JEE Advanced Model Paper 14 with solutions for online practice

f (x) =

1

3{x}−9{x}+1

=

1

t−t2+1

=

1

f(t)

where 3{x} = t
0 ≤ {x} < 1 ⇒ 1 ≤ 3{x} < 3 ⇒ 1 ≤ t < 3
f (t) = t - t2 + 1 ⇒ f'(t) = 1 - 2t
f'(t) < 0 for t ∊ [1 , 3] and f(t) is decreasing
∴ max = f (1) = 1
f (t) > f (3) = 3 - 9 + 1 = - 5
- 5 < f (t) ≤ 1
⇒ range of

1

f(t)

= ( - ∞ , -1/5) ∪ [1 , ∞)