Board Answer Paper: MARCH 2014

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1 Board Answer Paper: MARCH 04 and Statistics SECTION I Q.. (A) Select and write the correct answer from the given alternatives in each of the following: i. (C) Let l 0, m 3, n be the direction cosines of a line. l + m + n (0) option (C) is the correct answer. [] ii. (B) A Since, A(adj A) A I. K [] iii. (C) tan θ tan π θ tan 4 Since, tan θ tan α implies θ nπ ± α, n Z. the required general solution is θ nπ ± 4 π, n Z. [] (B) Attempt any THREE of the following: i. a + 3 b 5 c 0.(given) 5c a + 3 b 3b + a c 5 3b + a c [] 3+ by the section formula, point C(c) lies on the line segment joining points A (a ) and B (b), dividing it internally in the ratio 3 :. [] ii. Cartesian equation of the given line is 6 y+ 4 z the line passes through the points A(6, 4, 5) and has direction ratios, 7, 3. Let a be the position vector of point A and b be the vector parallel to the line.

2 Board Answer Paper: March 04 a 6i ˆ 4j ˆ+ 5kˆ and b i ˆ+ 7j ˆ+ 3kˆ [] The vector equation of a line passing through a point with position vector a and parallel to b is r a + λ b. vector equation of the required line is 6i ˆ 4j ˆ 5kˆ i ˆ+ 7j ˆ+ 3kˆ. [] r ( + ) + λ ( ) iii. The given equation is r ( ˆ ˆ ˆ) r n 8, where n 3i ˆ 4j ˆ+ kˆ Now, 3i 4j + k 8 n 3 + ( 4) + 3 [] The equation r n 8 can be written as n 8 r n n i.e., r ˆi ˆj kˆ which is the normal form of the plane The length of the perpendicular from origin is 8 units. [] 3 iv. Let θ be the acute angle between the lines whose direction ratios are 5,, 3 and 3, 4, 5. aa + bb + cc Then cos θ a + b + c a + b + c θ cos (3) + ( 4) + ( 3)(5) ( 3) 3 + ( 4) [] [] v. a. (p q) F [] b. Madhuri has curly hair or brown eyes. [] Q.. (A) Attempt any TWO of the following: i. y y z z The lines a b c are intersecting, if shortest distance is zero. y y z z i.e., if a b c 0 a b c Equations of the given lines are y+ z 3 k z and y 3 4 y y z z and a b c []

3 Here,, y, z, 3, y k, z 0, a, b 3, c 4, a, b, c Since the given lines intersect. k [] (3 8) (k + )( 4) (4 3) k + 0 k 9 0 k 9 [] ii. Let a, b and c be co-planar. Case : Suppose that any two of a, b and c are collinear vectors, say a and b. There eist scalars, y at least one of which is non-zero such that a + y b 0. a + y b + z c 0 is a required non-zero linear combination, where z 0. [] Case : None of the two vectors a, b and c are collinear. As c is coplanar with a and b, scalars, y are such that c a + y b a + y b + z c 0, is a required non-zero linear combination, where z [] Conversely, suppose a + y b + z c 0, where one of, y, z is non-zero, say z 0, then y c a + b z z c is coplanar with a and b. a, b and c are coplanar vectors. [] iii p q ~p p q ~p q (p q) ~p T T F T F F T F F T F F F T T T T T F F T F F F The entries in the columns 5 and 6 are identical. ~p q (p q) ~p [] [ mark each for column 5 and column 6] 3

4 Board Answer Paper: March 04 (B) Attempt any TWO of the following: i. Consider that for ABC, B is in a standard position i.e., verte B is at the origin and the side BC is along positive X-ais. As B is an angle of a triangle, B can be acute or B can be obtuse. Y A (c cos B, c sin B) Y (c cos B, c sin B) A c b c b X (0, 0) B a C (a, 0) X (0, 0) B a C(a, 0) X [] Using the Cartesian co-ordinate system, we get B (0, 0), A (c cos B, c sin B) and C (a, 0) Now consider l(ca) b [] b (a c cos B) + (0 c sin B).(by distance formula) a ac cos B + c cos B + c sin B [] a ac cos B + c (sin B + cos B) a ac cos B + c b a + c ac cos B [] ii. 4 Comparing the given equation with a + hy + by + g + fy + c 0, we get a, h 3, b 5, g 5, f 7, c 9 Consider, a h g 3 5 h b f g f c (45 49) + 3( ) + 5( 5) 4 + 3(8) + 5( 4) a h g h b f 0 g f c The given equation represents a pair of lines. [] Now, tan θ h ab a+ b θ tan 3 [] hf bg gh af The point of intersection is, ab h ab h , (, ) [] []

