Maths-
General
Easy

Question

A tangent to the curve, Y=f(x) at P(x, Y) meets X‐axis at A and Y‐axis at B If AP:BP=1: 3 and f(1)=1, then the curve also passes through the point

  1. left parenthesis fraction numerator 1 over denominator 2 end fraction comma 4 right parenthesis    
  2. left parenthesis fraction numerator 1 over denominator 3 end fraction comma 24 right parenthesis    
  3. left parenthesis 2 comma fraction numerator 1 over denominator 8 end fraction right parenthesis    
  4. left parenthesis 3 comma fraction numerator 1 over denominator 28 end fraction right parenthesis    

The correct answer is: left parenthesis 2 comma fraction numerator 1 over denominator 8 end fraction right parenthesis


    We have
    fraction numerator left parenthesis gamma minus gamma subscript 2 end subscript right parenthesis over denominator left parenthesis x minus x subscript 1 end subscript right parenthesis end fraction equals f to the power of ´ end exponent left parenthesis x subscript 1 end subscript right parenthesis rightwards double arrow gamma minus gamma subscript 1 end subscript equals f to the power of ´ end exponent left parenthesis x subscript 1 end subscript right parenthesis left parenthesis x minus x subscript 1 end subscript right parenthesis
    When gamma equals 0 colon fraction numerator negative gamma subscript 1 end subscript over denominator f ’ left parenthesis x subscript 1 end subscript right parenthesis end fraction equals x minus x subscript 1 end subscript rightwards double arrow chi equals x subscript 1 end subscript minus fraction numerator gamma subscript 1 end subscript over denominator f ’ left parenthesis x subscript 1 end subscript right parenthesis end fraction.
    Therefore, point A is A left parenthesis x subscript 1 end subscript minus fraction numerator gamma subscript 1 end subscript over denominator f ’ left parenthesis x subscript 1 end subscript right parenthesis end fraction comma blank 0 right parenthesis ,
    When x equals 0 colon gamma minus gamma subscript 1 end subscript equals f to the power of ´ end exponent left parenthesis x right parenthesis times left parenthesis negative x subscript 1 end subscript right parenthesis rightwards double arrow gamma equals gamma subscript 1 end subscript minus x subscript 1 end subscript f to the power of ´ end exponent left parenthesis x subscript 1 end subscript right parenthesis Therefore, point B is B left parenthesis O comma blank gamma subscript 1 end subscript minus x subscript 1 end subscript f to the power of ´ end exponent left parenthesis x subscript 1 end subscript right parenthesis right parenthesis .
    Point P divides AB in the rratio 1: 3.
    chi subscript 1 end subscript equals fraction numerator left square bracket 3 left parenthesis x right ceiling comma negative fraction numerator gamma subscript 1 end subscript over denominator f left parenthesis x subscript 1 end subscript right parenthesis end fraction right parenthesis right square bracket over denominator 4 end fraction
    gamma subscript 1 end subscript equals fraction numerator gamma subscript 1 end subscript minus x subscript 1 end subscript f to the power of ´ end exponent left parenthesis x subscript 1 end subscript right parenthesis over denominator 4 end fraction
    Therefore,
    4 gamma subscript 1 end subscript equals gamma subscript 1 end subscript minus x subscript 1 end subscript f to the power of ´ end exponent left parenthesis x subscript 1 end subscript right parenthesis
    rightwards double arrow f to the power of ´ end exponent left parenthesis x subscript 1 end subscript right parenthesis equals to the power of stack negative 3 gamma subscript 1 end subscript with _ below end exponent rightwards double arrow f to the power of ´ end exponent left parenthesis x right parenthesis equals to the power of stack negative 3 gamma with _ below end exponent
    Now,
    fraction numerator d gamma over denominator d x end fraction equals fraction numerator negative 3 gamma over denominator chi end fraction rightwards double arrow fraction numerator d gamma over denominator gamma end fraction equals fraction numerator negative 3 d x over denominator chi end fraction
    On integrating, we get
    In gamma equals negative 3 I n x plus C rightwards double arrow gamma equals k x to the power of negative 3 end exponent
    gamma left parenthesis 1 right parenthesis equals 1 rightwards double arrow k equals 1
    gamma equals fraction numerator 1 over denominator chi to the power of 3 end exponent end fraction
    When we substitute the values from the given options, only option (C) ssatisfies the above equation.
    Hence, the correct answer is option (c).

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