10th Maths Study Material 2024-25 PDF

www.Padasalai.Net www.Trb Tnpsc.Com EDUCATION DEPARTMENT VILLUPURAM DISTRICT MATHEMATICS 10 et...

www.Padasalai.Net www.Trb Tnpsc.Com EDUCATION DEPARTMENT VILLUPURAM DISTRICT MATHEMATICS 10 et i.N la sa da Pa Material for Students w. 2024-25 ww Achieved State Level 10th Place in SSLC Public Examination - March 2024 BEST WISHES bu. mwptHfd;/ M.A., M.A., B.Ed., M.Phill., Chief Educational Officer, Villupuram District. jd;dk;gpf;if + tplhKaw;rp + fod ciHg;g[ = btw;wp "The Struggle you're in Today will definitely develop the strength you need for Tomorrow. A. SIVAMOORTHY, Government High School, Perumbakkam, Villupuram District. Mobile: 9750827997 Kindly Send Me Your Key Answer to Our email id - [email protected]  www.Padasalai.Net www.Trb Tnpsc.Com 2 bu. mwptHfd;/ M.A., M.A., B.Ed., M.Phill., Kjd;ikf; fy;tp mYtyu;/ tpGg;g[uk;. ************************************************************************* et MESSAGE TO TEACHERS i.N First and foremost I would like to express my hearty gratitude to all the maths handling teachers la who took much effort so as to bring the SSLC public sa result in the 10th Place in State Level last year. I am very happy to here that more than 85% of da the question were asked only from our District Level Maths Material and more over the question papers Pa used for the preparatory through out the Academic Year 2023-24 were more helpful for the students to w. show their victory without any hindrance. I need the same kind of co-operation cent percent ww from the teachers’ side in this academic year too. In my point of view a dedicated and service minded teacher is always blessed by the God ever. Hence the teachers are asked to devote more time for the upliftment of poor rural pupils as education is the process of acquiring knowledge and skills. Kindly Send Me Your Key Answer to Our email id - [email protected]  Details of the Public Questions which took place in our Maths Material. GOVT. QUESTION PAPER - MARCH 2024 GOVT. QUESTION PAPER - JUNE 2024 Pub. Mat. Pub. Mat. Pub. Mat. Total Pub. Mat. Pub. Mat. Pub. Mat. Total Mark Mark Mark Mark Mark Mark Q.No. P.No. Q.No. P.No. Q.No. P.No. Marks Q.No. P.No. Q.No. P.No. Q.No. P.No. Marks 1 58-1 1 15 14-1 2 29 15-2(i) 5 8 1 58-10 1 15 14-4 2 29 17-6(i) 5 8 2 58-11 1 16 67-1 2 30 19-13 5 8 2 58-8 1 16 - - 30 68-3 5 6 www.Padasalai.Net ww 3 59-5 1 17 20-1 2 31 26-9(ii) 5 8 3 59-6 1 17 - - 31 25-5(ii) 5 6 4 59-10 1 18 27-2i 2 32 32-1 5 8 4 60-6 1 18 27-1(ii) 2 32 32-2 5 8 5 60-8 1 19 29-10 2 33 35-1 5 8 5 60-17 1 19 85-13 2 33 26-11 5 8 w. 6 60-12 1 20 105-2 2 34 38-1 5 8 6 61-9 1 20 104-6 2 34 34-10 5 8 7 61-6 1 21 119-2 2 35 43-4i 5 8 7 61-11 1 21 113-3 2 35 38-1 5 8 Pa 3 8 61-12 1 22 117-24 2 36 125-17 5 8 8 62-3 1 22 118-27 2 36 43-4(ii) 5 8 9 62-1 1 23 44-3 2 37 130-4 5 8 9 62-10 1 23 128-6 2 37 125-18 5 8 da 10 63-7 1 24 - - 38 49-4 5 6 10 63-2 1 24 139-1 2 38 134-16 5 8 11 64-3 - 25 138-8 2 39 - - 2 11 63-10 1 25 47-3 2 39 144-20 - 8 sa 12 64-15 1 26 49-1 2 40 la 146-1 5 8 12 64-3 1 26 50-7 2 40 - - 3 13 - - 27 51-17 2 41 56-16 5 7 13 65-9 1 27 51-16 2 41 52-2 5 8 14 65-12 1 28 71-2 2 42 25-4 5 8 14 65-2 1 28 21-9 2 42 - - 3 i.N 12 26 43 (a) 66-4 8 8 14 24 43 (a) 7-12 8 8 43 (b) 6-9 8 8 43 (b) 66-5 8 8 et 44 (a) 66-12 8 8 44 (a) 12-8 8 8 Kindly Send Me Your Key Answer to Our email id - [email protected] www.Trb Tnpsc.Com 44 (b) 11 - 6 8 8 44 (b) 66 - 4 8 8 97 135 87 125  Details of the Public Questions which took place in our Previous Year District Level Question Papers. GOVT. QUESTION PAPER - MARCH 2024 GOVT. QUESTION PAPER - JUNE 2024 Pub. Vil. Dt. Pub. Vil. Dt. Pub. Vil. Dt. Pub. Vil. Dt. Pub. Vil. Dt. Pub. Vil. Dt. Mark Mark Mark Mark Mark Mark Q.No. Q.P.No. Q.No. Q.P.No. Q.No. Q.P.No. Q.No. Q.P.No. Q.No. Q.P.No. Q.No. Q.P.No. 1 15 I R - 15 2 29 I R - 30 5 1 I Mid - 3 1 15 Hly - 15 2 29 Hly - 29 5 2 16 Qly - 16 2 30 I R - 29 5 2 III R - 1 1 16 30 II R - 29 5 www.Padasalai.Net ww 3 17 I Mid - 11 2 31 3 1 Mid -5 1 17 S.T 3 - 4 2 31 Hly - 31 5 4 18 32 4 18 32 II R - 33 5 5 IR-8 1 19 33 III R - 34 5 5 II Mid-1 1 19 Q-9 2 33 I R - 32 5 w. 6 20 34 III R - 36 5 6 20 34 S.T 7-12 5 7 21 35 7 S.T4 - 2 1 21 35 I R - 36 5 Pa 4 8 III R - 7 1 22 I R - 24 2 36 Hly - 38 5 8 22 36 9 23 Hly - 23 2 37 I R - 39 5 9 23 37 da 10 24 I R - 26 2 38 III R - 40 5 10 24 38 11 25 39 II Mid - 7 5 11 25 II R - 26 2 39 II R - 40 5 sa 