FunctionshardFree

Functions — JEE Maths practice question

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

Question

If the range of

f(x) = \sqrt{24\{x\}-5-16\{x\}^2}

, where

\{\cdot\}

denotes fractional part, is

[a,b]

, then the value of

b^2-a

Solution

Answer: 4

Step 1: Set

t = \{x\} \in [0,1)

and find where

f

is defined

24t-5-16t^2 \geq 0 \implies 16t^2-24t+5 \leq 0 \implies (4t-1)(4t-5) \leq 0

\frac{1}{4} \leq t \leq \frac{5}{4}

Combined with

t \in [0,1)

t \in \left[\frac{1}{4},1\right)

. Step 2: Find the range Let

h(t) = 24t-5-16t^2

h'(t) = 24-32t = 0

t = \frac{3}{4}

h\!\left(\frac{1}{4}\right) = 6-5-1 = 0, \quad h\!\left(\frac{3}{4}\right) = 18-5-9 = 4

The maximum of

h

\left[\frac{1}{4},1\right)

is 4 at

t = \frac{3}{4}

, minimum is 0 at

t = \frac{1}{4}

. Range of

f = \sqrt{h(t)}

[0, 2]

. So

a = 0

b = 2

. Step 3: Evaluate

b^2-a = 4-0 = 4

Answer: 4

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