recategorized by
758 views
1 votes
1 votes

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 by

2 Answers

1 votes
1 votes

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 $x = -1$

0 votes
0 votes
as cosx=(e^ix+e^-ix)/2 so I got (B).

Related questions

1 votes
1 votes
2 answers
1
Arjun asked Sep 23, 2019
604 views
Let $\cos ^6 \theta = a_6 \cos 6 \theta + a_5 \cos 5 \theta + a_4 \cos 4 \theta + a_3 \cos 3 \theta + a_2 \cos 2 \theta + a_1 \cos \theta +a_0$. Then $a_0$ is$0$$1/32$$...
1 votes
1 votes
1 answer
2
Arjun asked Sep 23, 2019
593 views
The number of real solutions of the equations $(9/10)^x = -3+x-x^2$ is$2$$0$$1$none of the above
0 votes
0 votes
1 answer
4
Arjun asked Sep 23, 2019
569 views
If $f(x) = \dfrac{\sqrt{3}\sin x}{2+\cos x}$, then the range of $f(x)$ isthe interval $[-1, \sqrt{3}/2]$the interval $[- \sqrt{3}/2, 1]$the interval $[-1, 1]$none of the ...