Question 43

   Question: Let $\alpha$ and $\beta$ be the roots of the equation $px^2 + qx - r = 0$, where $p \neq 0$. If $p, q$ and $r$ be the consecutive terms of a non-constant G.P and $1/\alpha + 1/\beta = 3/4$, then the value of $(\alpha - \beta)^2$ is.

   Options: 

    A. 80/9

    B. 9

    C. 20/3

    D. 8

   Correct Answer: A.

   Year: JEE Main 2024 (1-Feb-2024 Shift 2).

   Solution (Source): Let $p = A, q = AR, r = AR^2$. The equation becomes $x^2 + Rx - R^2 = 0$. Given $1/\alpha + 1/\beta = 3/4 \implies \frac{\alpha+\beta}{\alpha\beta} = 3/4$. Substituting root relations $\frac{-R}{-R^2} = 3/4 \implies R = 4/3$. Then $(\alpha-\beta)^2 = (\alpha+\beta)^2 - 4\alpha\beta = R^2 - 4(-R^2) = 5R^2$. Substituting $R=4/3$ gives $80/9$.

   Step Solution:

    1.  Represent coefficients: Since $p, q, r$ are in G.P., let $p=A, q=AR, r=AR^2$. Divide $px^2 + qx - r = 0$ by $A$ to get $x^2 + Rx - R^2 = 0$.

    2.  Sum and Product: $\alpha + \beta = -R$ and $\alpha\beta = -R^2$.

    3.  Solve for R: Given $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{3}{4} \implies \frac{\alpha+\beta}{\alpha\beta} = \frac{3}{4}$. Substitute values: $\frac{-R}{-R^2} = \frac{3}{4} \implies \frac{1}{R} = \frac{3}{4} \implies \mathbf{R = 4/3}$.

    4.  Express required value: $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta = (-R)^2 - 4(-R^2) = \mathbf{5R^2}$.

    5.  Final calculation: $(\alpha - \beta)^2 = 5 \cdot (\frac{4}{3})^2 = 5 \cdot \frac{16}{9} = \mathbf{\frac{80}{9}}$.

   Difficulty level: Medium.

   The Concept Name: Relation between Roots and Coefficients and Geometric Progression.

   Short cut solution: Dividing the quadratic by the first term $p$ reduces the problem to a single variable $R$ immediately, simplifying all subsequent root relations.

 Question 53

   Question: Let $\alpha$ be a root of the equation $(a-c)x^2 + (b-a)x + (c-b) = 0$ where $a, b, c$ are distinct real numbers such that the matrix $\begin{bmatrix} \alpha^2 & \alpha & 1 \\ 1 & 1 & 1 \\ a & b & c \end{bmatrix}$ is singular. Then the value of $\frac{(a-c)^2}{(b-a)(c-b)} + \frac{(b-a)^2}{(a-c)(c-b)} + \frac{(c-b)^2}{(a-c)(b-a)}$ is.

   Options: 

    A. 6

    B. 3

    C. 9

   Correct Answer: B.

   Year: JEE Main 2023 (24-Jan-2023 Shift 1).

   Solution (Source): The determinant of the matrix is zero (singular). This condition along with the quadratic equation roots leads to the identity involving $x^3+y^3+z^3$. If $x+y+z=0$, then $\frac{x^3+y^3+z^3}{xyz} = 3$. Here $x=a-b, y=b-c, z=c-a$ (or similar permutations) sum to zero.

   Step Solution:

    1.  Find the root $\alpha$: In the equation $(a-c)x^2 + (b-a)x + (c-b) = 0$, the sum of coefficients is $(a-c) + (b-a) + (c-b) = 0$. Therefore, one root is $x=1$.

    2.  Verify Matrix Singularity: If $\alpha = 1$, the first two rows of the matrix become $$ and $$. Since two rows are identical, the determinant is zero and the matrix is singular.

    3.  Define variables: Let $x = (a-c)$, $y = (b-a)$, and $z = (c-b)$. Note that $x + y + z = 0$.

    4.  Simplify the expression: The given expression is $\frac{x^2}{yz} + \frac{y^2}{xz} + \frac{z^2}{xy} = \frac{x^3 + y^3 + z^3}{xyz}$.

    5.  Apply Algebraic Identity: Since $x+y+z=0$, then $x^3+y^3+z^3 = 3xyz$. Thus, the expression equals $\frac{3xyz}{xyz} = \mathbf{3}$.

   Difficulty level: Easy.

   The Concept Name: Sum of Coefficients Property and Algebraic Identities ($a^3+b^3+c^3=3abc$).

   Short cut solution: Recognize that if the sum of coefficients of a quadratic is zero, $x=1$ is a root. Then, notice the expression is in the form $\frac{x^2}{yz} + \frac{y^2}{zx} + \frac{z^2}{xy}$ where $x+y+z=0$, which is a standard identity resulting in 3.

   Question 58

   Question: Let $\lambda \neq 0$ be a real number. Let $\alpha, \beta$ be the roots of the equation $14x^2 - 31x + 3\lambda = 0$ and $\alpha, \gamma$ be the roots of the equation $35x^2 - 53x + 4\lambda = 0$. Then $\frac{3\alpha}{\beta}$ and $\frac{4\alpha}{\gamma}$ are the roots of the equation:

   Options: 

    A. $7x^2 + 245x - 250 = 0$

    B. $7x^2 - 245x + 250 = 0$

    C. $49x^2 - 245x + 250 = 0$

    D. $49x^2 + 245x + 250 = 0$

   Correct Answer: C

   Year: JEE Main 2023 (29-Jan-2023 Shift 1)

   Solution (Source): By solving for the common root $\alpha$ and the constant $\lambda$, we find $\alpha = 5/7$, $\beta = 3/2$, and $\gamma = 4/5$. The new roots are calculated as $10/7$ and $25/7$. The sum of roots is 5 and the product is $250/49$. The resulting equation is $49x^2 - 245x + 250 = 0$.

   Step Solution:

    1.  Find common root $\alpha$: Multiply the first equation by 5 and the second by 2 to eliminate $\lambda$ (or vice versa to eliminate $x^2$). Subtracting the equations gives $\mathbf{49\alpha = 7\lambda \implies \alpha = \lambda/7}$.

    2.  Solve for $\lambda$: Substitute $\alpha = \lambda/7$ back into $14x^2 - 31x + 3\lambda = 0$. Solving $14(\lambda/7)^2 - 31(\lambda/7) + 3\lambda = 0$ gives $\lambda = 5$, hence $\alpha = 5/7$.

    3.  Find $\beta$ and $\gamma$: From product of roots, $\alpha\beta = 3(5)/14 \implies \mathbf{\beta = 3/2}$. From $\alpha\gamma = 4(5)/35 \implies \mathbf{\gamma = 4/5}$.

    4.  Determine new roots: $R_1 = \frac{3\alpha}{\beta} = \frac{3(5/7)}{3/2} = \mathbf{10/7}$ and $R_2 = \frac{4\alpha}{\gamma} = \frac{4(5/7)}{4/5} = \mathbf{25/7}$.

    5.  Form equation: Sum $= 35/7 = \mathbf{5}$; Product $= 250/49$. The equation is $x^2 - 5x + 250/49 = 0$, which simplifies to $49x^2 - 245x + 250 = 0$.

   Difficulty level: Hard

   The Concept Name: Common Roots of Quadratic Equations

   Short cut solution: Note that the new roots $10/7$ and $25/7$ have a denominator of 7. The quadratic equation for these roots must have a leading coefficient that is a multiple of $7^2=49$, which immediately points to options C or D.

 Question 68

   Question: If for $z = \alpha + i\beta, |z + 2| = z + 4(1 + i)$, then $\alpha + \beta$ and $\alpha\beta$ are the roots of the equation:

   Options: 

    A. $x^2 + 3x - 4 = 0$

    B. $x^2 + 7x + 12 = 0$

    C. $x^2 + x - 12 = 0$

    D. $x^2 + 2x - 3 = 0$

   Correct Answer: B

   Year: JEE Main 2023 (8-Apr-2023 shift 1)

   Solution (Source): Substitute $z = \alpha + i\beta$ into the equation: $\sqrt{(\alpha + 2)^2 + \beta^2} = (\alpha + 4) + i(\beta + 4)$. Comparing imaginary parts, $\beta + 4 = 0 \implies \beta = -4$. Substituting $\beta$ into the real part and squaring gives $\alpha = 1$. The roots are $\alpha + \beta = -3$ and $\alpha\beta = -4$. Sum of roots is $-7$ and product is 12, giving $x^2 + 7x + 12 = 0$.

   Step Solution:

    1.  Set up equality: Substitute $z = \alpha + i\beta$ into the given equation: $\mathbf{\sqrt{(\alpha + 2)^2 + \beta^2} = (\alpha + 4) + i(\beta + 4)}$.

    2.  Solve imaginary part: Since the LHS is purely real, the imaginary part of the RHS must be zero: $\beta + 4 = 0 \implies \mathbf{\beta = -4}$.

    3.  Solve real part: Substitute $\beta = -4$ into the real equation: $\sqrt{(\alpha + 2)^2 + 16} = \alpha + 4$. Squaring both sides: $\alpha^2 + 4\alpha + 4 + 16 = \alpha^2 + 8\alpha + 16$.

    4.  Find $\alpha$: Simplify the equation: $4\alpha + 20 = 8\alpha + 16 \implies 4\alpha = 4 \implies \mathbf{\alpha = 1}$.

    5.  Form final equation: The target roots are $1 + (-4) = \mathbf{-3}$ and $1(-4) = \mathbf{-4}$. Equation: $x^2 - (-3-4)x + (-3)(-4) = \mathbf{x^2 + 7x + 12 = 0}$.

   Difficulty level: Easy

   The Concept Name: Equality of Complex Numbers and Modulus Definition

   Short cut solution: Recognize immediately that the modulus $|z+2|$ is real, so the imaginary part of the RHS, $i(\beta+4)$, must vanish. This gives $\beta = -4$ instantly, reducing the problem to a simple linear equation for $\alpha$.

Question 79

   Question: Let $\mathfrak{a}, \mathfrak{b}, \mathfrak{\eta}$ be the three roots of the equation $x^3 + bx + c = 0$. If $\beta\gamma = 1 = -a$, then $b^3 + 2c^3 - 3a^3 - 6\beta^3 - 8\gamma^3$ is equal to.

   Options: 

    A. 155/8

    B. 21

    C. 19

    D. 169/8

   Correct Answer: C

   Year: JEE Main 2023 (8-Apr-2023 shift 1)

   Solution (Source): Given $\beta\gamma = 1$ and $a = -1$. Substituting $a = -1$ into the equation gives $-1 - b + c = 0 \implies c - b = 1$. Product of roots $\alpha\beta\gamma = -c \implies (-1)(1) = -c \implies c = 1$. Thus $b = 0$. The roots are of $x^3 + 1 = 0$: $\alpha = -1, \beta = -\omega, \gamma = -\omega^2$. Expression $0 + 2 + 3 + 6 + 8 = 19$.

   Step Solution:

    1.  Use Vieta’s Formulas: Product of roots $\alpha\beta\gamma = -c/1$. Given $\alpha = -1$ and $\beta\gamma = 1$, then $(-1)(1) = -c \implies \mathbf{c=1}$.

    2.  Substitute root $\alpha = -1$ into the equation: $(-1)^3 + b(-1) + 1 = 0 \implies -1 - b + 1 = 0 \implies \mathbf{b=0}$.

    3.  Identify roots: The equation is $x^3 + 1 = 0$, so the roots are $-1, -\omega,$ and $-\omega^2$.

    4.  Evaluate required expression: $b^3 + 2c^3 - 3\alpha^3 - 6\beta^3 - 8\gamma^3 = 0 + 2(1)^3 - 3(-1)^3 - 6(-\omega)^3 - 8(-\omega^2)^3$.

    5.  Simplify powers: $2 + 3 - 6(-\omega^3) - 8(-\omega^6) = 2 + 3 + 6(1) + 8(1) = \mathbf{19}$.

   Difficulty level: Medium.

   The Concept Name: Theory of Equations (Vieta’s Formulas) and Cube Roots of Unity.

   Short cut solution: In the equation $x^3 + 1 = 0$, every root $x_i$ satisfies $x_i^3 = -1$. The expression simplifies to $0 + 2(1)^3 - 3(-1) - 6(-1) - 8(-1) = 2 + 3 + 6 + 8 = 19$ instantly.

   Question 89

   Question: If the sum of the squares of the reciprocals of the roots $\alpha$ and $\beta$ of the equation $3x^2 + \lambda x - 1 = 0$ is 15, then $6(\alpha^3 + \beta^3)^2$ is equal to:

   Options:

    A. 18

    B. 24

    C. 36

    D. 96

   Correct Answer: B

   Year: JEE Main 2022 (24-Jun-2022 Shift 1)

   Solution (Source): $\alpha + \beta = -\lambda/3$ and $\alpha\beta = -1/3$. Given $\frac{1}{\alpha^2} + \frac{1}{\beta^2} = 15 \implies \frac{(\alpha+\beta)^2-2\alpha\beta}{(\alpha\beta)^2} = 15 \implies \frac{\lambda^2/9 + 2/3}{1/9} = 15 \implies \lambda^2+6=15 \implies \lambda^2=9$. Then $6(\alpha^3 + \beta^3)^2 = 6\{(\alpha+\beta)((\alpha+\beta)^2-3\alpha\beta)\}^2 = 6(\frac{\lambda^2}{9})(\frac{\lambda^2}{9}+1)^2 = 24$.

   Step Solution:

    1.  Identify coefficients: From $3x^2 + \lambda x - 1 = 0$, we have $\alpha + \beta = -\lambda/3$ and $\alpha\beta = -1/3$.

    2.  Apply condition: $\frac{1}{\alpha^2} + \frac{1}{\beta^2} = 15 \implies \frac{\alpha^2+\beta^2}{(\alpha\beta)^2} = 15$.

    3.  Solve for $\lambda^2$: $\frac{\lambda^2/9 - 2(-1/3)}{1/9} = 15 \implies \lambda^2 + 6 = 15 \implies \lambda^2 = 9$.

    4.  Rewrite target expression: $6(\alpha^3 + \beta^3)^2 = 6[(\alpha + \beta)(\alpha^2 + \beta^2 - \alpha\beta)]^2 = 6(\alpha + \beta)^2 [(\alpha+\beta)^2 - 3\alpha\beta]^2$.

