{"id":17669,"date":"2022-02-22T17:01:20","date_gmt":"2022-02-22T11:31:20","guid":{"rendered":"https:\/\/apboardsolutions.com\/?p=17669"},"modified":"2023-03-29T15:38:31","modified_gmt":"2023-03-29T10:08:31","slug":"ap-board-10th-class-maths-solutions-chapter-3-exercise-3-4-in-telugu","status":"publish","type":"post","link":"https:\/\/apboardsolutions.com\/ap-board-10th-class-maths-solutions-chapter-3-exercise-3-4-in-telugu\/","title":{"rendered":"AP Board 10th Class Maths Solutions Chapter 3 \u0c2c\u0c39\u0c41\u0c2a\u0c26\u0c41\u0c32\u0c41 Exercise 3.4"},"content":{"rendered":"

SCERT AP 10th Class Maths Textbook Solutions<\/a> Chapter 3 \u0c2c\u0c39\u0c41\u0c2a\u0c26\u0c41\u0c32\u0c41 Exercise 3.4 Textbook Exercise Questions and Answers.<\/p>\n

AP State Syllabus 10th Class Maths Solutions 3rd Lesson \u0c2c\u0c39\u0c41\u0c2a\u0c26\u0c41\u0c32\u0c41 Exercise 3.4<\/h2>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 1.
\n\u0c15\u0c3f\u0c02\u0c26 \u0c07\u0c35\u0c4d\u0c35\u0c2c\u0c21\u0c3f\u0c28 \u0c2c\u0c39\u0c41\u0c2a\u0c26\u0c41\u0c32\u0c32\u0c4b p(x) \u0c2c\u0c39\u0c41\u0c2a\u0c26\u0c3f\u0c28\u0c3f g(x) \u0c2c\u0c39\u0c41\u0c2a\u0c26\u0c3f\u0c1a\u0c47 \u0c2d\u0c3e\u0c17\u0c3f\u0c02\u0c1a\u0c3f \u0c2d\u0c3e\u0c17\u0c2b\u0c32\u0c3e\u0c28\u0c4d\u0c28\u0c3f, \u0c36\u0c47\u0c37\u0c3e\u0c28\u0c4d\u0c28\u0c3f \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c02\u0c21\u0c3f.
\n(i) p(x) = x3<\/sup> – 3x2<\/sup> + 5x – 3, g(x) = x2<\/sup> – 2
\n(ii) p(x) = x4<\/sup> – 3x2<\/sup> + 4x + 5, g(x) = x2<\/sup> + 1 – x
\n(iii) p(x) = x4<\/sup> – 5x + 6, g(x) = 2 – x2<\/sup>
\n\u0c38\u0c3e\u0c27\u0c28.
\n(i) p(x) = x3<\/sup> – 3x2<\/sup> + 5x – 3, g(x) = x2<\/sup> – 2<\/p>\n

\"AP<\/p>\n

\u0c2d\u0c3e\u0c17\u0c2b\u0c32\u0c02\u0c32\u0c4b \u0c2e\u0c4a\u0c26\u0c1f\u0c3f \u0c2a\u0c26\u0c02 = \\(\\frac{x^{3}}{x^{2}}\\) = x<\/p>\n

\u0c2d\u0c3e\u0c17\u0c2b\u0c32\u0c02\u0c32\u0c4b \u0c30\u0c46\u0c02\u0c21\u0c35 \u0c2a\u0c26\u0c02 = \\(\\frac{-3 x^{2}}{x^{2}}\\) = – 3
\n\u0c2d\u0c3e\u0c17\u0c2b\u0c32\u0c02 q(x) = x – 3
\n\u0c36\u0c47\u0c37\u0c02 r (x) = 7x – 9<\/p>\n

\"AP<\/p>\n

2\u0c35 \u0c2a\u0c26\u0c4d\u0c27\u0c24\u0c3f :
\np(x) = x3<\/sup> – 3x2<\/sup> + 5x – 3,
\ng(x) = x2<\/sup> – 2 = x2<\/sup> + 0x – 2<\/p>\n

\"AP<\/p>\n

\u0c2d\u0c3e\u0c17\u0c2b\u0c32\u0c02\u0c32\u0c4b \u0c2e\u0c4a\u0c26\u0c1f\u0c3f \u0c2a\u0c26\u0c02 = \\(\\frac{x^{3}}{x^{2}}\\) = x<\/p>\n

\u0c2d\u0c3e\u0c17\u0c2b\u0c32\u0c02\u0c32\u0c4b \u0c30\u0c46\u0c02\u0c21\u0c35 \u0c2a\u0c26\u0c02 = \\(\\frac{-3 x^{2}}{x^{2}}\\) = – 3
\n\u0c2d\u0c3e\u0c17\u0c2b\u0c32\u0c02 q(x) = x – 3
\n\u0c36\u0c47\u0c37\u0c2e\u0c41 r(x) = 7x – 9<\/p>\n

