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Question: Answered & Verified by Expert

If fx=x32x2+11+3x3x2+22xx3+6x3x4x22 for all x, then 2f0+f'0 is equal to

MathematicsDeterminantsJEE Main
Options:
  • A 48
  • B 24
  • C 42
  • D 18
Solution:
1591 Upvotes Verified Answer
The correct answer is: 42

Given: fx=x32x2+11+3x3x2+22xx3+6x3-x4x2-2

f0=01120604-2

f0=0+4+8

f0=12

f'x=x32x2+11+3x3x2+22xx3+63x2-102x+x32x2+11+3x6x23x2x3-x4x2-2+3x24x33x2+22xx3+6x3-x4x2-2

f'0=011206-100+01102004-2+00320604-2

f'0=0-0+6+0+0+0+0+38

f'0=-6+24=18

2f0+f'0=42

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