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1 Half Yearly Examinations X Class Mathematics Paper - I Model Paper (English Version) Time : 2 2 Hours PARTS - A & B Max. Marks: 50 Time: 2 Hours PART - A Marks: 5 SECTION - I Note: ) Answer any FIVE of the following questions choosing atleast two from each group A and B. 2) Each question carries TWO Marks. 5 2 = 0 GROUP - A (Statements - Sets, Functions, Polynomials). Prove (p q ) p Λ q. 2. If A B, then prove A B = B.. If f(x) = x + 2, g(x) = x -, find (fog) x. 4. Solve x 2 6x + 8 > 0. GROUP - B (Linear Programming, Real Numbers, Progressions) 5. Draw the graph 2x + y 6. x+a 6. Evaluate Lt 2a x a x a. 2x 7. Solve Insert four A.M.'s in between and 2. SECTION - II Note: ) Answer any FOUR of the following questions. 2) Each question carries ONE mark. 4 = 4 9. Show that p V p is a Tautology. 0. f : R R, If f(x) = x + 5 a bijection, find f.. Define Remainders Theorem.

2 2. x y Find the value of objective function P = + at (0, 50) If a x = b, b y = c, c z = a, show that xyz = , x, are in G.P. then find the value of 'x'. 7 2 SECTION - III Note: ) Answer any FOUR of the following questions choosing atleast two from each group A and B. 2) Each question carries FOUR marks. 4 4 = 6 GROUP - A 5. Prove that A (B C) = (A B) (A C) for any three sets A, B, C. 6. Let f, g, h be real functions defined as f(x) = x, g(x) = x, h(x) = x +, then show that ho(gof) = (hog)of. g() + g(2) + g() 7. If f(x) = x + 2, g(x) = x 2 x 2 then find the value of. f( 4) + f( 2) + f(2) 8. Factorize 4x 4 2x + 7x 2 + x 2 by using Remainders theorem. GROUP - B 9. A sweetshop makes gift packet of sweets combines two special types of sweets A and B which weight 7 kg. Atleast kg of A and no more than 5 kg of B should be used. The shop makes a profit of Rs. 5 on A and Rs. 20 on B per kg. Determine the product mix so as to obtain maximum profit (write inequalities only). 20. If y = + then s.t. y 9y = The AM, GM, HM of two numbers are A, G, H respectively. Show that A G H. 22. If the sum of the first 'n' natural numbers is S and that of their squares S 2 and cubes S, show that 9S 2 2 = S ( + 8S ). SECTION - IV Note: ) Answer any ONE of the following question. 2) This question carries FIVE marks. 5 = 5 2. Draw the graph of y = x 2 5x Maximise f = 5x + 7y subject to the constraints 2x + y 2, x + y 2, x 0, y 0.

3 Time: 2 Hour PART - B Marks: 5 Note: ) Answer all the questions in the paper itself. 2) Each question carries mark. 2 I. Choose the correct answer and write its letter in the brackets. 0 2 = 5. Converse of p q is ( ) A) p q B) q p C) p ^ q D) q p 2. If A = {p, q, r, s} then the number of subsets of A is ( ) A) 8 B) 6 C) 4 D) 2. If I(x) = x then I is... function (I : A A, x A) ( ) A) Inverse B) Identity C) Onto D) Constant 4. Number of elements in the range of Constant function ( ) A) B) O C) 2 D) Infinite 5. If nc = nc 7 then n = ( ) A) 7 B) C) 20 D) x = my 2 (m > 0) Parabola lies in ( ) A) I & II B) I & III C) I & IV D) II & IV 7. Any such line parallel to the line represented by f(x) = K is called ( ) A) Parallel lines B) Perpendicular lines C) Iso profit lines D) None 8. The value of ( ) A) 2 B) 64 C) 6 D) The relation between AM, GM and HM is ( ) A) A 2 = GH B) G 2 = AH C) H 2 = AG D) AG = H 0. HM of a and b is a + b a b 2ab A) B) C) D) ab 2 2 a + b II. Fill in the blanks with suitable answers. 0 2 = 5. (p v q) = If n(a) = 4, n(b) = 5, n(a B) = 2 then n(a B) =.... If a function is... then only its inverse also a function.

4 4. If n(a) = m, n(b) = n then number of relations from A to B is ( 9 The last term in the expansion of x + ) is... x 6. If (x - 2) is exactly divisible by x x 2 + 4x + k then k = Any point in the feasible region is called Lt x 2 + 5x =... x 0 x 9. If 'n' G.M.'s in between a and b, then common ratio is If K a, K b, K c are in G.P. then a, b, c are in... III. Match the following. 5 (i) GROUP - A GROUP - B 2. If x < 0, y < 0 then (x, y) lies in [ ] A) (-, 2) 22. A point in x + y > 6 [ ] B) 0 2. If A = {5, 6, 7}, B = {, 2} then n(a B) = [ ] C) If (x + y, ) = (, y - x) then x = [ ] D) (, 2) 25. Discriminent of x 2 + 2x + = 0 is [ ] E) F) 2 G) Q = H) Q 4 5 = (ii) GROUP A GROUP B 26. If x+ = 9 x+ then x = [ ] I) (25) [ ] J) 28. n th term of AP = [ ] K) x, 6, 9 are in G.P. then x = [ ] L) A.M. of 5 and 25 is [ ] M) - N) a + (n )d O) 0 n P) [2a + (n ) d] 2