\"AP<\/p>\n

(ii) p(x) = x4<\/sup> – 3x2<\/sup> + 4x + 5,
\ng(x) = x2<\/sup> + 1 – x
\n\u0c38\u0c3e\u0c27\u0c3e\u0c30\u0c23 \u0c30\u0c42\u0c2a\u0c02\u0c32\u0c4b p(x) = x4<\/sup> + 0x3<\/sup> – 3x2<\/sup> + 4x + 5
\n\u0c05\u0c35\u0c30\u0c4b\u0c39\u0c23 \u0c15\u0c4d\u0c30\u0c2e\u0c02\u0c32\u0c4b g(x) = x2<\/sup> – x + 1<\/p>\n

\"AP<\/p>\n

\u0c2d\u0c3e\u0c17\u0c2b\u0c32\u0c02\u0c32\u0c4b \u0c2e\u0c4a\u0c26\u0c1f\u0c3f \u0c2a\u0c26\u0c02 = \\(\\frac{x^{4}}{x^{2}}\\) = x<\/p>\n

\u0c2d\u0c3e\u0c17\u0c2b\u0c32\u0c02\u0c32\u0c4b \u0c30\u0c46\u0c02\u0c21\u0c35 \u0c2a\u0c26\u0c02 =\\(\\frac{x^{3}}{x^{2}}\\) = x<\/p>\n

\u0c2d\u0c3e\u0c17\u0c2b\u0c32\u0c02\u0c32\u0c4b \u0c2e\u0c4a\u0c26\u0c1f\u0c3f \u0c2a\u0c26\u0c02 = \\(\\frac{-3 x^{2}}{x^{2}}\\) = – 3<\/p>\n

\u0c2d\u0c3e\u0c17\u0c2b\u0c32\u0c02 q(x) = x2<\/sup> + x – 3
\n\u0c36\u0c47\u0c37\u0c02 r (x) = 8.<\/p>\n

\"AP<\/p>\n

(iii) p(x) = x4<\/sup> – 5x + 6, g(x) = 2 – x2<\/sup>
\n\u0c38\u0c3e\u0c27\u0c3e\u0c30\u0c23 \u0c30\u0c42\u0c2a\u0c02\u0c32\u0c4b
\np(x) = x4<\/sup> + 0x3<\/sup> + 0x2<\/sup> – 5x + 6
\n\u0c38\u0c3e\u0c27\u0c3e\u0c30\u0c23 \u0c30\u0c42\u0c2a\u0c02\u0c32\u0c4b g(x) = – x2<\/sup> + 0x + 2<\/p>\n

\"AP<\/p>\n

\u0c2d\u0c3e\u0c17\u0c2b\u0c32\u0c02\u0c32\u0c4b \u0c2e\u0c4a\u0c26\u0c1f\u0c3f \u0c2a\u0c26\u0c02 = \\(\\frac{x^{4}}{-x^{2}}\\) = – x2<\/sup><\/p>\n

\u0c2d\u0c3e\u0c17\u0c2b\u0c32\u0c02\u0c32\u0c4b \u0c30\u0c46\u0c02\u0c21\u0c35 \u0c2a\u0c26\u0c02 = \\(\\frac{2 x^{2}}{-x^{2}}\\) = – 2<\/p>\n

\u0c2d\u0c3e\u0c17\u0c2b\u0c32\u0c02 q(x) = – x2<\/sup> – 2;
\n\u0c36\u0c47\u0c37\u0c02 r (x) = – 5x + 10.<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 2.
\n\u0c15\u0c3f\u0c02\u0c26\u0c3f \u0c2c\u0c39\u0c41\u0c2a\u0c26\u0c41\u0c32\u0c32\u0c4b \u0c30\u0c46\u0c02\u0c21\u0c35 \u0c2c\u0c39\u0c41\u0c2a\u0c26\u0c3f\u0c28\u0c3f, \u0c2e\u0c4a\u0c26\u0c1f\u0c3f \u0c2c\u0c39\u0c41\u0c2a\u0c26\u0c3f\u0c1a\u0c47 \u0c2d\u0c3e\u0c17\u0c3f\u0c02\u0c1a\u0c3f \u0c2a\u0c4d\u0c30\u0c24\u0c3f \u0c38\u0c02\u0c26\u0c30\u0c4d\u0c2d\u0c02\u0c32\u0c4b \u0c2e\u0c4a\u0c26\u0c1f\u0c3f \u0c2c\u0c39\u0c41\u0c2a\u0c26\u0c3f \u0c15\u0c3e\u0c30\u0c23\u0c3e\u0c02\u0c15\u0c02 \u0c05\u0c17\u0c41\u0c28\u0c4b, \u0c15\u0c3e\u0c26\u0c4b \u0c38\u0c30\u0c3f\u0c1a\u0c42\u0c21\u0c02\u0c21\u0c3f.
\n(i) t3<\/sup> – 3, 2t4<\/sup> + 3t3<\/sup> – 2t2<\/sup> – 9t – 12
\n(ii) x2<\/sup> + 3x + 1, 3x4<\/sup> + 5x3<\/sup> – 7x2<\/sup> + 2x + 2
\n(iii) x3<\/sup> – 3x + 1, x5<\/sup> – 4x3<\/sup> + x2<\/sup> + 3x + 1
\n\u0c38\u0c3e\u0c27\u0c28.
\n(i) t2<\/sup> – 3, 2t4<\/sup> + 3t3<\/sup> – 2t2<\/sup> – 9t- 12<\/p>\n

