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Suppose that for infinitely many primes p the polynomial pf g has a rational root Prove that f has a rational root A5 Find all functions f R→ Rthat satisfy the conditions f(1xy)− f(xy) = f(x)f(y) for all x,y ∈ R and f(−1) 6= 0 A6 Let f N → N be a function, and let fm be fO n e c o p y o f a g f i r i P B o f e m a i l s (3 p a g e s ) f g i n c l u d e an e m a il fro m FBIHQ , C T D , d allefl L W /2 1 /2 0 0 5 to A SC , e t a l, b l b 6 b 7 C b 2 b 7 E D e ta ils (R' sclL aiL x i C fc S > b 6 b 7 C A L L I N F O R M A T I O N C O N T A I N E DJun 07, 19 · f B A ̒ ɂ͂ ̃y g n E X 1 7000 X N G A E t B g( 1579 g ) A 12 x b h A Ǝ ۂ y ɑ傫 Ƃ 낪 ܂ A ۂɂ̓t A E v ɂ A ̖ʐς 1 79 X N G A E t B g( 936 g ) A e X v ܂߂ O ̖ʐς 5730 X N G A E t B g( 5325 g ) ŁA 5 x b h A5 o X A3 p E _ E ŁA ̃ B O E ƃ_ C j O E G A i 镨 B ̂ 1912 N Ă ꂽ j ̂ ɁA V Ɍ z ꂽ R
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Y e t , d e s p i t e t h e l a r g e d rop in required balances, Figure 2 shows 2 t h a t f u n d s r a t e v o l a t i l i t y i n c r e a s e d o n l y s l i g h t l y , a n d remained well below its level of early 1991 W h y was the low level of required reserve balances in 1991 accompanied by a sharpP r a y f o r r e v iv al R E A D P s a lm 85 6 Wh ere do y o u n eed r evival in y o ur s p iritu al l ife?F L(P;f) >0 But this is a contradiction to R b a f(x)dx= 0Therefore f= 0 for all x2a;b #4 If f(x) g(x) on a;b, then R b a f(x)dx R b a g(x)dx Let P be a partition of a;b then L(f;P) L(g;P) Since f is integrable we know R b a f = b a f = supfL(f;P)g supfL(g;P)g= R b a = R b a g
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> @ < a b / @ 0 c d e f a / / g h i i j k k j h l m n o l p g q o j n ( / " $ 0 " $ 1 2 3( '3 a) truth table b) sop y0 = (a'b'c'd)(a'b'cd')(a'bc'd')(a'bcd)(ab'c'd')(ab'cd)(abc'd)(a bcd') y1= (a'b'cd)(a'bc'dThis implies x ∈ A∩B and y ∈ C, so p = (x,y) ∈ (A∩B)×C This proves (A×C)∩(B ×C) ⊆ (A∩B)×C Together the two inclusions prove the claimed equality 1 2 114 (e) Prove that A∩B and A\B are disjoint, and that A = (A∩B)∪
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Then there exists a bijection f∶A→ B De ne a function g∶P(A) → P(B) by g(X) ={f(x)Sx∈X} I claim gis a bijection To see that this is a bijection, it is enough to write down an inverse De ne h∶P(B) → P(A) by h(Y) ={f−1(y)Sy∈Y} This de nition makes sense because fis a bijection, so f−1 actually exists For any X∈P(A) we haveFYBA / 6 The structure will be as under Bachelor of Arts (BA) I The student joining the First Year BA Course shall offer six subjects as follows ( i ) The student can offer not more than one subject from one group ( ii) Subject in Group "A" is compulsory (iii) The student has to offer at least one languageThe length S of the curve y x from a to b is given by dx dx dy S b a 2 1 where 7 2 b a giving dx dx dy S 7 2 2 1 2 3 2 x x dx dy y x x x x Thus, S x x dx 7 2 1 2 2
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