12 26 I R - 27 2 40 12 26 II R - 27 2 40 13 27 41 III R - 43 5 13 II R - 14 1 27 41 S.T 6-16 5 la 14 I R - 13 1 28 42 II R - 32 5 14 I R - 14 1 28 ST 2 - 4 2 42 43 (a) Hly - 43 8 43 (a) i.N 43 (b) 43 (b) Hly - 43 8 44 (a) S.T4 - 4 8 44 (a) Hly - 44 8 et Kindly Send Me Your Key Answer to Our email id - [email protected] 44 (b) 44 (b) www.Trb Tnpsc.Com 3 14 66 7 12 61 Total = 83 Marks Total = 80 Marks  www.Padasalai.Net www.Trb Tnpsc.Com Minimum Material 5 4 GEOMETRY 1. Construct a triangle similar to a given 3. Construct a triangle similar to a given 2 triangle PQR with its sides equal to 3 of triangle PQR with its sides equal to 5 3 of the corresponding sides of the the corresponding sides of the triangle PQR 2 triangle PQR (scale factor < 1). (scale factor 3 < 1) P 3 5 Solution: Rough diagram P' et Q R' R Q1 Q2 i.N  Q3 4. Construct a triangle similar to a given 4 triangle LMN with its sides equal to of the la 5 corresponding sides of the triangle LMN 4 L sa (scale factor < 1). L' 5 da M 2. Cons Construct a triangle similar to a given N' N M1 7 triangle PQR with its sides equal to of M2 4 the corresponding sides of the triangle PQR M3 Pa 7 M4 (scale factor > 1) M5 4 Solution: Rough diagram 5. Construct a triangle similar to a given w. 6 triangle ABC with its sides equal to of the corresponding sides of the 5 6 ww triangle ABC (scale factor > 1). SEP-20 5  A' C C' A B1 B2 B3 B4 B5 B6 Kindly Send Me Your Key Answer to Our email id - [email protected]  www.Padasalai.Net www.Trb Tnpsc.Com 10th Std - Mathematics 6 6. Construct a triangle similar to a given 7 triangle PQR with its sides equal to 3 of the corresponding sides of the triangle 7 PQR (scale factor > 1). 3 P' P Q R R' Q1 et Q2 Q3 Q4 i.N Q5 9. Draw a circle of diameter 6 cm from a point Q6 Q7 P, which is 8 cm away from its centre. Draw the two tangents PA and PB to the circle and 7. Draw a circle of radius 3 cm. Take a point P measure their lengths. on this circle and draw a tangent at P. Solution: Given, radius r = 3 cm la Verification: In the right angle triangle OAP. PA2 – OA2 = 64 – 9 = 55 sa Rough diagram PA = 55 = 7.4 cm Solution: Rough diagram da Pa w. ww 8. Draw a circle of radius 4 cm. At a point L on it draw a tangent to the circle using the alternate segment. Solution: Rough diagram A. SIVAMOORTHY, BT. Asst. GHS, Perumpakkam, Villupuram Dt. Kindly Send Me Your Key Answer to Our email id - [email protected]  www.Padasalai.Net www.Trb Tnpsc.Com Minimum Material 7 10. Draw a circle of radius 4.5 cm. Take a point P on the circle. Draw the tangent at that point using the alternate segment theorem. cm 8.6 10 cm 8.6 Solution: cm M B A m m 5c 5c O N O 4.5 cm T et Proof: L In ∆OPA i.N PA2 = OP2 – OA2 = 102 – 52 = 100 – 25 = 75 T' PA = 75 = 8.6 cm (approx) 11. Draw a circle of radius 4.5 cm. Take a point la on the circle. Draw the tangent at that point using the alternate segment theorem. 13. Take a point which is 11 cm away from the centre of a circle of radius 4 cm and draw the two tangents to the circle from that point. sa Solution: M Rough Diagram A 4cm da O 11cm P O B N Pa 4.5 cm T L w. O 4cm B 4cm A 4cm T' ww 12. Draw the two tangents from a point which is 10 cm away from the centre of a circle of M radius 5 cm. Also, measure the lengths of the 11cm 10.3 tangents. m SEP-20 c 10.3 c Solution: m Rough Diagram A P Verification: 3cm 5cm O P In ∆OPA AP2 = OP2 – OA2 = 112 – 42 = 121 – 16 = 105 B AP = 105 = 10.2 cm Kindly Send Me Your Key Answer to Our email id - [email protected]  www.Padasalai.Net www.Trb Tnpsc.Com 10th Std - Mathematics 8 14. Draw the two tangents from a point which 15. Draw a tangent to the circle from the point P is 5 cm away from the centre of a circle of having radius 3.6cm, and centre at O. Point diameter 6 cm. Also, measure the lengths of P is at a distance 7.2 cm from the centre. the tangents. SEP-21 Solution: Rough Diagram Solution: A Rough Diagram m 3.6c A 7.2cm O P 3cm O 5cm P B B P A et 6.2c 4c m cm m 5 cm 4c m 3.6 i.N O M 7.2cm P B A cm 3cm A 3.6 la B O sa Verification Verification: In ∆OPA AP2 = OP2 – OA2 In ∆OPA, PA2 = OP2 – OA2 da = 52 – 32 = 25 – 9 = 16 = 7.22 – 3.62 AP = 16 = 4 cm = 51.84 – 12.96 = 38.88 Pa PA = 38.88 = 6.2 cm (approx) *** w. ww A. SIVAMOORTHY, BT. Asst. GHS, Perumpakkam, Villupuram Dt. Kindly Send Me Your Key Answer to Our email id - [email protected]  www.Padasalai.Net www.Trb Tnpsc.Com Minimum