It's immediate that P must be linear: p(y) = ay + b. So P(x^2) = a x^2 + b, and P(P(x^2) + 1) = a[P(x^2) + 1] + b = aP(x^2) + a + b = a^2x^2 + ab + a + b. a = 3, b = 2.
Hi mr.syber, can u pls help me solve, from l.h.s. If alpha + beta + gamma = 180 deg show that cos alpha/2 + cos(beta/2) + cos(gamma/2) = 4cos((beta + gamma)/4) * cos((alpha + gamma)/4) * cos((alpha + beta)/4)
Ok, what I did was differentiated. I got that P’(P(x^2)+1)(P’(x^2)2x)=18x This means that P’(P(x^2)+1)P’(x^2)=9 We can see that P’(x)=m, where m is some number. We can see this since p’(p(x^2)+1)=c0+c1x…cnx^n. and p’(x^2)=d0+d1x…dnx^k When we multiply those two together, we see that all have x terms except d0c0 which are actually the same value m. This must therefore mean c1…cn and d1…dn =0. So p’(x)=m We can solve for m immediately m^2=9, so m=3 or -3 We can solve for c by letting x=0 in our original equation. P(P(0)+1)=11. m(c+1)+c=0. We get that… c=-7 when m=-3, and c=2 when m=3. We therefore have P(x)=3x+2, and P(x)=-3x-7
It's immediate that P must be linear: p(y) = ay + b. So P(x^2) = a x^2 + b, and P(P(x^2) + 1) = a[P(x^2) + 1] + b = aP(x^2) + a + b = a^2x^2 + ab + a + b. a = 3, b = 2.
Using x^2 as a variable is a red herring: you could just as well say P[P(t)+1]=9*t + 11 for all (non-negative?) t.
Got 'em both!
Hi mr.syber, can u pls help me solve, from l.h.s.
If alpha + beta + gamma = 180 deg show that cos alpha/2 + cos(beta/2) + cos(gamma/2) = 4cos((beta + gamma)/4) * cos((alpha + gamma)/4) * cos((alpha + beta)/4)
yes Ican help you
@@Aymen-bt1ly Thanks. I am waiting for ur reply, i can solve from right side, but from left side, it is difficult to prove.
Dear@@waiphyoemg1668
Please order the steps
Yay! I solved this in 3 minutes by myself.
Yessssss !!!!
P(x)=3x+2...P(x)=-3x-7....mah??
We can sub y=x*2 to simplify the problem. 😋😋😋😋😋😋
good thinking
easy
Ok, what I did was differentiated. I got that P’(P(x^2)+1)(P’(x^2)2x)=18x
This means that P’(P(x^2)+1)P’(x^2)=9
We can see that P’(x)=m, where m is some number. We can see this since p’(p(x^2)+1)=c0+c1x…cnx^n.
and
p’(x^2)=d0+d1x…dnx^k
When we multiply those two together, we see that all have x terms except d0c0 which are actually the same value m.
This must therefore mean c1…cn and d1…dn =0. So p’(x)=m
We can solve for m immediately
m^2=9, so m=3 or -3
We can solve for c by letting x=0 in our original equation.
P(P(0)+1)=11.
m(c+1)+c=0. We get that…
c=-7 when m=-3, and c=2 when m=3. We therefore have
P(x)=3x+2, and P(x)=-3x-7
Wow!