5 SOLUTIONS SECTION - I GROUP - A. Show that (p q) p Λ q. Sol: p q p q (p q) q p Λ q T T T F F F T F F T T T F T T F F F F F T F T F (p q) p Λ q. 2. If A B then show that A B = B. Sol: x A B x A or x B x B (... A B) A B B... () x B x Aor x B (... A B) x A B B A B... (2) From () and (2) A B = B.. If f(x) = x + 2, g(x) = x - then find (fog) - (x). Sol: fog(x) = f(x - ) = x = x - y = x - x = y + (fog) - (x) = x Solve x 2 6x + 8 > 0. Sol: x 2-4x - 2x + 8 > 0 x(x - 4) -2(x - 4) > 0 (x - 4)(x - 2) > 0 If (x - α)(x - β) > 0 then x value not lies between α, β x value not lies between 2 and 4.

6 GROUP - B 5. Draw the graph 2x + y 6. Sol: 2x + y = 6 x 0 y 2 0 [ x + a - 2a ] 6. Evaluate x a Lt x - a [ x + a - ] 2a Sol: Lt x - a x a Y- axis X - axis [ x + a - 2a x + a + ] 2a = Lt x a x - a x + a + 2a [ ( x + a ) 2 - ( 2a ) 2 ] = Lt x a ( x - a)( x + a + 2a ) [ x + a - 2a ] = Lt x a ( x - a)( x + a + 2a ) x - a [ ] = Lt x a ( x - a)( x + a + 2a ) [ = Lt ] x a x + a + 2a = a + a + 2a = 2a + 2a =. 2 2a 2x - 7. Solve 5. 2x - Sol: 5 2x (... x a -a x a )

7 -5 2x x x 6-7 x Insert four A.M.'s between and 2. Sol: Let, x, x 2, x, x 4, 2 a =, t 6 = a + 5d = 2 + 5d = 2 5d = d = 20 d = = 4 5 x = t 2 = a + d = + 4 = 7 x 2 = t = t 2 + d = = x = t 4 = t + d = + 4 = 5 x 4 = t 5 = t 4 + d = = 9. SECTION - II 9. Show that p v ~ p is a Tautology. Sol: p ~ p p v ~ p T F T F T T p v ~ p is a tautology. 0. f : R R, If f(x) = x + 5 is a bijection, find f -. Sol: y = f(x) y = x + 5 x = y 5 y 5 x = y - 5 f - (y) = x =.. Remainder theorem. Sol: Remainder theorem: If a rational integral function of x, say f(x) is divided by (x a) then the remainder is f(a).

8 x y 2. Find the value of objective function P = + at (0, 50) (50) Sol: At (0, 50), P = = = 60 2 = 0. If a x = b, b y = c, c z = a then s.t. xyz =. Sol: a x = b (c z ) x = b c zx = b (b y ) zx = b b xyz = b xyz = , x, are in G.P. then find the value of x x 2 Sol: = -2 x x 2 = 2 7 x 2 = ; x = = ±. SECTION - III GROUP - A 5. Show that A- (B C) = (A- B) (A- C). Sol: (i) Prove that A- (B C) (A- B) (A- C) (ii) (A- B) (A- C) A- (B C) (i) x A- (B C) x A and x (B C) x A and (x B or x C) (x A and x B) or (x A and x C) x (A- B) (A- C) A- (B C) (A- B) (A- C)... ()

9 (ii) x (A- B) (A- C) (x A and x B) or (x A and x C) x A and (x B or x C) x A and x (B C) x A- (B C) (A- B) (A- C) A- (B C)... (2) From () and (2) A- (B C) = (A- B) (A- C). 6. Let f, g, h be real functions defined as follows f(x) = x, g(x) = - x, h(x) = x + show that ho(gof) = (hog)of. Sol: (i) ho(gof)(x) gof(x) = g[f(x)] = g[x] = - x ho(gof)(x) = h[gof(x)] = h[ - x] = - x + = 2 - x... () (ii) (hog)of(x) hog(x) = h[g(x)] = h[ - x] = - x + = 2 - x (hog)of(x) = hog[f(x)] = hog(x) = 2 - x... (2) From () and (2) ho(gof) = (hog)of. 7. If f(x) = x + 2, g(x) = x 2 - x - 2, then find g() + g(2) + g(). f(-4) + f(-2) + f(2) Sol: f(x) = x + 2 g(x) = x 2 - x- 2 f(-4) = = -2 g() = = -2 f(-2) = = 0 g(2) = = 4-4 = 0 f(2) = = 4 g() = = 9-5 = 4 g() + g(2) + g() = = =. f(-4) + f(-2) + f(2)

