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The real root of the equation 5x – 2 cosx – 1 = 0 (up to two decimal accuracy) is _______.
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- 0.5424
- 0.5423
- 1.5424
- 1.5423
Correct Option: A
Let f(x) =5x – 2 cos x – 1
⇒ f'(x) = 5 + 2 sin x
f(0) = – 3; f(1) = 2.9
By intermediate value theorem roots lie between 0 and 1.
Let x0 = 1 rad = 57.32º
By Newton Raphson method,
Xn+1 = Xn - | f'(xn) |
⇒ xn+1 | 5 + 2Sin xn |
⇒ x1 = 0.5632
⇒ x2 = 0.5425
⇒ x3 = 0.5424