Algebra (Mathematics Chapterwise Solved Papers for Teaching)Theory of Equation and InequationsTotal Questions: 5041. A geometric progression has first term 3 and the sum of all its 2n terms is three times the sum of its odd place terms, the common ratio is(a) 2(b) 3(c)(d) √2Correct Answer: (a) 2Solution:42. If the equation 𝓍² + a𝓍 + b = 0 has roots α and β B, then value of (1 + α + α²) (1 + β + β²) is- [RPSC PGT 2022](a) a²+b²-ab-a-b(b) a²+b²-ab-a-b- 1(c) a²+b²-ab-a-b+1(d) a²+b²+ab-a-b+1Correct Answer: (c) a²+b²-ab-a-b+1Solution:43. The roots of the quadratic equation are real and equal if the discriminant is _______ zero. [DSE TGT 2022](a) ≥(b) ≤(c) ≠(d) =Correct Answer: (d) =Solution:We know that if discriminant of quadratic equation will equal to zero, then roots will real and equal.44. One non zero root of the equation (2√2)ˣ² = 8³ˣ [DSE TGT 2022](a) 4(b) 0(c) 6(d) 5Correct Answer: (c) 6Solution:45. One solution of the quadratic equation 2𝓍² - 7x = 39 is [DSE TGT 2022](a) 2(b) -3(c) 5(d) 4Correct Answer: (b) -3Solution:46. Minimum possible number of real roots of a quadratic equation is [DSE TGT 2022](a) 1(b) 0(c) 2(d) 3Correct Answer: (b) 0Solution:We known that any quadratic equation will have maximum two roots. But if discriminant will be negative, then roots will be imaginary and no real roots. Hence, minimum possible number of real roots of a quadratic equation is 0.47. What is the maximum number of distinet roots a quadratic equation can have?(a) 1(b) 0(c) 2(d) 3Correct Answer: (c) 2Solution:We know that a quadratic equation can have maximum two roots. If discriminant will be positive and greater than zero, then roots of eqⁿ will be real and distinct. Hence, a quadratic equation will have maximum two distinct roots.48. How many distinct real roots are there for the equation 𝓍²-6𝓍+5=0? [DSE TGT 2022](a) 0(b) 1(c) 2(d) InfiniteCorrect Answer: (c) 2Solution:Given; 𝓍² - 6𝓍 + 5 = 0 𝓍² - 5𝓍 - 𝓍 + 5 = 0 ⇒ 𝓍 (𝓍-5) −1 (𝓍-5) = 0 ⇒ (𝓍-5) (𝓍-1) =0 ⇒ 𝓍 = 5, 1 Hence, eqⁿ have two distinct roots.49. In which of the following interval k lies so that the equation k𝓍² - 8𝓍 + k = 0 possess real solution? [DSE TGT 2022](a) (-3,3)(b) [3,3](c) (-4,4)(d) [-4,4]Correct Answer: (d) [-4,4]Solution:Given; k𝓍² - 8𝓍 +k = 0 If eqⁿ have real solution, then discriminant; D≥0 ⇒ b² - 4ac ≥ 0 ⇒ (-8) 2 - 4× k × k ≥ 0 ⇒ 64 - 4k² ≥ 0 ⇒ 64 ≥ 4k² ⇒ k² ≤ 16 ⇒k ∈ [4, 4]50. Maximum value of the expression 5+4𝓍 - 4𝓍² is [DSE TGT 2022](a) 5(b) 6(c) 1(d) 2Correct Answer: (b) 6Solution:Submit Quiz« Previous12345