If the set of all a in mathbbR - \1\$a \in \mathbb{R} - \{1\}$, for which the roots of the equation (1 - a)x^2 + 2(a - 3)x + 9 = 0$(1 - a)x^2 + 2(a - 3)x + 9 = 0$ are positive is (-infty, -alpha] cup [beta, gamma)$(-\infty, -\alpha] \cup [\beta, \gamma)$, then 2alpha + beta + gamma$2\alpha + \beta + \gamma$ is equal to ____________.
Numerical Answer Type:
Enter a numerical valueAnswer: 7 to 7+4 marks
Solution & Explanation
### Related Formula
textFor a quadratic equation A x^2 + B x + C = 0 text to have two positive real roots:$$\text{For a quadratic equation } A x^2 + B x + C = 0 \text{ to have two positive real roots:}$$text1. Real roots: D = B^2 - 4AC ge 0$$\text{1. Real roots: } D = B^2 - 4AC \ge 0$$text2. Sum of roots: -fracBA > 0$$\text{2. Sum of roots: } -\frac{B}{A} > 0$$text3. Product of roots: fracCA > 0$$\text{3. Product of roots: } \frac{C}{A} > 0$$
### Core Logic
We write down three systems of inequalities based on real and positive root conditions, find their intersection, and map the boundaries to solve for the parameters.
### Step 1: Apply the discriminant condition (Real roots)
For real roots, the discriminant D ge 0$D \ge 0$:
D = left[ 2(a - 3) right]^2 - 4(1 - a)(9) ge 0$$D = \left[ 2(a - 3) \right]^2 - 4(1 - a)(9) \ge 0$$4(a^2 - 6a + 9) - 36(1 - a) ge 0$$4(a^2 - 6a + 9) - 36(1 - a) \ge 0$$(a^2 - 6a + 9) - 9(1 - a) ge 0$$(a^2 - 6a + 9) - 9(1 - a) \ge 0$$a^2 - 6a + 9 - 9 + 9 a ge 0$$a^2 - 6a + 9 - 9 + 9 a \ge 0$$a^2 + 3a ge 0 implies a(a + 3) ge 0$$a^2 + 3a \ge 0 \implies a(a + 3) \ge 0$$
Thus, the interval is:
a in (-infty, -3] cup [0, infty) quad text--- (1)$$a \in (-\infty, -3] \cup [0, \infty) \quad \text{--- (1)}$$
### Step 2: Apply the sum of roots condition (Positive sum)
For positive roots, the sum of roots must be positive:
-fracBA = frac-2(a - 3)1 - a = frac2(a - 3)a - 1 > 0$$-\frac{B}{A} = \frac{-2(a - 3)}{1 - a} = \frac{2(a - 3)}{a - 1} > 0$$
Using the wavy curve method for fraca-3a-1 > 0$\frac{a-3}{a-1} > 0$:
a in (-infty, 1) cup (3, infty) quad text--- (2)$$a \in (-\infty, 1) \cup (3, \infty) \quad \text{--- (2)}$$
### Step 3: Apply the product of roots condition (Positive product)
For positive roots, the product of roots must be positive:
fracCA = frac91 - a > 0 implies 1 - a > 0 implies a < 1$$\frac{C}{A} = \frac{9}{1 - a} > 0 \implies 1 - a > 0 \implies a < 1$$
Thus, the interval is:
a in (-infty, 1) quad text--- (3)$$a \in (-\infty, 1) \quad \text{--- (3)}$$
### Step 4: Find the intersection of all conditions
Intersecting equations (1), (2), and (3):
