The Gateway to Computer Science Excellence
0 votes
12 views

The number of real roots of the equation

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

  1. $0$
  2. $1$
  3. $2$
  4. infinitely many
in Numerical Ability by Veteran (425k points)
recategorized by | 12 views

1 Answer

0 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 $x = -1$

by Boss (13.4k points)
0

@ankitgupta.1729

Can you please check-out this solution?

Related questions

Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
50,644 questions
56,517 answers
195,583 comments
101,144 users