the maximum possible value of (6^x-36^x) over all real values of x is
options 1) infinity 2)0 3)1 4)1/4 5) 1/2

• 1/4
  (6^x-36^x)
  
  Differentiating (6^x-36^x) w.r.t x and equating to zero, we get
  
  6^x * log x - 36^x * log 36 = 6^x -(6^2x)*(2log 6) =0
  6^x =1/2
  putting this value of 6^x in (6^x-36^x)
  1/2-1/4= 1/4
• 12 years agoHelpfull: Yes(19) No(3)
• y = 6^x-36^x
  let p=6^x
  then y= p-p^2
  y= 1/4-(p^2-p+1/4)
  y= 1/4-(p+1/2)^2
  for max y, (p+1/2)=0
  Max y=1/4-0
  Max y=1/4
• 12 years agoHelpfull: Yes(12) No(0)
• 4) 1/4
  
  explanation -
  let 6^x = y (where y>=0), 0 y = 1/2
  again, by taking double derivative
  d^2f(y)/dy^2 = -2
  so, f(y) has max value at y = 1/2
  => so max value = 1/2-(1/2)^2 = 1/4
• 12 years agoHelpfull: Yes(2) No(3)
• ans. is 0..bcoz 0 is a real no.
• 12 years agoHelpfull: Yes(1) No(7)
• option no. 3
  by equating single derivative of given expression to zero ,we can get extreme values of the expression
  by substituting this in double derivative we can get max value
• 12 years agoHelpfull: Yes(1) No(1)
• let, y=(6^x-36^x)
  for maximum value of y
  dy/dx=0
  6^xln6-36^xln36=0
  x=log(12) with base 6
  after putting this in y we get y=1/2 ans hence option 5 is correct
• 12 years agoHelpfull: Yes(0) No(6)