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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
(-2x + 3y + 2z)2 का विस्तार ज्ञात कीजिए। |
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Answer» (-2x + 3y + 2x)2 = (-2x)2 + (3y)2 + (2z)2 + 2(-2x)(3y) + 2(3y) (2z) + 2(-2x)(2z) = 4x2 + 9y2 + 4z2 – 12xy + 12yz – 8xz |
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| 2. |
निम्न के विस्तार ज्ञात कीजिए।(i) (-3x + y + z)2 (ii) (-x + 2y + z)2 (iii) (3x + 2y – z)2 (iv) (2 + x – 2y)2 (v) (m + 2n – 5p)2 (vi) (ab + bc + ca)2 |
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Answer» (i) (-3x + y + x)2 = (–3x)2 + (y)2 +(z)2 + 2(–3x)(y) + 2(y) (z) + 2(-3x)(z) =9x2 + y2 +z2 – 6xy + 2yz – 6xz (ii) (-x + 2y + z)2 = (-x)2 + (2y)2 +(z)2 + 2(-x)(2y) + 2(2y)(z) + 2(-x)(z) = x2 + y2 + z2 – 4xy + 4yz – 2xz (iii) (3x + 2y – z)2 = (3x)2 + (2y)2 +(+z)2 + 2(3x)(2y) + 2(2y)(-z) + 2(3x)(-z) = 9x2 + 4y2 + z2 + 12xy – 4yz – 6xz (iv) (2 + x – 2y)2 = (2)2 + (x)2 + (-2y)2 + 2.2.x + 2. x(-2y) + 2.2(-2y) =4 + x2 + 4y2 + 4x – 4xy – 8y (v) (m + 2n – 5p)2 = (m)2 + (2n)2 + (-5p)2 + 2. m. 2n + 2. 2n(-5p) +2. m(-5p) = m2 + 4n2 +25p2 + 4mn – 20np – 10mp (vi) (ab + bc + ca)2 = (ab)2 + (bc)2 + (ca)2 + 2ab.bc + 2bc.ca + 2ab.ca =a2b2 + b2c2 +c2a2 +2ab2c + 2abc2 + 2a2bc |
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| 3. |
(4a – 2b – 3c)2 का विस्तार कीजिए। |
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Answer» (4a – 2b – 3c)2 = (4a)2 + (-2b)2 + (-3c)2 – 16ab + 12bc – 24ac = 16a2 + 4b2 + 9c2 – 16ab + 12bc – 24ac |
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| 4. |
(3a + 4b + 5c)2 का विस्तार कीजिए। |
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Answer» (3a + 4b + 5c)2 = 9a2 + 16b2 + 25c2 + 24ab + 40 bc + 30ac |
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| 5. |
यदि x + y + z = 9 व xy + yz + zx = 23, तब x2 + y2 + z2 =(a) 25(b) 35(c) 45(d) 305 |
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Answer» विकल्प (b) 35 सही है। x + y + z =9 दोनों पक्षों का वर्ग करने पर, (x + y + z)2 = 92 ⇒ x2 + y2 + z2 + 2(xy + yz + zx) = 81 ⇒ x2 + y2 + z2 = 81 – 2(xy + yz + zx) = 81 – 2 × 23 (∵ xy + yz + zx = 23) = 81 – 46 = 35 |
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| 6. |
यदि x + y + z = 6 व xy + yz + zx = 11, तब सिद्ध कीजिए x3 + y3 + z3 -3xyz = 18 |
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Answer» ∵ x + y + z = 6 ……………. (1) x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx) ……..(2) समी० (1) का वर्ग करने पर x2 + y2 + z2 + 2(xy + yz + zx) = 36 x2 + y2 + z2 + 2(11) = 36 x2 + y2 + z2 = 36 – 22 = 14 समी० (2) में x2 + y2 + z2 का मान रखने पर x3 + y3 + z3 – 3xyz = (6) (14 – 11) = (6) (3) = 18 |
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| 7. |
सर्वसमिकाओं का प्रयोग कर सरल कीजिए। 322 x 322 - 2 x 322 x 22 + 22 x 22 |
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Answer» 322 × 322 – 2 × 322 × 22 + 22 × 22 माना 322 = a तथा 22 = b a × a – 2 × a × b + b× b =a2 – 2ab + b2 = (a – b)2 = (322 – 22)2 = (300)2 = 90000 |