Material 9 GRAPH 1. Varshika drew 6 circles with different sizes. Draw a graph for the relationship between the diameter and circumference of each circle as shown in the table and use it to find the circumference of a circle when its diameter is 6 cm. Diameter (x) cm 1 2 3 4 5 Circumference (y) cm 3.1 6.2 9.3 12.4 15.5 Solution: 1. Table: Diameter(x) cm 1 2 3 4 5 6 Circumference 3.1 6.2 9.3 12.4 15.5 18.6 (y) cm et 2. Variation: Direct Variation 3. Equation: i.N y = kx k = y = 3.1 ==6.2 3.1= 9.3 = 12.4 =.... = 3.1 x 1 2 3 4 y = 3.1x 4. Points: la (1, 3.1) (2, 6.2) (3, 9.3), (4, 12.4), sa (5, 15.5), (6, 18.6) 5. Solution: From the graph, when diameter is 6 cm, its da circumference is 18.6 cm. 2. A bus is travelling at a uniform speed of 50 km/hr. Draw the distance-time graph and hence find (i) the constant of variation (ii) how far will it travel in 1½ hr Pa (iii) the time required to cover a distance of 300 km from the graph. Solution: 1. Table Time taken x 60 120 180 240 300 360 (in minutes) w. Distance y 50 100 150 200 250 300 (in km) ww 2. Variation: Direct Variation. 3. Equation: y = kx y 50 100 5 5 5 k= = == = y= x x 60 120 6 6 6 III. Points: ( 60, 50), (120, 100), (180, 150), (240, 200), (300, 250) IV. Solution: From the graph, 5 (i) Constant of variation k = 6 (ii) The distance travelled in 90 mins = 75 km (iii) The time taken to cover 300 km = 360 minutes = 6 hours. Kindly Send Me Your Key Answer to Our email id - [email protected]  www.Padasalai.Net www.Trb Tnpsc.Com 10th Std - Mathematics 10 3. A company initially started with 40 workers to complete the work by 150 days. Later, it decided to fasten up the work increasing the number of workers as shown below. Number of workers (x) 40 50 60 75 Number of days (y) 150 120 100 80 (i) Graph the above data and identify the type of variation. (ii) From the graph, find the number of days required to complete the work if the company decides to opt for 120 workers? (iii) If the work has to be completed by 30 days, how many workers are required? Solution: 1. Table Number of workers (x) 40 50 60 75 100 120 Number of days (y) 150 120 100 80 60 50 2. Variation:  Indirect Variation. 3. Equation: et  xy = k xy = 40 × 150 = 6000 i.N xy = 6000 4. Points: (30, 200), (40, 150) (50, 120) (60, 100), (75, 80) 5. Solution: From the graph, la sa (i) Type of variation = Indirect variation (ii) T he required number of days to complete the work when the company decides to work with 120 workers = 50 days. (iii) If the work has to be completed by 200 days, the number of workers required = 30 workers da 4. Nishanth is the winner in a Marathon race of 12 km distance. He ran at the uniform speed of 12 km/hr and reached the destination in 1 hour. He was followed by Aradhana, Ponmozhi, Jeyanth, Sathya and Swetha with their respective speed of 6 km/hr, 4 km/hr, 3 km/hr and 2 km/hr. And, Pa they covered the distance in 2 hrs, 3 hrs, 4 hrs and 6 hours respectively. Draw the speed-time graph and use it to find the time taken to Kaushik with his speed of 2.4 km/hr. Solution: 1. Table: Speed x (km/hr) 12 6 4 3 2 w. Time y (hours) 1 2 3 4 6 2. Variation: ww Indirect Variation 3. Equation xy = k xy = 12 × 1 = 12 xy = 12 4. Points: (12, 1), (6, 2), (4, 3), (3, 4), (2, 6) 5. Solution: From the graph, The time taken by Kaushik to go at a speed of 2.4 km/hr = 5 hours. A. SIVAMOORTHY, BT. Asst. GHS, Perumpakkam, Villupuram Dt. Kindly Send Me Your Key Answer to Our email id - [email protected]  www.Padasalai.Net www.Trb Tnpsc.Com Minimum Material 11 5. A garment shop announces a flat 50% discount on every purchase of items for their customers. Draw the graph for the relation between the Marked Price and the Discount. Hence find (i) the marked price when a customer gets a discount of ₹ 3250 (from graph) (ii) the discount when the marked price is ₹ 2500 Solution: 1. Table (Given) Y Marked Price Scale 1000 2000 3000 4000 5000 6000 7000 ` (x) x-axis – 1 cm – 1000 units y-axis – 1 cm – 500 units Discounted 500 1000 1500 2000 2500 3000 3500 5500 Price ` (y) 2. Variation: 5000  Direct variation. 