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 2025Let S1={100,105,110,115,...}S_1 = \{100, 105, 110, 115, ... \}S1={100,105,110,115,...} and S2={100,95,90,85,...}S_2 = \{100, 95, 90, 85, ... \}S2={100,95,90,85,...} be two series in arithmetic progression. If aka_kak and bkb_kbk are the kkk-th terms of S1S_1S1 and S2S_2S2, respectively, then ∑k=120akbk\sum_{k=1}^{20} a_k b_k∑k=120akbk equals __________.IPMAT Indore 2020If 112+122+132+…\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \ldots121+221+321+… up to ∞=π26\infty = \frac{\pi^2}{6}∞=6π2, then the value of 112+132+152+…\frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \ldots121+321+521+… up to ∞\infty∞ 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 _______