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Functions question

2023 · 31 Jan · Shift 2 · Q32
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Functions question

2023 · 31 Jan · Shift 2 · Q32

JEE MainMathematicsFunctionsMCQ+4 / −1
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]f(x)=​x2−x+1​+[x2−x+1], where [t][t][t] denotes the greatest integer function, in the interval [−1,2][-1,2][−1,2], is :
  1. A
    34\frac{3}{4}43​
  2. B
    32\frac{3}{2}23​
  3. C
    14\frac{1}{4}41​
  4. D
    54\frac{5}{4}45​
View written solutionFree

Correct answer: A

  1. Let t=x2−x+1.t=x^2-x+1.t=x2−x+1. Then the given function becomes f(x)=∣t∣+[t],f(x)=|t|+[t],f(x)=∣t∣+[t], where [t][t][t] is the greatest integer function.

  2. First, find the range of t=x2−x+1t=x^2-x+1t=x2−x+1 for x∈[−1,2]x\in[-1,2]x∈[−1,2].

Complete the square: x2−x+1=(x−12)2+34.x^2-x+1=\left(x-\frac12\right)^2+\frac34.x2−x+1=(x−21​)2+43​. So, t≥34,t\ge \frac34,t≥43​, and the minimum occurs at x=12.x=\frac12.x=21​.

Now check the maximum on [−1,2][-1,2][−1,2]: t(−1)=(−1)2−(−1)+1=3,t(-1)=(-1)^2-(-1)+1=3,t(−1)=(−1)2−(−1)+1=3, t(2)=22−2+1=3.t(2)=2^2-2+1=3.t(2)=22−2+1=3. Hence, t∈[34,3].t\in\left[\frac34,3\right].t∈[43​,3].

  1. Since t≥34>0t\ge \frac34>0t≥43​>0, we have ∣t∣=t.|t|=t.∣t∣=t. Therefore, f(x)=t+[t].f(x)=t+[t].f(x)=t+[t]. So we now minimize f=t+[t]f=t+[t]f=t+[t] for t∈[34,3].t\in\left[\frac34,3\right].t∈[43​,3].

  2. Consider intervals for ttt according to the greatest integer function.

  • If 34≤t<1\frac34\le t<143​≤t<1, then [t]=0[t]=0[t]=0, so f=t.f=t.f=t. Its minimum on this interval is attained at the smallest ttt, namely t=34t=\frac34t=43​, giving f=34.f=\frac34.f=43​.

  • If 1≤t<21\le t<21≤t<2, then [t]=1[t]=1[t]=1, so f=t+1≥2.f=t+1\ge 2.f=t+1≥2.

  • If 2≤t<32\le t<32≤t<3, then [t]=2[t]=2[t]=2, so f=t+2≥4.f=t+2\ge 4.f=t+2≥4.

  • If t=3t=3t=3, then f=3+[3]=3+3=6.f=3+[3]=3+3=6.f=3+[3]=3+3=6.

  1. Comparing all cases, the absolute minimum value is 34.\boxed{\frac34}.43​​.

  2. This corresponds to option A.

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