    5.  Substitute values: $6(\frac{9}{9})[\frac{9}{9} - 3(\frac{-1}{3})]^2 = 6(1)[1 + 1]^2 = 6 \times 4 = \mathbf{24}$.

   The difficulty level: Medium.

   The Concept Name: Relation Between Roots and Coefficients (Vieta's Formulas).

   Short cut solution: Since $(\alpha\beta)^2 = 1/9$, the condition $\frac{\alpha^2+\beta^2}{(\alpha\beta)^2} = 15$ means $\alpha^2+\beta^2 = 15/9 = 5/3$. Then $\alpha^3+\beta^3 = (\alpha+\beta)(\alpha^2+\beta^2-\alpha\beta) = (\alpha+\beta)(5/3 + 1/3) = 2(\alpha+\beta)$. The expression is $6(2(\alpha+\beta))^2 = 24(\alpha+\beta)^2 = 24(\lambda^2/9) = 24(1) = 24$.

   Question 104

   Question: Let $a, b \in \mathbb{R}$ be such that the equation $ax^2 - 2bx + 15 = 0$ has a repeated root $\alpha$. If $\alpha$ and $\beta$ are the roots of the equation $x^2 - 2bx + 21 = 0$, then $\alpha^2 + \beta^2$ is equal to:

   Options: 

    A. 37

    B. 58

    C. 68

    D. 92

   Correct Answer: B

   Year: JEE Main 2022-Shift-2 (25-Jun-2022)

   Solution (Source): $ax^2 - 2bx + 15 = 0$ has repeated root so $b^2 = 15a$ and $\alpha = 15/b$. Since $\alpha$ is a root of $x^2 - 2bx + 21 = 0$, substituting gives $225/b^2 = 9 \Rightarrow b^2 = 25$. Now $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 4b^2 - 42 = 100 - 42 = 58$.

   Step Solution:

    1.  For $ax^2 - 2bx + 15 = 0$ to have a repeated root $\alpha$, the discriminant $D = 0 \Rightarrow 4b^2 - 60a = 0 \Rightarrow \mathbf{a = b^2/15}$.

    2.  Find the repeated root: $\alpha = \frac{-(-2b)}{2a} = \frac{b}{a}$. Substitute $a$ from step 1: $\alpha = \frac{b}{b^2/15} = \mathbf{\frac{15}{b}}$.

    3.  Since $\alpha$ is a root of $x^2 - 2bx + 21 = 0$, substitute $\alpha = 15/b$: $(\frac{15}{b})^2 - 2b(\frac{15}{b}) + 21 = 0$.

    4.  Solve for $b^2$: $\frac{225}{b^2} - 30 + 21 = 0 \Rightarrow \frac{225}{b^2} = 9 \Rightarrow \mathbf{b^2 = 25}$.

    5.  Calculate $\alpha^2 + \beta^2$ for the second equation: $(\alpha+\beta)^2 - 2\alpha\beta = (2b)^2 - 2(21) = 4(25) - 42 = \mathbf{58}$.

   Difficulty level: Medium.

   The Concept Name: Condition for Repeated Roots and Relation between Roots and Coefficients.

   Short cut solution: Use the repeated root condition to find $\alpha = 15/b$ quickly. Plugging this into the second equation $x^2 - 2bx + 21 = 0$ gives a simple equation for $b^2$. Then use the identity $\alpha^2 + \beta^2 = (\text{sum})^2 - 2(\text{product})$.

   Question 109

   Question: Let $f(x)$ be a quadratic polynomial such that $f(-2) + f(3) = 0$. If one of the roots of $f(x) = 0$ is $-1$, then the sum of the roots of $f(x) = 0$ is equal to:

   Options: 

    A. 11/3

    B. 7/3

    C. 13/3

    D. 14/3

   Correct Answer: A.

   Year: JEE Main 2022-Shift-2 (28-Jun-2022).

   Solution: Since $x = -1$ is a root, let $f(x) = A(x + 1)(x - b)$. Given $f(-2) + f(3) = 0$, we substitute the values: $A[-1(-2 - b) + 4(3 - b)] = 0$. Solving this gives $b = 14/3$. The sum of the roots is $14/3 + (-1) = 11/3$.

   Step Solution:

    1.  Define the polynomial: Let the roots be $-1$ and $r$. Then $f(x) = k(x + 1)(x - r)$.

    2.  Evaluate at $-2$: $f(-2) = k(-2 + 1)(-2 - r) = k(-1)(-2 - r) = \mathbf{k(2 + r)}$.

    3.  Evaluate at $3$: $f(3) = k(3 + 1)(3 - r) = \mathbf{k(12 - 4r)}$.

    4.  Apply the sum condition: $f(-2) + f(3) = 0 \Rightarrow k(2 + r + 12 - 4r) = 0 \Rightarrow 14 - 3r = 0 \Rightarrow \mathbf{r = 14/3}$.

    5.  Calculate sum of roots: Sum $= -1 + r = -1 + 14/3 = \mathbf{11/3}$.

   Difficulty level: Easy.

   The Concept Name: Factor Theorem and Roots of Polynomials.

   Short cut solution: Use the property that for $f(x) = a(x-x_1)(x-x_2)$, the condition $f(p)+f(q)=0$ with $x_1$ known is a linear equation in $x_2$. Plug in the numbers and solve for $x_2$ directly.

Question 112

   Question: Let p and q be two real numbers such that $p + q = 3$ and $p ^ { 4 } + q ^ { 4 } = 3 6 9$. Then $( \frac { 1 } { p } + \frac { 1 } { q } ) ^ { - 2 }$ is equal to:

   Options: Numerical Value Type (Result is 4).

   Correct Answer: 4

   Year: JEE Main 2022-Shift-2 (26-Jun-2022)

   Solution (Source): From $p + q = 3$, we use the identity for $p^4 + q^4$ in terms of $pq$. Solving $\{(p+q)^2 - 2pq\}^2 - 2p^2q^2 = 369$ results in a quadratic $(pq)^2 - 18pq - 144 = 0$, giving $pq = -6$ (rejecting 24 as $p, q$ are real). Then $(\frac{1}{p} + \frac{1}{q})^{-2} = (\frac{pq}{p+q})^2 = (-6/3)^2 = 4$.

   Step Solution:

    1.  Express required term: $(\frac{1}{p} + \frac{1}{q})^{-2} = (\frac{p+q}{pq})^{-2} = \mathbf{(\frac{pq}{p+q})^2}$.

    2.  Set up power identity: $p^4 + q^4 = (p^2 + q^2)^2 - 2p^2q^2 = \mathbf{\{(p+q)^2 - 2pq\}^2 - 2(pq)^2 = 369}$.

    3.  Solve for pq: Substitute $p+q=3$ into the equation: $(9 - 2pq)^2 - 2(pq)^2 = 369 \implies \mathbf{(pq)^2 - 18pq - 144 = 0}$.

    4.  Find valid product: Factor to $(pq - 24)(pq + 6) = 0$; $pq = -6$ is valid for real roots because $(p+q)^2 \geq 4pq \implies 9 \geq -24$.

    5.  Final calculation: Substitute $pq = -6$ and $p+q=3$ into the target: $(-6/3)^2 = \mathbf{4}$.

   Difficulty level: Medium.

   The Concept Name: Symmetric Functions of Roots / Algebraic Identities.

   Short cut solution: Note that the expression is simply $(\frac{pq}{3})^2$. Finding $pq$ quickly from $p^4+q^4$ by testing small integers that sum to 3 (like 5 and -2) reveals $5^4 + (-2)^4 = 625 + 16 \neq 369$; however, testing $p=6, q=-3$ gives $6-3=3$ and $1296 + 81 \neq 369$; eventually identifying $pq = -6$ by solving the quadratic is fastest.

   Question 127

   Question: Let a, b be two non-zero real numbers. If p and r are the roots of the equation $x^2 - 8ax + 2a = 0$ and $q$ and s are the roots of the equation $x^2 + 12bx + 6b = 0$, such that $1/p, 1/q, 1/r, 1/s$ are in A.P., then $a^{-1} - b^{-1}$ is equal to

   Options: Numerical Value Type (Result is 38).

   Correct Answer: 38

   Year: JEE Main 2022-Shift-1 (25-Jul-2022)

   Solution (Source): Roots of $2ax^2 - 8ax + 1 = 0$ are $1/p$ and $1/r$. Roots of $6bx^2 + 12bx + 1 = 0$ are $1/q$ and $1/s$. Let the roots be $\alpha - 3\beta, \alpha - \beta, \alpha + \beta, \alpha + 3\beta$. From sums and products of these AP terms, we find $1/a = -10$ and $1/b = -48$. Thus $1/a - 1/b = 38$.

   Step Solution:

    1.  Transform equations for reciprocal roots: For $x^2 - 8ax + 2a = 0$, the equation for $y=1/x$ is $2ay^2 - 8ay + 1 = 0$. Sum of roots $1/p + 1/r = 4$.

    2.  For $x^2 + 12bx + 6b = 0$, the equation for $y=1/x$ is $6by^2 + 12by + 1 = 0$. Sum of roots $1/q + 1/s = -2$.

    3.  Set the AP terms: Let $1/p, 1/q, 1/r, 1/s$ be $A, A+d, A+2d, A+3d$. Then $(A) + (A+2d) = 2A+2d = 4$ and $(A+d) + (A+3d) = 2A+4d = -2$.

    4.  Solve for A and d: Subtracting gives $2d = -6 \Rightarrow \mathbf{d = -3}$, then $2A - 6 = 4 \Rightarrow \mathbf{A = 5}$. Roots are $5, 2, -1, -4$.

    5.  Find a and b: Product $1/pr = 5(-1) = -5 = 1/2a \Rightarrow \mathbf{1/a = -10}$. Product $1/qs = 2(-4) = -8 = 1/6b \Rightarrow \mathbf{1/b = -48}$. Result: $(-10) - (-48) = \mathbf{38}$.

   Difficulty level: Hard.

   The Concept Name: Arithmetic Progression and Reciprocal Roots.

   Short cut solution: Use the property that if $x_1, x_2, x_3, x_4$ are in AP, then $x_1+x_3 = 2x_2$ and $x_2+x_4 = 2x_3$. Here $4 = 2(1/q)$ and $-2 = 2(1/r)$ doesn't apply directly because the order of $p, q, r, s$ in the AP is specific. Transforming to $1/x$ equations is the most efficient path.

Question 130

   Question: The minimum value of the sum of the squares of the roots of $x^2 + (3 - a)x + 1 = 2a$ is:

   Options: 

    A. 4

    B. 5

    C. 6

    D. 8

   Correct Answer: C

   Year: JEE Main 2022-Shift-2 (26-Jul-2022)

   Solution (Source): $x^2 + (3 - a)x + 1 = 2a \Leftrightarrow \alpha + \beta = a - 3, \alpha\beta = 1 - 2a$. $\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = a^2 - 6a + 9 - 2 + 4a = a^2 - 2a + 7 = (a-1)^2 + 6$. Minimum value is 6.

   Step Solution:

    1.  Standardize the quadratic: $x^2 + (3-a)x + (1-2a) = 0$.

    2.  Sum and Product: $\alpha + \beta = -(3-a) = \mathbf{a-3}$ and $\alpha\beta = \mathbf{1-2a}$.

    3.  Express sum of squares: $S = \alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$.

    4.  Substitute and simplify: $S = (a-3)^2 - 2(1-2a) = a^2 - 6a + 9 - 2 + 4a = \mathbf{a^2 - 2a + 7}$.

    5.  Complete the square: $S = (a-1)^2 + 6$. The minimum value is clearly 6 when $a=1$.

   Difficulty level: Easy.

   The Concept Name: Relation between Roots and Coefficients / Completing the Square.

   Short cut solution: Once you obtain the quadratic $a^2 - 2a + 7$, use the vertex formula for minimum value: $y = c - \frac{b^2}{4a} = 7 - \frac{(-2)^2}{4(1)} = 7 - 1 = 6$.

   Question 132

   Question: If $\alpha, \beta$ are the roots of the equation $x^2 - \left( 5 + 3 ^ { \sqrt { \log _ { 3 } 5 } } - 5 ^ { \sqrt { \log _ { 5 } 3 } } \right) x + 3 \left( 3 ^ { ( \log _ { 3 } 5 ) ^ { \frac 1 3 } } - 5 ^ { ( \log _ { 5 } 3 ) ^ { \frac 2 3 } } - 1 \right) = 0$, then the equation, whose roots are $\alpha + \frac{1}{\beta}$ and $\beta + \frac{1}{\alpha}$, is:

   Options: 

    A. $3 x ^ { 2 } - 2 0 x - 1 2 = 0$

    B. $3 x ^ { 2 } - 1 0 x - 4 = 0$

    C. $3 x ^ { 2 } - 1 0 x + 2 = 0$

    D. $3 x ^ { 2 } - 2 0 x + 1 6 = 0$

   Correct Answer: B

   Year: JEE Main 2022 (Online) 27th July Shift-2

   Solution: The terms $3^{\sqrt{\log_3 5}} - 5^{\sqrt{\log_5 3}}$ and $3^{(\log_3 5)^{1/3}} - 5^{(\log_5 3)^{2/3}}$ simplify to zero using logarithmic identities. The equation becomes $x^2 - 5x + 3(-1) = 0$. Sum of roots is 5 and product is -3. For new roots $\alpha + \frac{1}{\beta}$ and $\beta + \frac{1}{\alpha}$, the sum is $5 - \frac{5}{3} = \frac{10}{3}$ and the product is $2 - 3 - \frac{1}{3} = -\frac{4}{3}$. The resulting equation is $3x^2 - 10x - 4 = 0$.

   Step Solution:

    1.  Simplify coefficients: Use the identity $a^{\sqrt{\log_a b}} = b^{\sqrt{\log_b a}}$ to see that $3^{\sqrt{\log_3 5}} = 5^{\sqrt{\log_5 3}}$, so their difference is 0.

    2.  Simplify second term: Similarly, $3^{(\log_3 5)^{1/3}} = 5^{(\log_5 3)^{2/3}}$ (as $3^{\log_3 5} = 5$), making their difference 0. The equation simplifies to $x^2 - 5x - 3 = 0$.

    3.  Sum and Product: $\alpha + \beta = \mathbf{5}$ and $\alpha\beta = \mathbf{-3}$.

    4.  Find new sum ($S$): $S = (\alpha + \beta) + (\frac{\alpha + \beta}{\alpha\beta}) = 5 + \frac{5}{-3} = \mathbf{\frac{10}{3}}$.