10 8. Factorize 4x 4-2x + 7x 2 + x - 2 by using Remainders theorem. Sol: f(x) = 4x 4-2x + 7x 2 + x - 2 f() = = 4-4 = 0 (x - ) is a one factor. f(2) = 4(2) 4-2(2) + 7(2) 2 + (2) - 2 = = = 0 (x - 2) is another factor. By Horner's method of synthetic division x 2 - = (2x + )(2x - ) Factors of f(x) = (x - )(x - 2)(2x + )(2x - ). GROUP - B 9. A sweetshop makes gift packet of sweets combines two special types of sweets A and B which weight 7 kg. Atleast kg of A and no more than 5 kg of B should be used. The shop makes a profit of Rs. 5 on A and Rs. 20 on B per kg. Determine the product mix so as to obtain maximum profit (write inequalities only). Sol: Let A type of sweet packet = x kg B type of sweet packet = y kg x 0, y 0... () In one gift packet two types of sweets not more than 7 kgs x + y 7... (2) A type of sweet be atleast kgs x... () B type of sweet be no more than 5 kgs y 5... (4) Profit Rs. 5 on A & Rs. 20 on B per kg Profit function f = 5x + 20y.

11 20. If y = + then s.t. y - 9y = 0. Sol: y = / + / Cubing both sides ( y = / + ) / y = ( / ) ( + ) +. / ( / + ) / / ( / y = + + / + ) [... (a + b) = a + b + ab(a + b) ] / y = + + y y = 9+ +9y y = y; y = 0 + 9y y - 9y = The AM, GM, HM of two numbers are A, G, H respectively, then s.t. A G H. Sol: Let be two numbers x, y. x + y A= 2 G = xy 2xy H = x + y (i) A G x + y A- G = - xy 2 ( x ) 2 + ( y) 2-2 xy ( x - y ) 2 = = A G... () But G 2 = AH A G = G H

12 A G From () G H G H... (2) From () and (2) A G H. 22. If the sum of the first 'n' natural numbers is S and that of their squares S 2 and cubes S, show that 9S 2 2 = S ( + 8S ). n(n + ) Sol: S = Σn = 2 n(n + )(2n + ) S 2 = Σn 2 = [ 6 n(n + ) ] 2 n 2 (n + ) 2 S = Σn = 2 = 4 9[ n(n + )(2n + ) ] 2 LHS: 9S 2 2 = 6 [ n 2 (n + ) 2 (2n + ) 2 ] = 9 6 n 2 (n + ) 2 (2n + ) 2 =... () 4 n 2 (n + ) 2 n(n + ) 4 2 n 2 (n + ) 2 ( + 4n 2 + 4) = 4 n 2 (n + ) 2 (2n + ) 2 =... (2) 4 From () and (2) LHS = RHS. RHS: S ( + 8S ) = [ + 8 ] SECTION - IV 2. Draw the graph of y = x 2-5x + 6. Sol: x x x y

13 Scale: on X - axis cm = unit on Y - axis cm = unit t e n. a h b i t a r p u d a n e e. w w w From the graph x = 2 or 24. Sol: Maximise f = 5x + 7y subject to the constraints 2x + y 2, x + y 2, x 0, y 0. 2x + y = 2 x + y = 2 x y x y

Scale: on X - axis cm = unit on Y - axis cm = unit t e n. a h b i t a r p u d a n e e. w w w At A(4, 0) f = 5x + 7y = 5(4) + 7 (0) = 20 B (.4,.7) f = 5(.4) + 7(.7) = 7 +.9 = 28.

14 Scale: on X - axis cm = unit on Y - axis cm = unit t e n. a h b i t a r p u d a n e e. w w w At A(4, 0) f = 5x + 7y = 5(4) + 7 (0) = 20 B (.4,.7) f = 5(.4) + 7(.7) = = 28.9 C (0, 4) f = 5(0) +7(4) = = 28 At B (.4,.7) f is maximum. ) D ; 2) B; 9)B; 0) C; 4) 2mn; 5) x9 ; 20) A.P.; 2) G; ) B; ) p ^ 28) N; 0) K. 29) J; PART - B ANSWERS 4) A; 5) C; q; 2) 7; 6) C; 7) C; ) Bijection; 8) A; b n+ 6) 4; 7) Feasible solution; 8) 5; 9) a 22) D; 2) C; 24) E; 25) B; 26) M; 27) I; () Prepared by P. Venu Gopal

Final Exam A Name. 20 i C) Solve the equation by factoring. 4) x2 = x + 30 A) {-5, 6} B) {5, 6} C) {1, 30} D) {-5, -6} -9 ± i 3 14

Final Exam A Name. 20 i C) Solve the equation by factoring. 4) x2 = x + 30 A) {-5, 6} B) {5, 6} C) {1, 30} D) {-5, -6} -9 ± i 3 14 Final Exam A Name First, write the value(s) that make the denominator(s) zero. Then solve the equation. 1 1) x + 3 + 5 x - 3 = 30 (x + 3)(x - 3) 1) A) x -3, 3; B) x -3, 3; {4} C) No restrictions; {3} D)