1 votes 1 votes The number of real roots of the equation $2 \cos \big(\frac{x^2+x}{6}\big)=2^x+2^{-x}$ is $0$ $1$ $2$ $\infty$ Quantitative Aptitude isi2016-mmamma trigonometry quadratic-equations roots + – go_editor asked Sep 13, 2018 edited Nov 19, 2019 by Lakshman Bhaiya go_editor 492 views answer comment Share Follow See 1 comment See all 1 1 comment reply Shiva Sagar Rao commented May 7, 2021 reply Follow Share 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 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes 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) Snorlax answered Sep 15, 2020 Snorlax comment Share Follow See 1 comment See all 1 1 comment reply NastyBall commented Jun 17, 2021 reply Follow Share Your answer is wrong. Since at $x=-1$, we have $e^1+e^{-1}>2=2\cos\left(\frac{(-1)^2+(-1)}{6}\right)$. 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes 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. arvin answered Sep 13, 2018 arvin comment Share Follow See all 0 reply Please log in or register to add a comment.