| I. If \[x,\,\,y\] and z are all different real numbers, then \[\frac{1}{{{(x-y)}^{2}}}+\frac{1}{{{(y-z)}^{2}}}+\frac{1}{{{(z-x)}^{2}}}=\,{{\left( \frac{1}{x-y}+\frac{1}{y-z}+\frac{1}{z-x} \right)}^{2}}\] |
| II. \[{{\log }_{3}}\,x\cdot \,{{\log }_{4}}\,x\,\cdot \,{{\log }_{5}}\,\,x\] \[=\,\,({{\log }_{3}}\,x\cdot \,{{\log }_{4}}\,x)\] \[+\,({{\log }_{4}}\,x\cdot \,{{\log }_{5}}\,x)+\,({{\log }_{5}}\,x\cdot {{\log }_{3}}\,x)\] is true for exactly for one real value of x. |
| III. A matrix has 12 elements. Number of possible orders it can have a six. |
A) exactly one statement is correct
B) exactly two statements are correct
C) all statements are incorrect
D) all statements are correct
Correct Answer: B
Solution :
I. RHS \[{{K}_{a}}=\frac{[HCO{{O}^{-}}]\,[{{H}^{+}}]}{[HCOOH]}\] \[[{{K}_{a}}=\] \[(-3.7)=1.995\times {{10}^{-4}}]\] \[1.995\times {{10}^{-4}}=\frac{x(0.1)}{0.15}\] \[x=\frac{1.995\times {{10}^{-4}}\times 0.15}{0.1}\] = LHS II. \[[HCO{{O}^{-}}]=2.99\times {{10}^{-4}}\] \[\therefore \] \[=\frac{2.99\times {{10}^{-4}}}{0.15}\times 100\] \[I=\frac{pV}{RT}=\frac{5\times 9}{RT}\] \[II=\frac{10\times 6}{RT}\] \[\therefore \] \[=\frac{(45+60)}{RT}=\frac{105}{RT}\] \[\therefore \] \[\frac{nRT}{{{V}_{Total}}}\] \[=\frac{105}{RT}=\frac{RT}{15}\]or \[=7\,\text{atm}\] \[\frac{{{p}_{0}}-{{p}_{s}}}{{{p}_{s}}}=\frac{{{W}_{B}}}{{{W}_{B}}}\times \frac{{{M}_{A}}}{{{W}_{A}}}\] \[{{W}_{B}}=\] or \[{{M}_{B}}=\] So, \[{{W}_{A}}=\] and 60 are the required solution. III. Possible orders are \[{{M}_{B}}=\]
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