The number of solutions of the equation 2 cos | vedclass

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(A) We know that for any real $x$,by the $A.M. \ge G.M.$ inequality,$\frac{5^x + 5^{-x}}{2} \ge \sqrt{5^x \cdot 5^{-x}} = 1$. This implies $5^x + 5^{-

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(A) We know that for any real $x$,by the $A.M. \ge G.M.$ inequality,$\frac{5^x + 5^{-x}}{2} \ge \sqrt{5^x \cdot 5^{-x}} = 1$. This implies $5^x + 5^{-x} \ge 2$. Also,we know that the range of the cosine function is $[-1, 1]$,so $\cos(e^x) \le 1$. This implies $2 \cos(e^x) \le 2$. For the equation $2 \cos(e^x) = 5^x + 5^{-x}$ to hold,both sides must be equal to $2$. This requires $5^x + 5^{-x} = 2$,which occurs only at $x = 0$. Substituting $x = 0$ into the left side,we get $2 \cos(e^0) = 2 \cos(1)$. Since $\cos(1) \approx 0.54$,$2 \cos(1) \approx 1.08 
eq 2$. Therefore,there is no value of $x$ that satisfies the equation.

The number of solutions of the equation $2 \cos(e^x) = 5^x + 5^{-x}$ is:

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