4500 3. Equation: 4000 et y = kx 3500 Discount y 500 1 3250 k = = = i.N x 1000 2 3000 (6000, 3000) 1 2500 y = x (5000, 2500) 2 2000 (4000, 2000) 4. Points:  la (1000, 500), (2000, 1000), (3000, 1500), (4000, 1500 (3000, 1500) 1250 2000), (5000, 2500), (6000, 3000), (7000, 3500) 1000 (2000, 1000) sa 5. Solution: 500 (1000, 500)  rom the graph, F (i) If the customer gets a discount of ` 3250, then X’ O 2500 X 6500 1000 2000 3000 4000 5000 6000 7000 8000 da the Marked price = ` 6500 (ii) If the marked price is ` 2500, then the discount Y’ Market Price = ` 1250 6. Draw the graph of xy = 24, x, y > 0. Using the graph find, Pa (i) y when x = 3 and (ii) x when y = 6. Y Scale Solution: 1. Table: 28 x-axis – 1 cm – 2 units y-axis – 1 cm – 2 units x 12 8 6 4 3 2 26 w. y 2 3 4 6 8 12 24 (1, 24) 22 2. Variation: ww 20  Indirect variation. 18 3. Equation: 16 xy = k 14 xy = 12 × 2 = 24 12 (2, 12) xy = 24 10 4. Points: (3, 8) 8 (12, 2), (8, 3), (6, 4), (4, 6), (3, 8), (2, 12) 6 (4, 6) (6, 4) 5. Solution: 4 (8, 3) (12, 2) From the graph, 2 (24, 1) (i) If x = 3 then y = 8 (ii) If y = 6 then, x = 4 X’ 2 4 6 8 10 12 14 16 18 20 22 24 X Y’ Kindly Send Me Your Key Answer to Our email id - [email protected]  www.Padasalai.Net www.Trb Tnpsc.Com 10th Std - Mathematics 12 7. Graph the following linear function y =1/2 x. Identify the constant of variation and verify it with the graph. Also (i) find y when x = 9 (ii) find x when y = 7.5. Solution: 1. Table: x 2 4 6 8 10 12 14 16 y 1 2 3 4 5 6 7 8 2. Variation: Y Scale Direct Variation 9 x-axis – 1 cm – 1 units y-axis – 1 cm – 1 units 3. Equation: 8 y = kx 7.5 y 1 7 k = = x 2 6 1 y = x et 5 2 4.5 4. Points: 4  (2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, i.N 3 6), (14, 7), (16, 8) 2 5. Solution: 1 From the graph, la (i) If x = 9 then y = 4.5 (ii) If y = 7.5 then x = 15 X’ O 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 X Y’ sa 8.  The following table shows the data about the number of pipes and the time taken to till the same tank. No. of pipes (x) 2 3 6 9 Time Taken (in min) (y) 45 30 15 10 da Draw the graph for the above data and hence Y (i) find the time taken to fill the tank when five 46 Scale pipes are used 44 x-axis – 1 cm – 1 units y-axis – 1 cm – 2 units (ii) find the number of pipes when the time is 9 42 Pa minutes. 40 38 Solution: 36 1. Table: w. 34 No. of pipes (x) 2 3 5 6 9 10 32 Time Taken (y) 30 45 30 18 15 10 9 Time Taken (mins) 28 ww 26 2. Variation: 24 Indirect variation 22 3. Equation: 20 xy = k 18 xy = 2 × 45 = 90 16 xy = 90 14 3. Points: 12 (2, 45), (3, 30), (5, 18), (6, 15), (9, 10), (10, 9) 10 4. Solution: 8 From the graph, 6 (i) Time taken to fill the tank if using 4 5 pipes = 18 minutes 2 (ii) Number of pipes used if the tank fills up in X’ O 1 2 3 4 5 6 7 8 9 10 11 12 X Y’ Number of Pipes 9 minutes = 10 pipes A. SIVAMOORTHY, BT. Asst. GHS, Perumpakkam, Villupuram Dt. Kindly Send Me Your Key Answer to Our email id - [email protected]  www.Padasalai.Net www.Trb Tnpsc.Com Minimum Material 13 9. A school announces that for a certain competitions, the cash price will be distributed for all the participants equally as show below No. of participants (x) 2 4 6 8 10 Amount for each participant in ₹ (y) 180 90 60 45 36 (i) Find the constant of variation. (ii) Graph the above data and hence, find how much will each participant get if the number of participants are 12. 190 Y Solution: 1. Table: 180 Scale x-axis: 1 cm = 1 units 170 No. of participants y-axis: 1 cm = 10 units 2 4 6 8 10 12 160 (x) 150 Amount for each 140 180 90 60 45 36 30 participant in ` (y) 130 Amount for each participants 2. Variation: 120 et 110  Indirect variation. 100 3. Equation: 90 i.N xy = k 80 xy = 2 × 180 = 360 70 xy = 360 60 4. Points: 50 5. Solution: la (2, 180), (4, 90), (6, 60), (8, 45), (10, 36), (12, 30) 40 30 20 sa (i) Constant of Variation: k = 360 10 (ii) Cash Price each participant will get if X 12 participants participate = Rs. 30 X’ O 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Y’ Number of Participants da 10. A two wheeler parking zone near bus stand charges as below. Time (in hours) (x) 4 8 12 24 Amount ₹ (y) 60 120 180 360 Check if the amount charged are in direct variation or in inverse variation to the parking time. Pa Graph the data. Also (i) find the amount to be paid when parking time is 6 hr; (ii) find the parking duration when the amount paid is ₹150. 380 Y ) 60 Solution: I. Table: 360 ,3 Scale x-axis – 1 cm – 2 units (24 w. 340 y-axis – 1 cm – 20 units Time ( in hours) (x) 4 6 8 10 12 24 320 Amount in ` (y) 60 90 120 150 180 360 300 2. Variation: 280 ww  Direct variation. 