9 th CBSE Mega Test - II

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9 th CBSE Mega Test - II Time: 3 hours Max. Marks: 90 General Instructions All questions are compulsory. The question paper consists of 34 questions divided into four sections A, B, C and D.

1 9 th CBSE Mega Test - II Time: 3 hours Max. Marks: 90 General Instructions All questions are compulsory. The question paper consists of 34 questions divided into four sections A, B, C and D. Section A comprises of 8 questions of 1 mark each, Section B comprises of 6 questions of marks each. Section C comprises of 10 questions of 3 marks each and Section D comprises of 10 questions of 4 marks each. Question numbers 1 to 8 in Section A are multiple choice questions where you are to select one correct option out of the given four. There is no overall choice. However, internal choice has been provided in 1 question of two marks each, 3 questions of three marks each and questions of four marks each. You have to attempt only one of the alternatives in all such questions, Use of calculators is not permitted. NAME OF THE CANDIDATE CONTACT NUMBER PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 1

2 Solution Visits: Section-A [Question number 1 to 8 carries 1 mark each] 1. If x is irrational number, then x is : (a) rational (b) Irrational (c) 0 (d) real (D) real. One of the factors of (16y 1) + (1 4y) is : (a) (4 + y) (b) (4 y) (c) (4y + 1) (d) 8y (D) 8y [ (16y 1) + (1 4y) 16y y 8y 3y 8y 8y (4y 1)] 3. The value of 4 3 equal to: (a) 1/6 (b) 6 (c) 1/6 (d) 6 (C) 1/6 [ 4 3 {( ) 1/3 } 1/4 1/3 1/4 1/6 ] 4. In which of the following (x + ) is a factor? (a) 4x 3 13x + 6 (b) x 3 + x + x + 4 (c) 4x x 5 (d) x 3 + x x 9 (A) 4x 3 13x + 6 [ Let f(x) 4x 3 13x + 6 f( ) 4( ) 3 13 ( ) + 6] 4 ( 8) ] 5. An exterior angle of a triangle is 10 0 and the two interior opposite angles are equal. Each of these angle is : PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH

(a) 60 0 (b) 100 0 (c) 10 0 (d) 180 0 (A) 60 0 [ Let each interior opposite angle be x. We know that exterior angle of a triangle is equal to the sum of two its interior opposite angles.

3 (a) 60 0 (b) (c) 10 0 (d) (A) 60 0 [ Let each interior opposite angle be x. We know that exterior angle of a triangle is equal to the sum of two its interior opposite angles. x + x 10 0 x 10 0 x 60 0 ] 6. Two sides of a triangle are 1 cm and 13 cm. The length of the third side cannot be: (a) 0.8 cm. (b) 5cm. (c) 4 cm. (d) 6cm. (A) 0.8 cm. [ The sum of any two side of a triangle is grater then its third side. Here, ( ) cm. 1.8 cm 13 cm] 7. In the given figure, if AB AC and AP AQ, then by which congruence criterion PBC QCB. (a) SSS (b) ASA (c) SAS (d) RHS (C) SAS [ AB AC and AP AQ AB AP AC AQ BP CQ Now, in PBC and QCB, we have AB AC B C BP CQ (proved above) PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 3

4 BC CB (common) By SAS congruency, we have PBC QCB 8. Heron s formula is : (a) s(s a)(s b)(s c) (b) (s a)(s b)(s c) (c) s(s a)(s b)(s c),s a + b + c (d) s s a s b s c, s a + b + c (D) s s a s b s c, s a + b +c Section-B [Question number 9 to 14 carries mark each] 9. Simplify: Factorise: x 4 + 4x + 3 x 4 + 4x + 3 y + 4y + 3 PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 4

5 y + 3y + y + 3 ( where x y) y(y + 3) +1 ( y + 3) ( y+ 3) ( y+ 1) (x + 3) (x +1) ( y x )] 11. Evaluate (104) 3 using suitable identity. (104) 3 ( ) (100+ 4) [ Using identity (a + b) 3 a 3 + b 3 + 3ab(a +b)] Define (a) a square (b) perpendicular lines. (a) A square: A square is a rectangle having same length and breadth, here, undefined terms are length, breadth and rectangle. (b) Perpendicular lines: Two coplanar (in a plane) lines are perpendicular, if the angle between them at the point of intersection is one right angle. Here, the term one right angle is undefined. 13. In the given figure, prove that AB EF OR In the given figure, lines l and m intersect each others at O. If x 40 0, Then find the value of y, z and w. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 5

