20 views

The number of real roots of the equation

$$2 \cos \left( \frac{x^2+x}{6} \right) = 2^x +2^{-x} \text{ is }$$

1. $0$
2. $1$
3. $2$
4. infinitely many

recategorized | 20 views

Answer: $A$

Minimum value of $2^x + 2^{-x} = 2$

$$\implies 2\cos\bigg(\frac{x^2+x}{6}\bigg) = 2^x + 2^{-x}$$

$$\implies 2\cos\bigg(\frac{x^2+x}{6}\bigg) \ge 2$$

Now, $\because \cos 0 = 1$

$$\implies \frac{x^2+x}{6} = 0 \implies x(x+1) = 0 \implies x = 0, x = -1$$



So, $x = 0$ is the only solution as the equation is not satisfying at $x = -1$

by Boss (19.1k points)
0

@ankitgupta.1729

Can you please check-out this solution?