Verified Solution • 2025–26 Session moderate

QJEEM23Madae0: JEE Main 2023 Mathematics - Class 11 JEEM23Madae0 Solution: The absolute minimum value, of the function , where denotes the …

The absolute minimum value, of the function

f(x)=∣x2−x+1∣+[x2−x+1]f(x)=\left|x^{2}-x+1\right|+\left[x^{2}-x+1\right] ,

where [t][t] denotes the greatest integer function, in the interval [−1,2][-1,2] , is :

Multiple Choice Options

  • A 34\frac{3}{4}
  • B 32\frac{3}{2}
  • C 14\frac{1}{4}
  • D 54\frac{5}{4}

Step-by-Step Verified Solution

f(x)=∣x2−x+1∣+[x2−x+1];x∈[−1,2]\mathrm{f}(\mathrm{x})=\left|\mathrm{x}^{2}-\mathrm{x}+1\right|+\left[\mathrm{x}^{2}-\mathrm{x}+1\right] ; \mathrm{x} \in[-1,2]

Let g(x)=x2−x+1g(x)=x^{2}-x+1

=(x−12)2+34

∵|x2−x+1| and [x2−x+1]

Both have minimum value at x=1/2\mathrm{x}=1 / 2

⇒\Rightarrow Minimum f(x)=34+0\mathrm{f}(\mathrm{x})=\frac{3}{4}+0

=34=\frac{3}{4}

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