Aditya Gupta
Last Activity: 4 Years ago
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 :))
