DSSSB TGT Mathematics Solved Question Papers PDF

Delhi Subordinate Service Selection Board DSSSB TGT/PGT Mathematics MALE & FEMALE SOLVED PAPERS CHIEF EDITOR A.K. Mahajan Complied & Written By YCT...

Delhi Subordinate Service Selection Board DSSSB TGT/PGT Mathematics MALE & FEMALE SOLVED PAPERS CHIEF EDITOR A.K. Mahajan Complied & Written By YCT Exam Expert Group Computer Graphics By Balkrishna Tripathi & Vinay Sahu Editorial Office 12, Church Lane Prayagraj-211002 Mob. : 9415650134 Email : [email protected] website : www.yctbooks.com/www.yctfastbooks.com  All rights reserved with Publisher Publisher Declaration Edited and Published by A.K. Mahajan for YCT Publications Pvt. Ltd. and printed by Roop Printing Press In order to Publish the book, full care has been taken by the Editor and the Publisher, still your suggestions and queries are welcomed. Rs. : 495/- In the event of any dispute, the judicial area will be Prayagraj.  CONTENT DSSSB TRAINED GRADUATE TEACHER (TGT) 2014 Solved Paper (Exam Date : 28.12.2014)......................................................................................... 3-15 DSSSB TRAINED GRADUATE TEACHER (TGT) 2018 Solved Paper (Exam Date : 22.09.2018, Shift-II (Female)......................................................... 16-30 DSSSB TRAINED GRADUATE TEACHER (TGT) 2018 Solved Paper (Exam Date : 23.09.2018, Shift-I (Male)............................................................... 31-46 DSSSB TRAINED GRADUATE TEACHER (TGT) 2018 Solved Paper (Exam Date : 23.09.2018, Shift-II (Male)............................................................. 47-63 DSSSB TRAINED GRADUATE TEACHER (TGT) 2018 Solved Paper (Exam Date : 18.11.2018, Shift-III (Female)........................................................ 64-80 DSSSB TRAINED GRADUATE TEACHER (TGT) 2021 Solved Paper (Exam Date : 02.09.2021, Shift-I (Male)............................................................... 81-93 DSSSB TRAINED GRADUATE TEACHER (TGT) 2021 Solved Paper (Exam Date : 02.09.2021, Shift-III (Male)......................................................... 94-104 DSSSB TRAINED GRADUATE TEACHER (TGT) 2021 Solved Paper (Exam Date : 04.09.2021, Shift-I (Male)........................................................... 105-117 DSSSB TRAINED GRADUATE TEACHER (TGT) 2021 Solved Paper (Exam Date : 10.09.2021, Shift-II (Female)..................................................... 118-130 DSSSB TRAINED GRADUATE TEACHER (TGT) 2021 Solved Paper (Exam Date : 10.09.2021, Shift-III (Female).................................................... 131-145 DSSSB TRAINED GRADUATE TEACHER (TGT) 2021 Solved Paper (Exam Date : 11.09.2021, Shift-III (Female).................................................... 146-160 DSSSB TRAINED GRADUATE TEACHER (TGT) 2021 Solved Paper (Exam Date : 13.09.2021, Shift-I (Male)........................................................... 161-172 DSSSB TRAINED GRADUATE TEACHER (TGT) 2021 Solved Paper (Exam Date : 13.09.2021, Shift-II (Female)..................................................... 173-184 DSSSB POST GRADUATE TEACHER (PGT) 2015 Solved Paper (Exam Date : 28.06.2015, Tier-II...................................................................... 185-228 DSSSB POST GRADUATE TEACHER (PGT) 2018 Solved Paper (Exam Date : 03.07.2018, (Male)....................................................................... 229-255 DSSSB POST GRADUATE TEACHER (PGT) 2018 Solved Paper (Exam Date : 04.07.2018, (Female)................................................................... 256-280 DSSSB POST GRADUATE TEACHER (PGT) 2021 Solved Paper (Exam Date : 31.07.2021, (Male)....................................................................... 281-320 2  DSSSB TRAINED GRADUATE TEACHER (TGT) 2014 MATH Solved Paper [Exam Date : 28/12/2014 1. Decimal expansion of an irrational number is- (a) Two real roots / oes JeemleefJekeâ cetue DeheefjcesÙe mebKÙee keâe obMeceerÙe efJemleejCe neslee nw (b) Four real roots /Ûeej JeemleefJekeâ cetue (a) Non-terminating recurring/ Demeerefcele, DeeJeleea (c) One real root /Skeâ JeemleefJekeâ cetue (b) Terminating / meerefcele (d) No real root /keâesF& JeemleefJekeâ cetue veneR (c) Non-terminating non-recurring Ans. (b) : Given equation Demeerefcele, DeveeJeleea x2 – 5 |x| + 6 = 0 (d) None of these/ FveceW mes keâesF& veneR Now since Ans. (c) : An irrational number is a number which x ; x ≥ 0 |x| =  p  − x; x < 0 cannot be written in the form with p and q being co- we have q prime integers and q ≠ 0. Therefore, decimal expansion  x 2 − 5x + 6 ; x ≥ 0 of an irrational number is non-terminating and non- x2 – 5|x| + 6 =  2 recurring.  x + 5x + 6 ; x < 0 Now x2 – 5x + 6 = 0 clearly has two distinct real roots 2. If x = 3 + 2 , then value of x–2 is namely 2 and 3 and x2 + 5x + 6 = 0 has two distinct real Ùeefo x = 3 + 2 nw, lees x–2 keâe cetuÙe roots as well namely –2 and –3. 