6 Here, BCD BCE ECD Also, ABC BCD 0 ABC 50 But these are the alternate interior angles of AB and CD. Further AB CD.. (1) 0 0 CEF DCE But these are the interior angles on the same side of CD and EF. EF CD From (1) and (), we have AB CD and EF CD AB EF ( If two lines are parallel to the same line, they are parallel to each other) Lines l and m intersect at O x 0 + y y y Or z x 40 0 (vertically opposite angles) w y (vertically opposite angles) (Linear pair) PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 6

7 14. A point lies on the x-axis at a distance of 9 units from y-axis. What are its coordinates? What will be its coordinates if it lies on y-axis at a distance of 9 units from x-axis? 15. If x If a point lies on x-axis at a distance of 9 units from y-axis, then its abscissa is 9 and ordinate is 9. Coordinates of the point are (9, 0) If a point lies on y-axis at a distance of 9 units from x-axis, then its abscissa is 0 and ordinate is 0. Coordinates of the point are (0, 9) Section-C [Question number 15 to 4 carries 3 mark each] 3 3 and y 3 3, then find the value of x + y 10xy. OR Find the square root of 4.7 geometrically. x Similarly, we can find y 5 6 x + y 10xy (x + y xy) 8xy (x y) 8xy [(5 + 6) (5 6 )] 8(5 + 6) (5 6) [ ] 8(5 ( 6) ) PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 7

8 (4 6) 8(5 4) Or First of all draw a line of length 4.7 units such that AB 4.7 units. Now, from point B, mark a distance of 1 unit. Let this point be 'C. Let 'O' be the mid-point of the distance AC. Now, draw a semicircle with centre 'O' and radius OC. Let us draw a line perpendicular to AC passing through the point 'B' and intersecting the semicircle at point 'D'. The distance BD 4.7 Now, to represent 4.7 on the number line. Let us take the line BC as number line and point 'B' as zero, point 'C' as 1 and so on. Draw an arc with centre B and radius BD, which intersects the number line at point 'E'. Then, the point 'E' represents If x , find the value of x 3 1 x 3. OR If 5.36 and , find the value of x x PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 8

9 7 4 3 We have 1 x 14 x 1 x x On cubing both sides, we have x 3 x x 14 x x x x 3 x x 3 x If a + b + c 6, then find the value of ( a) 3 + ( b) 3 + ( c) 3 3( a) ( b) ( c). Sol. Let a x, b y and c z x + y+ z a + b + c 6 (a + b + c) Now, ( a) 3 + ( b) 3 + ( c) 3 3( a) ( b) ( c) x 3 + y 3 + z 3 3xyz (x + y + z) (x + y + z xy yz zx) 0 (x + y + z xy yz zx) 0 PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 9

10 18. Find the remainder when 3x 3 4x + 7x 5 is divided by (x-3) and (x + 3). Let f(x) 3x 3 4x +7x 5 When f(x) is divided by (x 3) Remainder f(3) 3(3) 3 4(3) 7(3) Also, when f(x) is divided by (x + 3) Remainder f( 3) 3( 3) 3 4( 3) + 7( 3) 5 3 ( 7) In the given figure, prove that x a + b + c. OR The sides BA and DC of a quadrilateral ABCD are produced as shown in figure. Show that x y a b. Sol. Let reflex ADC y Now, ADC + reflex ADC 360 PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 1

11 x + y (i) In quadrilateral ABCD, we have a + b + c + reflex ADC 360 a + b + c + y (ii) From (i) and {ii), we have x + y a + b + c + y x a + b + c Or Join BD a is the exterior angle of BDC D a CBD + CDB...(i) [ ext. sum of two interior opp. s] Also, b is the exterior angle of ABD b ABD ADB (ii) [ ext. sum of two interior opp. s] Adding (i) and (ii), we have a b CBD CBD ABD ADB CBD ABD CDB ADB x y 0. In the given figure, if AB CD, APQ 50 and PRD 17, find x and y. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 1