- First, intersect (2) and (3):
( (-infty, 1) cup (3, infty) ) cap (-infty, 1) = (-infty, 1)$$( (-\infty, 1) \cup (3, \infty) ) \cap (-\infty, 1) = (-\infty, 1)$$
- Next, intersect with (1):
( (-infty, -3] cup [0, infty) ) cap (-infty, 1) = (-infty, -3] cup [0, 1)$$( (-\infty, -3] \cup [0, \infty) ) \cap (-\infty, 1) = (-\infty, -3] \cup [0, 1)$$
Comparing this with (-infty, -alpha] cup [beta, gamma)$(-\infty, -\alpha] \cup [\beta, \gamma)$:
- alpha = 3$\alpha = 3$
- beta = 0$\beta = 0$
- gamma = 1$\gamma = 1$
Now calculate the target sum:
2alpha + beta + gamma = 2(3) + 0 + 1 = 7$$2\alpha + \beta + \gamma = 2(3) + 0 + 1 = 7$$
### Pattern Recognition
Location of roots: When both roots are positive, checking sum and product signs along with D ge 0$D \ge 0$ is the standard and fastest set of inequalities, avoiding complex vertex projections.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Practical Chemistry
Keywords:#location of roots positive quadratics#JEE Main 2025 Evening Q75#discriminant product sum inequalities#algebraic intervals wavy curve
More Complex Numbers and Quadratic Equations Previous-Year Questions — Page 4
Q54jee_main_2025_28_jan_eveningComplex Roots of Quadratic Equations
If alpha+ibeta$\alpha+i\beta$ and gamma+idelta$\gamma+i\delta$ are the roots of x^2-(3-2i)x-(2i-2)=0,$x^{2}-(3-2i)x-(2i-2)=0,$i=sqrt-1$i=\sqrt{-1}$ then alphagamma+betadelta$\alpha\gamma+\beta\delta$ is equal to :
A.6$6$
B.2$2$
C.-2$-2$
D.-6$-6$
Solution
### Related Formula
For a quadratic equation Ax^2 + Bx + C = 0$Ax^2 + Bx + C = 0$, roots can be obtained via the quadratic formula:
x = frac-B pm sqrtB^2 - 4AC2A$$x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$$
### Core Logic
Given quadratic equation:
x^2-(3-2i)x-(2i-2)=0$$x^{2}-(3-2i)x-(2i-2)=0$$
Using the quadratic formula where A=1, B=-(3-2i), C=-(2i-2)$A=1, B=-(3-2i), C=-(2i-2)$:
x = frac(3-2i) pm sqrt(3-2i)^2 - 4(1)(-(2i-2))2$$x = \frac{(3-2i) \pm \sqrt{(3-2i)^2 - 4(1)(-(2i-2))}}{2}$$
### Step 1: Simplify the Discriminant
textDiscriminant D = (3-2i)^2 + 4(2i-2)$$\text{Discriminant } D = (3-2i)^2 + 4(2i-2)$$D = (9 - 4 - 12i) + (8i - 8)$$D = (9 - 4 - 12i) + (8i - 8)$$D = 5 - 12i + 8i - 8 = -3 - 4i$$D = 5 - 12i + 8i - 8 = -3 - 4i$$
We need to find sqrt-3-4i$\sqrt{-3-4i}$. Let it be written as a perfect square:
-3-4i = 1 - 4 - 4i = 1^2 + (2i)^2 - 2(1)(2i) = (1-2i)^2$$-3-4i = 1 - 4 - 4i = 1^2 + (2i)^2 - 2(1)(2i) = (1-2i)^2$$
Thus, sqrtD = pm(1-2i)$\sqrt{D} = \pm(1-2i)$.
