| X Polynomial MCQ Assignments in Mathematics Class X (Term I) |
| 1. If α,β are zeroes of the polynomial f(x) = x2 + px + q, then polynomial having 1/α and 1/β as its zeroes is |
| (a) x2 + qx + p (b) x2 – px + q (c) qx2 + px + 1 (d) px2 + qx + 1 |
| 2. If α and β are zeroes of x2 – 4x + 1, then 1/α + 1/β – αβ is |
| (a) 3 (b) 5 (c) –5 (d) –3 |
| 3. The quadratic polynomial having zeroes as 1 and –2 is : |
| (a) x2 – x + 2 (b) x2 – x – 2 (c) x2 + x – 2 (d) x2 + x + 2 |
| 4. If α, β are zeroes of x2 – 6x + k, what is the value of k if 3α+2β=20 ? |
| (a)–16 (b) 8 (c) 2 (d) –8 |
| 5. If one zero of 2x2 – 3x + k is reciprocal to the other, then the value of k is |
| (a) 2 (b) −23 (c) −32 (d) –3 |
| 6. The quadratic polynomial whose sum of zeroes is 3 and product of zeroes is –2 is |
| (a) x2 + 3x – 2 (b) x2 – 2x + 3 (c) x2 – 3x + 2 (d) x2 – 3x – 2 |
| 7. If (x + 1) is a factor of x2 – 3ax + 3a – 7, then the value of a is : |
| (a)1 (b) –1 (c) 0 (d) –2 |
| 8. The number of polynomials having zeroes –2 and 5 is : |
| (a)1 (b) 2(c)3 (d) more than 3 |
| 9. The quadratic polynomial p(y) with –15 and–7 as sum and one of the zeroes respectively is : |
| (a) y2 – 15y – 56 (b) y2 – 15y + 56 (c) y2+ 15y + 56 (d) y2 + 15y – 56 |
| 10.The value of p for which the polynomial x3 + 4x2 – px + 8 is exactly divisible by (x – 2) is : |
| (a) 0 (b) 3 (c) 5 (d) 16 |
| 11. If 1 is a zero of the polynomial p(x) = ax2– 3(a – 1)x – 1, then the value of a is : |
| (a) 1 (b) –1 (c) 2 (d) –2 |
| 12. If –4 is a zero of the polynomial x2 – x – (2 + 2k), then the value of k is : |
| (a) 3 (b) 9 (c) 6 (d) –9 |
| 13.The degree of the polynomial (x + 1)(x2 – x – x4 + 1) is : (a) 2 (b) 3 (c) 4 (d) 5 |
| 14. If (x + 1) is a factor of x2– 3ax + 3a – 7, then the value of a is : |
| (a) 1 (b) –1 (c) 0 (d) –2 |
| 15. If sum of the squares of zeroes of the quadratic polynomial f(x) = x2 – 8x + k is 40, the value of k is |
| (a) 10 (b) 12 (c) 14 (d) 16 |
Class 09 Atoms and Molecules Numerical Problem based on Law of chemical Combination Law of conservation of mass Law of constant proportion Empirical formula 1. If 10 grams of CaCO 3 on heating gave 4.4g of CO 2 and 5.6g of CaO, show that these observations are in agreement with the law of conservation of mass.(Based on Law of conservation of mass) Solution: Mass of the reactants = 10g ; Mass of the products = 4.4 + 6.6g = 10g Since the mass of the reactants is equal to the mass of the products, the observations are in agreement with the law of conservation of mass. 2. 1.375 g of cupric oxide was reduced by heating and the weight of copper that remained was 1.098g. In another experiment 1.179 g of copper was dissolved in nitric acid and the resulting copper nitrate was converted into cupric oxide by ignition . The weight of cupric oxide formed was 1.476 g. which law of chemical combinations does this data state? Solution: in first experiment: Copper oxide = 1....
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