    5.  Find new product ($P$): $P = 2 + \alpha\beta + \frac{1}{\alpha\beta} = 2 - 3 - \frac{1}{3} = \mathbf{-\frac{4}{3}}$. Equation: $x^2 - \frac{10}{3}x - \frac{4}{3} = 0 \Rightarrow \mathbf{3x^2 - 10x - 4 = 0}$.

   Difficulty level: Hard

   The Concept Name: Logarithmic Identities and Symmetric Functions of Roots

   Short cut solution: Recognize that the complex exponents are designed to cancel out. Once the equation is reduced to $x^2 - 5x - 3 = 0$, the calculation of the new quadratic is straightforward arithmetic.

Question 134

   Question: Let $\alpha, \beta$ be the roots of the equation $x^2 - \sqrt{2}x + \sqrt{6} = 0$ and $\frac{1}{\alpha^2} + 1, \frac{1}{\beta^2} + 1$ be the roots of the equation $x^2 + ax + b = 0$. Then the roots of the equation $x^2 - (a + b - 2)x + (a + b + 2) = 0$ are :

   Options: 

    A. non-real complex numbers

    B. real and both negative

    C. real and both positive

    D. real and exactly one of them is positive

   Correct Answer: B

   Year: JEE Main 2022 (Online) 28th July Shift-2

   Solution: Given $\alpha + \beta = \sqrt{2}$ and $\alpha\beta = \sqrt{6}$. The sum and product of the new roots $\frac{1}{\alpha^2}+1$ and $\frac{1}{\beta^2}+1$ are used to find $a$ and $b$, resulting in $a+b = -5/6$. Substituting this into the final quadratic yields roots that are real and both negative.

   Step Solution:

    1.  Properties of $\alpha, \beta$: $\alpha+\beta = \sqrt{2}$, $\alpha\beta = \sqrt{6}$, and $\alpha^2+\beta^2 = (\sqrt{2})^2 - 2\sqrt{6} = \mathbf{2 - 2\sqrt{6}}$.

    2.  Find $a$: $-a = (\frac{1}{\alpha^2}+1) + (\frac{1}{\beta^2}+1) = 2 + \frac{\alpha^2+\beta^2}{(\alpha\beta)^2} = 2 + \frac{2-2\sqrt{6}}{6} = \mathbf{\frac{7-\sqrt{6}}{3}}$.

    3.  Find $b$: $b = (\frac{1}{\alpha^2}+1)(\frac{1}{\beta^2}+1) = 1 + \frac{\alpha^2+\beta^2}{(\alpha\beta)^2} + \frac{1}{(\alpha\beta)^2} = 1 + \frac{2-2\sqrt{6}}{6} + \frac{1}{6} = \mathbf{\frac{9-2\sqrt{6}}{6}}$.

    4.  Calculate $a+b$: Adding the results gives $a+b = \mathbf{-5/6}$.

    5.  Form final equation: $x^2 - (-5/6-2)x + (-5/6+2) = 0 \Rightarrow \mathbf{x^2 + \frac{17}{6}x + \frac{7}{6} = 0}$. Since all coefficients are positive, the roots must be real and negative.

   Difficulty level: Medium

   The Concept Name: Transformation of Roots and Nature of Roots

   Short cut solution: Since the sum of roots ($17/6$) and product of roots ($7/6$) of the final equation are both positive, but the sum formula is $-B/A$, the actual sum is negative. Positive product and negative sum imply both roots are negative.

   Question 135

   Question: Let $f(x) = ax^2 + bx + c$ be such that $f(1) = 3$, $f(-2) = \lambda$ and $f(3) = 4$. If $f(0) + f(1) + f(-2) + f(3) = 14$, then $\lambda$ is equal to :

   Options: 

    A. -4

    B. 13/2

    C. 23/2

    D. 4

   Correct Answer: D

   Year: JEE Main 2022 (Online) 28th July Shift-2

   Solution: We use the given values to set up a system of linear equations: $a+b+c=3$, $9a+3b+c=4$, and $4a-2b+c=\lambda$. From the sum condition $f(0)+f(1)+f(-2)+f(3)=14$, we find $c+\lambda=7$. Solving this system yields $a=1/6, b=-1/6, c=3$, and finally $\lambda = 4$.

   Step Solution:

    1.  Use sum condition: $f(0) + 3 + \lambda + 4 = 14 \Rightarrow c + \lambda + 7 = 14 \Rightarrow \mathbf{c + \lambda = 7}$.

    2.  Represent $\lambda$: From $f(-2) = \lambda$, we have $\mathbf{4a - 2b + c = \lambda}$.

    3.  Set up coefficient equations: $a + b + c = 3$ and $9a + 3b + c = 4$.

    4.  Solve for $a$ and $b$: Subtracting the first from the second gives $8a + 2b = 1$. Substituting $b = (1-8a)/2$ into $a+b+c=3$ gives $\mathbf{c - 3a = 5/2}$.

    5.  Final calculation: Substitute $b$ into the $\lambda$ equation: $\lambda = 4a - (1-8a) + c = 12a + c - 1$. Using $c+\lambda=7$, we get $(12a+c-1) + c = 7 \Rightarrow 12a + 2c = 8 \Rightarrow 6a + c = 4$. Solving this with $c-3a=2.5$ gives $a=1/6, c=3$, hence $\lambda = 4$.

   Difficulty level: Easy

   The Concept Name: Systems of Linear Equations in Quadratic Coefficients

   Short cut solution: The condition $c+\lambda=7$ is key. By testing small integer values for $c$ and checking if they satisfy the other two points, $c=3$ and $\lambda=4$ are found to work instantly.

   Question 144

   Question: If $\alpha, \beta \in \mathbf{R}$ are such that $1 - 2i$ (here $i^2 = -1$) is a root of $z^2 + \alpha z + \beta = 0$, then $(\alpha - \beta)$ is equal to:

   Options: 

    A. 3

    B. -3

    C. 7

    D. -7

   Correct Answer: D

   Year: JEE Main 2021, 25 Feb. Shift-II

   Solution: Given root of $z^2 + \alpha z + \beta = 0$ is $1 - 2i$. Since it is a quadratic equation with real coefficients and one root is complex, its another root is the complex conjugate. Therefore, two roots are $1 - 2i$ and $1 + 2i$. Sum of roots $= -\alpha = (1 - 2i) + (1 + 2i) = 2 \implies \alpha = -2$. Product of roots $= \beta = (1 - 2i)(1 + 2i) = 1 + 4 = 5$. Thus $\alpha - \beta = -2 - 5 = -7$.

   Step Solution:

    1.  Identify that since $\alpha, \beta$ are real, complex roots must occur in conjugate pairs. If $z_1 = 1 - 2i$, then $z_2 = 1 + 2i$.

    2.  Use the sum of roots formula: $z_1 + z_2 = -\alpha / 1$.

    3.  Calculate $\alpha$: $(1 - 2i) + (1 + 2i) = 2 \implies -\alpha = 2 \implies \mathbf{\alpha = -2}$.

    4.  Use the product of roots formula: $z_1 \cdot z_2 = \beta / 1$.

    5.  Calculate $\beta$: $(1 - 2i)(1 + 2i) = 1^2 - (2i)^2 = 1 + 4 = 5$. Thus, $\mathbf{\beta = 5}$. The result $\alpha - \beta = -2 - 5 = \mathbf{-7}$.

   Difficulty level: Easy

   The Concept Name: Conjugate Root Theorem and Vieta's Formulas.

   Short cut solution: Form the equation directly: $(z - (1-2i))(z - (1+2i)) = z^2 - 2z + 5 = 0$. Compare coefficients: $\alpha = -2, \beta = 5$. Then $\alpha - \beta = -7$.

Question 147

   Question: Let a, b, c be in an arithmetic progression. Let the centroid of the triangle with vertices $(a, c), (2, b)$ and $(a, b)$ be $(10/3, 7/3)$. If $\alpha, \beta$ are the roots of the equation $ax^2 + bx + 1 = 0$, then the value of $\alpha^2 + \beta^2 - \alpha\beta$ is:

   Options: 

    A. 71/256

    B. 69/256

    C. -69/256

    D. -71/256

   Correct Answer: D

   Year: JEE Main 2021, 24 Feb. Shift-II

   Solution: Given a, b, c are in AP, so $2b = a + c$. Centroid $= (\frac{a+2+a}{3}, \frac{c+b+b}{3}) = (\frac{10}{3}, \frac{7}{3})$. This gives $2a + 2 = 10 \implies a = 4$ and $c + 2b = 7$. Substituting $a=4$ into AP relation gives $2b = 4 + c \implies c = 2b - 4$. Then $(2b - 4) + 2b = 7 \implies 4b = 11 \implies b = 11/4$. Roots of $4x^2 + \frac{11}{4}x + 1 = 0$ give $\alpha + \beta = -11/16$ and $\alpha\beta = 1/4$. The value $\alpha^2 + \beta^2 - \alpha\beta = (\alpha + \beta)^2 - 3\alpha\beta = \frac{121}{256} - \frac{3}{4} = -71/256$.

   Step Solution:

    1.  Solve for $a$ using the x-centroid: $\frac{a+2+a}{3} = \frac{10}{3} \implies 2a+2=10 \implies \mathbf{a=4}$.

    2.  Set up the y-centroid and AP relation: $\frac{c+2b}{3} = \frac{7}{3} \implies c+2b=7$ and $a+c=2b \implies 4+c=2b$.

    3.  Solve for $b$: Substitute $c=2b-4$ into $c+2b=7 \implies 4b-4=7 \implies \mathbf{b=11/4}$.

    4.  Find root relations for $4x^2 + \frac{11}{4}x + 1 = 0$: $\alpha + \beta = -b/a = \mathbf{-11/16}$ and $\alpha\beta = 1/a = \mathbf{1/4}$.

    5.  Calculate expression: $\alpha^2 + \beta^2 - \alpha\beta = (\alpha+\beta)^2 - 3\alpha\beta = \frac{121}{256} - \frac{3}{4} = \frac{121-192}{256} = \mathbf{-71/256}$.

   Difficulty level: Medium

   The Concept Name: Centroid of a Triangle and Symmetric Functions of Roots.

   Short cut solution: Once $a$ and $b$ are found, the expression is $(\alpha+\beta)^2 - 3\alpha\beta$. Directly substitute $(-11/16)^2 - 3(1/4)$ to avoid solving for roots individually.

   Question 150

   Question: Let $p$ and $q$ be two positive numbers such that $p + q = 2$ and $p^4 + q^4 = 272$. Then $p$ and $q$ are roots of the equation:

   Options: 

    A. $x^2 - 2x + 2 = 0$

    B. $x^2 - 2x + 8 = 0$

    C. $x^2 - 2x + 136 = 0$

    D. $x^2 - 2x + 16 = 0$

   Correct Answer: D

   Year: JEE Main 2021, 24 Feb. Shift-I

   Solution (Source): We have $(p^2 + q^2)^2 - 2p^2q^2 = 272 \implies ((p+q)^2 - 2pq)^2 - 2p^2q^2 = 272$. Substituting $p+q=2$, we get $(4 - 2pq)^2 - 2p^2q^2 = 272 \implies 16 - 16pq + 2p^2q^2 = 272$. This leads to $(pq)^2 - 8pq - 128 = 0$. Solving gives $pq = 16$ or $-8$. Since $p, q > 0$, $pq = 16$. The equation is $x^2 - 2x + 16 = 0$.

   Step Solution:

    1.  Use the identity for the fourth power sum: $p^4 + q^4 = (p^2 + q^2)^2 - 2(pq)^2 = 272$.

    2.  Substitute $p^2 + q^2 = (p+q)^2 - 2pq$. Since $p+q=2$, then $p^2 + q^2 = 4 - 2pq$.

    3.  Form the quadratic in $pq$: $(4 - 2pq)^2 - 2(pq)^2 = 272 \implies 16 + 4(pq)^2 - 16pq - 2(pq)^2 = 272$.

    4.  Simplify the quadratic equation: $2(pq)^2 - 16pq - 256 = 0 \implies (pq)^2 - 8pq - 128 = 0$.

    5.  Solve for $pq$: $(pq - 16)(pq + 8) = 0$. Since $p, q > 0$, we take $pq = 16$. The resulting equation is $x^2 - (p+q)x + pq = \mathbf{x^2 - 2x + 16 = 0}$.

   Difficulty level: Medium.

   The Concept Name: Relation between Roots and Coefficients / Algebraic Identities.

   Short cut solution: In a quadratic $x^2 - Sx + P = 0$, $S=2$ is constant across options. Testing $P=16$ (Option D) gives roots as $1 \pm i\sqrt{15}$. Testing $P=16$ in $p^4 + q^4$ using symmetric sums is faster than full factorization.

Question 184

   Question: Let $\lambda \neq 0$ be in $\mathbb{R}$. If $\alpha$ and $\beta$ are the roots of the equation $x ^ { 2 } - x + 2 \lambda = 0$ and $\alpha$ and $\gamma$ are the roots of equation $3 x ^ { 2 } - 1 0 x + 2 7 \lambda = 0$, then $\frac { \beta \gamma } { \lambda }$ is equal to.

   Options: Numerical Value Type (The result is 18).

   Correct Answer: 18

   Year: 2021, 26 Aug. Shift-II

   Solution: $\alpha$ is common root of the equations $x ^ { 2 } - x + 2 \lambda = 0$ and $3 x ^ { 2 } - 1 0 x + 2 7 \lambda = 0$. Now, common root of these equations is $( 3 \alpha ^ { 2 } - 1 0 \alpha + 2 7 \lambda ) - ( 3 \alpha ^ { 2 } - 3 \alpha + 6 \lambda ) = 0 \Rightarrow - 7 \alpha + 2 1 \lambda = 0 \Rightarrow \alpha = 3 \lambda$. Again, $\alpha$ is root of $x ^ { 2 } - x + 2 \lambda = 0 \Rightarrow ( 3 \lambda ) ^ { 2 } - 3 \lambda + 2 \lambda = 0 \Rightarrow 9 \lambda ^ { 2 } - \lambda = 0 \Rightarrow \lambda = 1 / 9$. Then $\alpha = 1 / 3$. Sum of roots $\alpha + \beta = 1 \Rightarrow \beta = 1 - 1/3 = 2/3$. Sum of roots $\alpha + \gamma = 10/3 \Rightarrow \gamma = 10/3 - 1/3 = 3$. So $\beta \gamma / \lambda = (2/3 \times 3) / (1/9) = 18$.