6. The equation |x + 4| = – 4 has (a) 3− 2 (b) 5 + 2 6 |x + 4| = – 4 meceerkeâjCe kesâ nw~ (c) 3+ 2 (d) 5 − 2 6 (a) Unique solution / Skeâ cee$e meeOeve Ans. (d) : x = 3 + 2 (b) Two solutions/ oes meeOeve 1 1 (c) Many solutions /keâesF& meeOeve ⇒ = = 3− 2 (d) No solution /keâesF& meeOeve veneR x 3+ 2 Ans. (d) : Clearly |x + 4| = –4 has no solution 1 ( ) 2 ⇒ 2 = 3 − 2 = 3+ 2−2 6 = 5− 2 6 7. Sun of two irrational number is x oes DeheefjcesÙe mebKÙeeDeeW keâes peesÌ[ nw~ 0.05 (a) Rational /heefjcesÙe 3. The value of the expression 0.8 × is equal 0.004 (b) Irrational /DeheefjcesÙe to (c) Real /JeemeleefJekeâ 0.05 (d) Non of these /FveceW mes keâesF& veneR 0.8 × JÙebpekeâ keâe cetuÙe Fmekesâ yejeyej nw~ 0.004 Ans. (b) : Sum of two irrational numbers is always an (a) 10 (b) 0.1 irrational number. (c) 100 (d) 0.4 8. An odd degree polynomial equation has 0.05 efJe 5 is |2x+3|> 5 keâe meeOeve mecegÛÛeÙe nw~ Ans. (b) : Since n Cr + nCr+1 = n+1Cr+1 (a) (–∞, –4) we have (b) (1, ∞) 7 7 7 7 ( C0 + C1) + ( C1 + C2) +.... + ( C6 + C7) 7 7 (c) (–∞,–4) ∪(1, ∞) 8 8 = C1 + C2 +...... + C7 8 (d) None of these /FveceW mes keâesF& veneR = 28 – 2 Ans. (c) : We must have 37. For any two sets A and B the value of the set 2x + 3 < –5 or 2x+3 > 5 A∩(A∪B)c is Therefore, we must get A Deewj B efkeâmeer oes mecegÛÛeÙeeW kesâ efueS, mecegÛÛeÙe keâe x∈ (–∞, –4) ∪ (1,∞) A∩(A∪B)c cetuÙe nw~ x +1 (a) AC (b) BC 42. If f(x) = (x ≠ 1) is a real function then x –1 (c) φ (d) B f(f(f(2))) is equal to Ans. (a) : A ∩(A∪B) = A∩A ∩BC= φ c C x +1 Ùeefo f(x) = (x ≠ 1) Ùen JeemleefJekeâ heâueve nw, lees 38. If the area of a triangle having vertices (2,7), x –1 (5,1) and (x, 3) is 18 sq. units, then the value of f(f(f(2))) Fmekesâ yejeyej nw~ x is (a) 1 (b) 2 Ùeefo (2,7), (5,1) Deewj (x, 3) efMejesefyevog Jeeues Skeâ (c) 3 (d) 4 ef$ekeâesCe keâe #es$eheâue 18 sq. FkeâeF& nw, lees x keâe cetuÙe Ans. (c) : We have (a) 10 (b) 2 f(f(f(2))) = f(f(3)) = f(2) = 3 (c) –2 (d) –10 43. Which of the following limit does not exist ? Ans. (a) : efvecve ceW mes keâewve meer meercee DeefmlelJe ceW veneR nw? 1 ar (∆) = [x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 + y2)] |x| 2 (a) lim x →0 x with vertices (x1, y1), (x2, y2) and (x3, y3). Therefore, we get sin x (b) lim x →0 x 1 [ 2(−2) + 5(−4) + 6x ] = 18 x 2 (c) lim x → 0 cos x ⇒ 6x = 60 ⇒ x = 10 (d) None of these / FveceW mes keâesF& veneR 39. The coordinates of the point which divide the Ans. (a) : We have line segment joining the points (8,9) and (–7, 4) −x x internally in the ratio 2:3 is (a) lim = −1 and lim+ = 1 x → 0− x x →0 x Ùeefo (8,9) Deewj (–7, 4) keâes 2:3 kesâ Devegheele mes efYelej mes sin x peesÌ[ves Jeeues jsKee KeC[ keâes efJeYeeefpele keâjves Jeeues efyevog (b) lim =1 x → 0+ x keâe efveo&WMeebkeâ nw x (a) (7,2) (b) (2,7) (c) lim =0 x → 0+ cos x (c) (–7,2) (d) (2,–7) DSSSB (TGT-Math) 2014 (28.12.2014) 7 YCT  x  3 Ans. (c) : Let the two digit number be xy. Then we have 44. The value lt  1 +  is xy = 12 x →∝  x 12 ⇒ x  3 y= lt  1 +  keâe cetuÙe nw~ x x →∝  x Now 10x + y + 36 = 10y + x 1 (a) e 3 (b) e3 12 120 ⇒ 10x + + 36 = +x 1 x x (c) (d) 1 e ⇒ 10x2 +12 +36 x = 120 + x2 x ⇒ 9x2 + 36x –108 = 0  3 Ans. (b) : Let y = lim  1 +  ⇒ x2 +4x–12 = 0 x →∞  x ⇒ (x+6)(x–2) = 0  3 ⇒ ⇒ logey = lim x log e 1 +  x=2 x →∞  x Therefore the number is 26.  3 48. If the roots of the quadratic equation log e 1 +   x (4+m)x2+(m+1)x +1 = 0 are equal than the = lim x value of m is. x →∞ 1 x Ùeefo (4+m)x2+(m+1)x +1 = 0 Fme efÉIeeleer meceerkeâjCe 1  3  kesâ cetue Skeâ meceeve nw, lees m keâe cetuÙe – 2 3 (a) 5 1+  x  (b) 4 = lim x x =3 x →∞ 1 (c) 2 − 2 (d) None of these/FveceW mes keâesF& veneR x Therefore the required limit is e3. Ans. (a) :We must have x +1 −1 (m+1)2– 4(4+m) = 0 45. Value of the limit lim is ⇒ m2+1+2m –16–4m = 0 x →0 x ⇒ m2 –2m –15 = 0 x +1 −1 meercee lim x →0 keâe cetuÙe nw ⇒ m = 5, –3 x (a) 1 3 49. If sin (A+B) = cos (A–B) = , acute angles of (b) 2 2 1 A and B are (c) 2 3 (d) None of these / FveceW mes keâeF& veneR Ùeefo sin (A+B) = cos (A–B) = , nw, lees A Deewj B 2 x +1 −1 1 1 Ûeehe kesâ vÙetvekeâesCe nQ Ans. (c) : lim = lim = x →0 x x →0 2 x +1 2 (a) 45° and 15° /45° Deewj 15° sin28° sin28° (b) 30° and 60° /30° Deewj 60° 46. Value of is / keâe cetuÙe nw~ (c) 15° and 60° / 15° Deewj 60° cos62° cos62° (a) 1 (d) None of these / FveceW mes keâesF& veneR (b) 0 3 π 1 Ans. (a) : sin(A + B) = = sin (c) 2 3 2 (d) None of these / FveceW mes keâesF& veneR 3 π and cos (A – B) = = cos sin 28° sin ( 90° − 62° ) cos 62° 2 6 Ans. (a) : = = =1 π π cos 62° cos 62° cos 62° ⇒ A+B = and A – B = 47. A two-digit number is such that the product of its 3 6 digits 12. When 36 is added to this number, the π digits interchange their places. The number is ⇒ 2A = 2 oes DebkeâeW Jeeueer mebKÙee Ssmeer nw efpemeceW DebkeâeW keâe π iegCeveheâue 12 nw~ Fme mebKÙee ceW 36 peeWÌ[ves mes Debkeâes keâer ⇒ A= Deheveer peien mes Deouee yeoueer nesleer nw Jen mebKÙee nw~ 4 π (a) 62 (b) 36 ⇒ B= (c) 26 (d) 48 12 DSSSB (TGT-Math) 2014 (28.12.2014) 8 YCT  1 + 2 + 3 +... + n Ans. (c) : f(x) = beax + aebx 50. Value of lim , n ∈ N is equal to ⇒ f'(x) = abeax + abebx x →∞ n2 1 + 2 + 3 +... + n ⇒ f''(x) = a2beax + ab2ebx lim , n ∈ N keâe cetuÙe Fmekesâ yejeyej ⇒ f''(0) = a2b+ab2 = ab(a+b) x →∞ n2 nw  log x (a) 0 (b) 1  