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| 8. |
यदि x – y = 5 व xy = 12, तब x2 + y2 =(a) 49(b) 25(c) 144(d) इनमें से कोई नहीं |
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Answer» विकल्प (a) 49 सही है। x – y = 5 दोनों पक्षों का वर्ग करने पर, (x – y)2 = 52 ⇒ x2 + y2 – 2xy = 25 ⇒ x2 + y2 – 2(12) = 25 ⇒ x2 + y2 = 49 |
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| 9. |
यदि 3x +2y = 20 व xy = 14/9 तब 27x3 + 8y3 का मान ज्ञात कीजिए। |
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Answer» 3x + 2y = 20 …………… (1) वर्ग करने पर 9x2 + 4y2 + 12xy = 400 9x2 +4y2 = 400 – 12xy 27x3 + 8y3 = (3x)3 + (2y)3 = (3x + 2y) (9x2 + 4y2 – 6xy) = (3x + 2y)(400 – 12xy – 6xy) = (3x + 2y)(400 – 18xy) = (20)( 400 – 18 ×14/9 = (20) (400 – 28) = (20) (372) = 7440 |
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| 10. |
यदि 3x – 2y =11 व xy = 12, तब 27x3 – 8y3 का मान ज्ञात कीजिए। |
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Answer» ∵ 3x – 2y = 11 …………….. (1) वर्ग करने पर 9x2 + 4y2 – 12xy = 121 9x2 + 4y2 -12 × 12 = 121 9x2 + 4y2 = 121 + 144 = 265 27x3 – 8y3 = (3x)3 – (2y)3 = (3x – 2y)(9x2 + 4y2 + 6xy) = (11)(265 + 6 × 12) = (11)(265 + 72) = (11) (337) = 3707 |
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| 11. |
यदि x = 4, y =3, z = 2 तब सिद्ध कीजिए कि 4x2 + y2 + 25z2 + 4xy – 10yz – 20xz =1 |
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Answer» L.H.S. = 4x2 + y2 + 25z2 + 4xy -10yz – 20xz = (-2x)2 + (-y)2 + (58)2 + 2(-2x)(-y) + 2(-y) (5z) + 2(5z)(-2x) = (-2x – y + 5z)2 = (-2 × 4 – 3 + 5 × 2)2 = (-1)2 = 1 = R.H.S. |
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| 12. |
निम्न के मान ज्ञात कीजिए।(i) (46)3 + (34)3(ii) (23)3 – (17)3(iii) (111)3 – (89)3 |
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Answer» (i) (46)3 + (34)3 = (40 + 6)3 + (40-6)3 = (40)3 + (6)3 + 3 × 40 × 6(40 + 6) + (40)3 – (6)3 – 3 × 40 × 6(40 – 6) = 2(40)3 + 3 × 40 × 6[46 – 34] = 2(40)3 + 720 × 12 = 128000 + 8640 = 136640 (ii) (23)3 – (17)3 = (20 + 3)3 – (20 – 3)3 = (20)3 + (3)3 + 3 × 20 × 3(20 + 3) – (20)3 + (3)3 + 3 × 20 × 3(20 – 3) = 2(3)3 + 3 × 20 × 3(23 + 17) = 2 × 27 + 180(40) = 54+ 7200 = 7254 (iii) (111)3 – (89)3 = (100 + 11)3 – (100 – 11)3 = (100)3 + (11)3 + 3 × 100 × 11 (100 + 11) – (100)3 + (11)3 + 3 × 100 × 11(100 – 11) = 2(11)3 + 3 × 100 × 11(111 + 89) = 2(1331) + 3 × 100 × 11 × 200 = 2662 + 660000 = 662662 |
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| 13. |
निम्न को सरल कीजिए।(i) (4x + 2y)3 + (4x – 2y)3(ii) (4x + 2y)3 – (4x – 2y)3(iii) (a + 3)3 + (a – 3)3 |
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Answer» (i) (4x + 2y)3 = (4x)3 + (2y)3 + 3. (4x) . (2y) (4x + 2y) = 64x3 + 8y3 + 24xy (4x + 2y) = 64x3 + 8y3 + 96x2y + 48xy2 (4x – 2y)3 = (4x)3 + (-2y)3 + 3. (4x) (-2y) (4x -2y) = 64x3 – 8y3 – 24xy (4x -2y) = 64x3 – 8y3 – 96x2y + 48xy2 इसलिए (4x + 2y)3 + (4x – 2y)3 = 128x3 +96xy2 (ii) (4x + 2y)3 – (4x – 2y)3 = 64x3 + 8y3 + 96x2y + 98xy2 – (64x3 – 8y3 – 96x2y + 48xy2) = 64x3 + 8y3 + 96x32y + 48xy2 – 64x3 + 8y3 + 96x2y – 48xy2 = 16y3 + 192x2y (iii) (a + 3)3 = a3 + 27 + 3a . 3(a + 3) = a3 + 27 + 9a2 + 27a (a – 3)3 = a3 – 27 – 3a – 3(a – 3) = a3 – 27 – 9a2 + 27a ∴ (a + 3)3 + (a – 3)3 = 2a3 +54a |
