Sample Paper-05 Mathematics Class XII. Time allowed: 3 hours Answers Maximum Marks: 100. Section A. Section B

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1 Sample Paper-05 Mathematics Class XII Time allowed: hours Answers Maimum Marks: 00. No. (, ) R but (, ) R r. a () + ( ) + ( 5) 8 5 l, m, n [0, ]. A A ( 8) 8 ( 6) A 8 Hence Prove tan cos sin sin 6 tan cos 6 tan cos tan. tan tan tan - 5 A ( A) ' A' Hence proved Section A Section B

2 sin( a + b) 7. y cos( c + d ) d d cos( c + d ) sin( a + b) sin( a + b) cos( c + d ) d d d cos ( c + d ) cos( c + d )cos( a + b). a + sin( a + b)sin( c + d ). c d cos ( c + d ) 8. y (tan - ) (given) Differentiation both side w.r. to y tan. + ( + ) y tan Again differentiation both side w.r. to ( + ) y + y.. + ( + ) y + ( + ) y 9. Put f ( a + b) t f '( a + b). a d dt n+ n t t dt. a a n + n+ + c [ f ( a + b) ]. + c a n + + e + y e d 0. ( + y ) e - + e - - tan (y) + tan e c when 0, y c d - - tan y + tan e uuur. AB $ i $ j k$ uuur BC 6$ i $ j k$ uuur CA 9$ i $ j 6k$ uuur uuur AB, BC uuur and AC uuur uuur uuur AC AB + BC Hence points A, B, C are collinear.

3 0. Required probability 5 9 Section C 0. A A A A Now, A 5A + I Now given A 5A + I + X X 0 5 X 0 5 X 0 5 sin cos. f + cos sin ( + cos ) + cos sin ( + cos ) + cos ( + cos ) sin f + cos cos cos f ' + cos [ Q cos ]

4 Hence, cos ( cos ) > 0 0, and, + cos ( + cos ) cos cos < 0, d 5. a sin θ a( θ.cosθ sin θ ) dθ + + d aθ.cosθ dθ aθ.sinθ dθ tanθ d Slope of normal tanθ Equation of normal y y ( ) d cosθ y ( a sin θ a. θ cos θ ) ( a cos θ a. θ sinθ ) sinθ cosθ + y sinθ a length of from origin 0cosθ + 0sinθ a aproved cos θ + sin θ d 6. I cos( + a).cos( + b) sin[( + a) ( + b)] sin( a b) cos( + a).cos( + b) sin( + a).cos( + b) cos( + a).sin( + b) sin( a b) cos( + a).cos( + b) [ tan( a) tan( b) ] d sin( a b) + + [ log sec( + a) log sec( + b) ] + c sin( a b) sec( + a) log + c sin( a b) sec( + b) 7. Let E,F and A three events such that E selection of Bag A and Fselection of bag B A getting one red and one black ball of two Here, p(e)p(getting or in a throw of die) 6

5 p(f) Also, P(A/E)P (getting one red and one black if bag A is selected) 6 C C and 0 C 5 P(A/F)P(getting one red and one black if bag Black if bag B is selected) C 7C 0 C 5 Now, by theorem of total probability, p(a)p(e).p(a/e)+p(f).p(a/f) 8 + p(a) Let number of head be random variable X in four tosses of a coin.x may have values 0,,, or obviously repeated tosses of a coin are Bernoulli trials and thus X has binomial distribution with n and p probability of getting head in one toss qprobability of getting tail (not head) in one toss - since, we know that P(Xr) n r n r r therefore, P(X0) 0 0 C0 6 P( X ) C C Ρ q, r 0,,...n 6 6 C 6 P X P( X ) C P( X ) C 6 Now required probability distribution of X is 0 P() Required mean µ ipi variance σ ipi ipi X ipi µ