if x = 1 55. If f(x) =  x − 1 is continuous at x = 1 1 1  k if x = 1 (c) (d) 2 4 the value of k is 1 + 2 + 3 +... + n = lim n ( n + 1) = 1  log x if x = 1 Ùeefo f(x) =  x − 1 Ans. (c) : lim n →∞ n 2 n →∞ 2n 2 2 Ùen x = 1 hej efvejblej nw 51. The number of points at which the function  k if x = 1 1 , lees k keâe cetuÙe nw~ f(x) = is not continuous is x – [ x] (a) 0 (b) –1 (c) 1 (d) e 1 f(x) = heâueve Fve efyevogDeeW hej efvejblej Ans. (c) : We must have x – [ x] log x (a) 1 lim =k x →1 x − 1 (b) 2 (c) 3 1 ⇒ lim = 1 = k (d) None of these / Fveces mes keâesF& veneR x →1 x 1 56. If f(x) is an even function, then f'(x) is Ans. (d) : f(x) = is discontinuous at infinitely Ùeefo f(x) Ùen mece heâueve nw, lees f'(x) Ùen nw x − (x) many points since [x] is discontinuous at every integral (a) An even function /mece heâueve point. (b) An odd function /efJe0 (b) 2 (c) f(x) = 0 and f'(x) >0 (c) 1 (d) f'(x) = 0 and f'(x) n ( a12 + a 22 +... + a 2n ) du du dθ Since, we know that =. (c) (a1 + a 2 +a 3...+ a n )2 > ( n –1) ( a12 + a 22 +... + a 2n ) dv dθ dv 1 (d) (a1 + a 2...+ a n ) 2 ≤ ( n –1) ( a12 + a 22 +... + a 2n ) = 2× =1 2 Ans. (a) : (Chebyshev) Let a1,..., an be a sequence of Thus, the required derivative is 1. real numbers. Assume that a1 < a2 72, f(2) = 1, then f (n) is : ⇒ n – 1 = 1+ 6 3 n ( n + 1) n 2 – 3n + 2 (a) n(n – 1) (b) ⇒ n – 1 = 1+ 2 2 6 n(n – 1) 3 (c) (d) n ( n + 1) ⇒ 6n – 6 = 6 + n 2 – 3n + 2 2 2 ⇒ n 2 – 9n + 14 = 0 Ans. (c) : f ( n ) = ( n − 1) + f ( n − 1) : n > 72 ⇒ ( n – 2 )( n – 7 ) = 0 & f (2) = 1 Therefore, n = 7. Now, we have 1 – tanx f (3) = 3, f(4) = 6, f (5) = 10 and so on 9. Evaluation of ∫ 1 + tanx dx is: n ( n − 1) (a) log (sinx – cos x)+ c Therefore, f ( n ) = 2 (b) log (sinx – cot x)+ c 1 (c) log (sinx + cos x)+ c 12. If f ( x ) = then at x = 0 the function is: (d) log (cosx – cot x)+ c 1 + 21/x Ans. (c) : We have (a) Discontinuous because L lim f ( x ) ≠ R lim f ( x ) 1 – tan x x →0 x →0 ∫ 1 + tan x dx (b) Discontinuous because lim f ( 0 ) ≠ f ( 0 ) x →0 cos x – sin x  f '(x)  (c) Continuous =∫ dx ∵∫ dx = log f (x)  cos x + sin x  f (x)  (d) Discontinuous because R lim f ( x ) does not x →0 = log ( cos x + sin x ) + c exist. DSSSB (TGT-Math) 2018 (22.9.2018, Shift-II) 23 YCT  1 (a) 14 cm (b) 20 cm Ans. (a) : Given, f (x) = (c) 17 cm (d) 15 cm 1 + 21/ x We have, lim– f ( x ) = 1and lim+ f ( x ) = 0 Ans. (d) : x →0 x →0 Therefore, f(x) is discontinuous at x = 0. dy 13. If (sin x)y, then is ____ : dx y 2 cot x y 2 cot x (a) (b) By similarity of ∆ ACD and ∆ BCE we have 1 – y log (sin x) 1 – y log x AD AC y 2 cot x y 2 cot x = (c) (d) BE BC 1 + y log ( sin x ) 1 + y log x 3 x Ans. (a) : Let ⇒ = 5 x + 10 y = (sin x)y Taking log both side then, ⇒ 3x + 30 = 5x ⇒ log y = y log (sin x) {∵log(m) n = n log m} ⇒ x = 15 17. For every integer n > 2, the sum of the Differentiating, expansions 1 – n C1 + n C2 +... + ( –1) Cn is n n 1 dy dy y ⇒ = log(sin x) +.cos x y dx dx sin x (a) n (n – 1) 2n–2 dy (b) 0 ⇒ [1 – y log(sin x)] = y2 cot x (c) n2m–1 dx (2 C ) n 2 dy y 2 cot x (d) ⇒ = n dx 1 – y log(sin x) Ans. (b) : By Binomial Theorem, we have 14. If a, b, c, x, y, z are the real numbers such that (1 – x ) = 1 – n C1 x + n C2 x 2 +... + ( –1) n Cn x n n n...(i) a2 + b2 + c2 = 1, x2 + y2 + z2 = 1, then ax + by + cz Therefore, for sum of coefficients, putting x =1 in eqn (a) Less than or equal to 3 (i)- (b) Less than or equal to 1 1 – n C1 + n C2 +... + ( –1) n Cn = (1 –1)n = 0 n (c) Greater than or equal to 3 ⇒ 1 – n C1 + n C2 +... + ( –1) n Cn = 0. n (d) Greater than or equal to 1 Ans. (b) : By AM - GM inequality we have 18. If a, b, c be positive number such that a + b + c a 2 + b 2 + c 2 + x 2 + y 2 + z 2 ≥ 2 ( ax + by + cz ) = 4, then a3 + b3 + c3 is : ⇒ ax +by + cz ≤ 1 16 15. How many numbers from 1 to 1000 are not (a) Less than or equal to 19 divisible by 2, 3 and 5? (a) 262 (b) 268 16 (b) Greater than or equal to (c) 266 (d) 264 19 Ans. (c) : Numbers not divisible by any of 2, 3, and 5 64 (c) Less than or equal to are those whose residues modulo 30 are not divisible 9 by any of 2, 3, and 5. The relevant residues are 1, 7, 11, 64 13, 17, 19, 23 and 29 (eight). We count 8 × 33 = 264 (d) Greater than or equal to nonnegative whole numbers < 990 having these 9 residues, and two more for 991 and 997, leaving 734 Ans. (d) : We have a 3 + b3 + c3 = ( a + b + c ) ( a 2 + b 2 + c 2 – ab – bc – ca ) + 3abc with factors of 2, 3, or 5. 