### Step 2: Find the Roots
Boxedx = frac(3-2i) pm (1-2i)2$Boxed{x = \frac{(3-2i) \pm (1-2i)}{2}}$
Case 1 (+ sign):
x_1 = frac3 - 2i + 1 - 2i2 = frac4 - 4i2 = 2 - 2i$$x_1 = \frac{3 - 2i + 1 - 2i}{2} = \frac{4 - 4i}{2} = 2 - 2i$$
Case 2 (- sign):
x_2 = frac3 - 2i - 1 + 2i2 = frac22 = 1 + 0i$$x_2 = \frac{3 - 2i - 1 + 2i}{2} = \frac{2}{2} = 1 + 0i$$
Let the roots be alpha + ibeta = 2 - 2i implies alpha=2, beta=-2$\alpha + i\beta = 2 - 2i \implies \alpha=2, \beta=-2$
and gamma + idelta = 1 + 0i implies gamma=1, delta=0$\gamma + i\delta = 1 + 0i \implies \gamma=1, \delta=0$
### Step 3: Evaluate Target Expression
alphagamma + betadelta = (2)(1) + (-2)(0) = 2$$\alpha\gamma + \beta\delta = (2)(1) + (-2)(0) = 2$$
### Pattern Recognition
Always try to express the complex number under the square root in the form (a + bi)^2$(a + bi)^2$ by matching the imaginary part 2ab = -4i implies ab = -2$2ab = -4i \implies ab = -2$, and a^2 - b^2 = -3$a^2 - b^2 = -3$. This avoids long calculations.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Complex Numbers and Quadratic Equations
Qjee_main_2025_29_jan_morningGeometry of Complex Numbers
Let |z_1 - 8 - 2i| le 1$|z_1 - 8 - 2i| \le 1$ and |z_2 - 2 + 6i| le 2$|z_2 - 2 + 6i| \le 2$, z_1, z_2 in mathbbC$z_1, z_2 \in \mathbb{C}$. Then the minimum value of |z_1 - z_2|$|z_1 - z_2|$ is:
A. 3
B. 7
C. 13
D. 10
Solution
### Related Formula
textMinimum distance between two circles: d_min = C_1C_2 - r_1 - r_2$$\text{Minimum distance between two circles: } d_{\min} = C_1C_2 - r_1 - r_2$$
### Core Logic
The expressions define two circular disc fields in the complex plane:
Circle 1: Center C_1(8, 2)$C_1(8, 2)$, radius r_1 = 1$r_1 = 1$
Circle 2: Center C_2(2, -6)$C_2(2, -6)$, radius r_2 = 2$r_2 = 2$Geometry of Complex Numbers diagram for Q69 - JEE Main 2025 Morning
### Step 1: Calculate Center Distance
Using coordinate distance formulation:
C_1C_2 = sqrt(8 - 2)^2 + (2 - (-6))^2 = sqrt6^2 + 8^2 = 10$$C_1C_2 = \sqrt{(8 - 2)^2 + (2 - (-6))^2} = \sqrt{6^2 + 8^2} = 10$$
### Step 2: Find Minimum Separation
|z_1 - z_2|_min = C_1C_2 - r_1 - r_2 = 10 - 1 - 2 = 7$$|z_1 - z_2|_{\min} = C_1C_2 - r_1 - r_2 = 10 - 1 - 2 = 7$$
### Pattern Recognition
Always interpret modulus circle properties geometrically rather than algebraically. Disconnecting complex plane variables into simple 2D analytical geometry centers avoids calculation mistakes entirely.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Complex Numbers
Class 11 Mathematics: Coordinate Geometry
Q6jee_main_2024_01_february_morningGeometry of Complex Numbers
Let S=\zin C:|z-1|=1 text and (sqrt2-1)(z+overlinez)-i(z-overlinez)=2sqrt2\$S=\{z\in C:|z-1|=1 \text{ and } (\sqrt{2}-1)(z+\overline{z})-i(z-\overline{z})=2\sqrt{2}\}$. Let z_1, z_2in S$z_{1}, z_{2}\in S$ be such that |z_1|=max_zin S|z|$|z_{1}|=\max_{z\in S}|z|$ and |z_2|=min_zin S|z|$|z_{2}|=\min_{z\in S}|z|$. Then |sqrt2z_1-z_2|^2$|\sqrt{2}z_{1}-z_{2}|^{2}$ equals:
A.1$1$
B.4$4$
C.3$3$
D.2$2$
Solution
### Related Formula
For a complex number z = x + iy$z = x + iy$:
1. |z| = sqrtx^2 + y^2$|z| = \sqrt{x^2 + y^2}$
2. z + overlinez = 2x$z + \overline{z} = 2x$
3. z - overlinez = 2iy$z - \overline{z} = 2iy$
### Core Logic
Let z = x + iy$z = x + iy$.
