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Functions question

2023 · 30 Jan · Shift 2 · Q33
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  5. /2023 · 30 Jan · Shift 2 · Q33

Functions question

2023 · 30 Jan · Shift 2 · Q33

JEE MainMathematicsFunctionsMCQ+4 / −1
The range of the function f(x)=3−x+2+xf(x)=\sqrt{3-x}+\sqrt{2+x}f(x)=3−x​+2+x​ is :
  1. A
    [22,11][2 \sqrt{2}, \sqrt{11}][22​,11​]
  2. B
    [5,13][\sqrt{5}, \sqrt{13}][5​,13​]
  3. C
    [2,7][\sqrt{2}, \sqrt{7}][2​,7​]
  4. D
    [5,10][\sqrt{5}, \sqrt{10}][5​,10​]
View written solutionFree

Correct answer: D

  1. Find the domain

Given f(x)=3−x+2+xf(x)=\sqrt{3-x}+\sqrt{2+x}f(x)=3−x​+2+x​ For the square roots to be defined: 3−x≥0⇒x≤33-x\ge 0 \Rightarrow x\le 33−x≥0⇒x≤3 2+x≥0⇒x≥−22+x\ge 0 \Rightarrow x\ge -22+x≥0⇒x≥−2 So the domain is −2≤x≤3-2\le x\le 3−2≤x≤3

  1. Find the maximum/minimum of the function

Let f(x)=3−x+2+xf(x)=\sqrt{3-x}+\sqrt{2+x}f(x)=3−x​+2+x​ Differentiate: f′(x)=−123−x+122+xf'(x)=\frac{-1}{2\sqrt{3-x}}+\frac{1}{2\sqrt{2+x}}f′(x)=23−x​−1​+22+x​1​ Set f′(x)=0f'(x)=0f′(x)=0: 12+x=13−x\frac{1}{\sqrt{2+x}}=\frac{1}{\sqrt{3-x}}2+x​1​=3−x​1​ 2+x=3−x\sqrt{2+x}=\sqrt{3-x}2+x​=3−x​ 2+x=3−x2+x=3-x2+x=3−x 2x=12x=12x=1 x=12x=\frac12x=21​

  1. Check values at critical point and endpoints

  • At x=−2x=-2x=−2: f(−2)=5+0=5f(-2)=\sqrt{5}+0=\sqrt5f(−2)=5​+0=5​

  • At x=3x=3x=3: f(3)=0+5=5f(3)=0+\sqrt5=\sqrt5f(3)=0+5​=5​

  • At x=12x=\frac12x=21​: f(12)=52+52=252=10f\left(\frac12\right)=\sqrt{\frac52}+\sqrt{\frac52}=2\sqrt{\frac52}=\sqrt{10}f(21​)=25​​+25​​=225​​=10​

Thus,

  • minimum value =5=\sqrt5=5​
  • maximum value =10=\sqrt{10}=10​

Therefore the range is [5,10][\sqrt5,\sqrt{10}][5​,10​]

  1. Match with the options

Option D is [5,10][\sqrt5,\sqrt{10}][5​,10​] So the correct answer is D.

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