16. There are two circles of centers A and B of radii 3cm and 5cm respectively. The distance 64 & abc ≤ (by AM -GM inequality) between their centers AB = 10cm. A direct 9 common tangent is drawn that meets the line BA produced at C, then the length of CA is Therefore ( a 3 + b3 + c3 ) ≥ 64 equal to: 9 DSSSB (TGT-Math) 2018 (22.9.2018, Shift-II) 24 YCT 19. In the following figure, O is the centre of the 2. A park in the shape of a quadrilateral ABCD circle and D, E and F are the mid points of AB, has ∠C = 90º, AB = 9m, BC = 12 m, CD = 5 m BO and OA respectively. If ∠DEF = 30º, then and AD = 8m. The area it occupies is: find ∠ACB. (a) (30 + 4 35 ) m2 (b) (30 + 7 35 ) m2 (c) (30 + 5 35 ) m2 (d) (30 + 6 35 ) m2 Ans. (d) : Now, side BD = 13 m (by Pythagoras theorem) (a) 120º (b) 30º 5 × 12 ar (∆BCD) = = 30m 2 (c) 60º (d) 90º 2 Ans. (c) : Clearly ADEF is a parallelogram. Hence and ∠DEF = ∠DAF = 30º ar (∆ABD) = 15 (15 − 8 )(15 − 9 )(15 − 13) Also, in isosceles triangle AOB = 6 35m 2 ∠OAB = ∠OBA = 30º Therefore, ar (ABCD) = ar (∆BCD) + ar (∆ABD) and by ∠OAB + ∠OBA + ∠AOB = 180º We have ∠AOB = 120º ( = 30 + 6 35 m 2 ) Therefore ∠AOB = 2 × ∠ACB 3. A rectangular block 6 cm × 12 cm × 15 cm is ⇒ ∠ACB = 60º cut into exact number of equal cubes. The least 20. If a, b, c, d are four positive real numbers such possible number of cubes will be: that abcd = 1, then (1 + a) (1 + b)(1 + c) (1 + d) (a) 11 (b) 40 is : (c) 6 (d) 33 (a) Less than or equal to 8 6 ×12 × 15 Ans. (b) : Possible number of cubes = = 40 (b) Greater than or equal to 16 33 (c) Greater than or equal to 8 Note- in official answer key, option (c) is given as right (d) Less than or equal to 15 answer. Ans. (b) : For ai > 0 for i = 1,..., n and a1, a2,...an= 1 4. The value of 3 cosec 20° – sec20° is: (1 + a1) (1 + a2)... (1+ an) ≥ 2n (a) 0 (b) 1 Therefore, for given problem, we have- (c) 4 (d) –1 (1 + a) (1 + b) (1 + c) (1 + d) ≥ 24 = 16 Ans. (c) : We have Section : Discipline-4 ( 3 cos ec 20º – sec 20º ) 1. The ratio in which the line joining the points (1,  3 1  = 2  cos ec 20º – sec 20º  –1) and (4, 5) is divided by the point (2, 1) is:  2 2  (a) 3 : 1 (b) 2 : 1  3 (c) 1 : 2 (d) 1 : 3 ∵sin 60° =   sin 60º cos 60º  = 2 2  –    Ans. (c) : Let the ratio be k : 1 then we must have  sin 20º cos 20º  cos 60° = 1   2   sin 60º cos 20º – cos 60º sin 20º  k.4 + 1  m 2 x1 + m1 x 2 m y + m1 y 2  = 2× 2  ( 2sin A cos A = sin 2A ) ⇒2=  2sin 20º cos 20º   as x = &y = 2 1  k +1  m1 + m 2 m1 + m 2  sin 40º ∵ sin(A − B  =4 =4   ⇒ 2k + 2 = 4k + 1 sin 40º  = sin A cos B − cos A sin B ⇒ 2k = 1 cosA sinA 5. + is equal to: 1 1 – tanA 1 – cotA ⇒ k= 2 (a) sinA – cosA (b) sinA cosA Therefore, the required ratio is 1: 2 (c) cotA – tanA (d) sinA + cosA DSSSB (TGT-Math) 2018 (22.9.2018, Shift-II) 25 YCT Ans. (d) : 7. Let X ∼ N (µ,σ 2 ). If µ 2 = σ 2 ,(µ > 0), then the cos A sin A cos 2 A sin 2 A value of P(X 0) (–5, 2). The length of the altitude through A is: X–µ 9 13 Now, Z= (a) 53 (b) 106 σ 53 53 9 13 (c) 106 (d) 53 53 53 Ans. (b) : Since the Area of triangle formula is 1 Ar = [x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)] 2 gives Z = –2 when X = –1 & Z = 0 when X = 1 Therefore, P ( X < – µ X < µ ) = P ( X < –1 X < 1) = P ( Z < –2 Z < 0 ) 1 = P ( Z < –2 ∩ Z < 0 ) ⇒ Area of ∆ ABC is = [2(7 – 2) + 4 (2 – 3) + (–5) (3 2 P(Z < 0) – 7)] 1 – P ( Z < 2) 1 = = 2 1 – P ( Z ≤ 2 )  Area = [10 – 4 + 20] 1 2 2 Area = 13 square units...(i) 8. The coordinates of the point which is Now, line AD is the altitude from the point A on line equidistant from the vertices (8, –10), (7, –3) BC. Using distance formula and (0, –4) of a right angled triangle is: (x 2 − x1 ) 2 + ( y 2 − y1 ) = ( −5 − 4 ) + ( 2 − 7) 2 2 2 BC =  15 15  (a) (4, –7) (b)  , –  ⇒ BC = 81 + 25 = 106 units....(ii) 2 2 Now, we know that, 7 7 (c) (7, –6) (d)  , –  1 3 3 Area of triangle = × Base × height 2 Ans. (a) : Let (x, y) be the required point. Now 1 equidistance condition gives ⇒ 13 = × BC × AD from eqn. (i) (x – 8)2 + (y + 10)2 = (x – 7)2 + (y + 3)2 = x2 + (y + 4)2 2 1 ⇒ x2 + 64 – 16x + y2 + 100 + 20y 13 = × 106 × AD form eqn. (ii) = x2 + 49 – 14x + y2 + 9 + 6y 2 = x2 + y2 + 16 + 8y 26 26 106 ⇒ AD = ⇒ AD = × ⇒ x – 7y = 53 & 7x + y = 21 106 106 106 ⇒ x = 4 & y = –7 13 So, AD = 106 units 9. If the elevation of the top of the tree at point E 53 due east of the tree 60º and point S due west of 13 the tree is 30º, then elevation at the midpoint of Therefore, the length of the altitude from A is 106 53 ES is: units. (a) 90º (b) 45º Hence, option (b) is correct answer. (c) 30º (d) 60º DSSSB (TGT-Math) 2018 (22.9.2018, Shift-II) 26 YCT Ans. (d): ∵2 cos A cos B     = cos(A + B) + cos(A + B)  1 5 + 1 1  –1 5 +1  = × ×  +  2 4  2 2 4  5 + 1  –1 5 +1 =  +  16  2 4  1 = 16 We have 12. (sin α + cosec α)2 + (cos α + sec α) is equal to: SM = ME (M is the midpoint), ∠AFE = 90º, (a) tan2 α + cot2 α + 7 (b) sin2 α + cos2 α + 5 ∠ASM = 30º, ∠AEF= 60º (c) tan2 α + cos2 α + 7 (d) tan2 α + cot2 α + 5 In ∆AFE, Ans. (a) : AF tan 60º = ⇒ AF = 3 ( 50 – MF ) ( sin α + cos ecα ) + ( cos α + sec α ) 2 2 FE In ∆AFS = sin 2 α + cos ec 2 α + cos 2 α + sec 2 α + 4 AF 50 + MF = ( sin 2 α + cos 2 α ) + ( cot 2 α + tan 2 α ) + 6 tan 30º = ⇒ AF = SF 3 = tan 2 α + cot 2 α + 7 ⇒ 3 (50 – MF) = 50 + MF 13. The coordinates of A, B and C are (6, 3), (–3, 5) ⇒ MF = 25 and (4, –2) respectively. The area of ∆ABC is: Now in ∆AFM, (a) 51/2 square units (b) 49/2 square units AF 25 3 (c) 45/2 square units (d) 47/2 square units tan θ = = = 3 MF 25 Ans. (b) : ⇒ θ = 60º 6 3 1 1 10. The locus of the point that is equidistant from Ar( ∆ ABC) = –3 5 1 2 two points (2, 3) and (–4, 1) is : 4 –2 1 (a) 3x – y =1 (b) 2x + 3 y = 2 R 2 → R 2 – R1 , R 3 → R 3 – R 2 (c) 3x + y = –1 (d) x + 3y = 1 6 3 1 Ans. (c) : Let