STD. XII ISC - BOARD MATHEMATICS - SOLUTIONS

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Rao IIT Academ/XII-ISC-Board -05_Mathematics_SOLUTIONS Date: 6.0.05 Question STD. XII ISC - BOARD MATHEMATICS - SOLUTIONS SECTION A (i) 4 6 M 6 4 9 5 8 M 8 km k k k k M km I 0 5 8 k k 0 0 0 8 k k 0 0 0 4 k 0 4 k 8 k 0 0 8 k k 0 0 k 4 (ii) b a 8 b 4a e b a e a a 4 8 4a a 9

Rao IIT Academ/XII-ISC-Board -05_Mathematics_SOLUTIONS 9 a 8 a 4 b a 4a 4 8 Equation of ellipse a b 4 8 8 (iii) sin cos cos 6 sin cos sin cos sin cos cos (iv) lim 0 sin sin lim 0 cos cos sin lim 0 sin sin sin sin cos 4 cos 4 lim 0 4 cos lim 0 6

Rao IIT Academ/XII-ISC-Board -05_Mathematics_SOLUTIONS (v) I 4 d 8 8 4 d 4 d d 8 4 4 tan c (vi) f d cos d d 0 0 sin 0 sin sin 0 0 9 (vii) 4 0 9 4 b & 6 5 0 6 5 5 6 b b b b

b b r b b Rao IIT Academ/XII-ISC-Board -05_Mathematics_SOLUTIONS r (viii) P A (i) P B P A B 5 4 5 5. 4 P A B 5 5 5 P A B Taking LHS 6 5 a b c c a b 4 a b c c a b c a b c a b a b c a b c () Put t d d dt d d d dt d d d Now, sin 4 4

dt sin t d dt sin t d dt sin t Rao IIT Academ/XII-ISC-Board -05_Mathematics_SOLUTIONS d sin t sin t dt c sin t dt c cos t sec t dt tan t sect dt c tan tan Question t sec t c sec c (a) a b ab b ab a b a ( a b ) b a a b Let a b ab b ab a b a b a a b We shall tr to introduce zeros at a man places as possible keeping in mind that we have to introduce the factor a b. Appling C C bc and C C ac, we get a b 0 b 0 a b a b( a b ) a( a b ) a b 0 b ( a b ) 0 a b a a b [Taking ( a b ) common from both C and C ] 5 5

Rao IIT Academ/XII-ISC-Board -05_Mathematics_SOLUTIONS 0 b ( a b ) 0 a 0 0 a b [Appling R R br ar ] ( a b ) a 0 a b [Epanding along C ] ( a b ) (b) 5 4 AB 4 7 6 5 4 4 8 AB 4 7 5 4 4 8 6 4 5 4 6 8 0 0 AB 0 8 0 8 I 0 0 8 8 AB I A B I 8 A B 8 Let AX C 6 4 z AX C X A C we know that 5 4 6 8 z 7 6 66 60 4 6 8 z 4 6 A 8 B 6 6

8 6 8 z 4 z z Question Rao IIT Academ/XII-ISC-Board -05_Mathematics_SOLUTIONS (a) sin 5 sin 5 44 sin cos Let sin sin OPP HYP 44 adj cos HYP cos 44 5 44 5 44 69 69 7 7

(b) Rao IIT Academ/XII-ISC-Board -05_Mathematics_SOLUTIONS A B C A B C A B C L ABC ABC ABC ABC A A BC ABC BC AB B C B A BB B C L A B C Question 4 C A B (a) f sin sin on 0, f cos cos (i) f is differentiable on 0, (ii) Differentibilit Continuit f f is continuous on 0, LMVT is verified then there eist c c f b f a b a 0, such that C A cos c cos c sin sin sin 0 sin 0 0 cos c cos c 0 cosc cos c 0 cos c cos c 0 cos c cos c cos c 0 c c cos cos cos 0 8 8

c c Rao IIT Academ/XII-ISC-Board -05_Mathematics_SOLUTIONS cos cos 0 cosc, cos c c 0 0, c 0, c (b) foci 0, 0 be 0...() Equation of hperbola is b...() a equation passing through (, ) 9 4 b a We know that a b e b a 0 b 9 4 b 0 b Put b t 9 4 t 0 t 90 9 t 4t 0t t t t 90 0 t 8, t 5 b 8, b 5 b, b 5 When b then a 0 8 8 Neglected, a 8 is an imaginar number when b 5 9 9

then a 5 equation of hperbola Rao IIT Academ/XII-ISC-Board -05_Mathematics_SOLUTIONS 5 5 5 Question 5 (a) e m cos d e d mcos m d d m d d d d m a a d d d Dividing ; d d d m d d hence proved (b) AB r cos, AD r sin here r 0 AB 0 cos, AD 0 sin A r cos r sin Here r 0 D C A 4 r cos sin r sin cos 00 sin da 00 cos d A O r B da 0 d cos 0 4 da 400 sin d 0 0