\"AP<\/p>\n

\u0c2d\u0c3e\u0c17\u0c2b\u0c32\u0c02\u0c32\u0c4b \u0c2e\u0c4a\u0c26\u0c1f\u0c3f \u0c2a\u0c26\u0c02 = \\(\\frac{2 t^{4}}{t^{2}}\\) = 2t2<\/sup><\/p>\n

\u0c2d\u0c3e\u0c17\u0c2b\u0c32\u0c02\u0c32\u0c4b \u0c30\u0c46\u0c02\u0c21\u0c35 \u0c2a\u0c26\u0c02 = \\(\\frac{3 \\mathrm{t}^{3}}{\\mathrm{t}^{2}}\\) = 3t<\/p>\n

\u0c2d\u0c3e\u0c17\u0c2b\u0c32\u0c02\u0c32\u0c4b \u0c2e\u0c42\u0c21\u0c35 \u0c2a\u0c26\u0c02 = \\(\\frac{4 \\mathrm{t}^{2}}{\\mathrm{t}^{2}}\\) = 4<\/p>\n

\u0c36\u0c47\u0c37\u0c02 r (x) = 0
\n\u2234 t2<\/sup> – 3, 2t4<\/sup> + 3t3<\/sup> – 2t2<\/sup> – 9t – 12 \u0c15\u0c41 \u0c15\u0c3e\u0c30\u0c23\u0c3e\u0c02\u0c15\u0c2e\u0c41 \u0c05\u0c35\u0c41\u0c24\u0c41\u0c02\u0c26\u0c3f.<\/p>\n

\"AP<\/p>\n

(ii) x2<\/sup> + 3x + 1, 3x4<\/sup> + 5x3<\/sup> – 7x2<\/sup> + 2x + 2<\/p>\n

\"AP<\/p>\n

\u0c2d\u0c3e\u0c17\u0c2b\u0c32\u0c02\u0c32\u0c4b \u0c2e\u0c4a\u0c26\u0c1f\u0c3f \u0c2a\u0c26\u0c02 = \\(\\frac{3 x^{4}}{x^{2}}\\) = 3 x2<\/sup><\/p>\n

\u0c2d\u0c3e\u0c17\u0c2b\u0c32\u0c02\u0c32\u0c4b \u0c30\u0c46\u0c02\u0c21\u0c35 \u0c2a\u0c26\u0c02 = \\(\\frac{-4 x^{3}}{x^{2}}\\) = – 4 x<\/p>\n

\u0c2c\u0c3e\u0c17\u0c2b\u0c32\u0c02\u0c32\u0c4b \u0c2e\u0c42\u0c21\u0c35 \u0c2a\u0c26\u0c02 = \\(\\frac{2 x^{2}}{x^{2}}\\) = 2
\n\u0c36\u0c47\u0c37\u0c02 r (x) = 0
\n\u2234 x2<\/sup> + 3x + 1, 3x4<\/sup> + 5x3<\/sup> – 7x2<\/sup> + 2x + 2 \u0c15\u0c3e\u0c30\u0c23\u0c3e\u0c02\u0c15\u0c2e\u0c41 \u0c05\u0c35\u0c41\u0c24\u0c41\u0c02\u0c26\u0c3f.<\/p>\n

\"AP<\/p>\n

(iii) x3<\/sup> – 3x + 1, x5<\/sup> – 4x3<\/sup> + x2<\/sup> + 3x + 1<\/p>\n

\"AP<\/p>\n

\u0c2d\u0c3e\u0c17\u0c2b\u0c32\u0c02\u0c32\u0c4b \u0c2e\u0c4a\u0c26\u0c1f\u0c3f \u0c2a\u0c26\u0c02 = \\(\\frac{x^{5}}{x^{3}}\\) = x2<\/sup><\/p>\n