260 3. Equation: 240 y = kx, 220 y 60 k = = = 15 200 Amount x 4 ) 80 180 y = 15x ,1 (12 160 4. Points: 150 140  (4, 60), (6, 90), (8, 120), (10, 150), (12, 180), ) 120 0 12 (24, 320) (8 100 90 5. Solution: 80 From the graph, 60 ) 60 (4, (i) If the parking time is 6 hours, then the 40 parking charge = ` 90. 20 X (ii) If the amount ` 150 is paid, then the X’ 2 4 6 8 10 12 14 16 18 20 22 24 26 Parking time = 10 hours. Y’ Time Kindly Send Me Your Key Answer to Our email id - [email protected]  www.Padasalai.Net www.Trb Tnpsc.Com 10th Std - Mathematics 14 B × A = {2, 3, 5, 7} × {1, 2, 3} 1. Relations and Functions = {(2, 1), (2, 2), (2, 3), (3, 1), (3, 2), 2 Marks (3, 3), (5, 1), (5, 2), (5, 3), (7, 1), 1. I f A×B = {(3, 2), (3, 4), (5, 2), (5, 4)} then find (7, 2), (7, 3)} A and B. SEP-20 4. I f B × A = {(−2, 3), (−2, 4), (0, 3), (0, 4), Solution: (3, 3), (3, 4)} find A and B. A×B = {(3, 2), (3, 4), (5, 2), (5, 4)} then Solution: A = {Set of all first coordinates of elements of A = {Set of all second coordinates of elements A × B} A = {3, 5} of B × A} A = {3, 4} B = {Set of all second coordinates of elements B = {Set of all first coordinates of elements of of A × B} B = {2, 4} B × A} B = {–2, 0, 3} Thus A = {3, 5} and B = {2, 4} Thus, A = {3, 4} B = {–2, 0, 3} et 2. Find A × B, A × A and B × A 5. T  he arrow diagram shows a relationship i) A={2, −2, 3} and B = {1,−4} between the sets P and Q. Write the relation ii) A = B = {p, q} in i.N iii) A = {m, n} ; B = f Solution: i. A × B = {2, –2, 3} × {1, –4} (3, –4)} la = {(2, 1), (2, –4), (–2, 1), (–2, –4), (3, 1), A × A = {2, –2, 3} × {2, –2, 3} (i) Set builder form (ii) Roster form sa (iii) What is the domain and range of R. = {(2, 2), (2, –2), (2, 3), (–2, 2), (–2, –2), Solution: (–2, 3), (3, 2), (3, –2), (3, 3)} i.  et builder form of R = {(x, y) | y = x – 2, x P, S B × A = {1, –4} ×{2, –2, 3} da y Q} = {(1, 2), (1, –2), (1, 3), (–4, 2), (–4, –2), (–4, 3)} ii. Roster form R = {(5, 3), (6, 4), (7, 5)} iii. Domain of R = {5, 6, 7} and range of ii. Given A = B = {p, q} Pa A × B = {p, q} × {p, q} R = {3, 4, 5} = {(p, p), (p, q), (q, p),(q ,q)} 6. L  et X = {1, 2, 3, 4} and Y = {2, 4, 6, 8, 10} and A × A = {p, q} × p, q} R = {(1, 2), (2, 4), (3, 6), (4, 8)}. Show that R = {(p, p), (p, q), (q, p),(q ,q)} is a function and find its domain, co-domain w. B × A = {p, q} × {p, q} and range? = {(p, p), (p, q), (q, p),(q ,q)} Solution: ww iii. A = {m, n}, B = φ A × B = {(m, n) × { } = { } A × A = {(m, n) } × {m, n} = {(m, m), (m, n), (n, m), (n,n)} B × A = { } × {m, n} = { } Pictorial representation of R is given diagram,  et A = {1, 2, 3} and B = {x | x is a prime 3. L From the diagram, we see that for each number less than 10}. Find A × B and x X, there exists only one y Y. B × A. MAY-22 Thus all elements in X have only one image in Y. Solution: Therefor R is a function. A = {1, 2, 3} B = {2, 3, 5, 7} Domain X = {1, 2, 3, 4} A × B = {1, 2, 3} × {2, 3, 5, 7} Co-domain Y = { 2, 4, 6, 8, 10} = {(1, 2), (1, 3), (1, 5), (1, 7), (2, 2), (2, 3), (2, 5), (2, 7), Range of f ={2, 4, 6, 8} (3, 2), (3, 3), (3, 5), (3, 7)} A. SIVAMOORTHY, BT. Asst. GHS, Perumpakkam, Villupuram Dt. Kindly Send Me Your Key Answer to Our email id - [email protected]  www.Padasalai.Net www.Trb Tnpsc.Com Minimum Material 15 7. L  et A = {1, 2, 3, 4,..., 45} and R be the –6 x2 + 12x – 6 = 0 ÷ –6 relation defined as “is square of a number” x2 – 2x + 1 = 0 on A. Write R as a subset of A×A. Also, find (x – 1) (x – 1) = 0 x = 1, 1 the domain and range of R. SEP-21 11. Let A = {1, 2, 3, 4} and B = N. Solution: Let f : A→ B be defined by f(x) = x3 then, A = {1, 2, 3, …., 45} (i) find the range of f (ii) identify the type of R = {(1, 1), (2, 4), (3, 9), (4, 16), (5, 25), function. (6, 36)} Solution: R (A × A) A = {1, 2, 3, 4}, B = N Domain of R = {1, 2, 3, 4, 5, 6} f : A → B, f(x) = x3 Range of R = {1, 4, 9, 16, 25, 36} f(1) = (1)3 = 1; f(2) = (2)3 = 8;  Relation R is given by the set {(x, y) /y = x + 8. A f(3) = (3)3 = 27; f(4) = (4)3 = 64 et 3, x {0, 1, 2, 3, 4, 5}}. Determine its domain i) Range of f ={1, 8, 27, 64} and range. ii) It is one-one and into function. Solution: i.N 5 Marks x = {0, 1, 2, 3, 4, 5} f(x) = y = x + 3 1. If A = {1, 3, 5} and B = {2, 3} then f(0) = 3; f(1) = 4; f(2) = 5; (i) find A×B and B×A. f(3) = 6; f(4) = 7; la f(5) = 8 R = {(0, 3), (1, 4), (2, 5), (3, 6), (4, 7), (5, 8)} Domain of R = {0, 1, 2, 3, 4, 5} (ii) Is A×B = B×A? If not why? (iii) Show that n(A×B) = n(B×A) = n(A)× n(B)  