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| 14. |
(x + 4)(x + 10) का मान ज्ञात कीजिए। |
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Answer» (x + 4)(x + 10) = (x)2 + (4 + 10)x + 4 × 10 = x2 +14x + 40 |
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| 15. |
सिद्ध कीजिए कि (x + y + z)2 – (x – y – z)2 = 4x ( y + z) |
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Answer» L.H.S. = (x + y + z)2 – (x – y – z)2 = x2 + y2 + z2 + 2(xy + yz + zx) – (x2 + y2 + z2 – 2xy + 2yz – 2zx) = 4xy + 4xz = 4x(y + z) = R.H.S. |
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| 16. |
(3x + 4)(3x – 5) का मान ज्ञात कीजिए। |
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Answer» (3x + 4)(3x – 5) = (3x)2 + (4 – 5)3x – 4 × 5 = 9x2 – 3x – 20 |
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| 17. |
निम्न के मान ज्ञात कीजिए।(i) (99)3(ii) (9.9)3(iii) (10.4)3(iv) (598)3(v) (402)3(vi) (1002)3 |
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Answer» (i) (99)3 = (100 – 1)3 = (100) – (1)3 – 3 × 100 × 1(100 – 1) = 1000000 – 1 – 30000+ 300 = 970299 (ii) (9.9)3 = (10 – 0.1)3 = (10)3 – (0.1)3 – 3 × 10 × 0.1(10 – 0.1) = 1000 – 0.001 – 3(9.9) = 1000 – 0.001 – 29.7 = 970.299 (iii) (10.4)3 = (10 + 0.4)3 = (10)3 + (0.4)3 + 3 × 10 × 0.4(10 + 0.4) = 1000+ 0.064 + 12(10.4) = 1000 + 0.064 + 124.8 =1124.864 (iv) (598)3 = (600-2)3 = (600)3 – (2)3 – 3 × 600 × 2(600 – 2) = 216000000 – 8-21,60,000 + 7200 =21,38,47,192 (v) (402)3 = (400 + 2)3 = (400)3 + (2)3 + 3 × 400 × 2(400 + 2) = 64000000 + 8 + 960000 + 4800 = 64964808 (vi) (1002)3 = (1000 + 2)3 = (1000)3 + (2)3 + 3 × 1000 × 2(1000 + 2) = 1000000000 + 8+ 6000000 + 12000 = 1006012008 |
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| 18. |
(2x + 1)3 का मान ज्ञात कीजिए। |
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Answer» (2x + 1)3 = (2x)3 + (1)3 + 3 × 2x × 1(2x + 1) = 8x3 + 1 + 12x2 + 6x |
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| 19. |
यदि x = 5 और y = 7 तब 49x2 – 70xy + 25y2 का मान ज्ञात कीजिए। |
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Answer» x = 5 तथा y = 7 49x2 – 70xy + 25y2 = (7x)2 – 2 × 7x × 5y + (5y)2 = (7x – 5y)2 = (7 × 5 – 5 × 7)2 = (35 – 35)2 = 0 |
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| 20. |
यदि x = 15 व y = 27 तब 81x2 – 90xy + 25y2 का मान ज्ञात कीजिए। |
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Answer» 81x2 – 90xy + 25y2 = (9x)2 – 2 × 9x × 5y + (5y)2 = (9x – 5y)2 = (9 × 15 – 5 × 27)2 = (135 – 135) = 0 |
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| 21. |
यदि a2 + b2 + c2 = 20 व a + b+ c =0, तब ab+ bc + ca का मान ज्ञात कीजिए। |
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Answer» (a + b + c = 0 ) वर्ग करने पर a2 + b2 + c2 + 2(ab + bc + ca) = 0 20 + 2(ab + bc + ca) = 0 2(ab + bc + ca) = 0 – 20 ab + bc + ca = -20/2 = -10 |
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| 22. |
