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Differential Calculus

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23 Jul 2008 14:12:48 IST

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Prove that $x^2 = x \sin x + \cos x$ has exactly two real roots

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Aishwarya Vardhan Chaturvedi

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23 Jul 2008 14:23:15 IST

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squaring both sides

x^4=x^2(1+sin2x)
x4-x2(1+sin2x)=0
x2(x2-1+sin2x)=0

x=0 or
x=sinx-cosx
=root(2){sin(x+pi/4)}

Hari Shankar

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23 Jul 2008 14:31:56 IST

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check your working please

Conjurer

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23 Jul 2008 19:05:36 IST

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Consider the function f(x) = x^2 - xsinx - cosx

f'(x) = 2x -xcosx = x(2-cosx)

f'(x) is increasing for x>0 and decreasing for x<0

Now f(0) = -1

f(- infinity) = +infinity

This is a continuous function for obvious reasons.So from - inf to 0, f(x) crosses x axis once and since it increase after 0 it also crosses it again. Hence exactly 2 roots.

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