Maths-
General
Easy

Question

Let f left parenthesis x right parenthesis equals fraction numerator ln invisible function application x over denominator x end fraction for all x greater than 0 comma then
Statement-1: f open parentheses fraction numerator 5 over denominator 2 end fraction close parentheses greater than f open parentheses fraction numerator 9 over denominator 2 end fraction close parentheses because
Statement 2 : f open parentheses x subscript 1 end subscript close parentheses greater than f open parentheses x subscript 2 end subscript close parentheses for all x subscript 1 end subscript element of open parentheses 2 comma 4 close parentheses text  and  end text x subscript 2 end subscript element of open parentheses 0 comma 2 close parentheses union open parentheses 4 comma infinity close parentheses

  1. Statement-1is True, Statement-2 is True;Statement-2 is a correct explanation for Statement-1    
  2. Statement-1 is True, Statement-2 is True;Statement-2 is not a correct explanation    
  3. Statement -1 is True, Statement -2 is False    
  4. Statement -1 is False, Statement -2 is True    

The correct answer is: Statement-1is True, Statement-2 is True;Statement-2 is a correct explanation for Statement-1



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