\u0c2d\u0c3e\u0c17\u0c2b\u0c32\u0c02\u0c32\u0c4b \u0c30\u0c46\u0c02\u0c21\u0c35 \u0c2a\u0c26\u0c02 = \\(\\frac{-x^{3}}{x^{3}}\\) = – 1
\n\u0c36\u0c47\u0c37\u0c02 r (x) = 2
\n\u2234 x3<\/sup> – 3x + 1, x5<\/sup> – 4x3<\/sup> + x2<\/sup> + 3x + 1 \u0c15\u0c41 \u0c15\u0c3e\u0c30\u0c23\u0c3e\u0c02\u0c15\u0c2e\u0c41 \u0c15\u0c3e\u0c26\u0c41.<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 3.
\n3x4<\/sup> + 6x3<\/sup> – 2x2<\/sup> – 10x – 5 \u0c05\u0c28\u0c41 \u0c2c\u0c39\u0c41\u0c2a\u0c26\u0c3f\u0c15\u0c3f \u0c30\u0c46\u0c02\u0c21\u0c41 \u0c36\u0c42\u0c28\u0c4d\u0c2f\u0c3e\u0c32\u0c28\u0c41 \\(\\sqrt{\\frac{5}{3}}\\) \u0c2e\u0c30\u0c3f\u0c2f\u0c41 – \\(\\sqrt{\\frac{5}{3}}\\) \u0c05\u0c2f\u0c3f\u0c28 \u0c2e\u0c3f\u0c17\u0c3f\u0c32\u0c3f\u0c28 \u0c30\u0c46\u0c02\u0c21\u0c41 \u0c36\u0c42\u0c28\u0c4d\u0c2f\u0c3e\u0c32\u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c02\u0c21\u0c3f.
\n\u0c38\u0c3e\u0c27\u0c28.
\np(x) = 3x4<\/sup> + 6x3<\/sup> – 2x2<\/sup> – 10x – 5
\n\u0c05\u0c2f\u0c3f\u0c28 \u0c2e\u0c3f\u0c17\u0c3f\u0c32\u0c3f\u0c28 \u0c30\u0c46\u0c02\u0c21\u0c41
\np(x) = \\(\\sqrt{\\frac{5}{3}}\\) \u0c2e\u0c30\u0c3f\u0c2f\u0c41 – \\(\\sqrt{\\frac{5}{3}}\\) \u0c32\u0c41 \u0c30\u0c46\u0c02\u0c21\u0c41 \u0c36\u0c42\u0c28\u0c4d\u0c2f \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c32\u0c41 \u0c15\u0c3e\u0c35\u0c41\u0c28 (x – \\(\\sqrt{\\frac{5}{3}}\\)) (x + \\(\\sqrt{\\frac{5}{3}}\\))
\n= x2<\/sup> – \\(\\frac{5}{3}\\) p(x) \u0c15\u0c41 \u0c12\u0c15 \u0c15\u0c3e\u0c30\u0c23\u0c3e\u0c02\u0c15\u0c02 \u0c05\u0c35\u0c41\u0c24\u0c41\u0c02\u0c26\u0c3f.<\/p>\n

\"AP<\/p>\n

\u0c2d\u0c3e\u0c17\u0c2b\u0c32\u0c02\u0c32\u0c4b \u0c2e\u0c4a\u0c26\u0c1f\u0c3f \u0c2a\u0c26\u0c02 = \\(\\frac{3 x^{4}}{x^{2}}\\) = 3x2<\/sup><\/p>\n

\u0c2d\u0c3e\u0c17\u0c2b\u0c32\u0c02\u0c32\u0c4b \u0c30\u0c46\u0c02\u0c21\u0c35 \u0c2a\u0c26\u0c02 = \\(\\frac{6 x^{3}}{x^{2}}\\) = 6x<\/p>\n

\u0c2d\u0c3e\u0c17\u0c2b\u0c32\u0c02\u0c32\u0c4b \u0c2e\u0c42\u0c21\u0c35 \u0c2a\u0c26\u0c02 = \\(\\frac{3 x^{2}}{x^{2}}\\) = 3<\/p>\n

