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Functions: Range Denotes Fractional Part Value

JEE Maths question with a full step-by-step solution.

Question
If the range of f(x)=24{x}−5−16{x}2f(x) = \sqrt{24\{x\}-5-16\{x\}^2}, where {⋅}\{\cdot\} denotes fractional part, is [a,b][a,b], then the value of b2−ab^2-a.
Solution
Answer: 4
Step 1: Set t={x}∈[0,1)t = \{x\} \in [0,1) and find where ff is defined
24t−5−16t2≥0  ⟹  16t2−24t+5≤0  ⟹  (4t−1)(4t−5)≤024t-5-16t^2 \geq 0 \implies 16t^2-24t+5 \leq 0 \implies (4t-1)(4t-5) \leq 0
14≤t≤54\frac{1}{4} \leq t \leq \frac{5}{4}
 Combined with t∈[0,1)t \in [0,1): t∈[14,1)t \in \left[\frac{1}{4},1\right). Step 2: Find the range Let h(t)=24t−5−16t2h(t) = 24t-5-16t^2. h′(t)=24−32t=0h'(t) = 24-32t = 0 at t=34t = \frac{3}{4}.
h ⁣(14)=6−5−1=0,h ⁣(34)=18−5−9=4h\!\left(\frac{1}{4}\right) = 6-5-1 = 0, \quad h\!\left(\frac{3}{4}\right) = 18-5-9 = 4
 The maximum of hh on [14,1)\left[\frac{1}{4},1\right) is 4 at t=34t = \frac{3}{4}, minimum is 0 at t=14t = \frac{1}{4}. Range of f=h(t)f = \sqrt{h(t)}: [0,2][0, 2]. So a=0a = 0, b=2b = 2. Step 3: Evaluate
b2−a=4−0=4b^2-a = 4-0 = 4
 Answer: 4
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