the equidistant point be (x, y). Then we 1 have = –9 2 0 2 (x − 2) + (y − 3) = (x + 4) + (y − 1) 2 2 2 2 7 –7 0 1 x2 – 4x + 4 + y2 – 6y + 9 = x2 + 8x + 16 + y2 – 2y + 1 = (63 − 14) ⇒ 12x + 4y + 4 = 0 2 ⇒ 3x + y + 1 = 0 49 = Sq. units 3x + y = −1 2 14. A chord of a circle of radius 28 cm subtends an which describes the required locus. angle 90º at the centre of the circle. The area of 11. The value of expression cos60º cos36º cos42º the minor segment is: cos78º is (a) 224 cm2 (b) 248 cm2 2 1 3 (c) 184 cm (d) 290 cm2 (a) (b) Ans. (a) : 8 8 1 1 (c) (d) 16 4 Ans. (c) : cos 60º cos 36º cos 42º cos 78º 1 cos 36º cos 60º ( 2 cos 42º cos 78º ) = Area of circle = πr 2 = π ( 28 ) = 2464cm 2 2 2 1 90º = cos 36º cos 60º ( cos120º + cos 36º ) Area of the sector OACB = × 2464 = 616cm 2 2 360º DSSSB (TGT-Math) 2018 (22.9.2018, Shift-II) 27 YCT  1 18. If (3, 2), (6, 3), (x, y) and (6, 5) are the vertices & Ar (∆AOB) = × 28 × 28 of a parallelogram, the (x, y) is: 2 (a) (5, 6) (b) (9, 6) Ar (∆AOB) = 392cm2 (c) (9, 8) (d) (8, 7) Therefore, area of the minor segment ACB Ans. (b) : We must have = Ar of sector OACB – Ar (∆AOB)  3+ x 2+ y   6+ 6 3+5 = (616 –392) = 224 cm2  , = ,   2 2   2 2  15. The mean and variance of binomial (since diagonals of the parallelogram bisect each other) 4 distribution B(x:n, p) are 4 and respectively. 3+ x 2+ y 3 ⇒ = 6 and =4 2 2 What is the probability of getting 2 successes? ⇒ x = 9 and y = 6 1 20 (a) (b) 19. The radii of the ends of bucket of 16 cm height 243 243 are 20 cm and 8 cm. Find the outer curved 20 20 surface area of the bucket (c) (d) 423 342 (a) 880 cm2 (b) 3120 cm2 2 Ans. (b) : For a binomial distribution B(n, p). We have, (c) 1760 cm (d) 2240 cm2 mean = np = 4 Ans. (c) : Frustum of cone; hence total outer curved surface area = π(r + R) × l 4 & variance = np (1–p) = Now ; l = h 2 + ( R – r ) = 162 + ( 20 – 8 ) 2 2 3 2 ⇒ n = 6 and p = = 256 + 144 = 400 = 20cm 3 22 2 2  2 20 4 Therefore, CSA = ( 20 + 8) × 20 = 1760 cm2 Therefore, P(X = 2) = C 2    1 –  = 6 7    3 3  243 ∵CSA = πr1l  16. A cylinder has a diameter of 14 cm and its area   of curved surface is 220 cm2. The volume of the  where, r1 = R + r  cylinder is: 20. In the triangle ABC, ∠ABC = 90º, AB: BD: DC (a) 3080 cm3 (b) 1000 cm3 = 3: 1: 3. If AC = 20cm, then what is the length (c) 770 cm3 (d) 1540 cm3 of AD (in cm)? Ans. (c) : Curved surface area of cylinder = 2πrh 22 ⇒ 2 × × 7 × h = 220 7 ⇒ h = 5cm Now, volume of the cylinder (a) 4 5 (b) 5 2 22 (c) 4 10 (d) 6 3 = πr 2 h = × 49 × 5 = 770 cm3 7 Ans. (c) : 17. The volume of a hemisphere is 18π cm2. The total area of hemisphere is: (a) 21π cm2 (b) 18π cm2 2 (c) 27π cm (d) 24π cm2 2 3 Ans. (c) : Volume of hemi sphere = πr 3 Using Pythagoras theorem, we have- 2 3 AC2 = AB2 +BC2 ⇒ πr = 18π cm3 3 ⇒ (20)2 = (3x)2 + (4x)2 ⇒ r = 3 cm ⇒ 400 = 9x2 +16x2 Therefore, total surface area of the hemisphere ⇒ 400 = 25x2 = 3πr2 = 27π cm2 400 20 ⇒ x2 = ⇒x= =4 Note- in official answer key, option (b) is given as right 25 5 answer. ∴ AB = 12 cm and BD = 4 cm DSSSB (TGT-Math) 2018 (22.9.2018, Shift-II) 28 YCT Now in ∆ ABD, we have 3. The measure of the central tendency is given by ∵ (AD)2 = (AB)2 + (BD)2 the X-coordinate of the point of intersection of = (12)2 + (4)2 = 144 + 16 the more than ogive and less than ogive is: (AD)2 = 160 (a) Mode (b) Mean (c) All the above (d) Median AD = 4 10 cm Ans. (d) : The median can be determined by the Section : Discipline-5 intersection point of less than ogive and more than ogive. It is graphically shows bellow – 1. From the following results of two colleges A and B, find out which of the two is better? College A College B Exam Appeared Passed Appeared Passed M.A. 100 90 240 200 M.Sc. 60 45 200 160 B.A. 120 75 160 100 B.Sc. 200 150 200 140 Total 480 360 800 600 (a) college B (b) data is not sufficient (c) both are same (d) college A The value of the x-axis corresponding to the intersection Ans. (c) : Total number of students (appeared in exams) point indicates the median. of college A = 480 Total number of students (passed the exams) of college Hence, correct answer is median. Therefore, option (d) B = 360 is correct. 360 4. Consider the following distribution: ∴ Passing percentage of college A = ×100 = 75% Marks obtained No. of students 480 Total number of students of college B who appeared in More than or equal to zero 63 exams = 800 More than or equal to 10 58 Total number of students of college B who passed the More than or equal to 20 55 exams = 600 More than or equal to 30 51 600 ∴ Passing percentage of college B = ×100 = 75% More than or equal to 40 48 800 More than or equal to 50 42 Since, both colleges A and B have the same passing The frequency of the class 30 - 40 is : percentage. (a) 4 (b) 51 Therefore, option (c) is correct answer. (c) 48 (d) 3 If P ( A ∩ B ) = , P ( A ∩ B ) = and 2P (A) = p, Ans. (d) : We have 1 1 2. 2 2 P(B) = p, then the value of p is given by: (a) 1/3 (b) 1/4 (c) 2/3 (d) 1/2 Ans. (c) : ( ) P ( A ∩ B) = P A ∪ B = 1 – P ( A ∪ B) = 1 2 1 ⇒ P ( A ∪ B) = 2 ∴ P ( A ∪ B) = P ( A ) + P ( B) – P ( A ∩ B) Therefore, the required frequency is 3.  