26ssc Maths 1 Guess Paper

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S.S.C. – X Class

Model Paper – 2014

Time: 2 ½ Hours Mathematics: Paper – I max. Marks: 50

Instructions:1. Answer the questions under part ‘A’ on a separate answer book2. Write the answers to the questions under part ‘B’ on the question paper itself and attach it to the answer

book of part ‘A’.Time: 2 Hours Part – A marks: 35

Section – I ( Marks 5×2 = 10)

Note: 1. Answer any FIVE questions choosing at least TWO from each of the following two groups i.e. Group Aand Group B.

2. Each question carries 2 marks.

Group – A(Statements and sets, functions, polynomials)

1. Write the converse inverse and contra positive of the conditional “In a triangle ABC, if AB = AC, then B = C”2. Prove that ∩ = A – B

3. If function f: R ― {2}→ R is defined as f(x) = Then prove that = x

4. Find the value of K so tha − 3 + 4 + t is exactly divisible by (x – 2).

Group – B(Linear programming, Real Numbers, and Progressions)

5. Indicate the polynomial region represented by the system of inequation x ≥0, y≥0, x + y ≤ 1

6. If a + b + c = 0 then show that · · =

7. If a = x + √ + 1 then show that x = ( − )

8. Determine K, so that k + 2, 4K – 6, and 3K – 2 are the three consecutive terms of an A.P

Section – II (Marks: 4×1 = 4)

Note: 1. Answer ANY FOUR of the following six questions.

2. Each question carries 1 Mark.9. Show that the statement P ∩ (~ ) is Tantology.

10. When gof defined.11. If − 3 + 4 + is divisible by (x – 2), find the value of K12. Define the feasible region.

13. If / = then find P

14. Find the x so that , , are three consecutive terms in a G.P

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Section – III (Marks: 4×4 = 16)

Note: 1. Answer ANY FOUR questions, choosing at least TWO from each of the following groups. i.e. A and B2. Each question carries 4 marks.

Group – A(Statements and sets, functions, polynomials)

15. Prove that A - ( ∩ ) = (A – B) (A – C)

16. f(x) = x – 1, g(x) = x2

– 2, h(x) = x3

– 3 are defined then prove that fo(goh) = (fog)oh17. Let f be given by f(x) = x + 2 and f has the domain { :2 ≤ ≤5}: Find , its a domain and range.

18. Find the term independent of x in the expansion of 6 −

Group – B(Linear programming, Real Numbers, and Progressions)

19. A shopkeeper sells not more than 30 shirts of each colour. At least twice as many white ones are sold asgreen ones. If the profit on each of the white be Rs.20 and that of green be Rs.25, then find out how many of

each kind be sold to give him a maximum profit. (Graph need not be drawn)

20. If + + = 0 then show that ( + + ) = 27abc.21. The A.M, G.M and H.M of two numbers are A, G, H respectively. Show that A ≥ ≥ . 22. Find the sum of n terms of the progression 7, 77, 777……..

Section – C(Linear programming and Quadratic Equations)

23. Using graph of y = x 2, Solve the equation x 2 – 4x + 3 = 024. Maximize f = 3x + y, subject to the constraints 8x + 5y ≤ 40, 4x + 3y ≥ 12, x ≥0 y≥0

Part – BTime: 30 min. Marks: 15 Note:1. All questions are to be answered

2. Each question carries mark

3. Answers are to be written in the question paper only4. Marks will not be given for over-written, rewritten or erased answers.

