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let – 1≤ p ≤1 .Show that the equation 4x 3 -3x-p=0 has a unique root in the interval [-1/2, 1] and identify it. Ans: x= cos {1/3 cos -1 (p)} let – 1≤ p ≤1 .Show that the equation 4x3-3x-p=0 has a unique root in the interval [-1/2, 1] and identify it. Ans: x= cos {1/3 cos-1(p)}
First find f(-0.5) and f(1) in terms of p.And suit the domain for p lyng in [-1,1]and check relation in f(-0.5) and f(1).Next step is to proove the function is continuous and differntiable.I would be and then apply Mean Value theorem to it. simply.
Firstly the question is wrong , the interval in which root will be unique is [1/2, 1] instead of [-1/2, 1].The equation can be written as 4x3-3x = pDraw the graph of of function y = f(x) = 4x3-3x. by finding its maxima and minima.The given equation will have unique root in any interval if line y = p cuts the graph y=f(x) at exactly one point.From the graph it would be clear that that this will be true if interval of x is [1/2, 1] Let the root be a, so it will satisfy the given equation we have4a3-3a = plet a = cosr4cos3r-3cosr = pcos3r = pr = 1/3cos-1pa = cosr = cos(1/3cos-1p) will be the required root
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