badge image

Enroll For Free Now & Improve Your Performance.

×
User Icon
User Icon
User Icon
User Icon
User Icon

Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
Grade: 12

                        

Let A (a, 0) and B(b, 0) are the points on the curve y = x 2 –6x +5 in the cartesian coordinate system amd C is a point on arc AB , then the maximum value of | CA×CB| is

6 months ago

Answers : (1)

Aditya Gupta
2065 Points
							
obviously a and b are roots of eqn y = x2–6x +5 since y is 0 at these pts.
since y = x2–6x +5= (x – 5)(x – 1)
hence A(1, 0) and B(5, 0)
clearly |CA×CB|= |CA.CB.sin(t)| where t is angle ACB.
= 2*| ½ CA.CB.sin(t)|
but ½ CA.CB.sin(t) = A= area of triangle ABC (std formula)
so, |CA×CB|= 2*A. to maximise |CA×CB|, we need to maximise A. since only pt C is variable, while AB is constant (=4), so we can write A= ½ AB*h, where h is the height of perpendicular from C on base AB. since AB is constant, we need to maximise h in order to maximise A. You ll note after drawing the graph of curve y = x2–6x +5 that its min value occurs when y’ = 0 or at x=3. since h= magnitude of y coordinate of C, h is max when C’s magnitude of y coordinate is max, which clearly happens at x=3.
when x=3, y= – 4.
hence h|max= |-4|= 4
whence A|max= ½ *AB*h= ½ *4*4= 8
so, CA×CB|max= 2*A|max
= 2*8
16
KINDLY APPROVE :))
6 months ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

  • 731 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

  • 19 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details