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 2019The number of terms common to both the arithmetic progressions 2,5,8,11,...,1792, 5, 8, 11, ..., 1792,5,8,11,...,179 and 3,5,7,9,...,1013, 5, 7, 9, ..., 1013,5,7,9,...,101 isIPMAT Indore 2019Let α,β\alpha, \betaα,β be the roots of x2−x+p=0x^2 - x + p = 0x2−x+p=0 and γ,δ\gamma, \deltaγ,δ be the roots of x2−4x+q=0x^2 - 4x + q = 0x2−4x+q=0 where p and q are integers. If α,β,γ,δ\alpha, \beta, \gamma, \deltaα,β,γ,δ are in geometric progression then p+qp + qp+q isIPMAT Indore 2021It is given that the sequence {xnx_nxn} satisfies x1=0,xn+1=xn+1+2√(1+xn)x_1 = 0, x_{n+1} = x_n + 1 + 2√(1+x_n)x1=0,xn+1=xn+1+2√(1+xn) for n=1,2,...n = 1,2,...n=1,2,... Then x31x_{31}x31 is _______