ISI2016-MMA-3

Question

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

• A. $0$
• B. $1$
• C. $2$
• D. $\infty$

Comments

Same question asked in ISI 2015: https://gateoverflow.in/321864/isi2015-mma-13

https://math.stackexchange.com/questions/763481/no-of-real-solutions-of-the-equation-2-cos-fracx2-x6-2x-2-x

Answer 1

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)

Comments

Your answer is wrong. Since at $x=-1$, we have $e^1+e^{-1}>2=2\cos\left(\frac{(-1)^2+(-1)}{6}\right)$.

Answer 2

For real positive numbers a and b, the AM-GM Inequality for two numbers is: (a+b)/2 >= √ ab   => a+b >=  2√ ab 

so, 2^x+2^−x    ≥   2√(2^x *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.