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
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### Chapter Mix
Class 11 Chemistry: Practical Chemistry
More Complex Numbers and Quadratic Equations Previous-Year Questions — Page 5
Q21
jee_main_2024_29_january_evening
Roots of Quadratic Equations
Let alpha, beta$\alpha, \beta$ be the roots of the equation x^2 - sqrt6 x + 3 = 0$x^2 - \sqrt{6} x + 3 = 0$ such that operatornameIm(alpha) > operatornameIm(beta)$\operatorname{Im}(\alpha) > \operatorname{Im}(\beta)$. Let a, b$a, b$ be integers not divisible by 3 and n$n$ be a natural number such that fracalpha^99beta + alpha^98 = 3^n (a + ib), i = sqrt-1$\frac{\alpha^{99}}{\beta} + \alpha^{98} = 3^n (a + ib), i = \sqrt{-1}$. Then n + a + b$n + a + b$ is equal to
Numerical Answer. Answer: 49 to 49
Solution
### Related Formula
Using Euler's formula:
e^itheta = costheta + isintheta$$e^{i\theta} = \cos\theta + i\sin\theta$$
### Core Logic
Solving the quadratic root configurations for x^2 - sqrt6x + 3 = 0$x^2 - \sqrt{6}x + 3 = 0$:
x = fracsqrt6 pm sqrt6 - 122 = fracsqrt6 pm isqrt62 = fracsqrt62(1 pm i)$$x = \frac{\sqrt{6} \pm \sqrt{6 - 12}}{2} = \frac{\sqrt{6} \pm i\sqrt{6}}{2} = \frac{\sqrt{6}}{2}(1 \pm i)$$
Given operatornameIm(alpha) > operatornameIm(beta)$\operatorname{Im}(\alpha) > \operatorname{Im}(\beta)$, we set:
alpha = sqrt3 left(frac1+isqrt2right) = sqrt3 e^ipi/4$$\alpha = \sqrt{3} \left(\frac{1+i}{\sqrt{2}}\right) = \sqrt{3} e^{i\pi/4}$$
beta = sqrt3 left(frac1-isqrt2right) = sqrt3 e^-ipi/4$$\beta = \sqrt{3} \left(\frac{1-i}{\sqrt{2}}\right) = \sqrt{3} e^{-i\pi/4}$$
### Step 1: Simplify Target Expression
Let us factor out common variables:
fracalpha^99beta + alpha^98 = alpha^98 left( fracalphabeta + 1 right) = fracalpha^98(alpha + beta)beta$$\frac{\alpha^{99}}{\beta} + \alpha^{98} = \alpha^{98} \left( \frac{\alpha}{\beta} + 1 \right) = \frac{\alpha^{98}(\alpha + \beta)}{\beta}$$
Since alpha + beta = sqrt6$\alpha + \beta = \sqrt{6}$:
textValue = frac(sqrt3e^ipi/4)^98 cdot sqrt6sqrt3e^-ipi/4 = 3^49 e^i 98pi/4 cdot sqrt2 e^ipi/4 = 3^49 cdot sqrt2 e^i 99pi/4$$\text{Value} = \frac{(\sqrt{3}e^{i\pi/4})^{98} \cdot \sqrt{6}}{\sqrt{3}e^{-i\pi/4}} = 3^{49} e^{i 98\pi/4} \cdot \sqrt{2} e^{i\pi/4} = 3^{49} \cdot \sqrt{2} e^{i 99\pi/4}$$
Evaluate e^i 99pi/4$e^{i 99\pi/4}$:
99fracpi4 = 24pi + frac3pi4 implies e^i 99pi/4 = e^i 3pi/4 = frac-1+isqrt2$$99\frac{\pi}{4} = 24\pi + \frac{3\pi}{4} \implies e^{i 99\pi/4} = e^{i 3\pi/4} = \frac{-1+i}{\sqrt{2}}$$
Substituting this back:
textValue = 3^49 cdot sqrt2 left( frac-1+isqrt2 right) = 3^49(-1 + i)$$\text{Value} = 3^{49} \cdot \sqrt{2} \left( \frac{-1+i}{\sqrt{2}} \right) = 3^{49}(-1 + i)$$
### Step 2: Resolving Constants
Comparing with the given expression 3^n(a + ib)$3^n(a + ib)$:
n = 49, quad a = -1, quad b = 1$$n = 49, \quad a = -1, \quad b = 1$$
Therefore:
n + a + b = 49 - 1 + 1 = 49$$n + a + b = 49 - 1 + 1 = 49$$
### Pattern Recognition
Convert complex expressions into polar form r e^itheta$r e^{i\theta}$ early. Power scaling like alpha^98$\alpha^{98}$ becomes simple multiplication under Euler structures.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Complex Numbers and Quadratic Equations
Q29
jee_main_2024_29_january_evening
Integral Solutions
Let the set C = left\(x,y)mid x^2 -2^y = 2023, x,yin mathbbNright\$C = \left\{(x,y)\mid x^2 -2^{y} = 2023, x,y\in \mathbb{N}\right\}$. Then sum_(x,y)in C(x + y)$\sum_{(x,y)\in C}(x + y)$ is equal to
Numerical Answer. Answer: 46 to 46
Solution
### Related Formula
Analyze structural equations through modular constraints (e.g., modulo 3 or 4) to limit potential bounds.
