Board Pattern Maths Paper I Answers With Marking Skim

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SAFE HANDS Model marking schme and answers to Board Pattern Mathematics and statistics Paper I Hence to maximize z = 4x + 3y subjected to Q1.A (1) conditions 4x + 12y ≥ 18; 16x + 4y ≥ 24; 8x + 6y ≥ 16 and x≥ 0 & y ≥ 0 p T T F F

q T F T F

p→q T F T T

p∧(p→q) T F F F

p∧(p→q)→q T T T T

Since all truth values of last column are T the statement pattern is a “ Tautology” (2) p ≡ T, q ≡ T and r ≡F then (a) p ∨ (q ⇒ r) ≡ T ∨ (T ⇒ F) ≡ T ∨ F ≡ T (b) ¬q ∧ (p⇒q) ≡ ¬T ∧ (T ⇒ T) ≡ F ∧ T ≡ F (c) (∼r↔p) → ∼q ≡ (∼F↔T) → ∼T ≡ T → F ≡ F (3) The logical form of the given switch circuit is (A∧B) ∧ ( C∨B’) its behavior is A B C B’ A∧B C∨B’ (A∧B) ∧ (C∨B’) T T T F T T T T T F F T F F T F T T F T F T F F T F T F F T T F F T F F T F F F F F F F T T F T F F F F T F T F Q1 B (1) Consider the lines 5x + 3y =15 ≡ x/3 + y/5 = 1 will pass through (3,0) & (0,5) x + 2y = 6 ≡x/6 + y/3 = 1 will pass through (6,0) & (0,3) line x = 6 is parallel to Y axis and y = 5 is parallel to X axis. (1M) Y C

B D A X

Shaded region ABCD is the feasible region with vertices A(6,0); B(6, 5);C(0, 5) and D(12/7, 15/7) (1M)

(2) Let there be x kg of food A and y kg of food B. The cost of A is 4x and of B is 3y total cost z = 4x + 3y is to be minimized. Food A (x) Food B (y) Min. req. Fats 4 12 18 Carbohydrate 16 4 24 Proteins 8 6 16 Constrain due to fats 4x + 12y ≥ 18 Constrain due to carbohydrates 16x + 4y ≥ 24 Constrain due to protein is 8x + 6y ≥ 16

Q2 A. (1) In ∆ABC, AP⊥BC and BQ ⊥ CA. AP and BQ intersect each other at H B join CH and extend it to meet AB in R. To show CR ⊥ P R AB. Consider position H vectors with respect to H means let HA = a , HB = b A C Q and HC = c As HA ⊥ BC

hence ⇒

a

•(

a

•

c

=

c

a

-

b

•

b

HB ⊥ CA hence ⇒

b

•

=

a

b

)=0 ----(I) •(

b

•

-

a

)=0

c

-----(II)

c

a I and II ⇒ • c = b • c hence ( a - b )• c =0 Which means CH ⊥ AB thus altitudes of triangles are concurrent ( or any equivalent valid proof) (2) Use distributive property and fact that [ a a c c b ] = [ a b b ] = [ b ] = 0 and [ c a b ]

=[ a b c ] to prove ( b + c )•[( c c ) (3) Let a , b ,

+

)x(

a

and

c

r

+

a

b

)] = 2

a

•(

b

x

be given vectors

represented by OA , OB , OC and OR respectively. Through point R draw a plane parallel to planes BOC and COA and AOB intersecting lines OA, OB and OC intersect at L,M and N respectively. It is clear that a and OL are parallel hence

OL = x.

similarly OM = y.

a

b

and ON = z

c

Hence OR = OL + OM + ON = x. a + y. b + z. c To show uniqueness let there be two distinct linear combinations representing r say α. a + β. b + γ c = x a + y b + z c ∴ ( α - x). a

=

= (y-β).

a

b

+ (z -γ).

c

hence

( y − β) (z − γ ) b+ c contradiction to fact ( α − x) ( α − x)

that a , b , c are non coplanar. Hence no such distinct different linear combination will exist. Q2 B (1) Check that [ AB , AC AD ] = 0

−4 −6 −2 −1 4 3 = 0 hence A, B, C and D are coplanar. − 8 −1 3 (2) 2

a

+3

b

-5

c

=0

⇒

=

c

2a + 3b 2a + 3b = 5 2+3

(1M)

this indicates that C divides AB internally in ratio 3:2 and hence A-C-B are collinear. (1M)

