JEE Class main Answered
suppose a is a positive real number such that 
then
a) 
<2
b)2< <3
c)3< <4
d)4<=
Asked by Girijeshpandey.rjil | 28 Mar, 2019, 06:20: PM
Expert Answer
f(a) = a5 -a3 +a -2
if a =1, f(1) = -1 ; if a=2, f(2) = 24 ; hence root of f(a) is between 1 and 2 , i.e., 1< a <2
let us check middle value for root. let a = 3/2, f(3/2) = (243/32) -(27/8) +(3/2) -2 = 119/32 or f(a) >3
let us check middle value between 1 and 1.5, let a = 1.25 = 5/4, f(5/4) = (3125/1024) -(125/64) +(5/4) -2 ≈ 0.4
hence the root is near to 1.25. Let us consider a = 1+Δa , where Δa ≈ 0.25
then a6 = (1+Δa)6 ≈ 1+ 6C1 Δa + 6C2 (Δa)2 ..................................(1)
[ powers of Δa3 and higher orders are neglected in eqn.(1) ]
substituting Δa in eqn.(1), we get, a6 ≈ (1+0.25)6 ≈ 1 + 6×0.25 + 15×(1/16) ≈ 3.4
hence 3 < a6 < 4
Answered by Thiyagarajan K | 29 Mar, 2019, 07:56: AM
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