JEE MAINS 2018 For the equation `40/[x-1]-160/[x-4]-200/[x-5]+320/[x-8]=6x^2-27x` (A) Number o...
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Channel: Doubtnut · 2018 · Education
To ask Unlimited Maths doubts download Doubtnut from - https://goo.gl/9WZjCW For the equation `40/[x-1]-160/[x-4]-200/[x-5]+320/[x-8]=6x^2-27x` (A) Number of real solutions of above equation is 3 (B) If E denotes the product of non-zero real or complex roots of the equation, then sum of divisors of E is 2904 (C) If S denotes the set of all real roots of the equation then, sum of elements of S taken two at a time is 81 (D) If `alpha_1,alpha_2 in R` be two roots of the equation such that `log_[alpha_2](2alpha_1)` is defined then it must be 1. JEE Main 2018,JEE mains 2018 solutions,JEE mains cut off marks,JEE mains answer key,JEE mains Questions, JEE Mains previous year questions,Cheat code for JEE mains,JEE main crash course, Jee Mains preparation in last 48 hours, Instructions for JEE main exam, JEE mains online,How to score good marks in JEE, Chemistry question of JEE mains, Physics questions of JEE Mains, Last 10 year Jee mains questions,IIT JEE, JEE advanced
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