### Core Logic
Given the equation: x^2 - 2^y = 2023$x^2 - 2^y = 2023$.
Let us inspect the numbers modulo 8:
2023 equiv 7 pmod 8$$2023 \equiv 7 \pmod 8$$
A perfect square x^2$x^2$ can only be congruent to 0, 1, 4 pmod 8$0, 1, 4 \pmod 8$.
* If y geq 3$y \geq 3$, then 2^y equiv 0 pmod 8 implies x^2 equiv 7 pmod 8$2^y \equiv 0 \pmod 8 \implies x^2 \equiv 7 \pmod 8$, which is impossible.
Therefore, y$y$ must be less than 3. Since y in mathbbN$y \in \mathbb{N}$, the only choices are y = 1$y = 1$ or y = 2$y = 2$.
### Step 1: Testing Small Exponent Valuations
* Case 1: y = 1$y = 1$
x^2 - 2^1 = 2023 implies x^2 = 2025 implies x = 45 quad (textsince 45^2 = 2025)$$x^2 - 2^1 = 2023 \implies x^2 = 2025 \implies x = 45 \quad (\text{since } 45^2 = 2025)$$
This gives a valid natural solution pair: (45, 1)$(45, 1)$.
* Case 2: y = 2$y = 2$
x^2 - 2^2 = 2023 implies x^2 = 2027$$x^2 - 2^2 = 2023 \implies x^2 = 2027$$
Since 2027 is not a perfect square, this yields no natural solutions.
Thus, the only valid point element inside set C$C$ is (45, 1)$(45, 1)$.
### Step 2: Sum Evaluation
Evaluating the required target accumulation:
sum (x + y) = 45 + 1 = 46$$\sum (x + y) = 45 + 1 = 46$$
### Pattern Recognition
Exponential Diophantine equations (equations with variables in exponents) are best analyzed using modular arithmetic constraints to quickly find small finite upper bounds.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Principle of Mathematical Induction / Number Theory
Q11
jee_main_2024_27_jan_morning
Modulus and Conjugate
If S=\zin C:|z-i|=|z+i|=|z-1|\$S=\{z\in C:|z-i|=|z+i|=|z-1|\}$, then n(S)$n(S)$ is:
- A. 1$1$
- B. 0$0$
- C. 3$3$
- D. 2$2$
Solution
### Related Formula
|z - z_1| = |z - z_2|$$|z - z_1| = |z - z_2|$$
This represents the perpendicular bisector of the line segment joining the points z_1$z_1$ and z_2$z_2$ in the complex plane.
### Core Logic
The given set defines a complex number z$z$ that is equidistant from three fixed points:
A equiv (0, 1)$A \equiv (0, 1)$ corresponding to i$i$
B equiv (0, -1)$B \equiv (0, -1)$ corresponding to -i$-i$
C equiv (1, 0)$C \equiv (1, 0)$ corresponding to 1$1$
The condition |z-i|=|z+i|=|z-1|$|z-i|=|z+i|=|z-1|$ implies that z$z$ is the point of intersection of the perpendicular bisectors of the sides of the triangle formed by A$A$, B$B$, and C$C$.
### Step 1: Finding the Circumcenter
The point of intersection of the perpendicular bisectors of a triangle is its circumcenter.
Since A(0,1)$A(0,1)$, B(0,-1)$B(0,-1)$, and C(1,0)$C(1,0)$ form a unique, non-degenerate triangle, they have exactly one unique circumcenter.
