Page 1 Question 1. If p(x) = x 2 â€“ 2 2 x + 1, then find p () 22 . Solution: Since, p(x) = x 2 â€“ 2 2 x + 1 Then p () 22 = () 2 22 ?? ?? â€“ () 22 22 1 ?? + ?? = 4 (2) â€“ 4 (2) + 1 = 8 â€“ 8 + 1 = 1 Question 2. If a + b + c = 9, and ab + bc + ca = 26, find a 2 + b 2 + c 2 Solution: (a + b + c) 2 =(a 2 + b 2 + c 2 ) + 2 (ab + bc + ca) ? [9] 2 =(a 2 + b 2 + c 2 ) + 2 (26) = (a 2 + b 2 + c 2 ) + 52 ? a 2 + b 2 + c 2 =9 2 â€“ 52 = 81â€“ 52 = 29 Question 3. Factorise : 8p 3 + 12 5 p 2 + 6 25 p + 1 125 Solution: ? 8p 3 = (2p) 3 , and 1 125 = 3 1 5 ?? ?? ?? ? 8p 3 + 12 5 p 2 + 6 25 p + 1 125 = (2p) 3 + 3 (2p) 2 1 5 ?? ?? ?? + 3 (2p) 23 11 55 ?? ?? + ?? ?? ?? ?? = 3 1 2p + 5 ?? ?? ?? [? a 3 + 3a 2 b + 3ab 2 + b 3 = (a + b) 3 ] = 1 2p + 5 ?? ?? ?? 1 2p + 5 ?? ?? ?? 1 2p + 5 ?? ?? ?? Question 4. If a, b, c are all non-zero and a + b + c = 0, prove that 22 2 ab c 3 bc ca ab ++ = Solution: ? a + b + c = 0 ? a 3 + b 3 + c 3 â€“ 3abc = 0 or a 3 + b 3 + c 3 = 3abc ? 33 3 abc ab c 3 aa a a bc bc bc bc ++ = ? 22 2 ab c 3 bc ca ab ++ = Question 5. Prove that (a + b + c) 3 â€“ a 3 â€“ b 3 â€“ c 3 = 3(a + b) (b + c) (c + a) Solution: L.H.S. = (a + b + c) 3 â€“ a 3 â€“ b 3 â€“ c 3 = [(a + b + c) 3 â€“ a 3 ] â€“ [b 3 + c 3 ] ...(1) (a + b + c) 3 â€“ a 3 = (b + c) [3a 2 + b 2 + c 2 + 3ab + 2bc + 3ca] ...(2) [using x 3 â€“ y 3 = (x â€“ y) (x 2 + y 2 + xy)] and b 3 + c 3 = (b + c) [b 2 + c 2 â€“ bc] [using x 3 + y 3 = (x + y) (x 2 + y 2 â€“ xy)]...(3) From (1), (2) and (3), we get L.H.S. = (b + c) (3a 2 + b 2 + c 2 + 3ab + 2bc + 3ca) â€“ (b + c) (b 2 + c 2 â€“ bc) = (b + c) [3a 2 + b 2 + c 2 + 3ab + 2bc + 3ca â€“ b 2 â€“ c 2 + bc] = (b + c) [3a 2 + (b 2 â€“ b 2 ) + (c 2 â€“ c 2 ) + 3ab + (2bc + bc) + 3ca] = (b + c) [3a 2 + 0 + 0 + 3ab + 3bc + 3ca] = (b + c) [3 (a 2 + ab + bc + ca)] = 3 (b + c) [(a + b) (c + a)] = 3 (a + b) (b + c) (c + a) = RHSRead More

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