12 Sol. AB CD and PQ is a transversal PQR APQ (alternate interior angles) x 50 PRD is the exterior angle of PQR PRD PQR + QPR [ ext. sum of two interior 17 x + y opposite angles] y y Prove that the angles opposite to equal sides of an isosceles triangle are equal. Is the converse true? Given : A ABC in which AB AC To Prove : B C Construction : Draw the bisector of BAC which meets BC in D. Proof : In ABD and ACD, we have AB AC (given) BAD CAD (by construction) AD AD (common) So, by SAS congruence axiom, we have ABD ACD B C (c. p. c. t.) The converse of the above theorem is also true i.e., the sides opposite to equal angles of an isosceles triangle are equal. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 1

13 . In the given figure, D and E are points on the base BC of a ABC such that BD CE and AD AE. Prove that ABE ACD. BE BD + DE EC + DE ( BDCE) DC BE DC Also, AD AE AED ADE (angles opposite to equal sides are equal) In ABE and ACD, we have AD AE (given) AEB ADC ( AED ADE) BE CD (proved above) By SAS congruence axiom, we have ABE ACD 3. Prove that if two lines intersect each other, then the vertically opposite angles are equal. Given : Two lines AB and CD intersect each other at O. To Prove : (i) AOD COB (ii) AOC BOD PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 1

14 Proof : Since, ray OA stands on the line CD at O. AOC + AOD 180 (i) Also, ray OC stands on the line AB at O. AOC + COB (ii) From (i) and(ii), we have AOC + AOD AOC + COB AOD COB Similarly, we can prove AOC BOD 4. Find the area of a triangular park, two sides of which are 18 cm and 10 cm and the perimeter is 4 cm. Sol. Let a, b and c be the sides of a triangular park. Let a 18 cm, b 10 cm Perimeter of the triangular park 4 cm a + b + c 4cm 18 cm + 10 cm + c 4 cm 8 cm + c 4 cm c 4 cm 8 cm c 14 cm a b c 4 s 1cm. Area of the triangular park by Heron's formula s(s a)(s b)(s c) PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 1

15 cm Section-D [Question number 5 to 34 carries 4 mark each] 5. Using factor theorem, factorise the polynomial: x 4 + 3x 3 + x 3x 3. OR Show that x 3 + y 3 + z 3 3xyz 1 (x + y + z) [(x y) + (y - z) + (z x) ] Sol : Let f(x) x 4 + 3x 3 + x 3x 3 Put x 1 in f(x), we have f(1) (1) 3 + (1) 3(1) By factor theorem, x - 1 is a factor of f(x). Again, put x 1 in f(x), we have f ( 1) ( 1) 4 + 3( 1) 3 + ( 1) 3( 1) By factor theorem, x + 1 is a factor of f(x) Thus, (x + 1) and (x 1) are both factors of f(x) (x + 1) (x 1) x 1 is a factor of f(x) Now, divide f(x) by x 1, we have PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 1

16 4 3 x 1 x 3x x 3x 3(x 3x 3 x x 4 3 3x 3x 3x 3 3x 3 3x 3x 3 3x 3 0 We have, f(x) (x 1) (x + 3x + 3) (x 1)(x + 1)(x + 3x + 3) Or Here, R.H.S. 1 (x + y + z) [(x y) + (y z) + (z x) ] 1 (x + y + z) (x xy + y + y yz + z + z +zx + x ) 1 x y z x y z xy yz zx (x + y + z) (x + y + z xy yz zx) x 3 + y 3 +z 3 3xyz L.H.S. 6. Simplify: OR If a and b , find a b. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 1

17 Or We have a Similarly, we have b PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 1

18 Now, a b And a b Hence, a b a b a b Find the value of a, if (x + a) is a factor of x 4 a x + 3x a. Let p(x) x 4 a x + 3x a (x + a) is a factor of polynomial p(x) By factor theorem, we have p( a) 0 ( a) 4 a ( a) + 3( a) a 0 a 4 a 4 3a a 0 4a 0 a 0 8. The polynomial ax 3 + 4x + 3x 4 and x 3 4x + a when divided by (x 3) leave the same remainder. Find the value of a. Let f(x) ax 3 + 4x + 3x 4 And g(x) x 3 4x + 4 When f(x) and g(x) are divided by (x 3) leave the same remainder PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 1

