Mathematics - Sequence & Series Question with Solution | TestHub
Match each item in List I to all applicable entries in List II.
List I | List II |
|---|---|
(A) Find the largest positive term of the H.P. whose first two terms are 2/5 and 12/23. | (P) 2 |
(B) For a,b,c>0 with a+b+c=1, find the minimum of (1+a)(1+b)(1+c)/[(1-a)(1-b)(1-c)]. | (Q) 4 |
(C) Which listed integers belong to the range of f(x)=(2x²+2x+5)/(x²+x+1), for real x? | (R) 6 |
(D) Which listed values x satisfy (1/3)^(|x+6|/(1-|x|))>9? | (S) 7 |
| (T) 8 |
Options:
Answer:
Solution:
(A) Reciprocal terms form the A.P. (37-7n)/12, so the H.P. term is 12/(37-7n). The positive terms end at n=5, with largest value 6. (B) 1+a=(1-b)+(1-c)≥2√[(1-b)(1-c)]. Multiplying the three cyclic inequalities gives the ratio ≥8, attained at a=b=c=1/3. (C) f(x)=2+3/(x²+x+1), and x²+x+1≥3/4, giving range (2,6]. Listed values are 4,6. (D) The base 1/3 reverses the inequality, so |x+6|/(1-|x|)<-2. This requires |x|>1. Solving separately for x>1 and x<-1 gives x∈(-8/3,-1)∪(1,8), containing listed values 2,4,6,7. Thus A-R; B-T; C-Q,R; D-P,Q,R,S.