sa SEP-21 Range of R = {3, 4, 5, 6, 7, 8} Solution:  iven the function f: x → x2 − 5x + 6, evaluate 9. G Given that A = {1, 3, 5} and B = {2, 3} i. A × B = {1, 3, 5} ×{2 × 3} da (i) f (–1) (ii) f (2a) (iii) f (2) (iv) f (x −1) Solution: = {(1,2), (1,3), (3,2), (3,3), (5,2), (5,3)} Given: f: x → x2 – 5x + 6 ………(1) f(x) = x2 – 5x + 6 B × A = {2 × 3} × {1, 3, 5} Pa i. f(–1) = (–1)2 – 5(–1) + 6 = {(2,1), (2,3), (2,5), (3,1), (3,3), (3,5)} = 1+5+6 ………(2) = 12 ii. From (1) and (2) we conclude that w. A × B ≠ B × A as (1, 2) ≠ (2, 1) and ii. f(2a) = (2a)2 – 5(2a) + 6 (1, 3) ≠ (3, 1) etc = 4a2 – 10a + 6 iii. f(2) = (2)2 – 5(2) + 6 iii. n(A) = 3; n(B) = 2 ww = 4 –10 + 6 From (1) and (2) we observe that, = 0 n(A × B) = n(B × A) = 6; We see that, n(A) × n(B) = 3 × 2 = 6 iv. f(x – 1) = (x – 1)2 – 5(x – 1) + 6 Thus, n(A × B) = n(B × A) = n(A) × n(B). = x2 – 2x + 1 – 5x + 5 + 6 = x2 – 7x + 12 2. Let A = {x N | 1< x b 8.  If {(a, 8), (6, b)} represents an identity C) 0 ≤ r < b D) 0 < r ≤ b function, then the value of a and b are 2. Using Euclid’s division lemma, if the cube of respectively any positive integer is divided by 9 then the A) (8, 6) B) (8, 8) possible remainders are C) (6, 8) D) (6, 6) A) 0, 1, 8 B) 1, 4, 8 9. Let A = {1, 2, 3, 4} and B = {4, 8, 9, 10}. C) 0, 1, 3 D) 1, 3, 5 A function f : A → B given by f = {(1, 4), 3. If the HCF of 65 and 117 is expressible in the (2, 8), (3, 9), (4, 10)} is a form of 65m – 117, then the value of m is A) Many-one function B) Identity function A) 4 B) 2 C) One-to-one function D) Into function C) 1 D) 3 A. SIVAMOORTHY, BT. Asst. GHS, Perumpakkam, Villupuram Dt. Kindly Send Me Your Key Answer to Our email id - [email protected]  www.Padasalai.Net www.Trb Tnpsc.Com 1 Marks 59 4.  The sum of the exponents of the prime 13. The next term of the sequence 3 factors in the prime factorization of 1729 is A) 124 B) 127 A) 1 B) 2 C) 23 D) 181 C) 3 D) 4 14. If the sequence t1, t2, t3,......... are in A.P. then 5. The least number that is divisible by all the the sequence t6, t12, t18,......... is numbers from 1 to 10 (both inclusive) is A) a Geometric Progression A) 2025 B) 5220 B) an Arithmetic Progression C) 5025 D) 2520 C) neither an Arithmetic Progression nor a 6. 74k ≡ _____ (mod 100) Geometric Progression D) a constant sequence A) 1 B) 2 C) 3 D) 4 15. The value of (13 + 23 + 33 +........ + 153) − (1 + 2 + 3 +........ + 15) is et 7. Given F1 = 1, F2 = 3 and Fn = Fn–1 + Fn–2 then A) 14400 B) 14200 F5 is C) 14280 D) 14520 A) 3 B) 5 i.N C) 8 D) 11 3. Algebra 8. The first term of an arithmetic progression is 1. A system of three linear equations in three unity and the common difference is 4. Which la variables is inconsistent if their planes of the following will be a term of this A.P. A) intersect only at a point A) 4551 B) 10091 B) intersect in a line sa C) 7881 D) 13531 C) coincides with each other 9. If 6 times of 6th term of an A.P. is equal to 7 D) do not intersect times the 7th term, then the 13th term of the 2. The solution of the system x + y − 3z = −6, da A.P. is −7y + 7z = 7, 3z = 9 is A) 0 B) 6 A) x = 1, y = 2, z = 3 B) x = −1, y = 2, z = 3 C) 7 D) 13 C) x = −1, y = −2, z = 3 D) x = 1, y = −2, z = 3 Pa 10. An A.P. consists of 31 terms. If its 16th term 3. If (x – 6) is the HCF of x2 – 2x – 24 and is m, then the sum of all the terms of this A.P. x2 – kx – 6 then the value of k is is A) 3 B) 5 w. A) 16 m B) 62 m C) 6 D) 8 31 C) 31 m D) m 2 3y – 3 7y – 7 4. ÷ is 11. In an A.P., the first term is 1 and the common y 3y2 ww difference is 4. How many terms of the A.P. 9y 9 y3 A) B) must be taken for their sum to be equal to 7 (21y - 21) 120? C) 21y 2 - 42 y + 21 D) ( 7 y2 - 2 y + 1 ) A) 6 B) 7 3 y3 2 y C) 8 D) 9 1 12. If A = 265 and B = 264 + 263 + 262 +........ +20 5. y2 + is not equal to y2 which of the following is true? 2 y4 + 1 æ 1ö A) B is 264 more than A A) B) ç y + ÷ y2 è yø B) A and B are equal 2 2 C) B is larger than A by 1 æ 1ö C) æ y - 1 ö + 2 D) ç y + ÷ – 2 D) A is larger than B by 1 çè y ÷ø è yø Kindly Send Me Your Key Answer to Our email id - [email protected]  www.Padasalai.Net www.Trb Tnpsc.Com 10th Std - Mathematics 60 x 8 15. If A is a 2×3 matrix and B is a 3×4 matrix, 6. – 2 gives x – 25 x – 6x + 5 2 how many columns does AB have x 2 - 7 x + 40 x 2 + 7 x + 40 A) 3 B) 4 A) B) C) 2 D) 5 ( x - 5)( x + 5) ( x - 5)( x + 5)( x + 1) x 2 - 7 x + 40 x 2 + 10 16. If number of columns and rows are not equal C) D) in a matrix then it is said to be a ( ) x 2 - 25 ( x + 1) ( ) x 2 - 25 ( x + 1) A) diagonal matrix B) rectangular matrix 256x8y4z10 C) square matrix D) identity matrix 7. The square root of is equal to 25x6y6z6 17. Transpose of a column matrix is 16 x 2 z 4 y2 A) unit matrix B) diagonal matrix A) B) 16 2 4 5 y2 x z C) column matrix D) row matrix 16 xz 2 C) 16 y2 et D) æ 1 3ö æ 5 7ö 18. Find the matrix X if 2X + ç =ç 5 xz 5 y ÷ è 5 7ø è 9 5÷ø i.N 8. Which of the following should be added to æ -2 -2ö æ2 2 ö A) ç B) ç make x4 + 64 a perfect square è 2 -1÷ø è 2 -1÷ø A) 4x2 B) 16x2 æ1 2ö 2 1ö C) ç D) æç C) 8x2 D) –8x2 è2 2÷ø è 2 2÷ø A) –1 B) 2 la 9. The solution of (2x − 1)2 = 9 is equal to 19.  Which of the following can be calculated æ 1 2ö sa C) –1, 2 D) None of these from the given matrices A = ç 3 4÷ , ç ÷ 10. The values of a and b if 4x4 − 24x3 + 76x2 + ax æ 1 2 3ö è 5 6ø B = ç 4 5 6÷ da + b is a perfect square are ç ÷ A) 100, 120 B) 10, 12 è 7 8 9ø C) –120, 100 D) 12, 10 (i) A2 (ii) B2 Pa 11. If the roots of the equation q2x2 + p2x + r2 = (iii) AB (iv) BA 0 are the squares of the roots of the equation A) (i) and (ii) only B) (ii) and (iii) only qx2 + px + r = 0, then q, p, r are in _______ C) (ii) and (iv) only D) all of these A) A.P B) G.P w. æ1 0 ö C) Both A.P and G.P D) none of these æ 1 2 3ö 20. If A =A = ç , B = ç 2 -1÷ and 12. Graph of a linear equation is a _______. è 3 2 1÷ø ç ÷ è0 2 ø A) straight line B) circle æ 0 1ö ww C= ç. Which of the following C) parabola D) hyperbola è -2 5÷ø statements are correct? 13. The number of points of intersection of the æ 0 1ö quadratic polynomial x2 + 4x + 4 with the æ 5 5ö X-axis is (i) AB + C = ç (ii) BC = ç 2 -3÷ è 5 5÷ø ç ÷ A) 0 B) 1 è -4 10 ø C) 0 or 1 D) 2 æ 2 5ö æ -8 20ö (iii) BA + C = ç ÷ (iv) (AB)C = ç è 3 0ø è -8 13÷ø æ1 3 5 7 ö 14. For the given matrix A = ç 2 4 6 8 ÷ A) (i) and (ii) only B) (ii) and (iii) only ç ÷ è 9 11 13 15ø C) (iii) and (iv) only D) all of these the order of the matrix AT is A) 2×3 B) 3×2 C) 3×4 D) 4×3 A. SIVAMOORTHY, BT. Asst. GHS, Perumpakkam, Villupuram Dt. Kindly Send Me Your Key Answer to Our email id - [email protected]  www.Padasalai.Net www.Trb Tnpsc.Com 1 Marks 61 9. Two poles of heights 6 m and 11 m stand 4. Geometry vertically on a plane ground. If the distance AB BC between their feet is 12 m, what is the 1. If in triangles ABC and EDF, = then DE FD they will be similar, when distance between their tops? A) B = E B) A = D A) 13 m B) 14 m C) B = D D) A = F C) 15 m D) 12.8 m 2. In ΔLMN, L = 60°, M = 50°. 10. In the given figure, PR = 26 cm, QR = 24 If ΔLMN ~ ΔPQR then the value of R is cm, PAQ = 90° , PA = 6 cm and A) 40° B) 70° QA = 8 cm. Find PQR C) 30° D) 110° A) 80° B) 85° 3. If ΔABC is an isosceles triangle with C) 75° D) 90° et C = 90° and AC = 5 cm, then AB is A) 2.5 cm B) 5 cm 11. A tangent is perpendicular to the radius at C) 10 cm D) 5 2 cm the i.N 4. In a given figure ST || QR, PS 10= 62 cm and SQ A) centre B) point of contact = 3 cm. Then the ratio of the area 3 of ΔPQR to C) infinity D) chord the area of ΔPST is  1 1 la A) 25 : 4 12. How many tangents can be drawn to the 3 3 B) 25 : 7 circle from an exterior point? sa C) 25 : 11 A) one B) two D) 25 : 13 C) infinite D) zero 5. The perimeters of two similar triangles 13. The two tangents from an external points P da ΔABC and ΔPQR are 36 cm and 24 cm to a circle with centre at O are PA and PB. If respectively. If PQ = 10 cm, then the length of AB is 2 APB = 70° then the value of AOB is 2 10 6 A) 100° B) 110° Pa A) 6 cm B) cm 3 3 C) 120° D) 130° 2 C) 66 cm D) 151 cm1 14. In figure CP and CQ are tangents to a circle 3 3 3 with centre at O. ARB is another tangent w. 6. If in ΔABC , DE || BC. AB = 3.6 cm, AC = 2.4 cm and AD = 2.1 cm then the length of AE is touching the circle at R. If CP = 11 cm and A) 1.4 cm B) 1.8 cm BC = 7 cm, then the length of BR is ww C) 1.2 cm D) 1.05 cm A) 6 cm 7. In a ΔABC, AD is the bisector of BAC. If B) 5 cm AB = 8 cm, BD = 6 cm and DC = 3 cm. The C) 8 cm length of the side AC is D) 4 cm A) 6 cm B) 4 cm C) 3 cm D) 8 cm 15. In figure if PR is tangent to the circle at P and O is the centre of the circle, then POQ 8. In the adjacent figure BAC = 90° and AD BC then is A) BD. CD = BC2 A) 120° B) AB. AC = BC2 B) 100° C) BD. CD = AD2 C) 110° D) AB. AC = AD2 D) 90° Kindly Send Me Your Key Answer to Our email id - [email protected]  www.Padasalai.Net www.Trb Tnpsc.Com 10th Std - Mathematics 62 9. If A is a point on the Y axis whose ordinate 5. Coordinate Geometry is 8 and B is a point on the X axis whose 1. The area of triangle formed by the points abscissae is 5 then the equation of the line (−5, 0), (0, −5) and (5, 0) is AB is A) 0 sq.units B) 25 sq.units A) 8x + 5y = 40 B) 8x − 5y = 40 C) 5 sq.units D) none of these C) x = 8 D) y = 5 10. The equation of a line passing through the 2.  