ACJC NJC RVHS ACJC-NJC-RVHS H2 FM Prelim 2022 P2 Solutions MS

1. Let Pn be the statement ( )( )( ) ( ) ( )11 1 2 1 3 1 1 2 nnx x x nx x ++ + + +  + for all positive integers n. When 1,n= LHS 1 x=+ RHS (1)(2)1 1 (or ) RHS2 xx= + = +  = Thus 1P is true. Suppose Pk is true for some positive integer k. Then LHS of 1Pk+ ( )( )( ) ( ) ( )( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) 2 2 2 2 1 1 1 2 1 3 1 1 1 11 1 1 2 111 1 1 22 11 1 1 1 22 12 11 1 0 22 RHS of Pk x x x kx k x kk x k x k k k kk x x x k x kk x k k x kk x k k x + = + + + + + + + + + +  ++ = + + + + +   = + + + + + ++  + +   = Thus Pk is true implies that 1Pk+ is true. Since 1P is true and Pk is true implies that 1Pk+ is true, Pn is true for all positive integers n by mathematical induction. ( ) ( ) 1 ln 1 1 2 3ln 1 ln 1 ln 1 ln 1 1 2 3ln 1 1 1 1 1 11ln 1 (use in first part)2 1ln 1 2 3ln 2 ln 3 n k k n n n n n n n n n n n nn xnn n n n = +        = + + + + + + + +                    = + + + +           + +  =  +=+  +=  = + −  ln 2 (shown) 2. (a) det( ) 0A A is invertible but 00 00   is non- invertible.

OR 00det 000   00 00   is not in the set. Since the set does not contain the zero matrix, it is not a subspace of ( )22M . (b) Let U denote the set of all 22 matrices B such that T =−BB . 0 0 0 0 0 0 0 0 0 0 0 0 T      = =−           Hence 00 00 U  and U is non-empty. Consider any two elements B , C U . ( ) ( ) ( ) T TT+ = + =− + − =− +B C B C B C B C U + BC (Closure under addition) Consider any element B U and k . ( ) ( ) ( ) T Tk k k k= = − =−B B B B kUB (Closure under scalar multiplication) Hence U is a subspace of ( )22M . 3. dsec sec tan .d xx   =  = d d d d sec tan .d d d d y y x y xx == ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 22 23 22 2 2 2 2 3 2 2 2 2 2 2 2 2 4 2 2 2 2 4 2 2 2 d d d d sec tan sec tan secd d d d dd sec tan sec tan secdd dd sec sec 1 sec sec 1 secdd dd sec sec sec 2sec 1dd dd 21dd y y x y xx yy xx yy xx yy xx yyx x x x xx                    = + + = + + = − + − + = − + − = − + − ( ) ( ) 22 32 22 1 d d d 2 1 .d d d y y y x x xx x x  = − + − Substituting into the differential equation in x and y:

2 23 2 22 2 22 2 2 2 1 d 2 d d2 d d2 d sec d 2cos (shown)d y ky x x x y ky x y ky y ky      += += += += Characteristic equation: ( ) 2 0 i 0 mk m k k += =  Complementary function: ( ) ( ) 11 cos sin cos sec sin sec . cy A k B k A k x B k x  −− =+ =+ Since 22cos cos 2 1=+ , particular integral: cos 2 sin 2py C D E = + + 2 sin 2 2 cos 2 4 cos 2 4 sin 2 . p p y C D y C D    =− + =− − Substituting into the new differential equation: ( ) ( ) ( ) 4 cos 2 4 sin 2 cos 2 sin 2 cos 2 1 4 cos 2 4 sin 2 cos 2 1 C D k