\u0c2d\u0c3e\u0c17\u0c39\u0c3e\u0c30 \u0c05\u0c32\u0c4d \u0c17\u0c3e\u0c30\u0c3f\u0c25\u0c2e\u0c4d \u0c2a\u0c4d\u0c30\u0c15\u0c3e\u0c30\u0c02
\n3x4<\/sup> + 6x3<\/sup> – 2x2<\/sup> – 10x – 5
\n= (x2<\/sup> – \\(\\frac{5}{3}\\)) (3x2<\/sup> + 6x + 3) + 0
\n= (x2<\/sup> – \\(\\frac{5}{3}\\)) (3x2<\/sup> + 6x + 3)
\np(x) = 0 \u0c05\u0c2f\u0c3f\u0c28
\n(x2<\/sup> – \\(\\frac{5}{3}\\)) (3x2<\/sup> + 6x + 3) = 0
\nx2<\/sup> – \\(\\frac{5}{3}\\) = 0 \u0c32\u0c47\u0c26\u0c3e 3x2<\/sup> + 6x + 3 = 0
\nx2<\/sup> = \\(\\frac{5}{3}\\) \u0c32\u0c47\u0c26\u0c3e 3(x2<\/sup> + 2x + 1) = 0
\nx = \u00b1 \\(\\sqrt{\\frac{5}{3}}\\) \u0c32\u0c47\u0c26\u0c3e 3(x + 1)2<\/sup> = 0
\n(x + 1)2<\/sup> = 0
\nx + 1 = 0
\n\u2234 x = – 1
\np(x) \u0c2f\u0c4a\u0c15\u0c4d\u0c15 \u0c36\u0c42\u0c28\u0c4d\u0c2f \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c32\u0c41 \\(\\sqrt{\\frac{5}{3}}\\), – \\(\\sqrt{\\frac{5}{3}}\\) \u0c2e\u0c30\u0c3f\u0c2f\u0c41 – 1.<\/p>\n

\"AP<\/p>\n

2\u0c35 \u0c2a\u0c26\u0c4d\u0c17\u0c24\u0c3f:
\np(x) = 3x4<\/sup> + 6x3<\/sup> – 2x2<\/sup> – 10x – 5 \\(\\sqrt{\\frac{5}{3}}\\) \u0c2e\u0c30\u0c3f\u0c2f\u0c41 – \\(\\sqrt{\\frac{5}{3}}\\) \u0c32\u0c41 p(x) \u0c2f\u0c4a\u0c15\u0c4d\u0c15 \u0c30\u0c46\u0c02\u0c21\u0c41 \u0c36\u0c42\u0c28\u0c4d\u0c2f \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c32\u0c41.
\n(x – \\(\\sqrt{\\frac{5}{3}}\\)) (x + \\(\\sqrt{\\frac{5}{3}}\\)) = x2<\/sup> – \\(\\frac{5}{3}\\)
\n= \\(\\frac{3 x^{2}-5}{3}\\)
\n= \\(\\frac{1}{3}\\) (3x2<\/sup> – 5), p(x) \u0c15\u0c41 \u0c15\u0c3e\u0c30\u0c23\u0c3e\u0c02\u0c15\u0c02 \u0c05\u0c35\u0c41\u0c24\u0c41\u0c02\u0c26\u0c3f. \u0c15\u0c3e\u0c2c\u0c1f\u0c4d\u0c1f\u0c3f p(x) \u0c15\u0c41 3x – 5 \u0c15\u0c3e\u0c30\u0c23\u0c3e\u0c02\u0c15\u0c2e\u0c41<\/p>\n

\"AP<\/p>\n

\u0c2d\u0c3e\u0c17\u0c2b\u0c32\u0c02\u0c32\u0c4b \u0c2e\u0c4a\u0c26\u0c1f\u0c3f \u0c2a\u0c26\u0c02 = \\(\\frac{3 x^{4}}{3 x^{2}}\\) = x2<\/sup><\/p>\n

\u0c2d\u0c3e\u0c17\u0c2b\u0c32\u0c02\u0c32\u0c4b \u0c30\u0c46\u0c02\u0c21\u0c35 \u0c2a\u0c26\u0c02 = \\(\\frac{6 x^{3}}{3 x^{2}}\\) = 2x<\/p>\n

\u0c2d\u0c3e\u0c17\u0c2b\u0c32\u0c02\u0c32\u0c4b \u0c2e\u0c42\u0c21\u0c35 \u0c2a\u0c26\u0c02 = \\(\\frac{3 x^{2}}{3 x^{2}}\\) = 1<\/p>\n

