Question

Let F :P3 + R2 be a linear transformation with three values given below. F(22) = 11 F(x) = 1 f(1) = 1 Find F((x – 2)2) B. Find an element of ker(F)| A.

Answer 1

Now, IR² be a with F(x²) F(x2) - I FI IP3 Let P(x) c Linear Transformation [3] , Fle) = [] F4-6 = ax+bx+c EP F (ax²+bx+ 2.F(x2)+b.F(x) +C. F(1) [as Fis Linear > Flax²+bx+c) = a [] +c.[1] → F (ax2+bx+c Now, 1 3 tb 1 +c) atabtc 3 at btc

a+2btc 3 atbtc F(P(x)) = B) Let P, (x) = ax²+bx+4% E Kerf. Then. F (Pilz)) = [:] → an+26+47-[:] 3a+byte, Then, at 2b + c = 0 34+b,+C; = 0 LTD by Cross multiplication in (1) 8 (11) we get bin C =k where KER (1) Now, by ai 1 2 -5 G= -5K Then, at (2 +22-5) Then, a=k , b = 2K , Pi(x) = kx? + 2k2 -5k Then, ker F = L {[x2+2x–5)} A) Now, F((x-2)2) as F is Linear = F(x2-4x+4) - F(x2) – 4-F(x) +4.F(1) [as 0] -4/} ] +4[1] 1 - 3 3