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The Value Of 1 1 Factorial 2 2 N Factorial Is

(1) (n + 1)! (2) (n + 1)! + 1 (3) (n + 1)! - 1 (4) None of these Solution: Let S = 1 . 1! + 2 . 2! + 3 . 3! + ...+ n . n! =... View Article

The Sum Of Infinity Of The Series 1 2 By 3 6 By 3 2 10 By 3 3 Is

(1) 3 (2) 4 (3) 6 (4) 2 Solution: Let S = 1 + 2/3 + 6/32 + 10/33 + 14/34 + … (i) S/3 = 1/3 + 2/32 + 6/33 + 10/34 + 14/35 + …(ii)... View Article

If X 1 Then The Sum Of The Series 1 2x 3×2 4×3 Infinity Will Be

(1) 1/(1-x) (2) 1/ (1+x) (3) -1/(1+x2) (4) 1/(1-x)2 Solution: S = 1 + 2x + 3x2 + 4x3 +... ∞ ..(i) Multiply by x on both sides Sx... View Article

If Am And Gm Of X And Y Are In The Ratio P Q Then X Y Is

(1) p - √(p2+q2) : p + √(p2+q2) (2) p + √(p2 - q2) : p - √(p2 - q2) (3) p : q (4) None of these Solution: AM of x and y = (x+y)/2... View Article

Let Alpa Beta Gamma And Delta Be Four Positive Real Numbers Such That Their Product Is Unity Then The Least Value Of

(1) 6 (2) 16 (3) 0 (4) 32 Solution: Product, αβγδ = 1 (1 +α) (1+β) (1 + γ) (1 +δ ) = (1+ α + β + αβ ) (1+ γ+ δ + δγ) = 1+ α + β... View Article

If P Q R Are In Gp And Tan 1 P Tan 1 Q Tan 1 R Are In Ap Then P Q R Satisfies The Relation

(1) p = q = r (2) p ≠ q ≠ r (3) p + q = r (4) None of these Solution: Since p, q r are in GP q2 = pr … (i) Since tan-1 p, tan-1... View Article

If The Am Of Two Numbers Is A And Gm Is G Then The Numbers Will Be

1) A ± (A2 -G2 ) (2) √A ± √(A2-G2) (3) A ± √(A+G)(A-G) (4) (A ± √(A+G)(A-G) )/2 Solution: Let a, b be the numbers. Then AM, A =... View Article

If A Is A Positive Number Such That The Arithmetic Mean Of A And 2 Exceeds Their Geometric Mean By 1 Then The Value Of A Is

(1) 3 (2) 5 (3) 9 (4) 8 Solution: AM = (a+2)/2 GM = √(2a) Given AM exceeds GM by 1. (a+2)/2 - √(2a) = 1 (a+2) - 2√(2a) = 2 a... View Article

If A B C Are In Ap B A C B And A Are In Gp Then A B C Is

(1) 1 : 2 : 3 (2) 1 : 3 : 5 (3) 2 : 3 : 4 (4) 1 : 2 : 4 Solution: Since a, b c are in AP (b - a) = ( c- b) … (i) b - a, c - b... View Article

Three Positive Numbers Form An Increasing Gp If The Middle Term In This Gp Is Doubled Then New Numbers Are In Ap Then The Common Ratio Of The Gp Is

(1) 2 + √3 (2) 3 + √2 (3) 2 - √3 (4) 2 + √2 Solution: Let a, ar, ar2 be the numbers in GP. 2(2ar) = a + ar2 4r = 1+r2... View Article

If Three Real Numbers A B And C Are In Harmonic Progression Then Which Of The Following Is True

(1) 1/a, b, 1/c are in AP (2) 1/bc, 1/ca, 1/ab in HP (3) ab, bc, ac are in HP (4) a/c, b/c, c/a are in HP Solution: Given a, b, c... View Article

The Value Of 0 Bar 37 Where 0 Bar 37 Stands For The Number 0 037373 Is

The value of (1) 37/1000 (2) 37/990 (3) 1/37 (4) 1/27 Solution: Let x = 0.0373737… 1000x = 37.3737…. 10x = 0.3737… 1000x -... View Article

If A B C D And P Are Distinct Real Numbers Such That A2 B2 C2 P2 2ab Bc Cd P B2 C2 D2 0 Then A B C And D

(1) are in AP (2) are in GP (3) are in HP (4) satisfy ab = cd Solution: Given (a2 + b2 + c2)p2 - 2(ab + bc + cd)p + (b2 + c2 + d2)... View Article

In A Geometric Progression The Ratio Of The Sum Of The First Three Terms And The First Six Terms Is 125 152 The Common Ratio

(1) 1/5 (2) 2/5 (3) 4/5 (4) 3/5 Solution: Sum of first 3 terms of GP, S3 = a(r3-1)/(r-1) Sum of first 6 terms of GP, S6 =... View Article

If X 1 A A2 Infinity And Y 1 B B2 Infinity Where A And B Are Proper Fractions Then 1 Ab Equals

(1) xy / (x+y-1) (2) xy/(x-y) (3) (x2+ y2)/(x-y) (4) none of these Solution: 1+ a+ a2 +... ∞ = 1/(1-a) = x 1+b+b2+...∞ = 1/(1-b)... View Article

In An Infinite Geometric Series The First Term Is A And The Common Ratio Is R If The Sum Of The Series Is 4 And The Second Term Is 3 4 Then A R Is

(1) (4/7, 3/7) (2) (2, 3/8) (3) (3/2, 1/2) (4) (3, 1/4) Solution: Given first term = a Common ratio = r Sum, S = a/(1-r) = 4... View Article

Let R Be A Reflexive Relation On A Set A And I Be The Identity Relation On A Then

1) R ⊂ I 2) I ⊂ R 3) R = 1 4) None of these Solution: (3) A = {1, 2, 3} I = {(1, 1), (2, 2), (3, 3)} R = {(1, 1), (2, 2), (3, 3)} R = I... View Article

Let A 1 2 3 B 1 3 5 If Relational R From A To B Is Given By R 1 3 2 5 3 3 Then R 1 Is

1) {(3,3), (3,1), (5,2)} 2) {(1, 3), (2, 5), (3,3)} 3) {(1,3), (5,2)} 4) None of these Solution: (1) A = {1, 2, 3} B = {1, 3, 5} R =... View Article

If A Is The Set Of Even Natural Numbers Less Than 8 And B Is The Set Of Prime Numbers Less Than 7 Then The Number Of Relations From A To B Is

1) 29 2) 92 3) 32 4) 29 – 1 Solution: (1) A = {2, 4, 6 B = {2, 3, 5} Number of relations = 2mn m = 3, n = 3 29 is the correct... View Article

If R X Y X Y Belongs To Z X 2 Y 2 Less Than Or Equal To 4 Is A Relation In Z Then Domain Of R Is

1) {0,1,2} 2) {0, –1, –2} 3) {–2, –1, 0, 1, 2} 4) None of these Solution: (3) x2 + y2 ≤ 4 x, y ∈ Z D : {-2, -1, 0, 1, 2}... View Article

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