Download the app

Questions  

The maximum value of y=(x3)2+x222x2+x212 is 

a
3
b
10
c
25
d
None of these

detailed solution

Correct option is B

y=f(x)=x2−22+(x−3)2−x2+x2−12 Note that the first term describes the distance between Px,x2 and A(3,2) where as the second term describes the  distance between Px,x2 and B(0,1) .  Now PA−PB≤AB for possible positions of P Hencef(x)max= distance between AB=9+1=10

Talk to our academic expert!

+91

Are you a Sri Chaitanya student?


Similar Questions

A rectangular billiard table has vertices at P0,0,Q0,7,R10,7, and S10,0. A small billiard ball starts at M3,4moves in a straight line to the top M3,4, moves in a straight line to the top of the table, bounces to the right side of the table, and then comes to rest at N7,1. The y-coordinate of the point where it is the right side is 

 


phone icon
whats app icon