यदि a2 + b2 + c2 = 16 व ab+ bc + ca = 10, तब a + b + c का मान ज्ञात कीजिए। |
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Answer» (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) = 16 + 2 × 10 = 16 + 20 = 36 ∴ a + b + c = √36 = 6 |
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| 23. |
यदि a + b + c = 9 व ab+ bc + ca = 40, तब a2 + b2 + c2 का मान ज्ञात कीजिए। |
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Answer» यदि a + b + c = 9 वर्ग करने पर a2 + b2 + c2 + 2(ab + bc + ca) = 81 a2 + b2 + c2 + 2 × 40 = 81 a + b + c = 81 – 80 = 1 |
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| 24. |
(2a – 3b) का मान ज्ञात कीजिए। |
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Answer» (2a – 3b)3 = (2a)3 + (-3b)3 + 3(2a) (-3b)(2a – 3b) = 8a3 – 27b3 – 18ab(2a – 3b) = 8a3 – 27b3 – 36a2b + 54ab2 |
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| 25. |
यदि x + y = 12 व xy = 32 , तब x2 + y2 का मान ज्ञात कीजिए। |
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Answer» x + y = 12 ………………….. (1) xy = 32 ……………… (2) समी० (1) का वर्ग करने पर x2 + y2 + 2xy = 144 x2 + y2 + 2 × 32 = 144 x2 + y2 = 144 – 64 = 80 |
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| 26. |
यदि x + y = 10 व xy = 16, तब x2 – xy + y2 व x2 + xy + y2 के मान ज्ञात कीजिए। |
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Answer» x2 – xy + y2 = x2 + y2 + 2xy – 3xy = (x + y)2 – 3xy = (10)2 – 48 = 100 – 48 = 52 x2 + xy + y2 = (x + y)2 – xy = (10)2 – 16 = 100 – 16 = 84 |
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| 27. |
यदि a + b = 8 व ab = 6, तब a3 + b3 का मान ज्ञात कीजिए। |
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Answer» a + b = 8 ……………… (1) घन करने पर a3 + b3 + 3ab (a + b) = 512 a3 + b3 + 3 × 6(8) = 512 a3 + b3 = 512 – 144 = 368 |
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| 28. |
निम्न के मान ज्ञात कीजिए।(x – 0.1)(x + 0.1) |
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Answer» (x – 0.1)(x + 0.1) = x2 - (0.1)2 = x2 - 0.01 |
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| 29. |
यदि x – y = 6 व xy = 20, तब x3 – y3 के मान ज्ञात कीजिए। |
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Answer» ∵ x2 + y2 + xy = (x – y)2 + 3xy = (6)2 + 3 × 20 = 36 + 60 = 96 ∵ x2 – y2 = (x – y)(x2 + y2 + xy) = (6)(96) = 576 |
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| 30. |
यदि x + y = 10 व xy = 21, तब x3 + y3 का मान ज्ञात कीजिए। |
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Answer» x + y = 10 ……………. (1) घन करने पर (x + y)3 = (10)3 x3 + y3 + 3xy (x + y) = 1000 x3 + y3 + 3. 21 (10) = 1000 x3 + y3 + 630 = 1000 x3 + y 3 = 1000 – 630 = 370 |
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| 31. |
यदि a – b = 4 व ab = 21, तब a3 – b3 का मान ज्ञात कीजिए। |
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Answer» a – b = 4 वर्ग करने पर a2 + b2 – 2ab = 16 a2 + b2 – 2 × 21 = 16 a2 + b2 = 16 + 42 = 58 अब a3 – b3 = (a – b)(a + b2 + ab) = (4)(58 + 21) = (4)(79) = 316 |
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| 32. |