5 iii. Matri form of the given system of equations is This is of the form AX B, where A 3, X 3 4 y, B z 8 3 y 3 4 z Applying R R, we get 3 8 y 3 4 z Applying R R R, R 3 R 3 3R, we get y z 3 [] Applying R 3 R 3 R, we get y z 8 Hence, the original matri A is reduced to an upper triangular matri. By equality of matrices, we get + y + 3z 8...(i) 5y 5z 5 i.e. y + z 3...(ii) 8z 8 [] i.e. z Substituting z in equation (ii), we get y + 3 y Substituting y and z in equation (i), we get + () + 3() 8, y, z is the required solution. [] Q.3. (A) Attempt any TWO of the following: i. Let a + hy + by 0.(i) be a homogeneous equation of degree in and y. Case I: If b 0 (i.e., a 0, h 0), then the equation (i) reduces to a + hy 0 i.e., (a + hy) 0 This represents two lines, 0 and a + hy 0, both passing through the origin. Case II: If a 0 and b 0 (i.e., h 0), then the equation (i) reduces to hy 0, i.e., y 0 which represents the coordinate aes and they pass through the origin. [] 8 5 []

6 Board Answer Paper: March 04 Case III: If b 0, Multiplying both sides of equation (i) by b, we get ab + hby + b y 0 b y + hby ab To make L.H.S. a complete square, we add h on both the sides. b y + hby + h ab + h (by + h) (h ab) (by + h) ( h ab ) (by + h) ( h ab ) ( b + h ) + ( h ab ) (b + h ) ( ) h ab ii. y 0 [] This is the joint equation of two lines (by + h) + ( h ab ) and (by + h) ( h ab ) i.e., ( h h ab ) and ( h h ab ) by 0 + by 0 These lines pass through the origin. From the above three cases, we conclude that the equation a + hy + by 0 represents a pair of lines passing through the origin, if h ab 0. [] Let p : The switch S is closed, q : The switch S is closed, ~p : the switch S is closed, ~q : The switch S is closed. The symbolic form of the given switching circuit is (p ~q) (~p q). [] The switching table is as follows: iii. p q ~p ~q p ~q ~p q (p ~q) ( ~p q) T T F F T F T T F F T T F T F T T F F T T F F T T T F T In the above truth table, all the entries in the last column are T. the given circuit represents a tautology. Irrespective of the status of the switches, the current will always flow through the circuit. [] Let a, b, c, d be the position vectors of points A, B, C, D respectively. a î+ĵ + ˆk,b î+ĵ + 3ˆk, c 3 î+ ĵ+ ˆk and d 3 î + 3ĵ + 4 ˆk Now, ˆ ˆ ˆ ˆi+j+k ˆ ˆ î + ˆk AB b a ( i + j+ 3k ) ( ) AC c a ( 3i ˆ+ j ˆ+ k ˆ) ( ˆ ˆ ˆ) i+j+k î + ĵ + ˆk [] 6

7 AD d a ( 3i ˆ+ 3j ˆ+ 4k ˆ) ( ˆi+j+k ˆ ˆ) î + ĵ + 3 ˆk [] Volume of parallelopiped AB AC AD 0 AB AC AD 3 (3 ) 0 + (4 ) +4 5 Volume of parallelopiped is 5 cubic units. [] (B) Attempt any TWO of the following: i. Equation of the plane passing through the line of intersection of the planes y + z 3 0 and 4 3y + 5z is ( y + z 3) + λ(4 3y + 5z + 9) 0.(i) [] ( + 4λ) + ( 3λ)y + ( + 5λ)z 3 + 9λ 0 Direction ratios of normal to the plane (i) are ( + 4λ), ( + 3λ), + 5λ. But, the required plane is parallel to the line + y + 3 z 3 whose direction ratios are 4 5, 4, 5. Normal of the plane is perpendicular to the given line. ( + 4λ) 4( + 3λ) + 5( + 5λ) 0 [] 4 + 8λ 4 λ λ 0 λ 5 5 λ 5 Putting λ in (i), we get ( y + z 3) 5 (4 3y + 5z + 9) 0 [] 4 y + z y 5z y 4z y z 54 0 [] ii. To draw feasible region, construct table as follows: Inequality 3 + y + y 5 4 y 4 Corresponding equation (of line) 3 + y + y 5 4 y 4 Intersection of line with X-ais (4, 0) (5, 0) (4, 0) Intersection of line with Y-ais (0, 6) (0, 5) (0, 4) Region Non-origin side Non-origin side Origin side Origin side [] 7 []