I. Write the capital letters of the correct answers in the brackets provided against each question × =

1. ~ [ ∩(~ )] = [ ] A) ~ ∩~ B) ~ ~ C) ~ ∩ D) ~

2. If A B, B C then A C. this property is called [ ] A) ~ ∩~ B) ~ ~ C) ~ ∩ D) ~







3) If f(x) = x2 – 3x - 4 then f(-2) = [ ] A) 6 B) 20 C) 5 D) −1

4) If f(x) = x 2 + x + 4, then 24 is the image of [ ] A) 1 B) 2 C) 3 D) 4

5) If the roots of the equation Px 2 + qx + r = 0 are equal, then [ ] A) 2 = B) 2 = 4 C) 2 = 4 D) 2 =

6) The graph of x = ―my 2 lies in [ ] A) Q 1 and Q 2 B) Q 2 and Q 3 C) Q 3 and Q 4 D) Q 4 and Q 5

7) The in equation for 1 ≤ ≤5 is [ ] A) x2 -5 x + 6 < 0 B) x2 -6 x + 5 < 0 C) x2 -5 x + 6 > 0 D) x2 -5x + 5 > 0

8) If √ 2 = 16 then x = [ ] A) 8 B) 4 C) 2 D) 10

9) lim → = [ ]

A) 0 B) 3 C) 27 D) 9

10) If there are ‘n’ arithmetic means between a and b then common difference (d) = [ ]

A) B) C) D)

II Fill in the blanks with suitable answers.

11. ( ) =

12. Let f: R →R, defined by f(x) = 3x – 4 then (x) =

13. f: B →B if f(x) = x then the function f is

14. Expand ∑ a2 (b – c) =

15. If = then n =

16. The expression ax + by which is sought to be is called the objective function.

17. The limiting position of secant of a circle is

18. The value of (0.001 ) is

19. The Geometric mean of 5 and 125 is







20. ∑ n = 55 then n =

III For the following questions under Group – A, choose the correct answer from the master list Group – B andwrite the letter of the correct answer in the brackets provided against each item.

(i)

Group - A Group - B

21. A = 1,2,3,4 then n(A) =[ ] A. 6

22. The graph y = mx 2 is[ ] B. 7

23. f(x) = x 2 -2 x + 4 find f(-2) =[ ] C. ( − ) ∩ ( − )

24. A - ( ) = [ ] D. Straight line

25. The product of the roots of x 2 -4 x + 7 = 0[ ] E. 4

F. ( − ) ( − )

G. 12

H. Parabola

(ii)

Group - A Group - B

26. ∑ =[ ] I.

27. ∑ n3 =[ ] J. a + b

28. | | ≤ =[ ] K. a - b

29. √ + √ √ − √ =[ ] L.

( )

30. Arithmetic mean of a and b is[ ] M. -a ≤ ≤

O.( )( )







Part – B (ANSWERS)

I II

1.D 11. ∩

2. C 12.

3. A 13. Identity function

4. D 14. ( − ) + ( − ) + ( − )

5. C 15. 20

6. B 16. Maximise (or) Minimise

7. B 17. Tangent

8. A 18. 0.1

9. C 19. 25

10. B 20. 10

III

(i) (ii)

21.E 26.

22. H 27. O

23. G 28. M

24. C 29. K

25.B 30. I







(1, 0)

(0, 1)1

0 X + y = 1 X

Y

KEY SHEET

Section – I

Group – A

1. Converse: In a triangle ABC, if B = C, then AB = ACInverse: In a triangle ABC, if A ≠AC then B ≠ C

Contra positive: In a triangle ABC, if B ≠ C then AB ≠

2. ∩ = { / } = { / } = ∩ = A - B

3. Given f(x) =

f =

=( ) ( )

( )( ) ( )

( )

=( )

× ( )

=

= x

4. Let f(x) = − 3 + 4 +

Given that f(x) is exactly divisible by (x – 2)i.e. f(2) = 0

f(2) = 2 − 3(2) + 4(2) + = 8 – 12 + 8 + K = 0=16 – 12 + K = 0=4 + K =0

K = -4

Group – B5. Given

x ≥0, y ≥0, i.e. the solution set lies in Q1. The equation for the in equation x + y ≤1 is x + y = 1

x 0 1y 1 0







6. L.H.S. = · ·

= · ·

=

=

a + b + c = 0 a3 + b3 + c 3 = 3abc

= =x3 = R.H.S

7.Given

a = x + √ + 1

a – x =

√ + 1

Squaring both the sides

( − ) = √ + 1

a2 – 2ax + x 2 = x2 + 1a2 – 2ax = 1

2ax = a 2 – 1

2x = – = a -

x = [ − ]