Rao IIT Academ/XII-ISC-Board -05_Mathematics_SOLUTIONS da 800 sin 800 0 d So, Area is ma. when AB 0 cos 0 4 AD 0 cos 0 4 AB AD 0 Question 6 (a) I sec d cosec I cos d sin sin I sin cos sin d I tan sin d I tan cos Put tan t sin d sec d dt t t t dt t t dt dt Let t I t t dt t t t dt I t t dt

Rao IIT Academ/XII-ISC-Board -05_Mathematics_SOLUTIONS I t dt t I t t log t t log t t c I tan tan log tan tan log tan sec c t I tan sec log tan sec log tan sec c tan I tan sec log tan sec c (b) Question is wrong ( Data is missing) Question 7 (a) 9 0 4 5 6 00 4 0 5 8 9 8 00 44 69 96 5 56 9 6 9 0 9 5 64 5 6 0 0 4 75 8 77 58 8.5 n n 9 8 b 8 00 9 n 8 9 00 n 8 8 8 900 9 0000 448 900 548.6 6 6 b 8 00 9 n 8 8 5 n 8

Rao IIT Academ/XII-ISC-Board -05_Mathematics_SOLUTIONS 548 0.596 99 Equation of line of on is.5.6.5.6 0.88.5.6 9.6 Equation of line on is.5 0.596.5 (b) 0.596 0.6705.5 =.80 at.5 9.6.6.5 =.7555 Judge A R JudgeB R D = R R D 0 6 8 4 5.5 6 0.5 0.5 5.5 6.5 6.5 8 9 7 6 6 6 8 4 4 4 4 5 9 0 8 64 8 7 4 9 4 6 0 6 64 4 5 5 7 6 C. F m m 8 R 6 d n n cf 84.5 6 0 0 6 85 0 99. 0.

Question 8 Rao IIT Academ/XII-ISC-Board -05_Mathematics_SOLUTIONS (a) W B P(W ) P(B ) W B W B P(W / ) W P(B / ) W P( W / B ) P(B / ) B 0 W B W B W B W B P(B /W W ) P(B /W B ) P(B /W B ) P(B /B B ) P( W ) ; P( B ) 4 4 P( W / W ) ; P( B / W ) P( W / B ) ; P( B / B ) 4 4 P( B / W B ) P( B / W B ) P( B / W B ) P( B / W B ) 4 4 P(Required) 4 4 4 4 4 4 4 8 8 8 6 48 64 4 6 9 6 9 6 8 9 8 7 49 7 4 4

Rao IIT Academ/XII-ISC-Board -05_Mathematics_SOLUTIONS 4 5 4 (b) P A ; P B ; P C P A ; P B ; PC 5 4 () P (Eactl two person hit the target) P A B C P A B C P A B C P A P B P C P A P B P C P A P B P C 4 4 5 4 5 4 5 4 8 6 60 60 60 6 60 0 () P (At least two person hit the target) = P (Two person hit the target) + P (All three hit the target) 6 P A B C 60 6 P A P B P C 60 6 4 60 5 4 6 4 60 60 50 5 60 6 () P (None hit the target) P A B C P A P B P C 5 4 60 5 5

Question 9 (a) z i Rao IIT Academ/XII-ISC-Board -05_Mathematics_SOLUTIONS w iz z i i i i i i i i i w i i w i i 4 4 4 4 4 8 0 8 0 Locus is a circle 4 with centre 0, Radius 6 5 9 d (b) e e 0 d Put v d d v dv d v v dv e v e v 0 d 6 6

Rao IIT Academ/XII-ISC-Board -05_Mathematics_SOLUTIONS dv v v v v e e v e d v dv v v v e e v e ve v d v dv v e v e d e dv v v e d v v e dv d v v e log c log v e log c v v e v v e c c v e v v c e / when 0, c 0 e...() c Put c in equation () e 7 7

Rao IIT Academ/XII-ISC-Board -05_Mathematics_SOLUTIONS SECTION B Question 0 (a) Let sec AB be a diameter of a circle with centre C and P be an point on the circle other than A and B. Then APB is an angle subtended on a semicircle. Let AC CB a and CP r. P Then a r...() r AP AC CP a r r a BP BC CP CB CP a r A a C a B AP BP ( r a) ( r a) r r r a a r a a r a 0... ( r a a r ) AP BP APB is a right angle. Hence, the angle subtended on a semicircle is the right angle. (b) Volume of parallelopiped [ a b c] 4 Question (4 ) ( ) 4 ( 6) 0 5 0 45 cubic units (a) Equation of required plane passing through the intersection of the planes z 0 and 5z 0 is ( z ) ( 5z ) 0...() It passes through (,, ); ( ) ( 6 5 ) 0 ( 5) 0 5 In equation () ( z ) ( 5z ) 0 5 5 5 5z 5 6 9 5 z 6 0 4 0z 0 8 8