सर्वसमिकाओं का प्रयोग कर गुणनफल ज्ञात कीजिए।10008 × 995 |
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Answer» 10008 × 995 = (1000 + 8) × (1000 – 5) = (1000)2 + (8 – 5) × 1000 – 8 × 5 = 1000000 + 3000 – 40 = 1002960 |
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| 33. |
सर्वसमिकाओं का प्रयोग कर गुणनफल ज्ञात कीजिए। 105 × 97 |
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Answer» 105 × 97 = (100 + 5) × (100 -3) = (100)2 + (5 – 3) × 100 – 5 × 3 = 10000 + 200 – 15 = 10185 |
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| 34. |
सर्वसमिकाओं का प्रयोग कर गुणनफल ज्ञात कीजिए। 111 × 102 |
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Answer» 111 × 102 = (100 + 11) × (100 + 2) = (100)2 + (11 + 2) × 100 + 11 × 2 = 10000 + 1300 + 22 = 11322 |
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| 35. |
यदि a + b = 10 व ab = 21, तब a3 + b3 का मान ज्ञात कीजिए। |
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Answer» a + b = 10 घन करने पर a3 + b3 + 3ab (a + b) = 1000 a3 + b3 + 3 × 21(10) = 1000 ∴ a3 + b3 = 1000 – 630 = 370 |
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| 36. |
निम्न के मान ज्ञात कीजिए।(i) (3x – 4y + 5z)(9x2 + 16y2 + 25z2 + 12xy – 15yz + 20yz)(ii) (3x + 2y + 2z)(9x2 + 4y2 + 4z2 -6xy – 4yz – 6xz)(iii) (2x – y + 3z)(4x2 + y2 + 9z2 + 2xy + 3yz -6xz) |
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Answer» (i) (3x – 4y + 5z)(9x2 + 16y2 + 25z2 + 12xy – 15xz + 20yz) (ii) (3x + 2y + 2z(9x2 + 4y2 + 4z2 – 6xy -4yz – 6zx) (iii) (2x – y + 3z)(4x2 + y2 – 9z2 + 2xy + 3yz – 6xz) |
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| 37. |
सर्वसमिकाओं का प्रयोग कर गुणनफल ज्ञात कीजिए।85 × 96 |
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Answer» 85 × 96 = (100 – 15) × (100 – 4) = (100)2 + (-15 – 4) × 100 + 15 × 4 = 10000 – 1900 + 60 = 8160 |
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| 38. |
सर्वसमिका का प्रयोग करके निम्न के मान ज्ञात कीजिए। (1) 103 × 107(ii) 95 × 96(iii) 104 × 96 |
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Answer» (i) 103 × 107 = (100 + 3) × (100 + 7) = (100)2 + (3 + 7) × 100 + 3 × 7 =10000 + 1000 + 21 = 11021 (ii) 95 × 96 = (100 – 5) × (100 – 4) = (100)2 (5 + 4) × 100 + 5 × 4 = 10000 – 900 + 20 = 9120 (iii) 104 × 96 = (100 + 4) × (100 – 4) = (100)2 – (4)2 = 10000 – 16 = 9984 |
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| 39. |
(104)3 का मान सर्वसमिका का प्रयोग करके ज्ञात कीजिए। |
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Answer» (104)3 = (100 + 4)3 (100)3 + (4)3 + 3 × 100 × 4(100 + 4) = 1000000 + 64 + 124800 = 1124864 |
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| 40. |
निम्न के मान ज्ञात कीजिए।(a) (25)3 – (75)3 + (50)3(b) (0.2)3 – (0.3)3 + (0.1)3 (c) (1.5)3 – (0.9)3 – (0.6)3(d) (-12)3 + 73 + 53 |