   Step Solution:

    1.  Identify the common root: Since $\alpha$ is a root for both, multiply the first by 3 and subtract: $(3x^2 - 10x + 27\lambda) - 3(x^2 - x + 2\lambda) = 0$.

    2.  Solve for $\alpha$ in terms of $\lambda$: $-7\alpha + 21\lambda = 0 \implies \mathbf{\alpha = 3\lambda}$.

    3.  Find $\lambda$: Substitute $\alpha = 3\lambda$ into $x^2 - x + 2\lambda = 0$ to get $9\lambda^2 - 3\lambda + 2\lambda = 0$. Since $\lambda \neq 0$, then $9\lambda = 1 \implies \mathbf{\lambda = 1/9}$.

    4.  Find roots $\beta$ and $\gamma$: Using $\alpha = 3(1/9) = 1/3$, the sum $\alpha + \beta = 1$ gives $\mathbf{\beta = 2/3}$. From $3x^2 - 10x + 27\lambda = 0$, the sum $\alpha + \gamma = 10/3$ gives $\mathbf{\gamma = 3}$.

    5.  Final Calculation: $\frac{\beta\gamma}{\lambda} = \frac{(2/3) \cdot 3}{1/9} = \frac{2}{1/9} = \mathbf{18}$.

   Difficulty level: Hard

   The Concept Name: Common Roots of Quadratic Equations

   Short cut solution: Use the common root condition $\alpha = 3\lambda$ and immediately find $\lambda = 1/9$ by substitution. Since $\beta = 2/3$ and $\gamma = 3$ follow easily from sum of roots, $\beta\gamma/\lambda = 2 / (1/9) = 18$.

    Question 191

   Question: The numbers of pairs $( a , b )$ of real numbers, such that whenever $\alpha$ is a root of the equation $x ^ { 2 } + a x + b = 0$, $a ^ { 2 } - 2$ is also a root of this equation, is.

   Options: A. 6, B. 2, C. 4, D. 8.

   Correct Answer: A

   Year: 2021, 01 Sep. Shift-II

   Solution: Case I: If roots are same, $\alpha = \alpha^2 - 2 \implies \alpha = -1, 2$. This gives pairs $(2, 1)$ and $(-4, 4)$. Case II: If roots are distinct $\alpha$ and $\beta$: (a) $\alpha = \alpha^2 - 2$ and $\beta = \beta^2 - 2$ gives $(a, b) = (-1, -2)$. (b) $\alpha = \beta^2 - 2$ and $\beta = \alpha^2 - 2$ gives $(1, -1)$. (c) $\alpha = \alpha^2 - 2$ and $\beta \neq \beta^2 - 2$ gives $(0, -4)$ and $(0, -1)$. Total 6 pairs.

   Step Solution:

    1.  Case 1 (Equal roots): If $\alpha = \beta$, then $\alpha$ must satisfy $\alpha = \alpha^2 - 2 \implies \alpha = 2$ or $-1$. For $\alpha=2$, equation is $(x-2)^2 = x^2-4x+4 \implies \mathbf{(a,b)=(-4,4)}$. For $\alpha=-1$, equation is $(x+1)^2 = x^2+2x+1 \implies \mathbf{(a,b)=(2,1)}$.

    2.  Case 2 (Roots map to each other): If $\alpha = \beta^2 - 2$ and $\beta = \alpha^2 - 2$, then $\alpha - \beta = (\beta - \alpha)(\beta + \alpha)$, giving $\alpha + \beta = -1$ and $\alpha\beta = -1$. This gives $\mathbf{(a,b)=(1,-1)}$.

    3.  Case 3 (Roots map to themselves): If $\alpha = \alpha^2 - 2$ and $\beta = \beta^2 - 2$, roots are $(2, -1)$, which gives $x^2-x-2=0 \implies \mathbf{(a,b)=(-1,-2)}$.

    4.  Case 4 (Mixed): If $\alpha = 2$ (fixed) and $\beta$ is not fixed, but $\beta = \alpha^2 - 2 = 2$. This is already covered. If $\alpha = \alpha^2 - 2$ but $\beta$ is a root of the equation and $\beta^2 - 2$ must be a root. If $\alpha=2$, $\beta^2-2=2$ or $\beta$, this leads to $\mathbf{(0,-4)}$ and $\mathbf{(0,-1)}$.

    5.  Count pairs: Summing up the unique pairs $(a, b)$ identified across all cases results in exactly 6 pairs.

   Difficulty level: Hard

   The Concept Name: Functional Relationship between Roots of a Quadratic Equation

   Short cut solution: Exhaustively test the scenarios where roots $\alpha, \beta$ are related by $f(x) = x^2 - 2$. The distinct $(a, b)$ outcomes are derived from roots sets $\{2,2\}$, $\{-1,-1\}$, $\{2,-1\}$, $\{ \alpha, \beta \}$ where $\beta = \alpha^2 - 2$, etc., totaling 6.

    Question 202

   Question: Let $a, b \in \mathbb{R}$, $a \neq 0$ be such that the equation, $ax^2 - 2bx + 5 = 0$ has a repeated root $\alpha$, which is also a root of the equation, $x^2 - 2bx - 10 = 0$. If $\beta$ is the other root of this equation, then $\alpha^2 + \beta^2$ is equal to:

   Options: 

    A. 25

    B. 26

    C. 28

    D. 24

   Correct Answer: A

   Year: Jan. 9, 2020 (II)

   Solution (Source): $ax^2 - 2bx + 5 = 0$ has repeated root so $\alpha = b/a$ and $\alpha^2 = 5/a$. This implies $b^2 = 5a$. For $x^2 - 2bx - 10 = 0$, $\alpha$ is a root, so $(b/a)^2 - 2b(b/a) - 10 = 0$. Substituting $b^2 = 5a$ gives $a = 1/4$ and $b^2 = 5/4$. Then $\alpha^2 = 20$ and $\beta^2 = 5$. $\alpha^2 + \beta^2 = 25$.

   Step Solution:

    1.  For repeated root $\alpha$ in the first equation, use Vieta's: $\alpha = \mathbf{\frac{b}{a}}$ and $\alpha^2 = \mathbf{\frac{5}{a}}$.

    2.  Combine these to find the relationship: $(\frac{b}{a})^2 = \frac{5}{a} \implies \mathbf{b^2 = 5a}$.

    3.  Substitute root $\alpha$ into the second equation: $\alpha^2 - 2b\alpha - 10 = 0 \implies \mathbf{\frac{b^2}{a^2} - \frac{2b^2}{a} - 10 = 0}$.

    4.  Substitute $b^2=5a$: $\frac{5}{a} - 10 - 10 = 0 \implies \mathbf{a = 1/4}$. Then $\mathbf{b^2 = 5/4}$ and $\mathbf{\alpha^2 = 20}$.

    5.  In the second equation, product $\alpha\beta = -10$, so $\alpha^2\beta^2 = 100 \implies \beta^2 = \frac{100}{20} = \mathbf{5}$. Result: $20 + 5 = \mathbf{25}$.

   Difficulty level: Medium.

   The Concept Name: Condition for Repeated Roots / Common Roots.

   Short cut solution: Use the condition $b^2=5a$ derived from the first equation to quickly reduce the second equation to a linear one in $1/a$.

Question 211

   Question: If $\alpha$ and $\beta$ be two roots of the equation $x^2 - 64x + 256 = 0$. Then the value of $(\frac{\alpha^3}{\beta^5})^{1/8} + (\frac{\beta^3}{\alpha^5})^{1/8}$ is:

   Options: 

    A. 2

    B. 3

    C. 1

   Correct Answer: A

   Year: Sep. 06, 2020 (I)

   Solution (Source): $\alpha + \beta = 64, \alpha\beta = 256$. Expression is $\frac{\alpha^{3/8}}{\beta^{5/8}} + \frac{\beta^{3/8}}{\alpha^{5/8}} = \frac{\alpha^{8/8} + \beta^{8/8}}{(\alpha\beta)^{5/8}} = \frac{64}{(2^8)^{5/8}} = \frac{64}{32} = 2$.

   Step Solution:

    1.  Note root relations for $x^2 - 64x + 256 = 0$: $\alpha + \beta = 64$ and $\alpha\beta = 256 = 2^8$.

    2.  Rewrite the expression: $\frac{\alpha^{3/8}}{\beta^{5/8}} + \frac{\beta^{3/8}}{\alpha^{5/8}}$.

    3.  Take a common denominator: $\frac{\alpha^{3/8} \cdot \alpha^{5/8} + \beta^{3/8} \cdot \beta^{5/8}}{(\alpha\beta)^{5/8}}$.

    4.  Simplify powers: $\frac{\alpha + \beta}{(\alpha\beta)^{5/8}}$.

    5.  Substitute known values: $\frac{64}{(2^8)^{5/8}} = \frac{64}{2^5} = \frac{64}{32} = \mathbf{2}$.

   Difficulty level: Easy.

   The Concept Name: Symmetric Functions of Roots / Properties of Exponents.

   Short cut solution: Combine fractions immediately to see the numerator is simply the sum of the roots, which is $64$. The denominator is $(\alpha\beta)^{5/8} = 32$. $64/32 = 2$.

    Question 212

   Question: If $\alpha$ and $\beta$ are the roots of the equation $2x(2x + 1) = 1$, then $\beta$ is equal to:

   Options: 

    A. $2\alpha(\alpha + 1)$

    B. $-2\alpha(\alpha + 1)$

    C. $2\alpha(\alpha - 1)$

    D. $2\alpha^2$

   Correct Answer: B

   Year: Sep. 06, 2020 (II)

   Solution (Source): Let $\alpha$ and $\beta$ be the roots of the given quadratic equation, $2x^2 + 2x - 1 = 0$. Then, $\alpha + \beta = -1/2 \Rightarrow -1 = 2\alpha + 2\beta$ and $4\alpha^2 + 2\alpha - 1 = 0$ [ $\because \alpha$ is root of eq. (i)] $\Rightarrow 4\alpha^2 + 2\alpha + 2\alpha + 2\beta = 0 \Rightarrow \beta = -2\alpha(\alpha + 1)$.

   Step Solution:

    1.  Rewrite the given equation into standard quadratic form: $4x^2 + 2x - 1 = 0$.

    2.  Apply the sum of roots formula: $\alpha + \beta = -2/4 = \mathbf{-1/2}$.

    3.  Identify the root identity for $\alpha$: Since it's a root, $4\alpha^2 + 2\alpha - 1 = 0$, which means $1 = 4\alpha^2 + 2\alpha$.

    4.  Rearrange the sum of roots to isolate $\beta$: $\beta = -1/2 - \alpha \Rightarrow \mathbf{2\beta = -1 - 2\alpha}$.

    5.  Substitute the identity for '1' from Step 3 into Step 4: $2\beta = -(4\alpha^2 + 2\alpha) - 2\alpha = -4\alpha^2 - 4\alpha$. Dividing by 2 gives $\beta = -2\alpha(\alpha + 1)$.

   Difficulty level: Medium.

   The Concept Name: Relation between Roots and Coefficients.

   Short cut solution: Use the sum of roots $2\beta = -1 - 2\alpha$. Replace $-1$ using the equation $4\alpha^2 + 2\alpha - 1 = 0 \Rightarrow -1 = -4\alpha^2 - 2\alpha$ to get $2\beta = -4\alpha^2 - 4\alpha$ instantly.

   Question 214

   Question: If $\alpha$ and $\beta$ are the roots of the equation, $7x^2 - 3x - 2 = 0$ then the value of $\frac{\alpha}{1 - \alpha^2} + \frac{\beta}{1 - \beta^2}$ is equal to :

   Options: 

    A. 27/32

    B. 1/24

    C. 3/8

    D. 27/16

   Correct Answer: D

   Year: Sep. 05, 2020 (II)

   Solution (Source): Let $\alpha$ and $\beta$ be the roots of the quadratic equation $7x^2 - 3x - 2 = 0 \therefore \alpha + \beta = 3/7, \alpha\beta = -2/7$. Now, $\frac{\alpha}{1 - \alpha^2} + \frac{\beta}{1 - \beta^2} = \frac{\alpha - \alpha\beta^2 + \beta - \beta\alpha^2}{(1 - \alpha^2)(1 - \beta^2)} = \frac{(\alpha + \beta) - \alpha\beta(\alpha + \beta)}{1 - (\alpha^2 + \beta^2) + (\alpha\beta)^2} = \frac{(\alpha + \beta)(1 - \alpha\beta)}{1 - [(\alpha + \beta)^2 - 2\alpha\beta] + (\alpha\beta)^2}$.

   Step Solution:

    1.  Determine root properties from the equation: $\alpha + \beta = 3/7$ and $\alpha\beta = -2/7$.

    2.  Simplify the target expression by finding a common denominator: $\frac{(\alpha + \beta)(1 - \alpha\beta)}{1 - (\alpha^2 + \beta^2) + (\alpha\beta)^2}$.

    3.  Find the sum of squares: $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = (3/7)^2 - 2(-2/7) = \mathbf{37/49}$.

    4.  Substitute values into the expression: $\frac{(3/7)(1 - (-2/7))}{1 - (37/49) + (-2/7)^2} = \mathbf{\frac{(3/7)(9/7)}{1 - 37/49 + 4/49}}$.

    5.  Final Calculation: $\frac{27/49}{(49 - 37 + 4)/49} = \frac{27/49}{16/49} = \mathbf{27/16}$.

   Difficulty level: Medium.

   The Concept Name: Symmetric Functions of Roots.

   Short cut solution: Simplify the numerator to $(\alpha+\beta)(1-\alpha\beta)$ and the denominator to $1 - (\alpha+\beta)^2 + 2\alpha\beta + (\alpha\beta)^2$ and substitute the fractions $3/7$ and $-2/7$ directly.

   Question 216

   Question: Let $\lambda \neq 0$ be in R. If $\alpha$ and $\beta$ are roots of the equation, $x^2 - x + 2\lambda = 0$ and $\alpha$ and $\gamma$ are the roots of the equation, $3x^2 - 10x + 27\lambda = 0$, then $\frac{\beta\gamma}{\lambda}$ is equal to

   Options: 

    A. 27

    B. 18

    C. 9

    D. 36

   Correct Answer: B

   Year: Sep. 04, 2020 (II)

   Solution (Source): Since $\alpha$ is common root of $x^2 - x + 2\lambda = 0$ and $3x^2 - 10x + 27\lambda = 0 \therefore 3\alpha^2 - 10\alpha + 27\lambda = 0$ and $3\alpha^2 - 3\alpha + 6\lambda = 0$. On subtract, we get $\alpha = 3\lambda$. Now, $\alpha\beta = 2\lambda \Rightarrow 3\lambda \cdot \beta = 2\lambda \Rightarrow \beta = 2/3$. $\Rightarrow \alpha + \beta = 1 \Rightarrow 3\lambda + 2/3 = 1 \Rightarrow \lambda = 1/9$ and $\alpha\gamma = 9\lambda \Rightarrow 3\lambda \cdot \gamma = 9\lambda \Rightarrow \gamma = 3$. ∴ $\beta\gamma/\lambda = 18$.

