P = |22x - x^2 – 100|
Minimum value of P is 0
-(x^2-22x+100) = 0
x^2-22x+100=0
\(x= 22/2 ± \sqrt{121-100}\)
x= 11± √21
We know 4.5^2 = (4*5).25=20.25
√21>4.5
x=11+4.5xyz
x=15.5xyz or 6.4abc
Now minimum integral value of P occurs at the nearest integer value of 15.5xyz or 6.4abc, which are 6 and 16.
At x=6, P =4
At x=16, P=4
Since we have to find maximum value of P, answer must be 16.
Ora+b=22(equal to sum of roots); We have to choose integer a and b in such a way that product of them is closest to 100.
Use options to choose a and b. Go from the bottom as we need the maximum value
Option E- a=16; b=6; a*b=96 (closest one)
Option D- a=15; b=7; a*b = 105
Option C- a=11; b=11; a*b=121
Don't have to calculate further, as we already did in option E and D
(E)
Kritisood
Bunuel nick1816 is there any theory/generalization on how one can approach such questions? i seem to get lost often with these.