Let a and b be two integers such that 10a+b=5 and P(x)= x+ax+b The integer n such that P(10) - Maths - Relations and Functions

Question

Let a and b be two integers such that 10a+b=5 and P(x)= x+ax+b The
integer n such that P(10) P(11)=P(n) is

Devesh Kumar · Accepted Answer

Let a and b be two integers such that 10a+b=5 and P(x)= x+ax+b
 The integer n such that P(10)
P(11)=P(n) is what?P(10)=10+10a+b=10+5=15  …1     10a+b=5P(11)=11+11a+b=11+a+5=16+a   …2     10a+b=5Now,P(10)
P(11)=P(n)⇒15×16+a =n+na+b=n+n-10a+5   10a+b=5⇒240+15a=n+na-10a+5⇒n+na-25a-235=0⇒n=235+25aa+1  …3We see here that 'n' is just a special case of 'x' such that P(n)=P(10)
P(11)so, we can put eq 3 in 110=235+25aa+110a+10=235+25a-15a=225a=-15using  10a+b ⇒b=155for solving for a & b we assumed n=10⇒P(n)=P(10)
P(11)⇒P(10)=P(10)
P(11)⇒P(11)=1Now, with these values of a & b if P(11)=1, this would mean n=10 P(11)=16+a                  Using eq 2  P(11)=16+-15           putting a=-15P(11)=1 Therefore, n=10

Let a and b be two integers such that 10a+b=5 and P(x)= x+ax+b. The integer n such that P(10).P(11)=P(n) is