   Step Solution:

    1.  Find common root $\alpha$: Multiply the first equation by 3 and subtract the second: $3(x^2 - x + 2\lambda) - (3x^2 - 10x + 27\lambda) = 0 \Rightarrow 7\alpha - 21\lambda = 0 \Rightarrow \mathbf{\alpha = 3\lambda}$.

    2.  Find root $\beta$: From the product $\alpha\beta = 2\lambda$, substitute $\alpha$: $(3\lambda)\beta = 2\lambda \Rightarrow \mathbf{\beta = 2/3}$.

    3.  Find $\lambda$: Use the sum of roots $\alpha + \beta = 1$: $3\lambda + 2/3 = 1 \Rightarrow 3\lambda = 1/3 \Rightarrow \mathbf{\lambda = 1/9}$.

    4.  Find root $\gamma$: From the product in the second equation $\alpha\gamma = 27\lambda/3 = 9\lambda$, substitute $\alpha$: $(3\lambda)\gamma = 9\lambda \Rightarrow \mathbf{\gamma = 3}$.

    5.  Calculate final value: $\frac{\beta\gamma}{\lambda} = \frac{(2/3) \cdot 3}{1/9} = \frac{2}{1/9} = \mathbf{18}$.

   Difficulty level: Hard.

   The Concept Name: Common Roots of Quadratic Equations.

   Short cut solution: Identify $\alpha = 3\lambda$ by elimination. Then $\beta = 2/3$ and $\gamma = 3$ are found instantly from product relations, making the target $\frac{2}{1/9} = 18$.

   Question 217

   Question: If $\alpha$ and $\beta$ are the roots of the equation $x^2 + px + 2 = 0$ and $1/\alpha$ and $1/\beta$ are the roots of the equation $2x^2 + 2qx + 1 = 0$ then $(\alpha - 1/\alpha)(\beta - 1/\beta)(\alpha + 1/\beta)(\beta + 1/\alpha)$ is equal to:

   Options: 

    A. $\frac{9}{4}(9 + q^2)$

    B. $\frac{9}{4}(9 - q^2)$

    C. $\frac{9}{4}(9 + p^2)$

    D. $\frac{9}{4}(9 - p^2)$

   Correct Answer: D

   Year: JEE Main Sep. 03, 2020 (I)

   Solution: $\alpha\beta = 2$ and $\alpha + \beta = -p$. Also $1/\alpha + 1/\beta = -q \Rightarrow p = 2q$. The expression $(\alpha - 1/\alpha)(\beta - 1/\beta)(\alpha + 1/\beta)(\beta + 1/\alpha) = [\alpha\beta + 1/(\alpha\beta) - (\alpha/\beta + \beta/\alpha)][\alpha\beta + 1/(\alpha\beta) + 2] = \frac{9}{4}(9 - p^2)$ after substituting root relations.

   Step Solution:

    1.  Identify root relations for $x^2 + px + 2 = 0$: $\alpha + \beta = -p$ and $\alpha\beta = 2$.

    2.  Identify relations for $2x^2 + 2qx + 1 = 0$: Roots are $1/\alpha, 1/\beta$. Sum $1/\alpha + 1/\beta = -2q/2 = -q$.

    3.  Rewrite the target expression: $(\alpha - 1/\alpha)(\beta - 1/\beta) \cdot (\alpha + 1/\beta)(\beta + 1/\alpha) = (\frac{\alpha^2-1}{\alpha})(\frac{\beta^2-1}{\beta}) \cdot (\frac{\alpha\beta+1}{\beta})(\frac{\alpha\beta+1}{\alpha})$.

    4.  Substitute $\alpha\beta = 2$ into the last two terms: $(\frac{2+1}{\beta})(\frac{2+1}{\alpha}) = \frac{9}{\alpha\beta} = \mathbf{9/2}$. The first two terms expand to $\frac{\alpha^2\beta^2 - (\alpha^2+\beta^2) + 1}{\alpha\beta} = \mathbf{\frac{5 - (\alpha^2+\beta^2)}{2}}$.

    5.  Substitute $\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = p^2 - 4$: Result $= \frac{9}{2} \cdot \frac{5 - (p^2 - 4)}{2} = \mathbf{\frac{9}{4}(9 - p^2)}$.

   The difficulty level: Hard

   The Concept Name: Relation between Roots and Coefficients (Vieta's Formulas)

   Short cut solution: Recognize $\alpha\beta=2$ immediately simplifies the terms $(\alpha+1/\beta)(\beta+1/\alpha)$ to $9/2$. Then simplify $(\alpha-1/\alpha)(\beta-1/\beta)$ to $\frac{1}{2}(5 - (p^2-4))$ and multiply.

Question 220

   Question: If $\lambda$ be the ratio of the roots of the quadratic equation in $x, 3m^2x^2 + m(m - 4)x + 2 = 0$, then the least value of $m$ for which $\lambda + 1/\lambda = 1$, is:

   Options: 

    A. $2 - \sqrt{3}$

    B. $4 - 3\sqrt{2}$

    C. $-2 + \sqrt{2}$

    D. $4 - 2\sqrt{3}$

   Correct Answer: B

   Year: JEE Main Jan. 12, 2019 (I)

   Solution: Let roots be $\alpha, \beta$. $\lambda = \alpha/\beta$. The condition $\lambda + 1/\lambda = 1 \Rightarrow \alpha/\beta + \beta/\alpha = 1 \Rightarrow (\alpha+\beta)^2 = 3\alpha\beta$. From equation, $\alpha+\beta = \frac{4-m}{3m}$ and $\alpha\beta = \frac{2}{3m^2}$. Substitution gives $(m-4)^2 = 18$. $m = 4 \pm 3\sqrt{2}$. Least value is $4 - 3\sqrt{2}$.

   Step Solution:

    1.  Simplify the condition: $\lambda + \frac{1}{\lambda} = 1 \Rightarrow \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = 1 \Rightarrow \frac{\alpha^2 + \beta^2}{\alpha\beta} = 1$.

    2.  Express using sum and product: $\frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta} = 1 \Rightarrow \mathbf{(\alpha + \beta)^2 = 3\alpha\beta}$.

    3.  Identify coefficients: $\alpha + \beta = \mathbf{\frac{m(4 - m)}{3m^2} = \frac{4 - m}{3m}}$ and $\alpha\beta = \mathbf{\frac{2}{3m^2}}$.

    4.  Substitute and solve for $m$: $(\frac{4 - m}{3m})^2 = 3(\frac{2}{3m^2}) \Rightarrow \frac{(m - 4)^2}{9m^2} = \frac{2}{m^2} \Rightarrow \mathbf{(m - 4)^2 = 18}$.

    5.  Find the least $m$: $m = 4 \pm \sqrt{18} = 4 \pm 3\sqrt{2}$. The smaller value is $4 - 3\sqrt{2}$.

   The difficulty level: Medium

   The Concept Name: Ratio of Roots and Symmetric Functions

   Short cut solution: Use the shortcut for root ratio $r$: $\frac{(r+1)^2}{r} = \frac{b^2}{ac}$. Here $\lambda+1/\lambda=1 \Rightarrow \frac{(\lambda+1)^2}{\lambda} = \frac{\lambda^2+2\lambda+1}{\lambda} = (\lambda+1/\lambda)+2 = 1+2=3$. Set $b^2/ac = 3$ and solve $\frac{m^2(m-4)^2}{3m^2 \cdot 2} = 3 \Rightarrow (m-4)^2 = 18$ immediately.

   Question 221

   Question: If one real root of the quadratic equation $81x^2 + kx + 256 = 0$ is cube of the other root, then a value of k is :

   Options: 

    A. -81

    B. 100

    C. 144

    D. -300

   Correct Answer: D

   Year: JEE Main Jan. 11, 2019 (I)

   Solution: Let $\alpha$ and $\beta$ be the roots of the equation. Given $\beta = \alpha^3$. Product of roots $\alpha\beta = \frac{256}{81} \Rightarrow \alpha^4 = (\frac{4}{3})^4 \Rightarrow \alpha = \pm \frac{4}{3}$. Taking $\alpha = \frac{4}{3}$, then $\beta = \frac{64}{27}$. Sum of the roots $\alpha + \beta = -\frac{k}{81} \Rightarrow \frac{4}{3} + \frac{64}{27} = -\frac{k}{81} \Rightarrow \frac{100}{27} = -\frac{k}{81} \Rightarrow k = -300$.

   Step Solution:

    1.  Identify root relations: Product $\alpha \cdot \beta = \frac{256}{81}$ and the condition $\beta = \alpha^3$.

    2.  Solve for $\alpha$: $\alpha \cdot \alpha^3 = \alpha^4 = \frac{256}{81} \implies \alpha = \pm \frac{4}{3}$.

    3.  Find $\beta$: Using $\alpha = \frac{4}{3}$, then $\beta = (\frac{4}{3})^3 = \frac{64}{27}$.

    4.  Apply sum of roots: $\alpha + \beta = -\frac{k}{81} \implies \frac{4}{3} + \frac{64}{27} = -\frac{k}{81}$.

    5.  Calculate $k$: Multiply both sides by 81: $k = -81 \left( \frac{36+64}{27} \right) = -3 \times 100 = \mathbf{-300}$.

   The difficulty level: Medium

   The Concept Name: Relation between Roots and Coefficients (Vieta's Formulas)

   Short cut solution: Directly substitute $\alpha = 4/3$ into $81(\alpha + \alpha^3) = -k$. Calculating $81(4/3) + 81(64/27) = 108 + 192 = 300$, thus $k = -300$.

   Question 223

   Question: The value of $\lambda$ such that sum of the squares of the roots of the quadratic equation, $x^2 + (3 - \lambda) x + 2 = \lambda$ has the least value is:

   Options: 

    A. 15/8

    B. 1

    C. 4/9

    D. 2

   Correct Answer: D

   Year: JEE Main Jan. 10, 2019 (II)

   Solution: The given quadratic equation is $x^2 + (3 - \lambda)x + (2 - \lambda) = 0$. Sum of roots $\alpha + \beta = \lambda - 3$. Product of roots $\alpha\beta = 2 - \lambda$. $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = (\lambda - 3)^2 - 2(2 - \lambda) = \lambda^2 - 4\lambda + 5 = (\lambda - 2)^2 + 1$. For least value, $\lambda = 2$.

   Step Solution:

    1.  Standardize the equation: $x^2 + (3 - \lambda)x + (2 - \lambda) = 0$.

    2.  Write sum and product: $\alpha + \beta = \lambda - 3$ and $\alpha\beta = 2 - \lambda$.

    3.  Express sum of squares: $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$.

    4.  Substitute and simplify: $(\lambda - 3)^2 - 2(2 - \lambda) = \lambda^2 - 6\lambda + 9 - 4 + 2\lambda = \mathbf{\lambda^2 - 4\lambda + 5}$.

    5.  Minimize: Completing the square gives $(\lambda - 2)^2 + 1$. The minimum occurs at $\lambda = 2$.

   The difficulty level: Easy

   The Concept Name: Symmetric Functions of Roots and Minimization of Quadratics

   Short cut solution: Find the vertex of the parabola $f(\lambda) = \lambda^2 - 4\lambda + 5$ using $\lambda = -b/2a = -(-4)/2 = 2$.

   Question 228

   Question: If 5, 5r, $5r^2$ are the lengths of the sides of a triangle, then r cannot be equal to:

   Options: 

    A. 3/4

    B. 5/4

    C. 7/4

    D. 3/2

   Correct Answer: C

   Year: Jan. 10, 2019 (I)

   Solution:  For sides of a triangle, the sum of any two sides must exceed the third, so $5 + 5r > 5r^2$, which simplifies to $1 + r > r^2$. This inequality $r^2 - r - 1 < 0$ implies $r$ must lie in the interval $(\frac{1 - \sqrt{5}}{2}, \frac{1 + \sqrt{5}}{2})$. Since $7/4 = 1.75$ is greater than $\frac{1 + \sqrt{5}}{2} \approx 1.618$, it is not a valid ratio for the triangle sides. Thus, $r$ cannot be equal to 7/4.

   Step Solution:

    1.  Apply the Triangle Inequality where the sum of the two shorter sides must be greater than the longest side: $5 + 5r > 5r^2$.

    2.  Divide the inequality by 5 and rearrange it into a standard quadratic inequality: $r^2 - r - 1 < 0$.

    3.  Find the roots of the quadratic $r^2 - r - 1 = 0$ using the quadratic formula: $r = \frac{1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{1 \pm \sqrt{5}}{2}$.

    4.  Identify the valid range for $r$ from the inequality as $r \in (\frac{1 - \sqrt{5}}{2}, \frac{1 + \sqrt{5}}{2})$, which is approximately $(-0.618, 1.618)$.

    5.  Check the options against this interval: since $7/4 = 1.75$ falls outside the upper bound of $1.618$, it is the impossible value.

   The difficulty level: Medium.

   The Concept Name: Triangle Inequality Theorem.

   Short cut solution: Recognize that the ratio $r$ in a geometric progression of triangle sides must satisfy $r < \phi$, where $\phi$ is the Golden Ratio ($\approx 1.618$). Since $7/4 = 1.75 > 1.618$, it is immediately identified as the correct answer.

   Question 244

   Question: Let $p, q \in R$. If $2 - \sqrt{3}$ is a root of the quadratic equation, $x^2 + px + q = 0$, then:

   Options: 

    A. $p^2 - 4q + 12 = 0$

    B. $q^2 - 4p - 16 = 0$

    C. $q^2 + 4p + 14 = 0$

    D. $p^2 - 4q - 12 = 0$

   Correct Answer: D

   Year: April 10, 2019 (II)

   Solution: Since $2 - \sqrt{3}$ is a root and coefficients are real, $2 + \sqrt{3}$ is the other root. The sum of roots $\alpha + \beta = 4 = -p$, so $p = -4$. The product of roots $\alpha\beta = (2 - \sqrt{3})(2 + \sqrt{3}) = 1 = q$, so $q = 1$. Substituting $p$ and $q$ into option D, we get $16 - 4 - 12 = 0$.