8. If t 1, t 2, t 3 are in A.P. then t 2 – t 1 = t 3 – t 2 K+ 2, 4K – 6. 3K – 2 are in A.P (4K– 6) – (K + 2) = (3K – 2) – (4K – 6)

4K – 6 – K – 2 = 3K – 2 – 4K + 63K – 8 = -K + 43K + K = 4 + 84K = 12

K =

K = 3

Section – II

9.~ ∩(~ ) ∩(~ )

T F F TF T T

F T F F TF F T F T







Observing the last column all the values are true. So the statement P∩(~ ) is tautology.

10. gof is defined when the co domain o ‘f’is equal to the domain of ‘g’

11. Let f(x) = − 3 + 4 + If f(x) is exactly divisible by (x – 2), then the remainder f(2) = 0

f(2) = 2 − 3(2) + 4(2) + 0 = 8 – 12 + 8 + K0 = 16 – 12 + K K = -4

12. Feasible region: The solution set of constraints of an Linear programming problem is called Feasibleregion.

13. = x 2 ∟

= x 2

= 2

P = 2×

p = 3

14. If a, b, c are the three consecutive terms of a G.P then b 2 = ac If , x, are the three consecutive terms of a G.P then

X2

= × = 1X =√ 1 = ± 1

Section – III

Group – A

15. To prove A - ( ∩ ) = (A – B) (A – C)We have to prove(i) A - ( ∩ ) (A – B) (A – C)(ii) (A – B) (A – C) A - ( ∩ )

(i) Let x [A − ( ∩ ) ]

x A and x ( ∩ ) 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) (A – C) i.e. all the elements in A - ( ∩ ) are in (A – B) (A – C)







A - ( ∩ ) (A – B) (A – C) (1)(ii) Let x [A − ( ∩ ) ]

x (A – B) or x (A – C) x A and x B or x A and x Cx A and (x B or and x C )x A and x ( ∩ ) x A − ( ∩ )

i.e. All the elements in (A – B) (A – C) are in A − ( ∩ ) (A – B) (A – C) A - ( ∩ ) (2)

From (1) and (2) by the basic theorem of equality of sets, A - ( ∩ ) = (A – B) (A – C)

16. Givenf(x) = x – 1g(x) = x2 – 1

h(x) = x 3 – 3

fog(x) = f [ ( )] = f(x2 – 2)= X2 – 2 – 1

= x2 – 3 ( f(x) = x – 1)goh(x) = g [ ( )] = g(x3 – 3)

= ( −3) - 2 ( g(x) = x2 – 2)= X6 – 6x 3 + 9 – 2= x6 - 6x3 + 7

fo(goh)(x) = f [(goh) ( )]

= f(x6 - 6x3 + 7 – 1)= x6 - 6x3 + 6

(fog)o h(x) = fog(h(x))= fog(x 3 – 3)

= f [(g( −3)] = f(x6 - 6x3 + 7)= x6 - 6x3 + 7 – 1= x6 - 6x3 + 6

fo(goh) = (fog)oh

17. Given

f(x) = x + 2Domain of = { :2 ≤ ≤5}

i.e. x {2,3,4,5 } Domain of f = {2,3,4,5 }

f(2) = 2 + 2 = 4f(3) = 3 + 2 = 5f(4) = 4 + 2 = 6







f(5) = 5 + 2 = 7 f = {( 2,4 ), (3,5 ) , (4,6 ) , (5,7) } = {( 4,2 ), (5,3 ), (6,4 ), (7,5) }

Domain of = {4,5,6,7 } = { :4 ≤ ≤7} Range of = {2,3,4,5 } = { :2 ≤ ≤5}

18. Given

6 −

General term in the expansion of ( + ) is = · ·

+1 term in the equation of 6 − is

= 12 · (6 ) ·

= 12 · 6 · · (−1) ·

= 12 · 6 · (−1) · 5 · · = 12 · 6 · (−1) · 5 · = 12 · 6 · (−1) · 5 · (1)