19 f(3) g(3) a(3) 3 +4(3) +3(3) (3) + a 7a a 7a a 7a a a 6 a 1 9. The polynomial p(x) x 4 x 3 + 3x ax + 3a 7 when divided by (x + 1) leaves the remainder19. Find the value of a. Also, find the remainder, when p(x) is divided by(x + ). We have p(x) x 4 x 3 + 3x ax + 3a 7..(1) When p(x) is divided by (x +1) leaves remainder 19 p( 1) 19 ( 1) 4 ( 1) 3 + 3( 1) a( 1) + 3a a + 3a a a 0 a 5 Putting the value of a in (1), we have p(x) x 4 x 3 + 3x 5x x 4 x 3 + 3x 5x + 8 Remainder p( ) ( ) 4 ( ) 3 + 3( ) 5( ) PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH 1

20 30. In the given figure, PQR is an equilateral triangle in the coordinate axes and coordinates of Q and R are (0, 4) and (0, 4). Find the coordinates of the vertex P. Here, Q is (0, 4) and R is (0, 4) QR QR 8 units PQR is an equilateral triangle PQ QR PR 8 units Now, in rt. OPQ, PQ 8 units and OQ 4 units OP PQ OQ units Abscissa of P is 4 3and it lies on x axis. It ordinate is 0. Hence, the coordinates of P are ( 4 3, 0). 31. In the given figure, AP and DP are bisectors of two adjacent angles A and D of a quadrilateral ABCD. Prove that APD B C. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH

21 In quadrilateral ABCD, we have 0 A B C D A B C D A D 180 B C As, AP and DP are the bisectors of A and D 0 And Now, In PAD PDA 1 1 APD, we have A D 0 1 PAD PDA 180 B C 0 APD PAD PDA APD 180 B C APD B C APD B C 0 0 (Using (1))..(1) 3. AD is an altitude of an isosceles triangle ABC in which AB AC. Show that (i) AD bisects BC (ii) AD bisects A. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH

22 In ABD and ACD, we have AB AC (given) 0 ADB ADC 90 AD BC (given) AD AD By RHS congruence axiom, we have ABD BD DC ACD AD bisects BC (Common) (by c. p. c. t.) Also, BAD CAD (by c. p. c. t.) AD bisects A. 33. AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see in figure). Show that A C and B D. Given ABCD is a quadrilateral AB is the shortest side and CD is the longest side. To prove : A C and B B. Construction :Join A and C, also B and D. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH

23 Proof : In ABC, AB is the smallest side. BC AB 1...(i) ( angle opposite to longer side is greater) In ADC, CD is the longest side. CD > AD ii Adding (i) and (ii), we have A C Again, in ADB, AB is the smallest side. AD >AB iii In BCD, CD is the greatest side. CD > BC iv Adding (iii) and (iv), we have B D Hence, A C and B D. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH

24 34. AD, BE and CF, the altitudes of ABC are equal. Prove that ABC is an equilateral triangle. Consider right BCE and right CBF, we have Hyp. BC Hyp. BC (Common) CF BE (Given) By RHS congruence axiom, we have BCE CBF B C (by c. p. c. t.) AC BC (i) ( sides opposite to equal angles are equal) Similarly, ABD BAE B C (by c. p. c. t.) AC BC. (ii) ( sides opposite to equal angles are equal) From (i) and (ii), we have ABD BEA AB BC AC Hence, ABC is an equilateral triangle. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 40 D, CHANDIGARH

SOLUTIONS SECTION A [1] = 27(27 15)(27 25)(27 14) = 27(12)(2)(13) = cm. = s(s a)(s b)(s c)

SOLUTIONS SECTION A [1] = 27(27 15)(27 25)(27 14) = 27(12)(2)(13) = cm. = s(s a)(s b)(s c) 1. (A) 1 1 1 11 1 + 6 6 5 30 5 5 5 5 6 = 6 6 SOLUTIONS SECTION A. (B) Let the angles be x and 3x respectively x+3x = 180 o (sum of angles on same side of transversal is 180 o ) x=36 0 So, larger angle=3x