IPMAT IndoreAlgebra > Hard(1−q)log(p)−(1−p)log(q)p−q\dfrac{(1-q)\log(p)-(1-p)\log(q)}{p-q}p−q(1−q)log(p)−(1−p)log(q)(1−q)log(q)−(1−p)log(p)p−q\dfrac{(1-q)\log(q)-(1-p)\log(p)}{p-q}p−q(1−q)log(q)−(1−p)log(p)(1−q)log(p)+(1−p)log(q)p−q\dfrac{(1-q)\log(p)+(1-p)\log(q)}{p-q}p−q(1−q)log(p)+(1−p)log(q)(1−q)log(q)+(1−p)log(p)p−q\dfrac{(1-q)\log(q)+(1-p)\log(p)}{p-q}p−q(1−q)log(q)+(1−p)log(p)✅ Correct Option: 2Related questions:IPMAT Indore 2023A person standing at the centre of an open ground first walks 32 meters towards the east, takes a right turn and walks 16 meters, takes another right turn and walks 8 meters, and so on. How far will the person be from the original starting point after an infinite number of such walks in this pattern?IPMAT Indore 2024The sum of a given infinite geometric progression is 80 and the sum of its first two terms is 35. Then the value of nnn for which the sum of its first nnn terms is closest to 100, isIPMAT Indore 2019There are numbers a1,a2,a3,…,ana_1, a_2, a_3, \ldots, a_na1,a2,a3,…,an each of them being +1+1+1 or −1-1−1. If it is known that a1a2+a2a3+a3a4+…an−1an+ana1=0a_1 a_2 + a_2 a_3 + a_3 a_4 + \ldots a_{n-1} a_n + a_n a_1 = 0a1a2+a2a3+a3a4+…an−1an+ana1=0 then