p p 1 1 ∵ 2P(A) = p ⇒ P(A) =  5. The regression coefficients are b2 and b1. Then gives p + – =  2 the correlation coefficient r is: 2 2 2  P(B) = p  b b ⇒ 3p = 2 (a) 1 (b) 2 b2 b1 2 ⇒ p= (c) ± ( b1b 2 ) (d) b1b2 3 DSSSB (TGT-Math) 2018 (22.9.2018, Shift-II) 29 YCT Ans. (c) : 8. X, Y and Z are three uncorrelated variables Coefficient of correlation = ± b yx × b xy = ± b1b 2 having variances σ 2x , σ 2y and σ 2z respectively, then the correlation coefficient between X + Y 6. The values of c and P (X > 1) for which the and Y + Z is:  c (a) 1/2  ; 0 1 be fixed and a, b, c, d be arbitrary Remainder is P –2 integers. If a ≡ b (mod n) and c ≡ d (mod n), Similarly, then : (P – 2) ! = 1 (mod P) (a) a + c b + d (mod n) and ac bd (mod n) Given 27! = X (mod 29)...(i) (b) a + c b + d (mod n) but ac bd (mod n) P = 29 (which is prime) (c) a + c ≡ b + d (mod n) and ac ≡ bd (mod n) Wilson theorem (d) a + c b + d (mod n) and ac bd (mod n) (P – 2) ! = 1 (mod P) Ans. (c) : For n > 1 fixed and a, b, c, d arbitrary integers On comparing (i) with this property if a ≡ b (mod n) and c ≡ d (mod n) then we have we get, a – b = k1n and c – d = k2n X=1 DSSSB (TGT-Math) 2018 (23.9.2018, Shift-I-Male) 31 YCT 5. Consider a non-homogeneous system of 5 linear (d) Transpose of a diagonalizable matrix is equations in 4 variables, such that there is at diagonalizable. Since if A is diagonalizable then we least one variable with a non-zero coefficient in have each equation. Then the possible smallest and D = Q −1AQ largest ranks for the corresponding augmented matrix are: for invertible matrices Q and Q–1 and diagonal matrix D (a) Smallest = 1, Largest = 5 then on taking transpose we get ( ) T (b) Smallest = 2, Largest = 5 D = DT = QT A T Q −1 (c) Smallest = 1, Largest = 4 (d) Smallest = 2, Largest = 4 since the transpose of diagonal matrix is diagonal and (Q ) = (Q ) T −1. Thus if we set P = ( QT ) we have −1 T −1 Ans. (a) : Augmented matrix of a non-homogeneous system of 5 linear equations in 4 variables, such that there is at least one variable with a non-zero coefficient P −1A T P = D , T in each equation, is a 5×5 square matrix whose smallest and so A is diagonalizable. and largest ranks can be 1 and 5 respectively. 8. Identify a subspace of ℝ 2 out of four given 6. The units digit of 3100 is : options : (a) 0 (b) 1 (a) A line in ℝ 2 passing through the points (0, 1) (c) 2 (d) 3 and (–1, 0) Ans. (b) : Since the Fermat's theorem is aP–1≅1(mod P ) (b) A line in ℝ 2 passing through the points (0, – Now, we can use Fermat's theorem for the unit digit of 1) and (–1, 0) 3100. (c) A line in ℝ 2 passing through the points (1, 3) ⇒ 34≅ 1(mod 5 ) and (–1, –3) So, (34)25 = 3100≅125≅1(mod 5) (d) A line in ℝ 2 passing through the points (0, – ⇒3100≅125≅ 1(mod2) 1) and (1, 0) Thus, 3100≅1(mod 10 ) Ans. (c) : A line is a subspace of R2 if it passes through Hence, the unit digit of 3100 is 1. origin. Clearly the line joining (1, 3) and (–1, –3) passes 7. Which of the given statements is true? through origin since (1, 3) and (–1, –3) are symmetric (a) Every square matrix is diagonalizable about origin. Therefore, line joining (1, 3) and (–1, –3) is a subspace of R2. (b) The sum of diagonalizable matrices is also diagonalizable. 9. Consider the subset : (c) The product of diagonalizable matrices is also S = {[3,1, −1][ 2, 2, −1][ −5, −2, 2][1,3,1]} of R 3. If diagonalizable. A is the matrix having columns as vectors of set (d) If A is a diagonalizable matrix, then AT is S then: diagonalizable. (a) the set S is linearly dependent and the matrix Ans. (d) : A has rank 3. 0 1  (b) the set S is linearly dependent and the matrix (a) Consider the 2×2 matrix   , which is clearly A has rank 2. 0 0  not diagonalizable. (c) the set S is linearly independent and the matrix A has rank 3. (b) The sum of two diagonalizable matrix need not be (d) the set S is linearly independent and the 1 1  matrix A has rank 2. diagonalizable. For example, if A =   and 0 2  Ans. (a) :  −1 0  Since S = {( 3,1, −1) , ( 2,2, −1)( −5, −2, 2 )(1,3,1)} ⊆ R 3 cont B=  then A and B are diagonalizable, but A +  0 −2  ains more than 3 elements, S is linearly dependent as dim (R3) = 3. Now 0 1  B =  is not diagonalizable.  3 2 −5 1  1 2 −2 3 0 0   1 2 −2 3  R1 ↔ R 2  3 2 −5 1  (c) The product of two diagonalizable matrices need not   →   be diagonalizable as a counter example consider  −1 −1 2 1  −1 −1 2 1  1 0  1 1   1 1   =   1 2 −2 3  1 0 0 0   0 −1 0 −1  0 1 R 2 → R 2 − 3R1  →  0 −4 1 −8  C2 → C2 − 2C1 →  0 −4 1 −8  R 3 → R 3 + R1   CC3 → C3 + 2C1   1 0   0 1 0 4  4 → C 4 − 3C1 0 1 0 4  which is non - diagonalizable but   is a diagonal  0 −1  1 0 0 0  1 0 0 0  matrix and    →  0 −4 1 0   →  0 0 1 0  C 4 → C 4 − 2C2 C 2 → C2 + 4C3 −1  1 1   −1 1  −1 0  −1 1   0 1 0 2   0 1 0 2   =     0 −1  2 0  0 1  2 0  This clearly shows that rank of A is 3. DSSSB (TGT-Math) 2018 (23.9.2018, Shift-I-Male) 32 YCT 10. The real values of x satisfying the inequality Ans. (d) : By AM ≥ GM inequality we have x( 10 ) 2 log x –3log10 x+1 > 1000 are: 1 1 x+ ≥ 2 x⋅ (a) x > 100 (b) x > 1000 x x (c) x > 1 (d) x > 10 1 Ans. (b) : We have ⇒ x+ ≥2 x x ( 10 ) 2 log x −3log x +1 10 > 1000 13. The number of positive divisors of 8 × 81 × 125 taking log from both side is : (a) 20 (b) 80 ⇒ ( log x )2 − 3log x + 1 log x > log 103  10 10  10 10 (c) 16 (d) 100 ⇒ ( log x ) − 3log x + 1 log x − 3 > 0 2 Ans. (b) : The number of factors P1n 2 , P2n 2...........Pkn k is  10 10  10 (n1+1) (n2+1) (n3+1)...... (nk+1) ⇒ (( log10 2 ) x ) + 1 ( log10 x − 3) > 0 where, P's are distinct primes. We have ⇒ log10 x − 3 > 0 8×81 ×125 = 23 × 34 × 53 Therefore, the number of positive divisors ⇒ x > 103 = ( 3 + 1)( 4 + 1)( 3 + 1) = 80 ⇒ x > 1000 11. The system of linear congruences: Consider the linear congruence 6x ≡ 3 (mod 9). 