\u0c2d\u0c3e\u0c17\u0c39\u0c3e\u0c30 \u0c05\u0c32\u0c4d\u0c17\u0c3e\u0c30\u0c3f\u0c25\u0c2e\u0c4d \u0c2a\u0c4d\u0c30\u0c15\u0c3e\u0c30\u0c02 p(x) = 3x4<\/sup> + 6x3<\/sup> – 2x2<\/sup> – 10x – 5
\n= (3x2<\/sup> – 5) (x2<\/sup> + 2x + 1) + 0
\n\u2234 p(x) = (3x2<\/sup> – 5) (x2<\/sup> + 2x + 1)
\np(x) = 0 \u0c05\u0c2f\u0c3f\u0c28 (3x2<\/sup> – 5) (x2<\/sup> + 2x + 1) = 0
\n3x2<\/sup> – 5 = 0 \u0c32\u0c47\u0c26\u0c3e x2<\/sup> + 2x + 1 = 0
\n3x2<\/sup> = 5 \u0c32\u0c47\u0c26\u0c3e (x + 1)2<\/sup> = 0
\nx2<\/sup> = , \u0c32\u0c47\u0c26\u0c3e x + 1 = 0
\nx = \u00b1\\(\\sqrt{\\frac{5}{3}}\\) \u0c32\u0c47\u0c26\u0c3e x = – 1
\n\u2234 p(x) \u0c2f\u0c4a\u0c15\u0c4d\u0c15 \u0c36\u0c42\u0c28\u0c4d\u0c2f \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c32\u0c41 \\(\\sqrt{\\frac{5}{3}}\\), – \\(\\sqrt{\\frac{5}{3}}\\) \u0c2e\u0c30\u0c3f\u0c2f\u0c41 – 1.<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 4.
\nx3<\/sup> – 3x2<\/sup> + x + 2 \u0c05\u0c28\u0c41 \u0c2c\u0c39\u0c41\u0c2a\u0c26\u0c3f\u0c28\u0c3f g(x) \u0c05\u0c28\u0c47 \u0c2c\u0c39\u0c41\u0c2a\u0c26\u0c3f\u0c1a\u0c47 \u0c2d\u0c3e\u0c17\u0c3f\u0c38\u0c4d\u0c24\u0c47 \u0c2d\u0c3e\u0c17\u0c2b\u0c32\u0c2e\u0c41 x – 2 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 \u0c36\u0c47\u0c37\u0c2e\u0c41 – 2x + 4 \u0c05\u0c2f\u0c3f\u0c28 g(x) \u0c28\u0c41 \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c02\u0c21\u0c3f.
\n\u0c38\u0c3e\u0c27\u0c28.
\np(x) = x3<\/sup> – 3x2<\/sup> + x + 2
\ng(x) = ?
\nq(x) = x – 2
\nr(x) = – 2x + 4
\n\u0c2d\u0c3e\u0c17\u0c39\u0c3e\u0c30 \u0c05\u0c32\u0c4d \u0c17\u0c3e\u0c30\u0c3f\u0c25\u0c2e\u0c4d \u0c28\u0c41\u0c02\u0c21\u0c3f
\np(x) = g(x) . q(x) + r(x)
\np(x) – r(x) = g(x) . q(x)
\nx3<\/sup> – 3x2<\/sup> + x + 2 – (- 2x + 4) = g(x) (x – 2)
\nx3<\/sup> – 3x2<\/sup> + x + 2 + 2x – 4 = g(x) (x – 2)
\nx3<\/sup> – 3x2<\/sup> + 3x – 2 = g(x) (x – 2)<\/p>\n

\"AP<\/p>\n

\u0c2d\u0c3e\u0c17\u0c2b\u0c32\u0c02\u0c32\u0c4b \u0c2e\u0c4a\u0c26\u0c1f\u0c3f \u0c2a\u0c26\u0c02 = \\(\\frac{x^{3}}{x}\\) = x2<\/sup><\/p>\n

\u0c2d\u0c3e\u0c17\u0c2b\u0c32\u0c02\u0c32\u0c4b \u0c30\u0c46\u0c02\u0c21\u0c35 \u0c2a\u0c26\u0c02 = \\(\\frac{-x^{2}}{x}\\) = – x<\/p>\n

