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The number of real roots of the equation $2 \cos \big(\frac{x^2+x}{6}\big)=2^x+2^{-x}$ is

  1. $0$
  2. $1$
  3. $2$
  4. $\infty$
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2 Answers

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2cos$(x^2+x)/6$ = $2^{x}+2^{-x}$          , min value of $2^{x}+2^{-x}   is  1+1=2,  when ( x=0 $)

2cos$(x^2+x)/6$ $\geq$ 2

cos$(x^2+x)/6$ $\geq$ 1 , cos0 =1 so, $(x^2+x)/6$ = 0

$x(x+1)$= 0    , x= 0 and -1 (two real roots)

 

Ans C (pls correct me if I am wrong)
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For real positive numbers a and b, the AM-GM Inequality for two numbers is: (a+b)/2 >= √ ab   => a+b >=  2√ ab 

so, 2x+2−x    ≥   2√(2x *2 -x) = 2

now from the question : 2cos((x2+x)/6) >= 2 

                                        cos((x2+x)/6) >=1

    as range of cosx = [-1,1] so value of  cos((x2+x)/6) cannot be >1 .

means value of  cos((x2+x)/6)=1  = one solution

so answer is option B

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