Theory of equation practice · Mixed difficulty

Option A (10) is correct because by expanding the polynomial with roots 1, 2, 3, and 4, and comparing coefficients, we can find the values of a, b, and c, and then calculate a + 2b + c = 10.

Option C ($\pm \;2$) is correct because by substituting $|x|$ into the equation and solving, we find that the real roots are $\pm 2$.

Option D, $(-2,2)$, is correct because the condition for three distinct real roots is that $f(1)f(-1) < 0$, which leads to $-2 < a < 2$.

Option B is correct because there are two possible sets of triplets (a1, a2, a3) that satisfy the given conditions of the quadratic equations having exactly one root in common.

Option C ($\frac{\alpha}{{\alpha + 1}},\frac{\beta}{{\beta + 1}}$) is correct because the roots of the transformed quadratic equation are related to the roots of the original equation in this specific way.

Option C (7) is correct because it involves finding the product of g(x) evaluated at the roots of f(x) and using Vieta's formulas.

Option D ($b = 24, c = -32$) is incorrect. The correct answer is D ($b = 24, c = 32$) because the roots are equal, allowing us to expand (x-2)⁴ and match coefficients.

Option A (6, 4, -1) is correct because the roots are in the ratio 3:2 and their sum is 9, satisfying Vieta's formulas.

Option A ($r = pq$) is correct because if two roots are equal in magnitude but opposite in sign, one root is $p$, and substituting this into the equation gives $r = pq$.

Option D (no real roots) is correct because the equation simplifies to a form where sin(x) must be greater than 1, which is impossible.

Option D (None of these) is correct because the determinant simplifies to 0, regardless of the values of $\alpha$ and $\beta$.

Option D ((5,4)) is correct because when ${x^2} - 3x + 2$ is a factor of ${x^4} - p{x^2} + q$, the roots of ${x^2} - 3x + 2 = 0$ must also be roots of ${x^4} - p{x^2} + q = 0$. Substituting these roots allows us to solve for p and q.

Option C ($a = b = c$ or $a = - 2b = - 2c$) is correct because it satisfies the divisibility condition of the polynomial by both $x-a$ and $x-b$.

Option C ($-2, 1, 4$) is correct because the sum of these roots is 3, which matches the coefficient of the x² term in the given cubic equation.

Option C (26) is correct because after finding the polynomial's coefficients using the given derivatives and value at x=2, the second derivative at x=1 is calculated to be 26.

Option B ($\left(-\frac{1}{2}, 0\right) \cup\left(0, \frac{1}{2}\right)$) is correct because by simplifying the given equation and analyzing the discriminant of the resulting quadratic, we can find the range of values for 'a' that satisfy the condition of having real and distinct roots.

Option C ($\pm \;3$) is correct because the common difference is found by using the properties of Arithmetic Progression and solving the resulting equations.

Option A ($\left(y^2+x\right)^4=C\left|\left(y^2+2 x\right)^3\right|$) is correct because it represents the general solution obtained after solving the given differential equation using appropriate substitutions and integration.

Option D (51) is correct because it is the sum of squares of a, b, and c, where a, b, and c are coefficients found by using the given conditions and derivatives.

Option A ($-r$) is correct because, given the sum of two roots is zero, $pq$ simplifies to $-r$ using Vieta's formulas.

Option A (11) is correct because it is the value of n that satisfies the given conditions for the quadratic equation to have two consecutive integral solutions.

Option D (-1-i) is correct because by using the sum of roots property for quadratic equations, the other root can be found.

The correct answer is D ($49 x^{2}-245 x+250=0$) because it is the quadratic equation formed with roots derived from the given conditions and relationships between the roots of the two equations.

Option D (5) is correct because using Vieta's formulas and substituting into the expression yields 5.

Option A ($\frac{{ - 15}}{4}$) is correct because it is the result of applying Vieta's formulas and algebraic manipulation to find the sum of the squares of the roots.

Option C (3) is the correct answer because it represents the sum of all real values of x that satisfy the given equation, considering all possible cases where the equation equals 1.

Option B (${f^{m - 1}}(x)$) is correct because if $(x-c)$ is a factor of order m of $f(x)$, then $x=c$ is a root of $f(x)$ and its first m-1 derivatives.

Option A ($p=-4, q=7$) is correct because if one root is a complex number, the other root is its conjugate, and we can use the sum and product of roots to find p and q.

Option B ($\pm \;3,\; \pm \;i$) is correct because these are the roots obtained by factoring the given quartic equation.

Option B (-6) is correct because the value of $(x-y)^3+(y-z)^3+(z-x)^3$ is -6, which is derived from the given equations.