2011 Λύσεις Σχ. β&I

Iv úâè 0 x 0 îè x 1 õú ùô âèô

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iv) ¶Ú¤ÂÈ ≠ 0 x ≥ 0 Î·È x ≠ 1. ÕÚ·, ÙÔ Âȉ›Ô ÔÚÈÛÌÔ‡ Ù˘ Û˘Ó¿ÚÙËÛ˘ Â›Ó·È ÙÔ Û‡ÓÔÏÔ [0, + ) – {1} = [0, 1) (1, + ). 3. ∂›Ó·È f(–5) = (–5) 3 = –125. f(0) = 2 Ø 0 + 3 = 3. f(6) = 2 Ø 6 + 3 = 15. 4. i) ŒÛÙˆ x Ô ˙ËÙÔ‡ÌÂÓÔ˜ Ê˘ÛÈÎfi˜ ·ÚÈıÌfi˜. ΔfiÙÂ, Ô Ù‡Ô˜ Ù˘ Û˘Ó¿ÚÙËÛ˘ ı· ÚÔ·„ÂÈ ˆ˜ ÂÍ‹˜: x 1 x – 1
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x x + 1 (x + 1) Ø 4 (x + 1) 4 + x 2 . EÔ̤ӈ˜, ı· ÂÈÓ·È f(x) = (x + 1)4 + x 2 = x 2 + 4x + 4 = (x + 2) 2 ‰ËÏ·‰‹ f(x) = (x + 2) 2 , x . (1) ŒÙÛÈ ı· ¤¯Ô˘Ì f(0) = 2 2 = 4, f(1) = 3 2 = 9, f(2) = 4 2 = 16 Î·È f(3) = 5 2 = 25. ii) ∂Âȉ‹ x > 0, ¤¯Ô˘ÌÂ: f(x) = 36 (x + 2) 2 = 6 2 x + 2 = 6 x = 4. f(x) = 49 (x + 2) 2 = 7 2 x = 5. f(x) = 100 (x + 2) 2 = 10 2 x = 8. f(x) = 144 (x + 2) 2 = 12 2 x = 10. 5. i) °È· x ≠ 1 ¤¯Ô˘ÌÂ: f(x) = 7 2 (x – 1) = 4 x – 1 = 2 x = 3. ii) °È· x ≠ 0, 4 ¤¯Ô˘ÌÂ: g(x) = 2 x + 4 = 2x x = 4, ·‰‡Ó·ÙË. iii) °È· x ¤¯Ô˘ÌÂ: h(x) = x 2 + 1 = 5 x 2 = 4 x = 2 ‹ x = –2. ¨ 6.2. °Ú·ÊÈ΋ ·Ú¿ÛÙ·ÛË Û˘Ó¿ÚÙËÛ˘ ∞ã √ª∞¢∞™ 1. T· ÛËÌ›· Â›Ó·È ·ÔÙ˘ˆÌ¤Ó· ÛÙÔ ‰ÈÏ·Ófi Û¯‹Ì·. 2. ¶Ú¤ÂÈ 2 < x < 5 Î·È 1 < y < 6. 1 x 2 + 1 = 1 5 1 5 x + 4 x = 2 (x – 4)(x + 4) x(x – 4) = 2 x 2 – 16 x 2 – 4x = 2, 4 x – 1 = 2 4 x – 1 + 5 = 7 ∫∂º∞§∞π√ 6: μ∞™π∫∂™ ∂¡¡√π∂™ Δø¡ ™À¡∞ƒΔ∏™∂ø¡ 80 + 1 Ø 4 + x 2
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3. ΔÔ Û˘ÌÌÂÙÚÈÎfi ÙÔ˘ ∞(–1, 3), i) ˆ˜ ÚÔ˜ ÙÔÓ ¿ÍÔÓ· xãx Â›Ó·È ÙÔ μ(–1, –3) ii) ˆ˜ ÚÔ˜ ÙÔÓ ¿ÍÔÓ· yãy Â›Ó·È ÙÔ ¢(1, 3) iii) ˆ˜ ÚÔ˜ ÙË ‰È¯ÔÙfiÌÔ Ù˘ ÁˆÓ›·˜ x y Â›Ó·È ÙÔ ∂(3, –1) iv) ˆ˜ ÚÔ˜ ÙËÓ ·Ú¯‹ ÙˆÓ ·ÍfiÓˆÓ Â›Ó·È ÙÔ °(1, –3). 4. ªÂ ‚¿ÛË ÙÔÓ Ù‡Ô Ù˘ ·fiÛÙ·Û˘ ÙˆÓ ÛË- Ì›ˆÓ ∞(x 1 , y 1 ) Î·È B(x 2 , y 2 ), ¤¯Ô˘Ì i) ii) iii) iv) 5. i) ∂›Ó·È ÕÚ·, (AB) = (A°), ÔfiÙ ÙÔ ÙÚ›ÁˆÓÔ AB° Â›Ó·È ÈÛÔÛÎÂϤ˜ Ì ÎÔÚ˘Ê‹ ÙÔ ∞. ii) ∂›Ó·È ÔfiÙ (∞μ) 2 = 8. ÔfiÙ (∞°) 2 = 18. ÔfiÙ (μ°) 2 = 26. ¶·Ú·ÙËÚԇ̠fiÙÈ (μ°) 2 = (∞μ) 2 + (∞°) 2 . ÕÚ· ÙÔ ÙÚ›ÁˆÓÔ ∞μ° Â›Ó·È ÔÚıÔÁÒÓÈÔ, Ì ÔÚı‹ ÁˆÓ›· ÙËÓ ∞. 6. ∂›Ó·È (∞μ) = (5 – 2) 2 + (1 – 5) 2 = 5. (μ°) = (4 + 1) 2 + (2 – 1) 2 = 26, (A°) = (4 – 1) 2 + (2 + 1) 2 = 2 3 2 = 3 2, (AB) = (–1 – 1) 2 + (1 + 1) 2 = 2 2 2 = 2 2, (B°) = (–3 – 4) 2 + (5 + 2) 2 = 2 7 2 + = 7 2. (A°) = (–3 – 1) 2 + (5 – 2) 2 = 4 2 + 3 2 = 5. (AB) = (4 – 1) 2 + (–2 – 2) 2 = 3 2 + 4 2 = 5. (AB) = 0 2 + (4 + 1) 2 = 5. (AB) = (1 + 3) 2 + 0 2 = 4. (∞B) = (3 + 1) 2 + (4 – 1) 2 = 4 2 + 3 2 = 25 = 5. (O∞) = 4 2 + (–2) 2 = 20 = 2 5. (∞μ) = (x 2 – x 1 ) 2 + (y 2 – y 1 ) 2 6.2. °Ú·ÊÈ΋ ·Ú¿ÛÙ·ÛË Û˘Ó¿ÚÙËÛ˘ 81
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ÕÚ· ÙÔ ÙÂÙÚ¿Ï¢ÚÔ ∞μ°¢ ¤¯ÂÈ fiϘ ÙȘ Ï¢ڤ˜ ÙÔ˘ ›Û˜, ÔfiÙÂ Â›Ó·È ÚfiÌ‚Ô˜.
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