Q4(A)(a) (1) The slope of line x + 2y -1 = 0 is -½ let the slope of line making an angle of 30o with above line be m then 1 m − (− ) 1 2m + 1 2 tan30o = 1 ⇒ 3 = 2 − m squaring both 1 + m( − ) 2

sides and replacing m by y/x we get

Q3. (A) (a) (1) Verify A2 -2A + I = 0 by obtaining

2i + 1

A2 =   2i

− 2i  2 + 2i − 2i  , 2A =   1 − 2i 1 − 2i  2i

1 0   1

&I=  0

(1M)

post multiply by A-1 to get A-1 = 2I – A hence

1 − i i  A-1 =    − i 1 + i

(1M)

 1 1 1 x   2       (2) Using A.X = B we get 2 − 1 3  y  =  9   1 5 1 z  − 2 reduce to lower triangular ( or upper triangular) by elementary row transformations the matrix equation becomes –

 1 1 1  x   2 0 − 3 1  y  = 5      0 0 4 z  8

∴x + y + z = 2 , -3y + z = 5; 4z = 8 thus z= 2, y = -1, x = 1 Q3 (A) (b)(1) Given equation is ax2 + 2hxy + by2 = 0 ---I auxiliary equation is bm2 + 2hm + a = 0 If m1 and m2 are slopes of two lines represented by I then m1 + m2 = -2h/b, m1m2 = a/b. If θ is angle between lines with slopes m1 & m2 then tanθ =

m1 − m2 = 1 + m1.m2

(m1 + m2 )2 − 4m1.m2 1 + m1.m2

(1M)

but m1 + m2 = -2h/b , m1m2 = a/b ∴ tanθ =

2 h2 − ab a+b

(1M)

(2) Solve to get x + (cosecα ± cotα) y= 0 Q3(B) (1) Comparing with standard equation of pair of lines ax2 + 2hxy + by2 +2gx + 2fy + c = 0 we get a = 3, h = 5, b = 3, g = 0, f = 8, c = k condition of pair of lines is abc + 2fgh –ab2 – bg2 – ch2 = 0 ∴ k = -12 (2) Obtain radius by perpendicular distance formula p=

4(3) + 3(2) − 8 42 + 3 2

=2

(1M)

hence equation of circle is (x-3)2 + (y-2)2 = 22 hence equation of required equation is is x2 + y2 -6x -4y + 9 = 0 (1M)

x2 -16xy – 11y2 = 0 is given equation hence the result is proved. (2) comparing with standard equation a = 2, h = k/2 b = -9 (1M) using m1 + m2 = -2h/b = k/9 and m1m2 = a/b = -2/9 ∴ k/9 = 5( -2/9) ⇒ k = -10 (1M) (b) (1) As A and B are independent events P(A∩B) = P(A).P(B) hence P(A).P(B) = 0.2 and by addition theorem P(A∪B) = P(A) + P(B) – P(A∩B) we get P(A) + P(B) = 0.9 solving the quadratic equation P(A) = ½, P(B) = 2/5. (2) Number of elements in sample space =

15

C2 = 105

A = both balls are black or white ∴number of 4 6 elements in event space A = C2 + C2 = 21 hence

P(A) = 21/105 = 1/5 Q4 (B) (1) slope of tangent is -1 using equation y = mx ± a 2m2 + b2 get the answer as y + x = ± 13 (2) Solve to get 4x -3y + 5 = 0

Q5.(A) (a) (1) comparing with standard result m = 1, c = k a = 1/√2and b = 1/√3 and using c=±

a2m2 + b2 K = ±

5 6

(2) Expected equation is of type y2 = 4ax as point (-10,25) lies on parabola substituting a = 1 hence equation is y2 =4x (b) (1) Given α = β = γ hence dc’s are equal and cos2α + cos2β + cos2γ = 1 hence direction cosines are ±

1 3

(2) The angle between planes is angle between their normal, normal to 2x –y + 3z + 4 = 0 is 2i –j + 3k and to 3x + 2y – 4z -1 = 0 is 3i + 2j – 4k hence if θ is angle between the planes then

8 (2)(3) + ( −1)(2) + (3)( −4) cosθ = ⇒ θ = cos-1 406 14 . 29 Q5 (B) (1) a2 = 5 and b2 = 2, using b2 = a2(1-e2) using e=

3 5

Foci are (± ae, 0) hence (√3,0) and (-√3,0)

(2) Conjugate hyperbolas are

y2 x2 x2 y2 − = 1 and 2 − 2 = 1 a 2 b2 b a

(1M)

a 2 + b2 a 2 + b2 2 hence e = and e2 = hence a2 b2 1 1 + 2 =1 (1M) 2 e1 e2 2 1