JEE Main 2023MathematicsFunctionsHardMCQ

JEE Main 2023 — Functions Question with Solution

JEE Main 2023 (31 Jan Shift 2)

Question

Let f:R-2,6→R be real valued function defined as fx=x+2x+1x2-8x+12. Then range of f is

Choose an option

Show full solutionCorrect option: B
Correct answer
B-∞,-214∪[0,∞)

Step-by-step explanation

Given,

f:R-2,6→R be real valued function defined as fx=x+2x+1x2-8x+12

Now let, y=x2+2x+1x2-8x+12=x+12x-2x-6       …1

Now differentiating the above function we get,

⇒dydx=-2x+15x-16x-22x-62

Now by wavy curve method we get,

So Graph of y=x+12x-2x-6 for given domain will be,

 

So, from graph range is y∈-∞,214∪[0,∞)

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About this question

This is a previous-year question from JEE Main 2023, covering the Functions chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.