The first condition |z - 1| = 1$|z - 1| = 1$ describes a circle:
(x-1)^2 + y^2 = 1 quad implies (1) $$(x-1)^{2} + y^{2} = 1 \quad \implies (1) $$
The second condition gives a straight line:
(sqrt2-1)(2x) - i(2iy) = 2sqrt2$$(\sqrt{2}-1)(2x) - i(2iy) = 2\sqrt{2}$$2(sqrt2-1)x + 2y = 2sqrt2$$2(\sqrt{2}-1)x + 2y = 2\sqrt{2}$$(sqrt2-1)x + y = sqrt2 quad implies (2) $$(\sqrt{2}-1)x + y = \sqrt{2} \quad \implies (2) $$
The set S$S$ contains the intersection points of this circle and line.
### Step 1: Solve for Intersection Points
From (2), y = sqrt2 - (sqrt2-1)x$y = \sqrt{2} - (\sqrt{2}-1)x$. Substitute this into (1):
(x-1)^2 + [sqrt2 - (sqrt2-1)x]^2 = 1$$(x-1)^2 + [\sqrt{2} - (\sqrt{2}-1)x]^2 = 1$$
Solving this quadratic equation gives two values of x$x$:
x = 1 quad textor quad x = frac12-sqrt2 $$x = 1 \quad \text{or} \quad x = \frac{1}{2-\sqrt{2}} $$
- **Case A**: x = 1 implies y = sqrt2 - (sqrt2-1)(1) = 1$x = 1 \implies y = \sqrt{2} - (\sqrt{2}-1)(1) = 1$.
So, one point is z_A = 1 + i$z_A = 1 + i$.
Its magnitude is |z_A| = sqrt1^2 + 1^2 = sqrt2$|z_A| = \sqrt{1^2 + 1^2} = \sqrt{2}$.
- **Case B**: x = frac12-sqrt2 = frac2+sqrt22 = 1 + frac1sqrt2$x = \frac{1}{2-\sqrt{2}} = \frac{2+\sqrt{2}}{2} = 1 + \frac{1}{\sqrt{2}}$.
Then y = sqrt2 - (sqrt2-1)left(1+frac1sqrt2right) = sqrt2 - left(sqrt2 + 1 - 1 - frac1sqrt2right) = frac1sqrt2$y = \sqrt{2} - (\sqrt{2}-1)\left(1+\frac{1}{\sqrt{2}}\right) = \sqrt{2} - \left(\sqrt{2} + 1 - 1 - \frac{1}{\sqrt{2}}\right) = \frac{1}{\sqrt{2}}$.
So, the other point is z_B = left(1 + frac1sqrt2right) + fracisqrt2$z_B = \left(1 + \frac{1}{\sqrt{2}}\right) + \frac{i}{\sqrt{2}}$.
Its magnitude is |z_B| = sqrtleft(1+frac1sqrt2right)^2 + left(frac1sqrt2right)^2 = sqrt1 + sqrt2 + frac12 + frac12 = sqrt2 + sqrt2$|z_B| = \sqrt{\left(1+\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2} = \sqrt{1 + \sqrt{2} + \frac{1}{2} + \frac{1}{2}} = \sqrt{2 + \sqrt{2}}$.