   Step Solution:

    1.  Apply the Conjugate Root Theorem: if $2 - \sqrt{3}$ is an irrational root of a quadratic with rational/real coefficients, then $2 + \sqrt{3}$ must also be a root.

    2.  Calculate the sum of the roots: $(2 - \sqrt{3}) + (2 + \sqrt{3}) = 4$.

    3.  Calculate the product of the roots: $(2 - \sqrt{3})(2 + \sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1$.

    4.  Relate roots to coefficients $p$ and $q$: according to Vieta, $p = -(\text{sum}) = -4$ and $q = \text{product} = 1$.

    5.  Check the relation in Option D: $p^2 - 4q - 12 = (-4)^2 - 4(1) - 12 = 16 - 4 - 12 = 0$.

   The difficulty level: Easy.

   The Concept Name: Conjugate Root Theorem and Vieta's Formulas.

   Short cut solution: Construct the equation from its roots directly: $(x - 2)^2 = (\sqrt{3})^2 \Rightarrow x^2 - 4x + 4 = 3 \Rightarrow x^2 - 4x + 1 = 0$. Comparing with $x^2 + px + q = 0$, we have $p = -4$ and $q = 1$, which satisfies $p^2 - 4q - 12 = 0$.

   Question 245

   Question: If m is chosen in the quadratic equation $(m^2 + 1)x^2 - 3x + (m^2 + 1)^2 = 0$ such that the sum of its roots is greatest, then the absolute difference of the cubes of its roots is:

   Options: 

    A. $10\sqrt{5}$

    B. $8\sqrt{3}$

    C. $8\sqrt{5}$

    D. $4\sqrt{3}$

   Correct Answer: C

   Year: April 09, 2019 (II)

   Solution (Source): Sum of roots $= \frac{3}{m^2 + 1}$. Since the sum of roots is greatest, $m = 0$. Hence the equation becomes $x^2 - 3x + 1 = 0$. Now, $\alpha + \beta = 3, \alpha\beta = 1 \implies |\alpha - \beta| = \sqrt{5}$. $|\alpha^3 - \beta^3| = \sqrt{(\alpha - \beta)(\alpha^2 + \beta^2 + \alpha\beta)} = \sqrt{5}(9 - 1) = 8\sqrt{5}$.

   Step Solution:

    1.  Identify Sum of Roots: For $(m^2 + 1)x^2 - 3x + (m^2 + 1)^2 = 0$, the sum $S = \frac{3}{m^2 + 1}$.

    2.  Maximize Sum: $S$ is maximum when $m^2+1$ is minimum. Since $m^2 \geq 0$, the minimum is at $m=0$, giving the equation $x^2 - 3x + 1 = 0$.

    3.  Find Root Relations: For the new equation, $\alpha + \beta = 3$ and $\alpha\beta = 1$.

    4.  Calculate Difference: $|\alpha - \beta| = \sqrt{(\alpha + \beta)^2 - 4\alpha\beta} = \sqrt{3^2 - 4(1)} = \mathbf{\sqrt{5}}$.

    5.  Final Calculation: $|\alpha^3 - \beta^3| = |\alpha - \beta|(\alpha^2 + \alpha\beta + \beta^2) = |\alpha - \beta|[(\alpha + \beta)^2 - \alpha\beta] = \sqrt{5}(3^2 - 1) = \mathbf{8\sqrt{5}}$.

   Difficulty level: Medium.

   The Concept Name: Relation between Roots and Coefficients and Maximization.

   Short cut solution: Once $m=0$ is identified, use the property $|\alpha^3 - \beta^3| = |(\alpha - \beta)| \cdot (\frac{D}{a^2} + \frac{c}{a})$ where $D$ is the discriminant. Here $\sqrt{5}(5 + 1)$ is incorrect due to the middle term being $\alpha\beta$, so use $\sqrt{5}(9-1) = 8\sqrt{5}$ directly.

   Question 250

   Question: Let p, q and r be real numbers $(p \neq q, r \neq 0)$, such that the roots of the equation $\frac{1}{x+p} + \frac{1}{x+q} = \frac{1}{r}$ are equal in magnitude but opposite in sign, then the sum of squares of these roots is equal to:

   Options: 

    A. $p^2 + q^2 + r^2$

    B. $p^2 + q^2$

    C. $2(p^2 + q^2)$

    D. $\frac{p^2 + q^2}{2}$

   Correct Answer: B

   Year: Online April 16, 2018

   Solution (Source): Rearranging the equation gives $x^2 + (p + q - 2r)x + pq - pr - qr = 0$. Since roots are equal in magnitude but opposite in sign, Sum $= 0 \implies p + q - 2r = 0$. Sum of squares $= (\alpha + \beta)^2 - 2\alpha\beta = 0 - 2(pq - pr - qr) = p^2 + q^2 + 2r(2r - p - q) = p^2 + q^2 + 0 = p^2 + q^2$.

   Step Solution:

    1.  Form Quadratic: Cross-multiply: $r(x+q + x+p) = (x+p)(x+q) \implies \mathbf{x^2 + (p+q-2r)x + pq-pr-qr = 0}$.

    2.  Apply Condition: Roots are equal in magnitude but opposite in sign ($\alpha = -\beta$), so sum of roots is zero: $p+q-2r = 0 \implies 2r = p+q$.

    3.  Express Target: Sum of squares $S = \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$.

    4.  Substitute Sum and Product: $S = 0^2 - 2(pq - r(p+q)) = \mathbf{-2pq + 2r(p+q)}$.

    5.  Simplify with step 2: Substitute $2r = p+q$: $S = -2pq + (p+q)(p+q) = (p+q)^2 - 2pq = \mathbf{p^2 + q^2}$.

   Difficulty level: Medium.

   The Concept Name: Relation between Roots and Coefficients.

   Short cut solution: If the sum of roots of $ax^2+bx+c=0$ is zero, then $b=0$ and $\alpha^2+\beta^2 = -2c/a$. Here $b = p+q-2r=0$ and $c = pq-r(p+q)$. Thus $S = -2(pq-r(p+q))$. Using $r = (p+q)/2$, $S = -2pq + (p+q)^2 = p^2+q^2$.

Question 255

   Question: If $\lambda \in \mathbf{R}$ is such that the sum of the cubes of the roots of the equation, $\mathbf{x}^2 + (2 - \lambda) \mathbf{x} + (10 - \lambda) = \mathbf{0}$ is minimum, then the magnitude of the difference of the roots of this equation is.

   Options: 

    A. 20

    B. $2\sqrt{5}$

    C. $2\sqrt{7}$

    D. $4\sqrt{2}$

   Correct Answer: B

   Year: Online April 15, 2018

   Solution: Let the roots be $\alpha$ and $\beta$. The roots are calculated as $\frac{\lambda - 2 \pm \sqrt{\lambda^2 - 36}}{2}$. The magnitude of the difference of the roots is $|\alpha - \beta| = \sqrt{\lambda^2 - 36}$. The sum of cubes $\alpha^3 + \beta^3$ is expressed as a function $f(\lambda) = (\lambda - 2)(\lambda^2 - \lambda - 26)$. This function attains its minimum value at $\lambda = 4$. Substituting $\lambda = 4$ into the difference formula gives $\sqrt{20}$, which is $2\sqrt{5}$.

   Step Solution:

    1.  Identify root relations: $\alpha + \beta = \lambda - 2$ and $\alpha\beta = 10 - \lambda$.

    2.  Expand sum of cubes: $S = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) = (\lambda - 2)^3 - 3(10 - \lambda)(\lambda - 2)$.

    3.  Simplify the cubic expression: $S = (\lambda - 2)[\lambda^2 - 4\lambda + 4 - 30 + 3\lambda] = \lambda^3 - 3\lambda^2 - 24\lambda + 52$.

    4.  Minimize $S$ by differentiation: $S' = 3\lambda^2 - 6\lambda - 24 = 0 \Rightarrow \lambda^2 - 2\lambda - 8 = 0$, giving $\lambda = 4$ (for minimum).

    5.  Calculate root difference: $|\alpha - \beta| = \sqrt{(\alpha + \beta)^2 - 4\alpha\beta} = \sqrt{(\lambda - 2)^2 - 4(10 - \lambda)} = \sqrt{\lambda^2 - 36}$. For $\lambda = 4$, $|diff| = \sqrt{-20}$. The magnitude (absolute value of the complex difference) is $\sqrt{20} = \mathbf{2\sqrt{5}}$.

   The difficulty level: Hard.

   The Concept Name: Optimization of Symmetric Functions of Roots.

   Short cut solution: Use the property that the magnitude of the difference of roots is $\sqrt{|D|}$. Since $D = \lambda^2 - 36$, finding the critical point of the sum of cubes algebraically ($\lambda = 4$) leads to $\sqrt{|16 - 36|} = 2\sqrt{5}$.

Question 257

   Question: If, for a positive integer n, the quadratic equation, $\mathbf{x}(\mathbf{x}+1) + (\mathbf{x}+1)(\mathbf{x}+2) + . . . . . + (\mathbf{x}+\mathbf{n}-1)(\mathbf{x}+\mathbf{n}) = \mathbf{10n}$ has two consecutive integral solutions, then n is equal to:

   Options: 

    A. 11

    B. 12

    C. 9

    D. 10

   Correct Answer: A

   Year: 2017

   Solution: The sum of the terms is $\sum_{r=1}^n (x+r-1)(x+r) = 10n$. This expands and simplifies to the quadratic $x^2 + nx + \frac{n^2 - 31}{3} = 0$. Let the consecutive integral roots be $\alpha$ and $\alpha + 1$. Using the relation for roots, we find $n^2 = 121$, thus $n = 11$.

   Step Solution:

    1.  Simplify the series: $\sum_{r=1}^n [x^2 + (2r - 1)x + r(r - 1)] = 10n \Rightarrow nx^2 + n^2x + \frac{(n-1)n(n+1)}{3} = 10n$.

    2.  Normalize the quadratic: Divide by $n$ to get $x^2 + nx + \frac{n^2 - 1}{3} - 10 = 0 \Rightarrow x^2 + nx + \frac{n^2 - 31}{3} = 0$.

    3.  Define consecutive roots: Let roots be $\alpha$ and $\alpha + 1$, so their difference is 1.

    4.  Apply discriminant property: For any quadratic with roots differing by 1, $D = a^2 \Rightarrow n^2 - 4(\frac{n^2 - 31}{3}) = 1$.

    5.  Solve for $n$: $3n^2 - 4n^2 + 124 = 3 \Rightarrow n^2 = 121 \Rightarrow \mathbf{n = 11}$.

   The difficulty level: Hard.

   The Concept Name: Summation of Series and Root Properties.

   Short cut solution: The condition "consecutive integral solutions" means the discriminant $D$ of the reduced quadratic $x^2 + nx + \frac{n^2-31}{3} = 0$ must be 1. Solving $n^2 - 4(\frac{n^2-31}{3}) = 1$ instantly gives $n = 11$.

   Question 278

   Question: If $\pmb{\mathrm{a}}$ and $\beta$ are roots of the equation, $\mathbf{x}^2 - 4\sqrt{2}\mathbf{kx} + 2\mathbf{e}^{4\ln\mathbf{k}} - \mathbf{1} = \mathbf{0}$ for some $\mathbf{k}$, and $\mathbf{a}^2 + \mathbf{\beta}^2 = 66$, then $\boldsymbol{\mathfrak{a}}^3 + \boldsymbol{\mathfrak{b}}^3$ is equal to:

   Options: 

    A. $248\sqrt{2}$

    B. 280√2

    C. −32√2

    D. −280√2

   Correct Answer: D

   Year: Online April 11, 2014

   Solution: The equation is $x^2 - 4\sqrt{2}kx + 2k^4 - 1 = 0$. From the sum and product of roots, $\alpha + \beta = 4\sqrt{2}k$ and $\alpha\beta = 2k^4 - 1$. Using $\alpha^2 + \beta^2 = 66$, we solve to find $k^4 - 8k^2 + 16 = 0$, so $k^2 = 4$. Setting $k = -2$ to satisfy the required sum of cubes leads to $\alpha^3 + \beta^3 = -280\sqrt{2}$.

   Step Solution:

    1.  Simplify the equation: $2e^{4\ln k} = 2k^4$, so the equation is $x^2 - (4\sqrt{2}k)x + (2k^4 - 1) = 0$.

    2.  Use the sum of squares: $(\alpha + \beta)^2 - 2\alpha\beta = 66 \Rightarrow (4\sqrt{2}k)^2 - 2(2k^4 - 1) = 66$.

    3.  Solve for $k$: $32k^2 - 4k^4 + 2 = 66 \Rightarrow k^4 - 8k^2 + 16 = 0 \Rightarrow (k^2 - 4)^2 = 0 \Rightarrow \mathbf{k^2 = 4}$.

    4.  Determine sum and product: $\alpha\beta = 2(4)^2 - 1 = 31$ (Error in source math, should be $2(4)-1 = 7$ if $k^2=4$? Re-checking source: it says $2k^4-1$ where $k^2=4 \Rightarrow 2(16)-1 = 31$). Thus $\alpha\beta = 31$ and $\alpha + \beta = 4\sqrt{2}k$.

    5.  Calculate sum of cubes: For $k = -2$, $\alpha + \beta = -8\sqrt{2}$. Then $\alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 + \beta^2 - \alpha\beta) = (-8\sqrt{2})(66 - 31) = \mathbf{-280\sqrt{2}}$.

   The difficulty level: Medium.

   The Concept Name: Logarithmic Identities and Newton's Sums.

   Short cut solution: Recognize $2e^{4\ln k} - 1$ as $2k^4 - 1$. The condition $\alpha^2 + \beta^2 = 66$ creates a quadratic in $k^2$ that yields $k^2 = 4$. Since the options contain $\sqrt{2}$, and $\alpha+\beta \propto k$, only $k = \pm 2$ is possible. Evaluating $(\alpha+\beta)(66-\alpha\beta)$ for $k = -2$ gives $-280\sqrt{2}$.

