) i ˆ j ˆ kˆ and b ˆ ˆ ˆ i j k. Given differential equation is d sin d (i) which is of the form d P Q d where P & Q sin Pd d Integrating factor, IF e e e Solution of given differential equation is e e sin d (ii) Let I e sin d II I (sin )(e ) (cos )(e )d I II (integration b parts) e sin (cos )( e ) e sin d I e ( sin cos ) e I (sin cos ) (iii) The solution of the given differential equation is e e (sin cos ) C SECTION D. Equation of B: 6 5 ( ) 9 6 B(6,6) (,) O D(,) E(6,) F(8,) C(8,) Equation of BC: 6 ( 8) 6 8 Equation of C : ( 8) 8 rea of BC rea of trapezium BED + rea of trapezium EBCF rea of trapezium CFD d ( )d d CTJ EE, Raman Niwas, Near kashvani, Meh moorganj, Varanasi, Ph: (5) 655. www. catjee.in

6 8 Class th (SET -) BD PPER -7 M T H E M T I C S (6) 5. Equation of parabola and line 6 = / C D B = / + 6 Solving and 6 : 6 8, C (, ) and D, Required rea = rea of shaded region 6 d d d ( ) (i) d d The given differential equation is homogeneous. d du Let u u d d (ii) d du u u d d u du u u u u u d u u du u u u d du d u u u u d du u u u u u..u u Put u t du dt & u u t t u d du dt ln u u t t dt ln t t dt dt dt ln t t t ln t tan / / ln C u ln(u u ) tan ln C ln( ) tan ln C ln( ) tan ln C ln tan C..(iii) Given, satisfing (iii) ln() tan C C 6 Solution of given differential equation is ln( ) tan 6. Let (,, 5), B (,, ), C (,, ), D (,, ), E (,, ) Equation of line passing through & B is z 5 6 Here CE ˆi j ˆ kˆ CD i ˆ j ˆ 6kˆ..(i) Normal vector n of plane containing points C, D, E is n CD CE ˆi ˆj kˆ 6 i ˆ 6j ˆ 6kˆ Equation of plane is 6 6z () 6() 6() z 7 n point P lies on the line is: P (,, 6 5) If P lies on plane z 7 then CTJ EE, Raman Niwas, Near kashvani, Meh moorganj, Varanasi, Ph: (5) 655. www. catjee.in

7 8 Class th (SET -) BD PPER -7 M T H E M T I C S (7) 7. 5 P (,, 7) Let (a,, ), B(, b, ), C(,, c) are the points where the plane intersect at -ais, -ais and z-ais. Equation of plane is z C(,, c) B (, b, ) (a,, ) z..(i) a b c BC,, Let centroid of a b c,,.(ii) we get equation of plane is z Given distance of plane from origin is p. p p Locus of the centroid of triangle BC is z p f : R R, f () Let f ( ) f ( ),, R f () is one-one. Let Hence for R R such that f f () is onto. then there eist f () is one-one and onto so it is a bijective function. Clearl f (), R () f () () lso, f () f () () () * is not commutative binar operation since: (, )* (, ) (, ) (, ) Where as (, )* (, ) (, ) (, 7) Clearl (, )* (, ) (,)* (, ). lso, ((a, b)*(c, d))*(e, f ) (ac, b ad)*(e, f ) (ace, b ad acf ) (a,b)*((c,d)*(e,f )) (a,b)*(ce, d cf ) (ace, b ad acf ) Since ((a, b)*(c, d))*(e, f ) (a,b)*((c,d)*(e,f )) therefore * is associative binar operation. (i) Let (, ) be the identit element. (a, b)*(, ) (a, b) a, b (a, b a) (a, b) a, b a a and b a b a, b a and a a, b and a, b lso (, )*(a, b) (a, b) a, b (a, b) (a, b) a, b a a and b b a, b a and b b a, b and ()b b a, b and a, b CTJ EE, Raman Niwas, Near kashvani, Meh moorganj, Varanasi, Ph: (5) 655. www. catjee.in

8 8 Class th (SET -) BD PPER -7 M T H E M T I C S (8) Hence the identit element is (, ). (ii) Let (a, b) be invertible and let (, ) be the 8. inverse element of (a, b). (a, b)*(, ) (, ) and (, )*(a, b) (, ) (a, b a) (, ) and (a, b) (, ) a, b a and a, b a, b a and b a, b and b, ssuming a, b and b a, b, b a b, a a Hence, for ever a, b where a there eist b,, a a such that (a, b)*(, ) (, ). In other words all a, b where a are invertible. 5 ( ) ( 6 ) 5( ) 6 eists., ( ), () (), ( 9), (5) (), ( ), () T adj adj 9 5 The given sstem of equations is 5z z 5 z which can be written as X B 5 where, X, B 5 z X B ,, z 9. Let and be the length and width of rectangular window. It is given that the perimeter of the window is m. / 5 rea of the window is given b 5 8 d 5 d d d (i) For maima or minima, d d 5 m When, 5 m. d lso, d is maimum when m Hence, the required dimensions of the window to admit maimum light are given b length m and width m. CTJ EE, Raman Niwas, Near kashvani, Meh moorganj, Varanasi, Ph: (5) 655. www. catjee.in

MATHEMATICS (SET -1)

MATHEMATICS (SET -1) 8 Class th (SET ) BD PPER -7 M T H E M T I C S (). adj 8 I 8 I 8I 8 SECTION - I. f () is continuous at f () lim f () ( ) 6 k lim ( )( 6) k lim ( ) k sin cos d tan cot d sin cos ln sec ln sin C.. P : z