### Step 2: Evaluate Min and Max Magnitude Expressions
Comparing magnitudes, |z_B| > |z_A|$|z_B| > |z_A|$, so:
z_1 = z_B = left(1 + frac1sqrt2right) + fracisqrt2 $$z_1 = z_B = \left(1 + \frac{1}{\sqrt{2}}\right) + \frac{i}{\sqrt{2}} $$z_2 = z_A = 1 + i $$z_2 = z_A = 1 + i $$
Now, calculate |sqrt2z_1-z_2|^2$|\sqrt{2}z_{1}-z_{2}|^{2}$:
sqrt2z_1 = sqrt2left(1 + frac1sqrt2right) + sqrt2left(fracisqrt2right) = (sqrt2 + 1) + i$$\sqrt{2}z_1 = \sqrt{2}\left(1 + \frac{1}{\sqrt{2}}\right) + \sqrt{2}\left(\frac{i}{\sqrt{2}}\right) = (\sqrt{2} + 1) + i$$sqrt2z_1 - z_2 = (sqrt2 + 1 + i) - (1 + i) = sqrt2 $$\sqrt{2}z_1 - z_2 = (\sqrt{2} + 1 + i) - (1 + i) = \sqrt{2} $$|sqrt2z_1 - z_2|^2 = |sqrt2|^2 = 2 $$|\sqrt{2}z_1 - z_2|^2 = |\sqrt{2}|^2 = 2 $$
### Pattern Recognition
Sees: Geometric constraint mapping to a circle and line intersection in the complex plane.
Shortcut: Rationalizing terms like frac12-sqrt2$\frac{1}{2-\sqrt{2}}$ immediately into standard form 1+frac1sqrt2$1+\frac{1}{\sqrt{2}}$ saves you from handling layered fraction algebra down the line.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Complex Numbers: Geometry
Class 10 Coordinate Geometry: Lines and Circles
Q27jee_main_2024_01_february_morningGeometry of Complex Numbers
Let P=\zinmathbbC:|z+2-3i|le1\$P=\{z\in\mathbb{C}:|z+2-3i|\le1\}$ and Q=\zinmathbbC:z(1+i)+overlinez(1-i)le-8\$Q=\{z\in\mathbb{C}:z(1+i)+\overline{z}(1-i)\le-8\}$. Let in P cap Q, |z-3+2i|$P \cap Q, |z-3+2i|$ be maximum and minimum at z_1$z_{1}$ and z_2$z_{2}$ respectively. If |z_1|^2+2|z_2|^2=alpha+betasqrt2$|z_{1}|^{2}+2|z_{2}|^{2}=\alpha+\beta\sqrt{2}$, where alpha, beta$\alpha, \beta$ are integers, then alpha+beta$\alpha+\beta$ equals
Numerical Answer.Answer: 36 to 36
Solution
### Related Formula
For a complex coordinate transformation, substituting z = x + iy$z = x + iy$ and its conjugate overlinez = x - iy$\overline{z} = x - iy$ maps a complex condition directly into rectangular Cartesian coordinates.
### Core Logic
Let's translate the complex set properties into Cartesian geometry:
- Set P$P$: |z - (-2 + 3i)| le 1 implies$|z - (-2 + 3i)| \le 1 \implies$ Interior and boundary of a circle with center C(-2, 3)$C(-2, 3)$ and radius r = 1$r = 1$.
- Set Q$Q$: (x+iy)(1+i) + (x-iy)(1-i) le -8 implies (x - y + ix + iy) + (x - y - ix - iy) le -8$(x+iy)(1+i) + (x-iy)(1-i) \le -8 \implies (x - y + ix + iy) + (x - y - ix - iy) \le -8$2(x - y) le -8 implies x - y + 4 le 0$$2(x - y) \le -8 \implies x - y + 4 \le 0$$
This defines a half-plane below or to the left of the boundary line L_2: x - y + 4 = 0$L_2: x - y + 4 = 0$.
### Step 1: Identify Extreme Points for Distance from Point A
We want to find points in the region P cap Q$P \cap Q$ that minimize and maximize the distance to the external point A(3, -2)$A(3, -2)$, corresponding to |z - (3 - 2i)|$|z - (3 - 2i)|$.
The line L_1$L_1$ connecting center C(-2, 3)$C(-2, 3)$ and point A(3, -2)$A(3, -2)$ has slope:
m = frac-2 - 33 - (-2) = frac-55 = -1$$m = \frac{-2 - 3}{3 - (-2)} = \frac{-5}{5} = -1$$
Equation of line L_1$L_1$: y - 3 = -1(x + 2) implies x + y - 1 = 0$y - 3 = -1(x + 2) \implies x + y - 1 = 0$.