6 9. Here Now, r r r ˆi r ˆj + y {(i ˆ + yj ˆ + zk) ˆ ˆi}.{(i ˆ + yj ˆ + zk) ˆ ˆj} + y ( ykˆ + zj).(k ˆ ˆ zi) ˆ + y (0i ˆ+ zj ˆ yk) ˆ ( zi ˆ+ 0ˆj + k) ˆ + y y + y 0 0. Let P ( α, β, γ) be the point of intersection of the given line (i) and plane (ii) y + z.. (i) and y + z 5. (ii) since,point P ( α, β, γ) lies on line (i)( therefore it satisfy(i) α β + γ λ α λ + ; β λ ; γ λ + Also point P ( α, β, γ) lie on plane (ii) α β + γ 5 (iii) putting the value of α, β, γin (iii) we get λ + λ + + λ + 5 λ λ 0 α ; β ; γ hence the coordinate of the point of intersect ion p is (-,-,) therefore,required distance d ( + ) + ( + 5) + ( + 0) units. Here sin[ cot ( + ) ] cos(tan ) let cot ( + ) θ cotθ + θ + θ cos ec cot sinθ θ sin cot ( + ) sin + + again tan α tan α secα + tan α + cos α α cos tan cos now equation (i) becomes

7 sin sin cos cos y tan tan tan tan tan Let sinθ sin ( ) θ putting the value of, we get + sin θ Q 0 tan sinθ sin 0 < sinθ < sin θ θ cos + cosθ tan tan 0 < θ < 0 < < sinθ θ θ sin cos θ 0 < < θ θ tan cot tan tan θ > > θ θ sin > > θ,, differentiating both sides with respect to, we get d d. Given a cosθ +bsinθ d a sin θ + bcos θ dθ Also, y a sin θ b cos θ a cos θ + bsin θ dθ

8 d a cos θ + bsin θ θ d d a sin θ + b cos θ dθ y. d y d d y d y d y d y y y y y 0 d + d d d + Section D. A N N and * is a binary operation defined on A. a, b * c, d a + c, b + d c + a, d + b c, d * a, b The operation is commutative a, b * c, d * e, f a + c, b + d * e, f a + c + e, b + d + f Again, And ( a, b) ( c, d )*( e, f ) ( a, b) *( c + e, e + f ) ( a + c + e, b + d + f ) Here, ( a, b) *( c, d ) *( e, f ) ( a, b) ( c, d )*( e, f ) The operation is associative., a, b * e, f a + e, b + f Let identity function be ( e f ), then For identity function a a + e e 0 And for b + f b f 0 As 0 N, therefore, identity-element does not eist. 5. The point of intersection of the curves y, y: y y ( 78) 0 0, The shaded area is the required area.

9 A y y d 0 d 0 / 5 / A y y d d / 7 / 6 Thus, ratio of the areas is 5:75:9 6. The vector equation of the line through the point A and B is r r $ i + $ j + k + λ[(5 ) $ i + ( ) $ j + (6 ) kˆ ] r r $ i + $ j + kˆ + λ($ i $ j + 5 kˆ )...( i) Let P be the point where the line AB crosses the XY plane. Then the position vector r r of the point P is the form i $ + y$ j i $ + yi $ ( + λ) $ i + ( λ) $ j + ( + 5 λ) kˆ + λ y λ, y 5 5 req. point is,, I cos log(sin ) d. sin sin log(sin ). cos. d sin 0 log sin cos. sin cos + cos log d log Let be the cost of kg onions, y be the cost of kg wheat, z be the cost of kg rice. Thus we get the following equations: d

10 +y+z60 +y+6z90 6+y+z70 60 Let A 6, b A 50 0, A 0 0 0, X A b , y 8, z 8 Thus, per kg cost of onions, wheat and rice is Rs.5,Rs.8, Rs.8 respectively. 9. Suppose is the number of pieces of model A and y is the number of pieces of model B. Then, Profit Z y The mathematical formulation of the problem is as follows: Ma Z y s.t 9 + y 80( fabricating constra int) + y 60 + y 0( finishing constra int) 0, y 0 We graph the above inequalities. The feasible region is as shown in the figure. The corner points are O,A,B and C. The co-ordinates of the corner points are (0,0),(0,0), (,6),(0,0). Corner Point Z y (0,0) 0 (0,0) 6000 (,6) 6800 (0,0) 000 Thus profit is maimized by producing units of A and 6 units of B and maimum profit is 6800.