\u0c2d\u0c3e\u0c17\u0c2b\u0c32\u0c02\u0c32\u0c4b \u0c2e\u0c42\u0c21\u0c35 \u0c2a\u0c26\u0c02 = \\(\\frac{x}{x}\\) = 1
\n\u2234 g(x) = x = x + 1.<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 5.
\n\u0c2d\u0c3e\u0c17\u0c39\u0c3e\u0c30 \u0c28\u0c3f\u0c2f\u0c2e\u0c2e\u0c41 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 \u0c26\u0c3f\u0c17\u0c41\u0c35 \u0c07\u0c35\u0c4d\u0c35\u0c2c\u0c21\u0c3f\u0c28 \u0c28\u0c3f\u0c2f\u0c2e\u0c3e\u0c32\u0c28\u0c41 \u0c24\u0c43\u0c2a\u0c4d\u0c24\u0c3f\u0c2a\u0c30\u0c3f\u0c1a\u0c47 \u0c35\u0c3f\u0c27\u0c02\u0c17\u0c3e p(x), g(x), q{x) \u0c2e\u0c30\u0c3f\u0c2f\u0c41 r(x) \u0c2c\u0c39\u0c41\u0c2a\u0c26\u0c41\u0c32\u0c15\u0c41 \u0c24\u0c17\u0c3f\u0c28 \u0c09\u0c26\u0c3e\u0c39\u0c30\u0c23\u0c32\u0c28\u0c41 \u0c07\u0c35\u0c4d\u0c35\u0c02\u0c21\u0c3f.
\n(i) p(x) \u0c2a\u0c30\u0c3f\u0c2e\u0c3e\u0c23\u0c2e\u0c41 = q(x) \u0c2a\u0c30\u0c3f\u0c2e\u0c3e\u0c23\u0c2e\u0c41
\n(ii) q(x) \u0c2a\u0c30\u0c3f\u0c2e\u0c3e\u0c23\u0c2e\u0c41 = r(x) \u0c2a\u0c30\u0c3f\u0c2e\u0c3e\u0c23\u0c2e\u0c41
\n(iii) r(x) \u0c2a\u0c30\u0c3f\u0c2e\u0c3e\u0c23\u0c2e\u0c41 = 0
\n\u0c38\u0c3e\u0c27\u0c28.
\n(i) p(x) \u0c2a\u0c30\u0c3f\u0c2e\u0c3e\u0c23\u0c02 = q(x) \u0c2a\u0c30\u0c3f\u0c2e\u0c3e\u0c23\u0c02 \u0c2d\u0c3e\u0c17\u0c39\u0c3e\u0c30 \u0c05\u0c32\u0c4d\u0c17\u0c3e\u0c30\u0c3f\u0c25\u0c2e\u0c4d
\np(x) = g(x) q(x) + r(x)
\n\u0c09\u0c26\u0c3e – 1: p(x) \u0c2a\u0c30\u0c3f\u0c2e\u0c3e\u0c23\u0c02 = q(x) \u0c2a\u0c30\u0c3f\u0c2e\u0c3e\u0c23\u0c02 \u0c15\u0c3e\u0c35\u0c3e\u0c32\u0c02\u0c1f\u0c47 g(x) \u0c2f\u0c4a\u0c15\u0c4d\u0c15 \u0c2a\u0c30\u0c3f\u0c2e\u0c3e\u0c23\u0c02 \u0c36\u0c42\u0c28\u0c4d\u0c2f\u0c02 \u0c15\u0c3e\u0c35\u0c3e\u0c32\u0c3f \u0c05\u0c28\u0c17\u0c3e g(x) \u0c38\u0c4d\u0c25\u0c3f\u0c30 \u0c2c\u0c39\u0c41\u0c2a\u0c26\u0c3f.
\n3x3<\/sup> – 15x2<\/sup> + 9x + 12
\n= 3(x3<\/sup> – 5x2<\/sup> + 3x + 4) + 0
\np(x) = 3x3<\/sup> – 15x2<\/sup> + 9x + 12
\ng(x) = 3
\nq(x) = x3<\/sup> – 5x2<\/sup> + 3x + 4
\nr(x) = 0<\/p>\n

\u0c09\u0c26\u0c3e – 2 : 15x2<\/sup> + 8x + 10
\n= 5 (3x2<\/sup> + \\(\\frac{8}{5}\\) x + 2) + 0
\np(x) = 15x2<\/sup> + 8x + 10
\ng(x) = 5
\nq(x) = 3x2<\/sup> + \\(\\frac{8}{5}\\) x + 2
\nr(x) = 0<\/p>\n

\"AP<\/p>\n

(ii) q(x) \u0c2a\u0c30\u0c3f\u0c2e\u0c3e\u0c23\u0c02 = r(x) \u0c2a\u0c30\u0c3f\u0c2e\u0c3e\u0c23\u0c02 \u0c2d\u0c3e\u0c17\u0c39\u0c3e\u0c30 \u0c05\u0c32\u0c4d \u0c17\u0c3e\u0c30\u0c3f\u0c25\u0c2e\u0c4d p(x) = g(x) q(x) + r(x) \u0c32\u0c46\u0c15\u0c4d\u0c15 \u0c2a\u0c4d\u0c30\u0c15\u0c3e\u0c30\u0c02
\nq(x) \u0c2a\u0c30\u0c3f\u0c2e\u0c3e\u0c23\u0c02 = r(x) \u0c2a\u0c30\u0c3f\u0c2e\u0c3e\u0c23\u0c02
\nr(x) \u0c2a\u0c30\u0c3f\u0c2e\u0c3e\u0c23\u0c02 < g(x) \u0c2a\u0c30\u0c3f\u0c2e\u0c3e\u0c23\u0c2e\u0c41 \u0c05\u0c28\u0c3f \u0c2e\u0c28\u0c15\u0c41 \u0c24\u0c46\u0c32\u0c41\u0c38\u0c41.
\n\u2234 q(x) \u0c2a\u0c30\u0c3f\u0c2e\u0c3e\u0c23\u0c02 < g(x) \u0c2a\u0c30\u0c3f\u0c2e\u0c3e\u0c23\u0c02 \u0c15\u0c3e\u0c35\u0c3e\u0c32\u0c3f<\/p>\n