   Question 279

   Question: If $\frac{1}{\sqrt{\alpha}}$ and $\frac{1}{\sqrt{\beta}}$ are the roots of the equation, $ax^2 + bx + 1 = 0$ ($a \neq 0, a, b, \in R$), then the equation, $x(x + b^3) + (a^3 - 3abx) = 0$ has roots:

   Options: 

    A. $\alpha^{3/2}$ and $\beta^{3/2}$

    B. $a\beta^{1/2}$ and $\alpha^{1/2}\beta$

    C. $\sqrt{\alpha\beta}$ and $\alpha\beta$

    D. $\alpha^{-3/2}$ and $\beta^{-3}$

   Correct Answer: A

   Year: JEE Main Online April 9, 2014

   Solution (Source): Let $\frac{1}{\sqrt{\alpha}}$ and $\frac{1}{\sqrt{\beta}}$ be the roots of $ax^2 + bx + 1 = 0$. Sum $\frac{1}{\sqrt{\alpha}} + \frac{1}{\sqrt{\beta}} = \frac{\sqrt{\alpha} + \sqrt{\beta}}{\sqrt{\alpha \beta}} = -\frac{b}{a}$. Product $\frac{1}{\sqrt{\alpha}\sqrt{\beta}} = \frac{1}{a} \Rightarrow a = \sqrt{\alpha \beta}$. $b = -(\sqrt{\alpha} + \sqrt{\beta})$. Rearranging the target equation gives $x^2 + (b^3 - 3ab)x + a^3 = 0$. Substituting $a$ and $b$ leads to roots $\alpha^{3/2}$ and $\beta^{3/2}$.

   Step Solution:

    1.  Find Root Relations: For $ax^2 + bx + 1 = 0$, the product of roots is $1/a = \frac{1}{\sqrt{\alpha\beta}} \implies \mathbf{a = \sqrt{\alpha\beta}}$.

    2.  Determine $b$: The sum of roots is $-b/a = \frac{\sqrt{\alpha}+\sqrt{\beta}}{\sqrt{\alpha\beta}}$. Substituting $a = \sqrt{\alpha\beta}$ gives $b = -(\sqrt{\alpha} + \sqrt{\beta})$.

    3.  Standardize Target Equation: $x(x + b^3) + (a^3 - 3abx) = 0 \implies \mathbf{x^2 + (b^3 - 3ab)x + a^3 = 0}$.

    4.  Evaluate New Coefficients: $b^3 - 3ab = -(\sqrt{\alpha}+\sqrt{\beta})^3 - 3(\sqrt{\alpha\beta})(-(\sqrt{\alpha}+\sqrt{\beta}))$. Using identity $(x+y)^3 - 3xy(x+y) = x^3+y^3$, this simplifies to $-(\alpha^{3/2} + \beta^{3/2})$.

    5.  Identify Roots: The constant term is $a^3 = (\alpha\beta)^{3/2}$. The equation is $x^2 - (\alpha^{3/2} + \beta^{3/2})x + (\alpha\beta)^{3/2} = 0$, so roots are $\alpha^{3/2}$ and $\beta^{3/2}$.

   Difficulty level: Hard

   The Concept Name: Relation between Roots and Coefficients (Vieta's Formulas)

   Short cut solution: Recognize that $b^3 - 3ab$ follows the cubic identity format for the roots of the original equation. Since the original roots are powers of $\alpha$ and $\beta$, and $a^3$ is a power of their product, the new roots must be the cubes of the inverse of the original roots.

   Question 281

   Question: If equations $ax^2 + bx + c = 0$ ($a, b, c \in R, a \neq 0$) and $2x^2 + 3x + 4 = 0$ have a common root, then $a:b:c$ equals:

   Options: 

    A. 1: 2: 3

    B. 2: 3: 4

    C. 4: 3: 2

    D. 3: 2: 1

   Correct Answer: B

   Year: JEE Main Online April 9, 2014

   Solution (Source): Let $\alpha, \beta$ be the common roots. For $2x^2 + 3x + 4 = 0$, $D = 3^2 - 4(2)(4) = -23$. Since $D < 0$, roots are complex conjugates. If one root is common, both must be common. Thus $\frac{a}{2} = \frac{b}{3} = \frac{c}{4} \implies a:b:c = 2:3:4$.

   Step Solution:

    1.  Analyze given equation: For $2x^2 + 3x + 4 = 0$, calculate the discriminant: $D = 3^2 - 4(2)(4) = 9 - 32 = -23$.

    2.  Determine Nature of Roots: Since $D < 0$ and coefficients are real, the roots are non-real complex conjugate pairs.

    3.  Apply Common Root Theorem: If a quadratic with real coefficients has one complex root in common with another quadratic, both roots must be common.

    4.  Relate Coefficients: Since both roots are common, the equations are proportional: $\frac{a}{2} = \frac{b}{3} = \frac{c}{4}$.

    5.  Final Ratio: $a:b:c = \mathbf{2:3:4}$.

   Difficulty level: Easy

   The Concept Name: Condition for Common Roots (Complex Roots)

   Short cut solution: Check the discriminant of the constant-coefficient equation ($2x^2+3x+4=0$) first. If $D < 0$, both roots are common, and the answer is simply the ratio of those coefficients: 2:3:4.

Question 286

   Question: If p and $q$ are non-zero real numbers and $\alpha^3 + \beta^3 = -p, \alpha\beta = q$, then a quadratic equation whose roots are $\frac{\alpha^2}{\beta}, \frac{\beta^2}{\alpha}$ is:

   Options: 

    A. $px^2 - qx + p^2 = 0$

    B. $qx^2 + px + q^2 = 0$

    C. $px^2 + qx + p^2 = 0$

    D. $qx^2 - px + q^2 = 0$

   Correct Answer: B

   Year: JEE Main Online April 25, 2013

   Solution (Source): Let the roots be $\alpha^2/\beta$ and $\beta^2/\alpha$. Sum of roots $= \frac{\alpha^3 + \beta^3}{\alpha\beta} = \frac{-p}{q}$. Product of roots $= \alpha\beta = q$. Required equation is $x^2 - (\frac{-p}{q})x + q = 0 \Rightarrow qx^2 + px + q^2 = 0$.

   Step Solution:

    1.  Identify new roots: $x_1 = \frac{\alpha^2}{\beta}$ and $x_2 = \frac{\beta^2}{\alpha}$.

    2.  Calculate new sum ($S$): $S = x_1 + x_2 = \frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha} = \frac{\alpha^3 + \beta^3}{\alpha\beta}$.

    3.  Substitute known values: $S = \frac{-p}{q}$.

    4.  Calculate new product ($P$): $P = x_1 \cdot x_2 = \frac{\alpha^2}{\beta} \cdot \frac{\beta^2}{\alpha} = \alpha\beta = \mathbf{q}$.

    5.  Form equation: $x^2 - Sx + P = 0 \Rightarrow x^2 - (\frac{-p}{q})x + q = 0$. Multiply by $q$: $qx^2 + px + q^2 = 0$.

   Difficulty level: Easy

   The Concept Name: Transformation of Quadratic Equations and Symmetric Functions

   Short cut solution: Use the fact that the sum of roots is $\frac{\alpha^3+\beta^3}{\alpha\beta} = -p/q$. In the standard form $Ax^2+Bx+C=0$, the sum is $-B/A$. For option B, $-p/q$ matches. The product $C/A$ is $q^2/q = q$, which also matches the given $\alpha\beta$.

   Question 287

   Question: If $\alpha$ and $\beta$ are roots of the equation $x^2 + px + \frac{3p}{4} = 0$ such that $|\alpha - \beta| = \sqrt{10}$, then $p$ belongs to the set:

   Options: 

    A. $\{2, -5\}$

    B. $\{-3, 2\}$

    C. $\{-2, 5\}$

    D. $\{3, -5\}$

   Correct Answer: C

   Year: Online April 22, 2013

   Solution: Given quadratic equation is $x^2 + px + \frac{3p}{4} = 0$. So, $\alpha + \beta = -p$ and $\alpha\beta = \frac{3p}{4}$. Now, given $|\alpha - \beta| = \sqrt{10} \Rightarrow (\alpha - \beta)^2 = 10 \Rightarrow (\alpha + \beta)^2 - 4\alpha\beta = 10 \Rightarrow p^2 - 4 \times \frac{3p}{4} = 10 \Rightarrow p^2 - 3p - 10 = 0 \Rightarrow p = -2, 5 \Rightarrow p \in \{-2, 5\}$.

   Step Solution:

    1.  From the equation $x^2 + px + \frac{3p}{4} = 0$, identify root relations: $\alpha + \beta = -p$ and $\alpha\beta = \frac{3p}{4}$.

    2.  Use the given condition $|\alpha - \beta| = \sqrt{10}$ and square both sides to get $(\alpha - \beta)^2 = 10$.

    3.  Apply the algebraic identity: $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta = 10$.

    4.  Substitute the values from step 1 into the identity: $(-p)^2 - 4\left(\frac{3p}{4}\right) = 10 \Rightarrow p^2 - 3p - 10 = 0$.

    5.  Factorize the quadratic in $p$: $(p-5)(p+2) = 0$, which gives $p = 5$ or $p = -2$.

   The difficulty level: Medium

   The Concept Name: Relation between Roots and Coefficients (Vieta's Formulas)

   Short cut solution: Use the formula $|\alpha - \beta| = \frac{\sqrt{D}}{|a|}$. Here, $\sqrt{p^2 - 4(1)(\frac{3p}{4})} = \sqrt{10} \Rightarrow p^2 - 3p = 10$. Solve for $p$ to get $\{5, -2\}$.

   Question 296

   Question: Let $p, q, r \in R$ and $r > p > 0$. If the quadratic equation $px^2 + qx + r = 0$ has two complex roots $\alpha$ and $\beta$, then $|\alpha| + |\beta|$ is

   Options: 

    A. equal to 1

    B. less than 2 but not equal to 1

    C. greater than 2

    D. equal to 2

   Correct Answer: C

   Year: Online May 19, 2012

   Solution: Given quadratic equation is $px^2 + qx + r = 0$. Since $\alpha$ and $\beta$ are two complex roots, they are conjugates, so $|\beta| = |\alpha|$. Consider product of roots $\alpha\beta = r/p$. Then $|\alpha||\beta| = |\alpha|^2 = r/p$. Since $r > p > 0$, $|\alpha|^2 > 1 \Rightarrow |\alpha| > 1$. Thus $|\alpha| + |\beta| = 2|\alpha| > 2$.

   Step Solution:

    1.  Recognize that for real coefficients, complex roots must be conjugates: $\beta = \bar{\alpha}$.

    2.  This implies that their magnitudes are equal: $|\alpha| = |\beta|$.

    3.  Find the product of the roots using coefficients: $\alpha\beta = \frac{r}{p}$.

    4.  Take the modulus of the product: $|\alpha\beta| = |\alpha|^2 = \frac{r}{p}$. Since $r > p$, then $|\alpha|^2 > 1 \implies |\alpha| > 1$.

    5.  Calculate the required sum: $|\alpha| + |\beta| = 2|\alpha|$. Since $|\alpha| > 1$, the sum is greater than 2.

   The difficulty level: Medium

   The Concept Name: Relation between Roots and Coefficients and Complex Conjugates

   Short cut solution: The product of roots is $r/p$. Since $r > p$, the product is $>1$. Because the roots are complex conjugates, each root must have a magnitude greater than 1. Therefore, their sum of magnitudes must be $>2$.

 Question 297

   Question: If the sum of the square of the roots of the equation $x^2 - (\sin a - 2)x - (1 + \sin a) = 0$ is least, then $a$ is equal to:

   Options: 

    A. $\frac{\pi}{6}$

    B. $\frac{\pi}{4}$

    C. $\frac{\pi}{3}$

    D. $\frac{\pi}{2}$

   Correct Answer: D

   Year: Online May 12, 2012

   Solution: Let $x_1$ and $x_2$ be two roots of quadratic equation. $\therefore x_1 + x_2 = \sin \alpha - 2$ and $x_1x_2 = -(1 + \sin \alpha)$. $(x_1 + x_2)^2 = (\sin a - 2)^2 = \sin^2 a + 4 - 4 \sin a \Rightarrow x_1^2 + x_2^2 = \sin^2 a + 4 - 4 \sin a - 2x_1x_2 = \sin^2 a + 4 - 4 \sin a + 2(1 + \sin a) = \sin^2 a - 2 \sin a + 6$. We get least value of $x_1^2 + x_2^2$ at $\pi/2$.

   Step Solution:

    1.  Identify root relations using Vieta's formulas: Sum $S = x_1 + x_2 = \mathbf{\sin a - 2}$ and Product $P = x_1x_2 = \mathbf{-(1 + \sin a)}$.

    2.  Express the sum of squares as $E = x_1^2 + x_2^2 = \mathbf{(x_1 + x_2)^2 - 2x_1x_2}$.

    3.  Substitute the values of $S$ and $P$: $E = (\sin a - 2)^2 - 2[-(1 + \sin a)]$.

    4.  Expand and simplify: $E = (\sin^2 a - 4\sin a + 4) + (2 + 2\sin a) = \mathbf{\sin^2 a - 2\sin a + 6}$.

    5.  Complete the square to find the minimum: $E = \mathbf{(\sin a - 1)^2 + 5}$. This is least when $\sin a = 1$, which gives $a = \pi/2$.

   Difficulty level: Medium

   The Concept Name: Relation between Roots and Coefficients and Optimization of Quadratics.

   Short cut solution: Recognize that the expression for the sum of squares is a quadratic in terms of $\sin a$: $f(k) = k^2 - 2k + 6$. The minimum value of this parabola occurs at the vertex $k = -b/2a = -(-2)/2 = 1$. Since $k = \sin a$, then $\sin a = 1 \Rightarrow a = \pi/2$.

   Question 301

   Question: Sachin and Rahul attempted to solve a quadratic equation. Sachin made a mistake in writing down the constant term and ended up in roots (4,3). Rahul made a mistake in writing down coefficient of $X$ to get roots (3,2). The correct roots of equation are:

   Options: 

    A. 6,1

    B. 4,3

    C. -6,-1

    D. -4,-3

   Correct Answer: A

   Year: 2011 RS

   Solution: Let the correct equation be $ax^2 + bx + c = 0$. Sachin's equation has roots 4 and 3 $\Rightarrow -b/a = 7$. Rahul's equation has roots 3 and 2 $\Rightarrow c/a = 6$. From these, roots of the correct equation $x^2 - 7x + 6 = 0$ are 6 and 1.

   Step Solution:

    1.  Sachin miswrote the constant term ($c$), so his sum of roots is correct: $S = 4 + 3 = 7$.

    2.  Rahul miswrote the linear coefficient ($b$), so his product of roots is correct: $P = 3 \times 2 = 6$.

    3.  Construct the correct quadratic equation using the formula $x^2 - Sx + P = 0$: $x^2 - 7x + 6 = 0$.

    4.  Factor the equation: $(x - 6)(x - 1) = 0$.

    5.  The correct roots are $x = 6, 1$.

   Difficulty level: Easy

   The Concept Name: Roots and Coefficients (Vieta's Formulas).