(b) Rao IIT Academ/XII-ISC-Board -05_Mathematics_SOLUTIONS The given lines are r ( iˆ ˆj kˆ ) (iˆ ˆj 4 kˆ ) and r (iˆ 4 ˆj 5 kˆ ) (iˆ ˆj 4 kˆ ) or r (iˆ 4 ˆj 5 kˆ ) (iˆ ˆj 4 kˆ ) replacing b a single parameter. These two lines pass through the points A and B having position vectors a ˆ ˆ ˆ i j k a iˆ 4 ˆj 5kˆ S. D. ( a a ) b b and Here, ( a ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ a ) (i 4 j 5 k) ( i j k) i j k ( a a ) b ( iˆ ˆj kˆ ) (iˆ ˆj 4 kˆ ) iˆ ˆj kˆ (8 6) iˆ (4 4) ˆj ( 4) kˆ iˆ 0 ˆj kˆ 4 ( a a ) b () 0 ( ) 5 and b 4 9 6 9 Question Substituting these values in the formula for S.D. we have S.D. = 5 5 units 9 9 (a) R Y G P(R) = P(G) = 6 P(Y) = I W B III W 4 B II 4 W B P(W/R) = 5 P(W/Y) = 4 5 P(W/G) = 7 4 P( Total) 5 5 6 7 4 5 5 4 9 9

Rao IIT Academ/XII-ISC-Board -05_Mathematics_SOLUTIONS 4 56 5 0 0 P( R / W ) P( R) P( W / R) P( Total) 5 0 0 5 4 (b) Success p 6 Failure q n r n r r P ( r) C p q Here n 5, r 0,,, p, q Maimum three success P(0) P() P() P() 5 5 5 5 5 5 5 5 C0 C C C 5 5 0 0 6. 6 0 0

Rao IIT Academ/XII-ISC-Board -05_Mathematics_SOLUTIONS SECTION C Question c i n (a) P i i 9 P 50, 000 i 0.0075 00 c.0075 50, 000 0.0075.0075 60 c 50, 000 0.0075.565680 0.565680.0075 (b) c 045.70 Let one tpe of food be and another tpe be Food Vitamin A Vitamin B Vitamin C unit unit unit unit unit unit According to the given condition Minimize z 4 6 Subject to 0 or 6 8 0 6 8 6 0 5 8 8 Corner points Objective function z 4 6 A (0, 0) z 4 0 6 0 40 B (, 4) z 4 6 4 9 C (, 5) z 4 6 5 04 D (0, 8) z 0 4 6 8 88 Minimum value of z is 9 at B (, 4)

Rao IIT Academ/XII-ISC-Board -05_Mathematics_SOLUTIONS

Rao IIT Academ/XII-ISC-Board -05_Mathematics_SOLUTIONS Question 4 (a) Bill drawn on : 08.0.0 Tenure : 7 month Nominal due date : 08.0.0 Legall due date :.0.0 Discounted on : 08.05.0 Unepired Number of das (b) Ma June 0 Jul Aug Sep 0 Oct 46 das 46 n r rs. 65 5 B. D Face Value Amount realised 7650 7497 5 B. D A n i 5 7650 i 5 5 5 765 i 0.05 7650 500 0 Hence, the required rate of interest is 5% per annum. Let C be the total cost function. Then C 6 Average cost AC C AC C 5 C 5 6 Let MC be the marginal cost function. Then, dc d MC 5 6 5 d d d d 6 6 AC 5 d d Now, d d AC 6 0 0 For AC to be increasing 6 0 6 6 0 6 or 6 6 0

Question 5 (a) Commodit Rao IIT Academ/XII-ISC-Board -05_Mathematics_SOLUTIONS Price in Rs. 00 04 p p 0 Weight w p p0 A 4 8 6 B 5 6 0 60 50 C 4 5 4 70 56 D 9 8 8 p = 00 p0 = 60 p 0 p 00 p 0 00 00 60 5 (b) 4 quartl 4 quartl 4 quartl moving ear Quarter Profit Moving Total Average Average Counted st 0 9 nd 47 rd 0 6 40.5 th 4 56 9 47.75 44.5 st 0 68 0 50.75 49.5 nd 59 49 66.5 56.5 rd 66 65 66.5 64.5 th 4 7 85 7.5 68.75 st 04 88 86 7.5 7.75 nd 60 80 70 70.75 rd 60 75 68.75 69.75 th 4 67 4 4

Rao IIT Academ/XII-ISC-Board -05_Mathematics_SOLUTIONS 90 80 70 60 50 40 0 0 0 0 0 0 40 50 60 70 80 90 00 0 5 5

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