8 Board Answer Paper: March 04 Y 4 X 6 5 C 4,4 3 B(4, 4) 4 y 4 3 D(, 3) A(4, ) O X Y 3 + y + y 5 [] Shaded portion ABCD is the feasible region, whose vertices are A, B, C and D. A is the point of intersection of the lines 4 and + y 5. Putting 4 in + y 5, we get 4 + y 5 y A (4, ) B is the point of intersection of the lines y 4 and 4. B (4, 4) C is the point of intersection of the lines y 4 and 3 + y. Putting y 4 in 3 + y, we get 3 + (4) C,4 3 D is the point of intersection of the lines + y 5 and 3 + y. Solving the above equations, we get D (, 3) [] Here, the objective function is Z 6 + 4y Z at A(4,) 6(4) + 4() 8 Z at B(4, 4) 6(4) + 4(4) 40 Z at C 4, (4) 4 3 Z at D(, 3) 6() + 4(3) 4 Thus, Z is minimized at every point along the line segment CD and its minimum value is 4. Therefore, Z has infinite number of optimal solutions. [] 8

9 iii. Let cos 4 5 cos 4 5 and 0 < < π sin > 0 Now, sin cos [] Let y cos 3 cos y 3 and 0 < y < π sin y > 0 Now, sin y cos y But, cos ( + y) cos cos y sin sin y y cos cos cos 3 33 cos 65 [] [] [] 9

10 Board Answer Paper: March 04 Q.4. (A) 0 SECTION II Select and write the correct answer from the given alternatives in each of the following: i. (A) y cos θ and sin θ d sin θ and cos θ dθ dθ d dθ d sin θ cosθ dθ π tan d π 4 [] θ 4 ii. (C) + y sec tan d Here, P sec Pd Integrating factor e sec d e log sec + tan e sec + tan [] iii. (B) y 3 + d d (,3) 6() 5 the equation of the tangent at (, 3) is y 3 5( ) y y 5 [] (B) Attempt any THREE of the following: i. f(0).(given) lim f() lim (sin cos ) 0 Since, lim 0 lim sin lim cos sin 0 cos 0 0 f(0) [] f() f(0), f is continuous at 0. [] ii. f() 5 + 9, [,4] As f() is a polynomial function, a. f() is continuous on [, 4] b. f() is differentiable on (, 4). f() () 5() f(4) (4) 5(4) f() f(4) Thus, all the conditions of Rolle s theorem are satisfied. [] The derivative of f() should vanish for at least one point c (, 4).

11 To obtain the value of c, f() f () 5 f (c) 0 c 5 0 c 5 (, 4) Thus, Rolle s theorem is verified. [] n iii. Let I sec tan d n sec sec tan d Put sec y Differentiating w.r.t., we get sec tan d [] n n y I y n + n yn + c I n secn + c [] iv. E(X ) P( i) i [] [] v. X ~ B(n 0, p) and E(X) 8, n 0.(given) But, E(X) np [] 8 0p p [] Q.5. (A) Attempt any TWO of the following: i. Let δ be a small increment in the value of. Since u is a function of, there should be a corresponding increment δu in the value of u. Also y is a function of u. there should be a corresponding increment δy in the value of y. δy δy δu Consider, δ δu δ Taking lim on both sides, we get δ 0 δy lim δ 0 δ δy δu lim lim δ 0δu δ 0δ.[ δ 0, δu 0] [] δy lim δy δu lim lim δ 0 δ δu 0δu δ 0δ.(i) δu But, lim δ 0 δ eists and is finite. d

12 Board Answer Paper: March 04 δy Also, lim δu 0δ y eists and is finite. u du limits on R.H.S. of (i) eist and are finite. [] Hence, limits on L.H.S. should also eist and be finite. δy lim δ 0 δ y eists and is finite. d d y du du d [] ii. y A cos (log ) + B sin (log ). (i) Differentiating w.r.t., we get d Asin(log ) Bcos ( log) + d y A sin (log ) + B cos (log ) d [] Again, differentiating w.r.t., we get d y + d y d d Acos(log ) Bsin ( log) d y + d y [Acos (log ) + Bsin (log )] d d [] d y + d y d d. [From (i)] d y + d y d d [] iii. Let I d ( + )( + ) Let ( + )( + ) A + B + + A( + ) + B( + ).(i) Putting in (i), we get A[( ) + ] 3A A 3 Putting in (i), we get B + B 3 3 [] ( + )( + )