   Short cut solution: Look for an option where the sum of roots is 7 (from Sachin) and the product is 6 (from Rahul). Option A ($6+1=7, 6\times1=6$) satisfies both immediately.

    Question 319

   Question: If the roots of the equation $x^2 - bx + c = 0$ be two consecutive integers, then $b^2 - 4c$ equals:

   Options: 

    A. -2

    B. 3

    C. 2

    D. 1

   Correct Answer: D

   Year: 2005

   Solution (Source): Let $a$, $a + 1$ be roots. Then $a + a + 1 = b = \text{sum of roots}$ and $a(a + 1) = c = \text{product of roots}$. $\therefore b^2 - 4c = (2a + 1)^2 - 4a(a + 1) = 1$.

   Step Solution:

    1.  Let the roots be $n$ and $n + 1$.

    2.  Write Vieta's relations: $b = n + (n + 1) = \mathbf{2n + 1}$ and $c = \mathbf{n(n + 1)}$.

    3.  Set up the expression for the discriminant: $b^2 - 4c = \mathbf{(2n + 1)^2 - 4(n(n + 1))}$.

    4.  Expand the terms: $\mathbf{4n^2 + 4n + 1 - 4n^2 - 4n}$.

    5.  Simplify the result: $\mathbf{1}$.

   The difficulty level: Easy

   The Concept Name: Relation between Roots and Coefficients (Vieta's Formulas)

   Short cut solution: The term $b^2 - 4c$ is the discriminant ($D$). The difference between roots $(\alpha - \beta)$ is given by $\sqrt{D}/|a|$. Since the roots are consecutive integers, $|\alpha - \beta| = 1$. Thus, $\sqrt{b^2 - 4c} = 1$, which implies $b^2 - 4c = 1$.

   Question 321

   Question: The value of $a$ for which the sum of the squares of the roots of the equation $x^2 - (a - 2)x - a - 1 = 0$ assume the least value is:

   Options: 

    A. 1

    B. 0

    C. 3

    D. 2

   Correct Answer: A

   Year: 2005

   Solution (Source): Given equation is $x^2 - (a - 2)x - a - 1 = 0 \Rightarrow \alpha + \beta = a - 2$; $\alpha\beta = -(a + 1)$. $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = a^2 - 2a + 6 = (a - 1)^2 + 5$. For min. value of $\alpha^2 + \beta^2$, $a - 1 = 0 \Rightarrow a = 1$.

   Step Solution:

    1.  Identify sum and product: $S = \alpha + \beta = \mathbf{a - 2}$ and $P = \alpha\beta = \mathbf{-(a + 1)}$.

    2.  Use the identity for sum of squares: $\alpha^2 + \beta^2 = \mathbf{S^2 - 2P}$.

    3.  Substitute the values: $\mathbf{(a - 2)^2 - 2(-(a + 1))}$.

    4.  Simplify to a quadratic in $a$: $a^2 - 4a + 4 + 2a + 2 = \mathbf{a^2 - 2a + 6}$.

    5.  Complete the square: $\mathbf{(a - 1)^2 + 5}$. This is least when $a = \mathbf{1}$.

   The difficulty level: Easy

   The Concept Name: Symmetric Functions of Roots and Minimization of Quadratics

   Short cut solution: Once you obtain the expression $a^2 - 2a + 6$, find the vertex of the parabola using $a = -B/2A = -(-2)/(2 \cdot 1) = 1$.

   Question 325

   Question: If one root of the equation $x^2 + px + 12 = 0$ is 4, while the equation $x^2 + px + q = 0$ has equal roots, then the value of $q$ is:

   Options: 

    A. 4

    B. 12

    C. 3

    D. 49/4

   Correct Answer: D

   Year: 2004

   Solution (Source): Given that 4 is a root of $x^2 + px + 12 = 0 \Rightarrow 16 + 4p + 12 = 0 \Rightarrow p = -7$. Now, the equation $x^2 + px + q = 0$ has equal roots. $\therefore D = 0 \Rightarrow p^2 - 4q = 0 \Rightarrow q = p^2/4 = 49/4$.

   Step Solution:

    1.  Substitute root $x = 4$ into the first equation: $\mathbf{4^2 + p(4) + 12 = 0}$.

    2.  Calculate $p$: $16 + 4p + 12 = 0 \Rightarrow 4p = -28 \Rightarrow \mathbf{p = -7}$.

    3.  For equal roots in the second equation, set the discriminant to zero: $\mathbf{p^2 - 4q = 0}$.

    4.  Substitute $p = -7$ into the discriminant: $\mathbf{(-7)^2 - 4q = 0}$.

    5.  Solve for $q$: $49 = 4q \Rightarrow \mathbf{q = 49/4}$.

   The difficulty level: Easy

   The Concept Name: Discriminant and Nature of Roots

   Short cut solution: In $x^2 + px + 12 = 0$, the product of roots is 12. If one root is 4, the other is 3. The sum is $4 + 3 = -p \Rightarrow p = -7$. For $x^2 + px + q = 0$ to have equal roots, the root must be $-p/2 = 7/2$. The product $q$ is $(7/2)^2 = 49/4$.

   Question 330

   Question: The value of ' a ' for which one root of the quadratic equation $( a ^ { 2 } - 5 a + 3 ) x ^ { 2 } + ( 3 a - 1 ) x + 2 = 0$ is twice as large as the other is.

   Options: 

    A. $- \frac { 1 } { 3 }$

    B. $\frac { 2 } { 3 }$

    C. $- \frac { 2 } { 3 }$

    D. $\frac { 1 } { 3 }$

   Correct Answer: B

   Year: 2003

   Solution: Let one root of the given equation be $\alpha$ and the second root be $2\alpha$. According to Vieta's formulas, the sum of roots is $3\alpha = \frac{1-3a}{a^2-5a+3}$ and the product is $2\alpha^2 = \frac{2}{a^2-5a+3}$. Substituting the value of $\alpha$ from the sum equation into the product equation leads to $\frac{(1-3a)^2}{a^2-5a+3} = 9$. Expanding this results in $9a^2 - 6a + 1 = 9a^2 - 45a + 27$, which simplifies to $39a = 26$, giving $a = 2/3$.

   Step Solution:

    1.  Define roots as $\alpha$ and $2\alpha$. Use the sum of roots: $3\alpha = -\frac{3a-1}{a^2-5a+3} \implies \alpha = \frac{1-3a}{3(a^2-5a+3)}$.

    2.  Use the product of roots: $2\alpha^2 = \frac{2}{a^2-5a+3} \implies \mathbf{\alpha^2 = \frac{1}{a^2-5a+3}}$.

    3.  Substitute the expression for $\alpha$ into the squared equation: $\left( \frac{1-3a}{3(a^2-5a+3)} \right)^2 = \frac{1}{a^2-5a+3}$.

    4.  Simplify by canceling one term of $(a^2-5a+3)$ from the denominators: $\frac{(1-3a)^2}{9(a^2-5a+3)} = 1 \implies (1-3a)^2 = 9(a^2-5a+3)$.

    5.  Expand and solve the linear equation in $a$: $1 - 6a + 9a^2 = 9a^2 - 45a + 27 \implies 39a = 26 \implies \mathbf{a = 2/3}$.

   The difficulty level: Medium.

   The Concept Name: Vieta's Formulas (Relation between Roots and Coefficients).

   Short cut solution: Use the formula for roots in ratio $n:1$, which is $nb^2 = (n+1)^2ac$ [Outside Source]. For $n=2$, $2(3a-1)^2 = 9(a^2-5a+3)(2) \implies (3a-1)^2 = 9(a^2-5a+3)$, leading directly to $a=2/3$.

  Question 335

   Question: If p and q are the roots of the equation $x^2 + px + q = 0$, then:

   Options: 

    A. $p = 1, q = -2$

    B. $p = 0, q = 1$

    C. $p = -2, q = 0$

    D. $p = -2, q = 1$

   Correct Answer: A

   Year: 2002

   Solution: From the sum of roots, $p + q = -p$, which implies $q = -2p$. From the product of roots, $pq = q$, which leads to $q(p - 1) = 0$. This gives two possibilities: $q = 0$ (which leads to $p = 0$) or $p = 1$ (which leads to $q = -2$).

   Step Solution:

    1.  Apply the sum of roots formula to $x^2 + px + q = 0$: $p + q = -p$.

    2.  Rearrange to find $q$ in terms of $p$: $q = -2p$.

    3.  Apply the product of roots formula: $pq = q$.

    4.  Solve the product equation: $pq - q = 0 \implies \mathbf{q(p - 1) = 0}$.

    5.  Check cases: If $p = 1$, then $q = -2(1) = \mathbf{-2}$. (The case $p=0, q=0$ is not among the distinct options provided).

   The difficulty level: Easy.

   The Concept Name: Vieta's Formulas (Relation between Roots and Coefficients).

   Short cut solution: Test Option A: If $p=1, q=-2$, the equation is $x^2 + x - 2 = 0$. Its roots are $1$ and $-2$, which are exactly the values of $p$ and $q$.

   Question 337

   Question: Difference between the corresponding roots of $\mathbf { x } ^ { 2 } + \mathbf { a x } + \mathbf { b } = \mathbf { 0 }$ and $\mathbf { x } ^ { 2 } + \mathbf { b } \mathbf { x } + \mathbf { a } = \mathbf { 0 }$ is same and $\mathbf { a } \neq \mathbf { b }$, then:

   Options: 

    A. $\mathbf{a} + \mathbf{b} + 4 = 0$

    B. $\mathbf{a} + \mathbf{b} - 4 = 0$

    C. $\mathbf{a} - \mathbf{b} - 4 = 0$

    D. $\mathbf{a} - \mathbf{b} + 4 = 0$

   Correct Answer: A

   Year: 2002

   Solution: Let $\alpha$ and $\beta$ are roots of the equation $x^2 + ax + b = 0$ and $\gamma$ and $\delta$ be the roots of the equation $x^2 + bx + a = 0$ respectively. $\therefore \alpha + \beta = -a, \alpha\beta = b$ and $\gamma + \delta = -b, \gamma\delta = a$. Given $|\alpha - \beta| = |\gamma - \delta| \Rightarrow (\alpha - \beta)^2 = (\gamma - \delta)^2 \Rightarrow (\alpha + \beta)^2 - 4\alpha\beta = (\gamma + \delta)^2 - 4\gamma\delta \Rightarrow a^2 - 4b = b^2 - 4a \Rightarrow (a^2 - b^2) + 4(a - b) = 0 \Rightarrow a + b + 4 = 0$ (since $a \neq b$).

   Step Solution:

    1.  Define roots and use Vieta's Formulas: For the first equation, $\alpha + \beta = -a$ and $\alpha\beta = b$; for the second, $\gamma + \delta = -b$ and $\gamma\delta = a$.

    2.  Set up the given condition (equal differences) by squaring: $(\alpha - \beta)^2 = (\gamma - \delta)^2$.

    3.  Convert differences to sum and product form: $(\alpha + \beta)^2 - 4\alpha\beta = (\gamma + \delta)^2 - 4\gamma\delta$.

    4.  Substitute coefficients: $a^2 - 4b = b^2 - 4a$.

    5.  Rearrange and factorize: $a^2 - b^2 + 4a - 4b = 0 \implies (a - b)(a + b) + 4(a - b) = 0 \implies (a - b)(a + b + 4) = 0$. Since $a \neq b$, then $a + b + 4 = 0$.

   Difficulty level: Medium

   The Concept Name: Relation between Roots and Coefficients (Vieta's Formulas)

   Short cut solution: The difference between roots of $Ax^2+Bx+C=0$ is $\frac{\sqrt{D}}{|A|}$. Since $|A|=1$ for both, the discriminants must be equal: $a^2 - 4b = b^2 - 4a \implies a^2 - b^2 = -4(a - b) \implies a + b = -4$.

   Question 338

   Question: If $\alpha \neq \beta$ but $\mathbf { a } ^ { 2 } = 5 \mathbf { a } - 3$ and $\beta ^ { 2 } = 5 \beta - 3$ then the equation having $\alpha / \beta$ and $\mathbf { \beta } / \mathbf { a }$ as its roots is:

   Options: 

    A. $3x^2 - 19x + 3 = 0$

    B. $3x^2 + 19x - 3 = 0$

    C. $3x^2 - 19x - 3 = 0$

    D. $x^2 - 5x + 3 = 0$

   Correct Answer: A

   Year: 2002

   Solution: Given that $\alpha^2 = 5\alpha - 3$ and $\beta^2 = 5\beta - 3$; $\Rightarrow \alpha$ and $\beta$ are roots of equation $x^2 = 5x - 3$ or $x^2 - 5x + 3 = 0$. $\therefore \alpha + \beta = 5$ and $\alpha\beta = 3$. Thus, the equation whose roots are $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$ is $x^2 - x(\frac{\alpha}{\beta} + \frac{\beta}{\alpha}) + \frac{\alpha\beta}{\alpha\beta} = 0 \Rightarrow x^2 - x(\frac{\alpha^2 + \beta^2}{\alpha\beta}) + 1 = 0$, which results in $3x^2 - 19x + 3 = 0$.

   Step Solution:

    1.  Identify the base equation: Since both $\alpha$ and $\beta$ satisfy $x^2 = 5x - 3$, they are the roots of $x^2 - 5x + 3 = 0$.

    2.  Find root relations: $\alpha + \beta = 5$ and $\alpha\beta = 3$.

    3.  Define the sum ($S$) and product ($P$) of the new roots: $S = \frac{\alpha}{\beta} + \frac{\beta}{\alpha}$ and $P = \frac{\alpha}{\beta} \cdot \frac{\beta}{\alpha} = 1$.

    4.  Calculate the new sum using symmetric functions: $S = \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta} = \frac{5^2 - 2(3)}{3} = \mathbf{19/3}$.

    5.  Construct the required quadratic equation: $x^2 - (19/3)x + 1 = 0$, which simplifies to $3x^2 - 19x + 3 = 0$.

   Difficulty level: Easy

   The Concept Name: Symmetric Functions of Roots and Formation of Quadratic Equations

   Short cut solution: Use the identity $\alpha^2 + \beta^2 = 5(\alpha + \beta) - 6$ (from the given square relations). This gives $\alpha^2 + \beta^2 = 5(5) - 6 = 19$. The sum of new roots is $19/3$ and product is $1$, directly leading to Option A.