Consider the function
$$ f(x) = 2\cos((x^2+x)/6) - 2^x - 2^{-x}. $$
Clearly $x= 0$ is a root of equation so number of real roots can not be 0 but how to figure out question how many ? From where does this observation start?
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What equation?? – John Doe May 09 '17 at 04:12
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Oops I forgot to write equation but now u can see . – Gopesh Thakur May 09 '17 at 04:21
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Should it be tagged as abstract algebra? – Zuriel May 09 '17 at 04:36
2 Answers
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The Rhs is symmetric about y axis and its minimum value is 2. The graph is increasing on both the sides . While the maximum of Lhs is $2$ as $-1\leq cos (...)\leq 1$ . Thus the equation has $1$ real root and that is at $x=0$
Archis Welankar
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Hint: $\cos a \le 1\,$ for any $\forall a \in \mathbb{R}\,$, and $b + \frac{1}{b} \ge 2\,$ for any $\forall b \gt 0\,$ with equality iff $b=1\,$. So:
$$ 2 \cos\left(\frac{x^2+x}{6}\right) \;\le\; 2 \;\le\; 2^x + \frac{1}{2^x} $$
For the equality to hold, both sides must equal $2\,$, therefore $\,b = 2^x = 1\,$, so $\,x=\,\cdots$
dxiv
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