To get the term independent of x in the given expansion, we have to find r so that12 – 3r = 0

12 = 3rR = = 4

i.e. r = 4 5 th term is the term independent of x

T5 = 12 : 6 8 : 5 4

Group – B

19. Let the green shirts be ‘x’White coloured shirts be ‘y’

Total number of shirts sold = 30 + ≤30

At least twice as many white shirts are sold as green shirts 2x ≤ y

The conditions are x ≥0, y ≥0, x + y ≤ 30 and 2x – y ≤ 0 The function to be maximized is p = 25x + 20y

Subject to the above constraints,Points (30, 0) and (0, 30) are the solution sets of x + y ≤ 30 which represents the line x + y = 30 and

the origin side of it.Similarly (5, 10), (3, 6) and (2, 4) are the solution sets of 2x – y ≤ 0 which represents the line 2x – y = 0

and the non origin side of it.Hence the maximum profit lies at one of the vertices (0, 0), (10, 20) and (0, 30)









≥ 1

G ≥ H (2) From (1) and (2)

A≥ G ≥ H

22. 7 + 77 + 777 + 7777 + ………… n terms

= 7[1+11+111+1111+

] = [9+99+999+9999+ ]

= [(10 −1) + (100 −1) + (1000 −1) + (10000 −1) + ]

= [(10 −1) + (10 −1) + (10 −1) + (10 −1) + ]

= [10+10 +10 + 10 + ] - [1 + 1 + 1 + 1 + ]

Here 10+ 10 +10 + 10 + up to n terms is in G.P.

Where a = 10, r = = 10

r > 1

= ( )

= ( )

( ) −

( )

[10 (10 − 1 ) − 9 ]

Section – iv

23. Givenx2 – 4x + 3 = 0x2 = 4x - 3

y = x2

Y = 4x – 3The roots of x 2 – 4x + 3 = 0 are the x – co-ordinates of the parabola y = x 2 and straight line Y = 4x – 3

Consider y = x 2

Consider: Y = 4x – 3

Choose the scale On x – axis 1 cm = 1 UnitOn y – axis 1 cm = 2 Unit

X 0 1 2 3 -1 -2 -3y 0 1 4 9 1 4 9

X 0 1 2 3 -1y -3 1 5 9 -7







The line Y = 4x – 3 intersects the parabola at (1, 1) and (3, 9)X coordinate are 1, 3

Solution set = {1,3 } Verification:

x2 – 4x + 3 = 0x2 – 3x - x + 3 = 0x(x – 3) - 1(x – 3) = 0(x – 3) (x – 1) = 0x – 3 = 0 or x – 1 = 0

x = 3 or x = 1

24. x ≥0 ; y≥0 ……… these in equations represents the Q 1 first quadrant including the lines and .To determine the region of 8x – 5y ≤ 40First we draw the straight line 8x – 5y = 40

(0, 8), (5, 0)

To determine the region of 4x + 3y ≥ 12

We draw a straight line 4x + 3y ≥ 12

8x +5y ≤ 40

Substituting (0, 0) in the inequation8(0)+5(0) ≤ 40

0 ≤ 40 is true 8x +5y ≤ 40 represents the region towards the origin.

4x + 3y ≥ 124(0) + 3(0) ≥ 120 ≥ 12 is false 4x + 3y ≥ 12 represents the region that does not contain the origin.

ABCD is closed Convex Polygon.A (3, 0); B (5, 0), C (0, 8); D (0, 4)Objective function: f = 3x + y

Value of f at A (3, 0) = 3(3) + 0 = 9 + 0 = 9Value of f at B (5, 0) = 3(5) + 0 = 15 + 0 =15Value of f at C (0, 8) = 3(0) + 8 = 0 + 8 = 8Value of f at D (0, 4) = 3(0) + 4 = 0 + 4= 4

f = 3x + y is maximum at the vertex B (5, 0)

X 0 5

y 8 0

X 0 3y 4 0




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26ssc Maths 1 Guess Paper