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 5 5 P : 5 5z z 8 Distance between P & P SECTION - B 5. is a skew - smmetric matri T T det() ( ) det() det() det() n ( det(k) k det()) MTHEMTICS (SET -) d d a b c 8 5 6. Given f (),, Since f () is a polnomial function hence it is continuous in,,. f f () Rolle's Theorem is applicable. and differentiable in f (c) c c, Clearl c, Hence, c 7. Let be the length of a side of cube, V is the volume and S is surface area. V & S 6, where is a function of time t. Given dv 9 dt d d d 9 9 dt dt dt lso S 6 ds d dt dt 6 6.6 cm / sec CTJ EE, Raman Niwas, Near kashvani, Meh moorganj, Varanasi, Ph: (5) 655. www. catjee.in 8. f() 6 / f () 6 6 ( ) / f () R Therefore, the function f is increasing on R. P,, 9. Given Q 5,, Equation of line passing through P and Q is z Substituting z z z z-coordinate is. Here sample space S {,,,, 5, 6} : Number obtained is even B : Number obtained is red. Clearl {,, 6} B,, B {} Here P(), P(B), 6 P( B) 6
8 Class th (SET ) BD PPER -7 M T H E M T I C S () Since P( B) P()P(B), therefore, and B are not independent events.. Let work for das and B works for das. Labour cost Z (to be minimized) The constraints are: 6 6 d d. Let I 5 8 8 5 d.. 5 d ( ) Put t d dt dt t. t log C t ( ) log C ( ) a d log C a a a SECTION - C tan tan tan tan tan tan tan tan 7 7 6 7 7. Let 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 b Hence, is a matri. Let c 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 5. Given a Let u, v b u v a Differentiating both sides with respect to : du dv (i) d d CTJ EE, Raman Niwas, Near kashvani, Meh moorganj, Varanasi, Ph: (5) 655. www. catjee.in
8 Class th (SET ) BD PPER -7 M T H E M T I C S () 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 ( ) lso, 6. Let I d d d ( ) d cos d ( sin )(5 cos ) cos d ( sin )( sin ) Put sin t cos d dt I dt ( t )( t ) dt dt ( t )( t ) (t )(t ) (t ) (t ) dt 5 (t )(t ) dt 5 (t ) (t ) t t tan tan C 5 / / t tan t tan C 5 sin tan (sin ) tan C 5 7. 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 CTJ EE, Raman Niwas, Near kashvani, Meh moorganj, Varanasi, Ph: (5) 655. www. catjee.in
8 Class th (SET ) BD PPER -7 M T H E M T I C S () ( )d d ( )d (5 )d ( )d 5 6 ( ) 5 9 6 8 9 6 8. The given differential equation is tan d d d tan d d tan d d This is of the form P Q d, where tan P, Q. d Pd tan IF e e e Solution of given differential equation is tan tan tan e e d Let tan t d dt tan tan t t t I e d t e dt te e u v (integration b parts) t tan I e (t ) tan e So, solution of given differential equation is tan tan e tan e C tan tan Ce 9. 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. Let O is origin and position vector of, B, C, D are i ˆ 6j ˆ 9k, ˆ ˆ i j ˆ k, ˆ i ˆ j ˆ kˆ and i ˆ 6j ˆ kˆ respectivel. B OB O i ˆ j ˆ 6kˆ C OC O ˆi j ˆ 8kˆ D OD O ˆi j ˆ ( 9)kˆ Normal n to the plane passing through the point, B, C is : n B C ˆi ˆj kˆ ˆi ˆj kˆ n 6 7i ˆ 5j ˆ kˆ 8 8 Hence, equation of the plane passing through the point, B, C is 7 5 z 7() 5() () 7 5 z If above plane passes through the point D then: 7() 5(6).. 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 CTJ EE, Raman Niwas, Near kashvani, Meh moorganj, Varanasi, Ph: (5) 655. www. catjee.in
8 Class th (SET ) BD PPER -7 M T H E M T I C S (5) 6 8 6 6 6 6 6 6 6 6 8 8 6 Variance of X, Var(X) X i pi X 6 6 6 6 6 6 6 6 6 6.667. 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() P(E )P / E P(E )P / E P(E )P / E (Baes' Theorem)..7..7.7. 7 Yes regularit in a school is important to understand the core concepts of the subjects properl which ensures good grade point in the eams.. Here Z and the given constraints are, (, )D (, 5) B(,) (, ) + = = C(5, ) + = Corner point corresponding value of Z = + (, 5) B(, ) C(5, ) 5 D(, ) Hence maimum value of Z, which occurs when and. SECTION - D. Let D 7 and 5 D 7 5 8 8 8 7 6 7 7 9 5 5 6 5 6 8 8 8 D 8I 5. D I D (i) 8 8 The given sstem of equation : z z 9 z The given sstem of equation can be written as X B, where, X, B 9 z X B DB 8 7 9 8 5 6 8 8,, z f : R R, f () CTJ EE, Raman Niwas, Near kashvani, Meh moorganj, Varanasi, Ph: (5) 655. www. catjee.in
8 Class th (SET ) BD PPER -7 M T H E M T I C S (6) Let f ( ) f ( ),, R 6 9 9 6 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 Hence the identit element is (, ). (ii) Let (a, b) be invertible and let (, ) be the 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. 6. Let length of the square base of the cuboid be & height be and let its volume be V. Volume of cuboid, V.(i) Surface area: S V V CTJ EE, Raman Niwas, Near kashvani, Meh moorganj, Varanasi, Ph: (5) 655. www. catjee.in
8 Class th (SET ) BD PPER -7 M T H E M T I C S (7) ds V d (ii) d S 8V d.(iii) For maima or minima ds d V V lso, for d S 8V d Hence, S is minimum when =, that is, cuboid becomes cube. 7. 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. 6 8 8 5 8 d ( )d d 6 6 8 8 5 9 8 6 5 5 6 9 6 6 6 96 7 8 6 6 6 8 8 8 5 8 7 9 6 6 8. 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 6 6 6 7 7 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 CTJ EE, Raman Niwas, Near kashvani, Meh moorganj, Varanasi, Ph: (5) 655. www. catjee.in
8 Class th (SET ) BD PPER -7 M T H E M T I C S (8) 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 9. 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 6 6 5 7 5 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 CTJ EE, Raman Niwas, Near kashvani, Meh moorganj, Varanasi, Ph: (5) 655. www. catjee.in