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Question: Answered & Verified by Expert
Let f:R-2,6R be real valued function defined as fx=x+2x+1x2-8x+12. Then range of f is
MathematicsFunctionsJEE MainJEE Main 2023 (31 Jan Shift 2)
Options:
  • A -,-214214,
  • B -,-214[0,)
  • C -,-2140,
  • D -,-214[1,)
Solution:
2980 Upvotes Verified Answer
The correct answer is: -,-214[0,)

Given,

f:R-2,6R 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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