UGCNET Questions

Solving 2x + 3 ≤ 11, we subtract 3 from both sides to get 2x ≤ 8. Dividing both sides by 2 yields x ≤ 4. On the real number line, this corresponds to all numbers less than or equal to 4. In interval notation, that set is written as (−∞, 4], with a square bracket at 4 indicating that 4 is included.

Option A:
Option A, (−∞, 4), excludes 4 itself and would represent x < 4, which is a strict inequality, not the given ≤ relation. Option B: Option B correctly captures x ≤ 4 by including all values up to and including 4, thus satisfying the inequality exactly. Option C: Option C, (−∞, 7], includes many values greater than 4, violating the upper bound imposed by 2x ≤ 8. Option D: Option D, [4, ∞), represents x ≥ 4, the opposite direction of the inequality, and only includes 4 and greater values.

When multiplying powers of the same base, we keep the base unchanged and add the exponents. The expression 3⁴ × 3² therefore becomes 3^(4 + 2). Adding the exponents gives 3⁶. This rule arises directly from expanding each power into repeated multiplication and then counting the total number of factors of 3.

Option A:
Option A, 3⁸, results from incorrectly multiplying the exponents instead of adding them. There is no exponent law that justifies multiplying 4 and 2 in this context.

Option B:
Option B correctly applies the exponent rule aᵐ × aⁿ = aᵐ⁺ⁿ. By adding 4 and 2, it expresses the product as a single power of 3 with exponent 6.

Option C:
Option C, 3², keeps only one of the original exponents and discards the contribution of the other factor. This contradicts the idea of combining powers.

Option D:
Option D, 3⁴, similarly ignores the factor 3² and so cannot be equivalent to the full product.

Equal real roots occur when the discriminant of a quadratic equation is zero. For the equation 2x² − kx + 8 = 0, the discriminant is b² − 4ac, which becomes k² − 4(2)(8). Setting this equal to zero gives k² − 64 = 0. Solving, we get k² = 64 and hence k = ±8, but the given options only include 8 as a valid choice. Therefore, k = 8 satisfies the equal roots condition using the standard discriminant rule.

Option A:
Option A, 4√2, when squared gives 32, and substituting into the discriminant k² − 64 yields 32 − 64 = −32, which is negative. A negative discriminant would produce complex, not equal real, roots.

Option B:
Option B leads to k² = 64, so the discriminant becomes 64 − 64 = 0. This matches the precise condition for equal real roots in a quadratic and is consistent with the formula b² − 4ac = 0.

Option C:
Option C, 16, gives k² = 256 and a discriminant of 256 − 64 = 192, which is positive and corresponds to two distinct real roots rather than equal ones.

Option D:
Option D, 16√2, squares to 512, making the discriminant 512 − 64 = 448. This again yields unequal real roots, not the required equal roots.

To solve the inequality 2x − 5 > 7, we first add 5 to both sides, giving 2x > 12. Then we divide both sides by 2, which is positive, so the inequality direction remains the same, yielding x > 6. Thus, the solution set consists of all real numbers greater than 6, and this is the equivalent inequality in terms of x.

Option A:
Option A captures the full result of correctly isolating x. It incorporates both the addition and division steps and expresses the final constraint directly as x > 6.

Option B:
Option B, x > 5, would follow only if the inequality were 2x − 5 > 5, not 2x − 5 > 7. It represents a weaker condition than required.

Option C:
Option C, x > 4, ignores more of the numerical shift and significantly broadens the solution set beyond what the original inequality allows.

Option D:
Option D, x > 3, is even looser and does not align with the operations needed to isolate x in the given inequality.