The graphic details the intersection region bounded by the circular locus and the line inequality, showing points z1 and z2 relative to external reference point P.
### Step 2: Calculate Coordinates for z1 and z2
- **Minimum Distance Point (z_2$z_2$):** By geometric observation, the minimum distance from A$A$ to the bounded region is the intersection point of lines L_1$L_1$ and L_2$L_2$:
x - y + 4 = 0 quad textand quad x + y - 1 = 0 implies 2x + 3 = 0 implies x = -frac32, \, y = frac52$$x - y + 4 = 0 \quad \text{and} \quad x + y - 1 = 0 \implies 2x + 3 = 0 \implies x = -\frac{3}{2}, \, y = \frac{5}{2}$$
So, z_2 = left(-frac32, frac52
ight)$z_2 = \left(-\frac{3}{2}, \frac{5}{2}
ight)$.
- **Maximum Distance Point (z_1$z_1$):** The maximum distance is at the far boundary edge of the circle along line L_1$L_1$. The vector path from C(-2, 3)$C(-2, 3)$ opposite to A$A$ has unit direction left(-frac1sqrt2, frac1sqrt2right)$\left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$:
z_1 = left(-2 - frac1sqrt2, \, 3 + frac1sqrt2right)$$z_1 = \left(-2 - \frac{1}{\sqrt{2}}, \, 3 + \frac{1}{\sqrt{2}}\right)$$
### Step 3: Evaluate Magnitudes and Sum Parameters
Calculate the squares of the moduli:
|z_1|^2 = left(-2 - frac1sqrt2right)^2 + left(3 + frac1sqrt2right)^2 = 4 + 2sqrt2 + frac12 + 9 + 3sqrt2 + frac12 = 14 + 5sqrt2$$|z_1|^2 = \left(-2 - \frac{1}{\sqrt{2}}\right)^2 + \left(3 + \frac{1}{\sqrt{2}}\right)^2 = 4 + 2\sqrt{2} + \frac{1}{2} + 9 + 3\sqrt{2} + \frac{1}{2} = 14 + 5\sqrt{2}$$|z_2|^2 = left(-frac32right)^2 + left(frac52
ight)^2 = frac94 + frac254 = frac344 = frac172$$|z_2|^2 = \left(-\frac{3}{2}\right)^2 + \left(\frac{5}{2}
ight)^2 = \frac{9}{4} + \frac{25}{4} = \frac{34}{4} = \frac{17}{2}$$
Now compute the total requested term:
|z_1|^2 + 2|z_2|^2 = (14 + 5sqrt2) + 2left(frac172right) = 14 + 5sqrt2 + 17 = 31 + 5sqrt2$$|z_1|^2 + 2|z_2|^2 = (14 + 5\sqrt{2}) + 2\left(\frac{17}{2}\right) = 14 + 5\sqrt{2} + 17 = 31 + 5\sqrt{2}$$
Matching with alpha + betasqrt2$\alpha + \beta\sqrt{2}$ gives alpha = 31$\alpha = 31$ and beta = 5$\beta = 5$. Thus:
alpha + beta = 31 + 5 = 36$$\alpha + \beta = 31 + 5 = 36$$
### Pattern Recognition
Sees: Locus intersection involving geometric complex inequalities.
Shortcut: Translating complex equations into standard 2D graphs reveals the geometry instantly, mapping extreme distances to line intersections or boundary nodes cleanly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Complex Numbers: Geometry
Class 11 Coordinate Geometry: Straight Lines
Q6jee_main_2024_29_january_eveningModulus and Argument of a Complex Number
Let r$r$ and theta$\theta$ respectively be the modulus and amplitude of the complex number z = 2 - i left(2 tan frac5 pi8right)$z = 2 - i \left(2 \tan \frac{5 \pi}{8}\right)$, then (r, theta)$(r, \theta)$ is equal to
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