13 I 3 d + 3 d + 3 d + ( ) tan 6 I (B) 3 tan 3 tan ( ) 6 tan Attempt any TWO of the following: d [] tan + c + c 3 tan ( ) + c [] i. Let be the side of each square removed from the piece of cardboard. The length of the side of the bo will be 8 cm and the height of the bo will be cm. Volume of bo area of base height (8 ) ( ) [] f() f () f () 4 44 [] For maima or minima, f () ( 9) ( 3) 0 9 or 3 For 3, f (3) 4(3) < 0 [] The volume will be maimum when 3. Maimum volume f(3) 4(3) 3 7(3) + 34(3) c.c. [] ii. Let I a a f( )d+ f(a )d I + I 0 0 For I, put a t d dt d dt When 0, t a and when a, t a [] I a f(a )d f() t dt 0 a a a a a b a f(t)dt... f( )d f( )d [] a b b b f( )d. f( )d f(t)dt a a a I I + I a a f( )d+ f( )d 0 a [] 3

14 Board Answer Paper: March 04 a b c b f( )d. f ( ) d f( ) d+ f ( ) d ;a < c< b a a c 0 a a a f( )d f( )d+ f(a )d [] iii. As f() is continuous in [, ], it is continuous at 0. lim f() lim f() f(0) lim sin a ( + ) [] sin a lim a 0 (0) + a a() a a 3 [] Also, f() is continuous at. lim f() f() f() Q.6. (A) lim ( + ) + lim + ( b + 3 ) [] () + b b() 3 + 4b b [] a 3, b i. Attempt any TWO of the following: d y+ + y y + + y d y + + y...(i) Put y u...(ii) y u Differentiating w.r.t., we get d u + du...(iii) d Substituting (ii) and (iii) in (i), we get u + du d u + + u [] du d + u du d + u Integrating on both sides, we get du d + log c + u log u+ + u log + log c [] 4

15 log u+ + u log c u + + u c y + + y c y+ + y c y + + y c [] ii. Let X be the number of heads out of 8 tosses. P(getting head) p, q p Given n 8 X ~ B (n, p) X ~ B 8, The p.m.f. of X is given by 8 8 P(X ) p() C, 0,,,.,8 [] P(getting at least one head) P(X ) 0 8 P(X ) P(no head) P(X < ) P(X 0) 8 C 0 [] ()() The probability of getting at least one head is [] iii. p y q ( + y) p+q Taking logarithm on both sides, we get p log + q log y (p + q) log ( + y) Differentiating w.r.t., we get p + q d y y d + y. + d [] q p+ q p+ q p y d + yd + y q p+ qd y (p+ q) p( + y) y + yd ( + y) [] q + qy py qy p+ q p py y ( + y) d ( + y) q py q py y d y d d y [] 5

16 Board Answer Paper: March 04 (B) Attempt any TWO of the following: i. Equation of the circle is + y 6.(i) Equation of line is y.(ii) Solving (i) and (ii), we get Y 6 ± y.[ area is in the first quadrant] y B (,) X O C X A(4, 0),. [] The point of intersection of curve (i) and (ii) in the first quadrant is ( ) Draw BC perpendicular to X-ais A Area of OCB + Area of region CABC 4 d + 6 d 0 Y [] 6 6 sin ( ) 4 8sin () 8 8sin + π π y 6 π sq.units. [] [] ii. Let I a d d ( ) a d a d d d a d a ( a ) + a a d a a a a + d a a a a d a d a [] [] I + [] a I a log a +c I a a log + a +c a c I a log + a + ad a c a log + a + c, where c [] 6

17 iii. a. Since, P(X) is the probability distribution of X. 6 0 P(X ) P(X 0) + P(X ) + P(X ) + P(X 3) + P(X 4) + P(X 5) + P(X 6) k + 3k + 5k + 7k + 9k + k + 3k 49k k 49 [] b. P(0 < X < 4) P(X ) + P(X ) + P(X 3) 3k + 5k + 7k 5k 5 49 P(0 < X < 4) 5 49 c. By definition of c.d.f., F() P(X ) [] F(0) P(X 0) P(X 0) k 49 F() P(X ) P(X 0) + P(X ) k + 3k 4k 4 49 F() P(X ) P(X 0) + P(X ) + P(X ) k + 3k + 5k 9k 9 49 F(3) P(X 3) P(X 0) + P(X ) + P(X ) + P(X 3) k + 3k +5k +7k 6k 6 49 F(4) P(X 4) P(X 0) + P(X ) + P(X ) + P(X 3) + P(X 4) k + 3k + 5k + 7k + 9k 5k 5 49 F(5) P(X 5) P(X 0) + P(X ) + P(X ) + P(X 3) + P(X 4) + P(X 5) k + 3k + 5k + 7k + 9k + k 36k F(6) P(X 6) P(X 0) + P(X ) + P(X ) + P(X 3) + P(X 4) + P(X 5) + P(X 6) k + 3k + 5k + 7k + k + 3k 49k c.d.f. of X is as follows: i F( i ) [] [] 7

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