14. x ≡ 1 (mod 5) Then the incongruent solutions modulo 9 of this congruence are: x ≡ 3 (mod 3) (a) 0, 3, 6 (b) no solution x ≡ 2 (mod 7) (c) 2, 5, 8 (d) 1, 4, 7 (a) has a simultaneous solution, which is not unique modulo 105 Ans. (c) : 6x ≡ 3(mod 9)...(i) (b) has no simultaneous solution. d = GCD (6, 9) = 3 (c) has a simultaneous solution, which is uniqueWe have exactly 3 incongruent solutions modulo 105. Now, dividing (i), with 3, we get (d) cannot be solved by Chinese Remainder 6x 3  9 ⇒ ≡  mod  ⇒ 2x ≡ 1 (mod 3) theorem. 3 3 3 Ans. (c) : Chinese Remainder Theorem:- x=2 if a1, a2..........ak are any integers and m1, m2........., mk 4 –1 = 3 and (3 mod 3) = 0 are relative prime pairwise integers then the one solution is x1 = 2 simultaneous congruences Now, find all the in congruent solutions , x ≅ a1 (mod m1 ) m x = x1 + P x ≅ a 2 (mod m 2 ) d where, 0 ≤ P ≤ (3 – 1) ⋮ ⋮ ⋮ Put, P = 0 x ≅ a k (mod m k ) X=2+0=2 Have a solution and the solution is unique modulo M. put, P = 1 where, M = m1,m2,.........,mk X=2+3×1=5 Sol:- Given that m1 = 5, m2 = 3, m3 = 7 put, P = 2, Since, m1, m2 and m3 are coprime, X=2+3×2=8 So, system of linear congruences will have The three in congruent solutions simultaneous solution with modulo modulo 9 of this congruence are 2, 5, 8 M = m1×m2×m3 15. If a1, a2,...,an and b1, b2,...,bn are non - negative ⇒ M = 5×3×7 real numbers such that a1 > b1, a2 > b2,...,an > bn ⇒ M = 105 then : Therefore, system of linear congruences simultaneous (a) a1 + a 2 +... + a n > b1 + b 2 +... + b n and solution, which is unique modulo 105,option (c) correct answer. a1a 2...a n > b1b 2...b n 12. For any real x ≠ 0, which of the inequality is (b) a1 + a 2 +... + a n < b1 + b2 +... + b n and true? a1a 2...a n < b1b 2...b n 1 1+ x 2 (c) a1a 2...a n > b1b 2...b n but (a) x + < 0 (b) b1 + b 2 +... + b n but (c) x + < 2 (d) x + ≥ 2 x x a1a 2...a n < b1b 2...b n DSSSB (TGT-Math) 2018 (23.9.2018, Shift-I-Male) 33 YCT Ans. (a) : If a1 , a 2 ,..., a n and b1 , b2 ,..., b n are non- 19. If a is any integer and a3 ≡ x (mod 9), then the negative real number such that a1 > b1 , a2 > b2,...,an > bn possible values of x are: then we must have (a) 5, 6, 7 (b) 0, 1, 8 a1 + a 2 +... + a n > b1 + b 2 +... + b n (c) 3, 5, 7 (d) 2, 4, 6 Ans. (b) : Given that a is any integer & a1a 2...a n > b1b 2...b n i.e. a = 1,2,3,4,5........ on adding and multiplying inequalities. To find the possible values of x when a3 ≡ x(mod9). 16. If x (≠ 1) is any real number, then If a = 1 then (1)3 ≡ 1(mod9) x 3 + 1 > x 2 + x if In this case we obtain x = 1 (a) x > –1 If a = 2 then (2)3 ≡ 8(mod9) (b) Inequality doesn't hold for any real x In this case we obtain x = 8 (c) x = –1 If a = 3 then (3)3≡ 9 (mod9) ≡ 0 (mod9) (d) x < –1 In this case we obtain x = 0 and so on..... Ans. (a) : We have, Therefore, as we continue this process we get the same x3 +1 > x2 + x ; x ≠ 1 remainders as 0,1,8 respectively, Hence, The possible values of x are 0,1,8. ⇒ x3 − x 2 − x + 1 > 0 Thus, option (b) is correct answer. ⇒ x 2 ( x − 1) − 1( x − 1) > 0 20. If p and q are distinct prime with ap ≡ a (mod ⇒ ( 2 ) x − 1 ( x − 1) > 0 q) and aq ≡ a (mod p), then: ⇒ x > –1 (a) apq ≡ –a (mod pq) (b) apq ≡ a (mod pq) 17. If W is a nonempty subset of real vector space (c) apq ≡ 0 (mod pq) (d) apq ≡ 1 (mod pq) 3 R using the same operations, then W is a Ans. (b) : We know that least common multiple of two subspace of R3 if and only if: relatively prime numbers is the product of the numbers. (a) W is not closed under scalar multiplication Now, if ap ≡ a (mod q) and aq ≡ a(mod p); p and q are but closed under vector addition in R3 (b) W is closed under vector addition and scalar primes; then using Fermat's Little theorem i.e. ap ≡ multiplication in R3 a(mod p) and a ≡ a (mod q) we get q ( ) (c) W is not closed under vector addition but q a pq ≡ a p ≡ a p ≡ a ( mod p ) closed under scalar multiplication in R3 ( ) (d) W is neither closed under vector addition nor p closed under scalar multiplication in R3 & a pq ≡ a q ≡ a q ≡ a ( mod q ) Ans. (b) : A non-empty subset W of a vector space V is ⇒ p a pq − a and q a pq − a a subspace of V if and only if W is closed under vector addition and scalar multiplication induced by vector ⇒ pq a pq − a addition and scalar multiplication defined for V i.e. αx + βy ∈ W since LCM (p, q) = pq for every x, y in W and α, β ∈ F. ⇒ apq ≡ a (mod pq). 