\u0c09\u0c26\u0c3e -1 : p(x) = (x2<\/sup> + 5x + 6) (x + 3) + x – 4
\n= x3<\/sup> + 5x2<\/sup> + 6x + 3x2<\/sup> + 15x + 18 + x – 4
\np(x) = x3<\/sup> + 8x + 22x + 14
\ng(x) = x2<\/sup> + 5x + 6
\nq(x) = x + 3
\nr(x) = x – 4<\/p>\n

\u0c09\u0c26\u0c3e – 2: p(x) = (3x3<\/sup> – 4x + 5) (x2<\/sup> – 2) + x2<\/sup> + 2x + 5
\n= 3x5<\/sup> – 4x3<\/sup> + 5x2<\/sup> – 6x3<\/sup> + 8x – 10 + x2<\/sup> + 2x + 5
\np(x) = 3x5<\/sup> – 10x3<\/sup> + 6x2<\/sup> + 10x – 5
\ng(x) = 3x3<\/sup> – 4x + 5
\nq(x) = x2<\/sup> – 2
\nr(x) = x2<\/sup> + 2x + 5<\/p>\n

\"AP<\/p>\n

(iii) r(x) \u0c2a\u0c30\u0c3f\u0c2e\u0c3e\u0c23\u0c02 = 0
\nr(x) \u0c2a\u0c30\u0c3f\u0c2e\u0c3e\u0c23\u0c02 = 0 \u0c05\u0c28\u0c17\u0c3e f(x) \u0c38\u0c4d\u0c25\u0c3f\u0c30 \u0c2c\u0c39\u0c41\u0c2a\u0c26\u0c3f \u0c05\u0c35\u0c41\u0c24\u0c41\u0c02\u0c26\u0c3f.
\n\u0c09\u0c26\u0c3e – 1: r(x) = 5
\ng(x) = x + 5
\nq(x) = x + 5x + 7
\n\u0c2d\u0c3e\u0c17\u0c39\u0c3e\u0c30 \u0c05\u0c32\u0c4d \u0c17\u0c3e\u0c30\u0c3f\u0c25\u0c2e\u0c4d
\np(x) = g(x) . q(x) + r(x) .
\n= (x + 5) (x2<\/sup> + 5x + 7) + 5
\np(x) = x3<\/sup> + 5x2<\/sup> + 7x + 5x2<\/sup> + 25x + 35 + 5
\np(x) = x3<\/sup> + 10x2<\/sup> + 32x + 40.<\/p>\n

\u0c09\u0c26\u0c3e – 2:
\nr(x) = 8
\ng(x) = x – 3
\nq(x) = x + 3
\np(x) = g(x) . q(x) + r(x)
\np(x) = (x – 3) (x + 3) +8
\np(x) = x2<\/sup> – 9 + 8
\np(x) = x2<\/sup> – 1<\/p>\n","protected":false},"excerpt":{"rendered":"

SCERT AP 10th Class Maths Textbook Solutions Chapter 3 \u0c2c\u0c39\u0c41\u0c2a\u0c26\u0c41\u0c32\u0c41 Exercise 3.4 Textbook Exercise Questions and Answers. AP State Syllabus 10th Class Maths Solutions 3rd Lesson \u0c2c\u0c39\u0c41\u0c2a\u0c26\u0c41\u0c32\u0c41 Exercise 3.4 \u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 1. \u0c15\u0c3f\u0c02\u0c26 \u0c07\u0c35\u0c4d\u0c35\u0c2c\u0c21\u0c3f\u0c28 \u0c2c\u0c39\u0c41\u0c2a\u0c26\u0c41\u0c32\u0c32\u0c4b p(x) \u0c2c\u0c39\u0c41\u0c2a\u0c26\u0c3f\u0c28\u0c3f g(x) \u0c2c\u0c39\u0c41\u0c2a\u0c26\u0c3f\u0c1a\u0c47 \u0c2d\u0c3e\u0c17\u0c3f\u0c02\u0c1a\u0c3f \u0c2d\u0c3e\u0c17\u0c2b\u0c32\u0c3e\u0c28\u0c4d\u0c28\u0c3f, \u0c36\u0c47\u0c37\u0c3e\u0c28\u0c4d\u0c28\u0c3f \u0c15\u0c28\u0c41\u0c17\u0c4a\u0c28\u0c02\u0c21\u0c3f. (i) p(x) = x3 – 3×2 + 5x – 3, g(x) = x2 … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":35294,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[2],"tags":[],"jetpack_featured_media_url":"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/02\/Chapter-3-\u0c2c\u0c39\u0c41\u0c2a\u0c26\u0c41\u0c32\u0c41-Exercise-3.4-1.png","_links":{"self":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/17669"}],"collection":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/comments?post=17669"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/17669\/revisions"}],"predecessor-version":[{"id":17670,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/17669\/revisions\/17670"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media\/35294"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media?parent=17669"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/categories?post=17669"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/tags?post=17669"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}