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GATE-2016
MATHEMATICS
CE/PAPER 1/Q1
Now f(x) = 3x –ex + sinx
So f’(x) = 3 – ex + cosx
Now f(x) = 3x –ex + sinx
So f’(x) = 3 – ex + cosx
Given that x0 = 0.333
Now f(x) = 3x –ex + sinx
So f’(x) = 3 – ex + cosx
Given that x0 = 0.333
So f(x0) = -0.069
Now f(x) = 3x –ex + sinx
So f’(x) = 3 – ex + cosx
Given that x0 = 0.333,f(x0) =-0.069
The Newton-Raphson formula is
as follows x1 = x0 – f’(x0)/f(x0)
Now f(x) = 3x –ex + sinx
So f’(x) = 3 – ex + cosx
Given that f(x0) = 0.333,f(x0)=--0.069
The Newton-Raphson formula is
as follows x1 = x0 – f’(x0)/f(x0)
We have f’(x0)=2.55
Now f(x) = 3x –ex + sinx
So f’(x) = 3 – ex + cosx
Given that f(x0) = 0.333,f(x0)=--0.069
The Newton-Raphson formula is
as follows x1 = x0 – f’(x0)/f(x0)
We have f’(x0)=2.55
So x1= 0.333+ 0.069/2.55=0.360
Gate 2016 ce paperi mathematicsq1
Gate 2016 ce paperi mathematicsq1

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Gate 2016 ce paperi mathematicsq1

  • 2.
  • 3. Now f(x) = 3x –ex + sinx So f’(x) = 3 – ex + cosx
  • 4. Now f(x) = 3x –ex + sinx So f’(x) = 3 – ex + cosx Given that x0 = 0.333
  • 5. Now f(x) = 3x –ex + sinx So f’(x) = 3 – ex + cosx Given that x0 = 0.333 So f(x0) = -0.069
  • 6. Now f(x) = 3x –ex + sinx So f’(x) = 3 – ex + cosx Given that x0 = 0.333,f(x0) =-0.069 The Newton-Raphson formula is as follows x1 = x0 – f’(x0)/f(x0)
  • 7. Now f(x) = 3x –ex + sinx So f’(x) = 3 – ex + cosx Given that f(x0) = 0.333,f(x0)=--0.069 The Newton-Raphson formula is as follows x1 = x0 – f’(x0)/f(x0) We have f’(x0)=2.55
  • 8. Now f(x) = 3x –ex + sinx So f’(x) = 3 – ex + cosx Given that f(x0) = 0.333,f(x0)=--0.069 The Newton-Raphson formula is as follows x1 = x0 – f’(x0)/f(x0) We have f’(x0)=2.55 So x1= 0.333+ 0.069/2.55=0.360