18. A transformation T: R2→ R2 first reflects Section : Discipline-2 points through the vertical axis (x2- axis) and 1. The general solution of recurrence relation then reflects points through the line x2 = x1. ar – 5ar-1 + 6ar-2 = 4r ,r ≥ 2 is: Then the standard matrix of T is: 0 −1  −1 0  1 (a) A1.2r + A 2.3r −.4r , where A1 and A2 are (a)   (b)   8 1 0   0 1 arbitrary constants. 1 0   0 1 (b) A1.5r + A 2.6r + 8.4r , where A1 and A2 are (c)   (d)   0 −1  −1 0  arbitrary constants. Ans. (d) : Linear transformation Tx 2 −axis : ℝ 2 → ℝ 2 and (c) A1.2r + A 2.3r + 8.4r , where A1 and A2 are arbitrary constants. Tx 2 = x1 : ℝ 2 → ℝ 2 are given by matrices 1 (d) A1.2r + A 2.4r +.3r , where A1 and A2 are  −1 0  0 1  8 TA=   and TB= 1 0  respectively. Therefore the arbitrary constants.  0 1    required linear transformation T: R2 → R2 is given by Ans. (c) : ar – 5ar–1 + 6ar–2= = 4r, r ≥ 2 the matrix Associate homogeneous solution is ar – 5ar–1 + 6ar–2  0 1   −1 0   0 1  T = TA×TB=   =  Characteristic equation 1 0  0 1   −1 0  x2 – 5x + 6 = 0 DSSSB (TGT-Math) 2018 (23.9.2018, Shift-I-Male) 34 YCT x = 3 or 2 sum of roots = p = 2 + 8 = 10 Solution - & product of roots = q = (–1) × (–9) = 9 an = 6an–2 – an–1 Therefore, x2 – 10x + 9 = 0 is the required quadratic and Ah = A1 (2)r + A2 (3)r its roots are 1 and 9. f(r) = 1 × 4r it is of the Form and 4 is not a root. 5. If there are m arithmetic progressions with the Therefore, it's particular solution is A4r General first term of each being unity and common solution of recurrence relation is differences as 1, 2, 3...m respectively, then the A1(2)r + A2 (3)r + A4r sum of their nth terms is: After Substitution partial solution in recurrence relation n (A4r) – 5(A.4r–1) + 6 (A.4r–2) = 4r (a) [mn + m – n + 1] 2 5A 6A m A– + =1 (b) [mn + m – n + 1] 4 16 2 ∴A=8 n Therefore General Solution is A1(2)r + A2 (3)r + 8.4r (c) [mn – m + n + 1] 2 2. If the point P(x, y) is equidistant from the points A (a + b, b – a) and B (a – b, a + b), then m (d) [mn – m + n + 1] (a) x = ay (b) y = bx 2 (c) ax = by (d) bx = ay Ans. (d) : nth terms of each arithmetic progression with Ans. (d) : If P (x, y) is equidistant from the points A (a first term unity and common difference 1,2,..,m + b, b – a) and B (a – b, a + b) then we have- respectively are PA =AB ⇒(PA)2 = (PB)2 n, 2n – 1, 3n – 2,... , mn – m + 1 respectively which is an arithmetic progression with  x − ( a + b )  +  y − ( b − a )  =  x − ( a − b )  +  y − ( a + b )  2 2 2 2 common difference n – 1 and m terms. Therefore, the ⇒ x2 + (a + b)2 – 2x(a + b) + y2 + (b – a)2 – 2y(b – a) required sum is = x2 + (a – b)2 – 2x(a – b) + y2 + (a + b)2 – 2y(a + b) m ⇒ x(a + b) + y(b – a) = x(a – b) + y(a + b) [ mn − m + n + 1] 2 ⇒ x(a + b) – x(a – b) = y(a + b) – y(b – a) 6. If A (6, 3), B (–3, 5), C (4, –2) and D (x, 3x) are ⇒ ax + bx – ax + bx = ay + by – by + ay Area(∆DBC) 3 ⇒ bx = ay four points with = , then the Area(∆ABC) 2 21  1 1 value of x is: 3. In the expansion of  5 3 + 2 7  , the sum of all 23 177   (a) (b)   8 56 rational terms is: (a) 78117 (b) 78133 173 25 (c) (d) (c) 78149 (d) 78125 56 8 Ans. (b) : The general term in the expansion of Ans. (d) : Given, A (6, 3), B (–3, 5), C (4, –2) and D (x, 3x)  1 1  21 By using area of triangle  53 + 2 7  using Binomial theorem is   1   =  x1 ( y 2 − y3 ) + x 2 ( y3 − y1 ) + x 3 ( y1 − y 2 )  21− r r 2 x ( 5 + 2 ) + ( −3)( −2 − 3x ) + 4 ( 3x − 5 ) 3 7 Tr+1 = 21Cr 5.2 −r r Area of ∆DBC = 2 = 21C r 57 ⋅ 5 3 ⋅ 2 7 7x + 6 + 9x + 12x − 20 28x − 14 which is clearly rational only when r = 0 and r = 21. = = Therefore, the required sum is 2 2 21 C0 57 + 21C 21 23 = 78125 + 8 = 78133 6 ( 5 + 2 ) + ( −3)( −2 − 3) + 4 ( 3 − 5) Area of ∆ABC = 4. Two candidates attempted the problem, 2 x 2 + px + q = 0. One started with a wrong value 42 + 15 − 8 49 of p and got the roots –1 and –9, whereas the = = 2 2 other started with an incorrect value of q and get the roots 2 and 8. Find the correct roots Area of ∆DBC 3 Given: = (a) 9 and –2 (b) 8 and –9 Area of ∆ABC 2 (c) –1 and 2 (d) 1 and 9 28x − 14 Ans. (d) : Given quadratic is 2 3 = = x 2 + px + q = 0 49 2 Now we must have 2 DSSSB (TGT-Math) 2018 (23.9.2018, Shift-I-Male) 35 YCT ⇒ 56x – 28 = 147 10. The sum of n terms of four arithmetical ⇒ 56x = 175 progressions are S1, S2, S3, and S4, with the first term of each as 2 and the common differences 28 ⇒ x=. as 1, 3, 5 and 7 respectively. Then: 5 (a) S4 + S3 = S1 + S2 (b) S4 + S2 = S1 + S3 7. Consider three quadratic equations, (c) S4 + S1+ S2 = 2S3 (d) S4 + S1 = S2 + S3 x 2 + ax + bc = 0 , x 2 + bx + ca = 0 , Ans. (d) : a = 2, d1 = 1 x 2 + cx + ab = 0. If each pair of these equations l1 = 2 + (n – 1)1 = 2+(n–1)1= 2 + n – 1 = n + 1 n has a common root, then the sum and product S1 = [ a + ℓ1 ] of these common roots are: 2 − (a + b + c) n (a) sum = , Product = a2b2c2 S1 =  2 + ( n + 1)  2 2 − (a + b + c) n = [ n + 3] ----(i) (b) sum = , Product = abc 2 2 a+b+c a = 2, d2 = 3, l2 = 2 + (n – 1)3 = 3n – 1 (c) sum = , Product = a2b2c2 n 2 S2 = [ a + ℓ 2 ] a+b+c