A man walks near a wall, such that the origin and perpendicular to the line distance between him and the wall is 10 7x − 3y + 4 = 0 is units. Consider the wall to be the Y axis. The A) 7x − 3y + 4 = 0 B) 3x − 7y + 4 = 0 path travelled by the man is C) 3x + 7y = 0 D) 7x − 3y = 0 A) x = 10 B) y = 10 11. Consider four straight lines et C) x = 0 D) y = 0 (i) l1 : 3y = 4x + 5; (ii) l2: 4y = 3x − 1 3. The straight line given by the equation (iii) l3 : 4y + 3x = 7 (iv) l4 : 4x + 3y = 2 i.N x = 11 is Which of the following statement is true? A) parallel to X axis A) l1 and l2 are perpendicular B) parallel to Y axis B) l1 and l4 are parallel C) passing through the origin la C) l2 and l4 are perpendicular D) passing through the point (0, 11) D) l2 and l3 are parallel sa 4. If (5, 7), (3, p) and (6, 6) are collinear, then 12. A straight line has equation 8y = 4x + 21. the value of p is Which of the following is true? A) 3 B) 6 A) The slope is 0.5 and the y intercept is 2.6 da C) 9 D) 12 B) The slope is 5 and the y intercept is 1.6 C) The slope is 0.5 and the y intercept is 1.6 5. The point of intersection of 3x − y = 4 and D) The slope is 5 and the y intercept is 2.6 x + y = 8 is Pa A) (5, 3) B) (2, 4) 13.  When proving that a quadrilateral is a trapezium, it is necessary to show C) (3, 5) D) (4, 4) A) Two sides are parallel 6. The slope of the line joining (12, 3), (4, a) is. B) Two parallel and two non-parallel sides 1 w.. The value of ‘a’ is C) Opposite sides are parallel 8 A) 1 B) 4. D) All sides are of equal length ww C) –5 D) 2 14.  When proving that a quadrilateral is a 7. The slope of the line which is perpendicular parallelogram by using slopes you must find to a line joining the points (0, 0) and (–8, 8) is A) The slopes of two sides B) The slopes of two pair of opposite sides A) –1 B) 1 C) 1 D) –8 C) The lengths of all sides 3 D) Both the lengths and slopes of two sides 8. If slope of the line PQ is 1 then slope of the 15. (2, 1) is the point of intersection of two lines. 3 A) x − y − 3 = 0; 3x − y − 7 = 0 perpendicular bisector of PQ is B) x + y = 3; 3x + y = 7 A) 3 B) – 3 C) 3x + y = 3; x + y = 7 C) 1 D) 0 D) x + 3y − 3 = 0; x − y − 7 = 0 3 A. SIVAMOORTHY, BT. Asst. GHS, Perumpakkam, Villupuram Dt. Kindly Send Me Your Key Answer to Our email id - [email protected]  www.Padasalai.Net www.Trb Tnpsc.Com 1 Marks 63 A) 45° B) 30° 6. Trigonometry C) 90° D) 60° 1 1. The value of sin2θ + is equal to 11. The electric pole subtends an angle of 30° 1 + tan2q at a point on the same level as its foot. At a A) tan2θ B) 1 second point ‘b’ metres above the first, the C) cot2θ D) 0 depression of the foot of the pole is 60°. The height of the pole (in metres) is equal to 2. tanθcosec2θ – tanθ is equal to A) secθ B) cot2θ A) 3b B) C) sinθ D) cotθ b b C) D) 3. If (sinα + cosecα)2 + (cosα + secα)2 = k + tan2α 2 3 + cot2α, then the value of k is equal to 12. A tower is 60 m heigh. Its shadow is x metres et A) 9 B) 7 shorter when the sun’s altitude is 45° than C) 5 D) 3 when it has been 30°, then x is equal to 4. If sinθ + cosθ = a and secθ + cosecθ = b, then A) 41.92 m B) 43.92 m i.N the value of b(a2 – 1) is equal to C) 43 m D) 45.6 m A) 2a B) 3a 13.  The angle of depression of the top and C) 0 D) 2ab bottom of 20 m tall building from the top 5. If 5x = secθ and equal to 5 x la 1 = tanθ , then x2 – 2 is x of a multistoried building are 30° and 60° respectively. The height of the multistoried building and the distance between two sa 1 buildings (in metres) is A) 25 B) 25 A) 20, 10 3 B) 30, 5 3 C) 5 D) 1 C) 20, 10 D) 30, 10 3 da 6. If sinθ = cosθ, then 2tan2θ + sin2θ − 1 is equal 14. Two persons are standing ‘x’ metres apart to from each other and the height of the first -3 3 person is double that of the other. If from the A) B) Pa 2 2 middle point of the line joining their feet an 2 C) D) -2 observer finds the angular elevations of their 3