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Answer» (a) (25)3 – (75)3 + (50)3 = (25)3 – (25 + 50)3 + (50)3 = (25)3 – (25)3 – (50)3 – 3 × 25 × 50(25 + 50) + (50)3 =-3 × 25 × 50 × 75 = -281250 (b) (0.2)3 – (0.3)3 + (0.1)3 = (0.2)3 – (0.2 + 0.1) + (0.1) 3 = (0.2)3 s – (0.2)3 – (0.1)3 – 3 × 0.2 × 0.1 × (0.2 + 0.1) + (0.1)3 =-3 × 0.2 × 0.1 × (0.3) = -0.018 (c) (1.5)3 – (0.9)3 – (0.6)3 = (0.9 + 0.6)3 – (0.9)3 – (0.6)3 = (0.9)3 + (0.6)3 + 3 × 0.9 × 0.6 × (0.9 + 0.6) – (0.9)3 – (0.6)3 = 3 × 0.9 × 0.6 × 1.5 = 2.430 (d) (-12)3 + 73 + 53 ⇒ 73 + (-7 – 5)3 + 53 ⇒ 73 + (-7)3 + (-5)3 + 3 × (-7)(-5)(-7 – 5) + 53 = 3 × -7 × -5(-12) = 105 × (-12) = -1260 |
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| 41. |
सिद्ध कीजिए कि x2 + y2 + z2 – xy – y – zx; x, y व z के सभी मानों के लिए सदैव धनात्मक होगा। |
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Answer» x2 + y2 + z2 – xy – yz – zx =1/2 [(x – y)2 + (y – z)2 + (z – x)2] ∵ (x – y)2,(y – z)2, (z – x)2 पूर्ण वर्ग है जो हमेशा धनात्मक होते हैं। अतः इसका योग सदैव धनात्मक होगा। |
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| 42. |
यदि 2x + 3y = 8 व xy = 2, तब सिद्ध कीजिए कि 4x2 + 9y2 = 40 |
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Answer» 2x + 3y = 8 वर्ग करने पर, 4x2 + 9y2 + 2(2x)(3y) = 64 ⇒ 4x2 + 9y2 = 64 – 12(xy) = 64 – 12 × 2 = 40 |
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| 43. |
यदि 3x – 7y = 10 व xy = -1, तब सिद्ध कीजिए कि 9x2 + 49y2 = 58 |
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Answer» 3x – 7y = 10 दोनों ओर का वर्ग करने पर 9x2 + 49y2 – 42xy = 100 9x2 + 49y2 – 42 × (-1) = 100 9x2 + 49y2 = 100 – 42 = 58 |
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| 44. |
यदि x + y + z = 9 और x2 + y2 + z2 = 35, तब x3 + y3 + z3 – 3xyz के मान ज्ञात कीजिए। |
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Answer» x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx) ……..(1) ∵ x + y + z = 9 वर्ग करने पर . x2 + y2 + z2 + 2(xy + y + zx) = 81 35 + 2(xy + yz + zx) = 81 2(xy + yz + zx) = 81 – 35 = 46 xy + yz + zx = 46/2 = 23 समी० (1) से x3 + y3 + z3 – 3xyz = 9(35 – 23) = 9 × 12 = 108 |
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| 45. |
यदि 4x2 + y2 = 40 व xy =6, तब सिद्ध कीजिए कि 2x + y = ± 8 |
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Answer» 4x2 + y2 = 40 ⇒ (2x)2 + (y)2 + 2(2x)(y) = 40 + 2(2x)(y) ⇒ (2x + y)2 = 40 + 4(6) = 40 + 24 ⇒ (2x + y)2 = 64 ⇒ (2x + y) = ± 8 |
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| 46. |
x3 + y3 + z3 – 3xyz का मान ज्ञात कीजिए, यदि x + y + z = 14 व x2 + y2 + z2 = 60 |
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Answer» x + y + z = 14 ………………(1) वर्ग करने पर, x2 + y2 + z2 + 2(xy + yz + zx) = 196 60 + 2(xy + yz + zx) = 196 2(xy + yz + zx) = 196 – 60 = 136 xy + yz + zx = 136/2 = 68 x3 + y3 + z3 – 3xyz = (x + y + z)[x2 + y2 + z2 – xy – yz – zx] = (14)[60 – 68] = 14 × (-8) = -112 |
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| 47. |
x3 – 8y3 – 36xy – 216 का मान ज्ञात कीजिए, यदि x = 2y + 6 |
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Answer» x = 2y + 6 का मान रखने पर, (2y + 6)3 – 8y3 – 36(2y + 6)y – 216 = 8y3 + 216 + 36y (2y + 6) – 8y3 – 72y2 – 216y – 216 = 72y2 + 216y – 72y2 – 216y = 0 |
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