### Step 2: Final Conclusion
Therefore, there is only one such complex number z$z$ that satisfies the condition.
n(S) = 1$n(S) = 1$
### Pattern Recognition
Recognize that |z - z_1| = |z - z_2| = |z - z_3|$|z - z_1| = |z - z_2| = |z - z_3|$ is geometrically identical to finding the circumcenter of a triangle with vertices at z_1$z_1$, z_2$z_2$, and z_3$z_3$. A non-collinear set of three points always yields exactly 1 circumcenter.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Maths: Complex Numbers
Class 11 Maths: Straight Lines
Q30
jee_main_2024_27_jan_morning
Cube Roots of Unity
If
alpha$\alpha$ satisfies the equation x^2+x+1=0$x^{2}+x+1=0$ and
(1+alpha)^7=A+Balpha+Calpha^2$(1+\alpha)^{7}=A+B\alpha+C\alpha^{2}$,
A, B, Cge 0$A, B, C\ge 0$, then
5(3A-2B-C)$5(3A-2B-C)$ is equal to:
Numerical Answer. Answer: 5 to 5
Solution
### Related Formula
1 + omega + omega^2 = 0$$1 + \omega + \omega^2 = 0$$
omega^3 = 1$\omega^3 = 1$
### Core Logic
The equation x^2 + x + 1 = 0$x^2 + x + 1 = 0$ is the standard identity whose roots are the non-real cube roots of unity, omega$\omega$ and omega^2$\omega^2$.
Let us assign alpha = omega$\alpha = \omega$.
We are given the expression (1+alpha)^7$(1+\alpha)^7$.
Substituting the root: (1+omega)^7$(1+\omega)^7$.
### Step 1: Simplify using Unity Properties
From the identity 1 + omega + omega^2 = 0$1 + \omega + \omega^2 = 0$, we extract:
1 + omega = -omega^2$$1 + \omega = -\omega^2$$
Substitute this into the expression:
(1+omega)^7 = (-omega^2)^7 = -omega^14$$(1+\omega)^7 = (-\omega^2)^7 = -\omega^{14}$$
### Step 2: Cyclical Reduction
Using omega^3 = 1$\omega^3 = 1$, reduce the exponent 14 modulo 3:
14 = 3(4) + 2 Rightarrow omega^14 = (omega^3)^4 cdot omega^2 = 1 cdot omega^2 = omega^2$$14 = 3(4) + 2 \Rightarrow \omega^{14} = (\omega^3)^4 \cdot \omega^2 = 1 \cdot \omega^2 = \omega^2$$
Thus, the expression reduces to -omega^2$-\omega^2$.
Rewrite this back to its linear form using 1 + omega + omega^2 = 0$1 + \omega + \omega^2 = 0$:
-omega^2 = 1 + omega = 1 + alpha$$-\omega^2 = 1 + \omega = 1 + \alpha$$
### Step 3: Finding Co-efficients
We compare 1 + alpha$1 + \alpha$ with A + Balpha + Calpha^2$A + B\alpha + C\alpha^2$.
Notice that 1 + alpha$1 + \alpha$ can be directly represented without any alpha^2$\alpha^2$ term (and we must keep A, B, C ge 0$\ge 0$).
So, A = 1$A = 1$, B = 1$B = 1$, C = 0$C = 0$.
### Step 4: Final Output Evaluation
Substitute these constants into the required equation 5(3A - 2B - C)$5(3A - 2B - C)$:
5(3(1) - 2(1) - 0)$$5(3(1) - 2(1) - 0)$$
5(3 - 2) = 5(1) = 5$$5(3 - 2) = 5(1) = 5$$
### Pattern Recognition
The roots of x^2+x+1=0$x^2+x+1=0$ are always omega, omega^2$\omega, \omega^2$. Expressions of the form (1+omega)^k$(1+\omega)^k$ rapidly collapse down to single variables via the 1+omega+omega^2=0$1+\omega+\omega^2=0$ rule, making multi-variable polynomial equations instantly trivial.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Maths: Complex Numbers
Q5
jee_main_2024_29_jan_morning
Modulus and Equality of Complex Numbers
If z=frac12-2i$z=\frac{1}{2}-2i$, is such that |z+1|=alpha z+beta(1+i)$|z+1|=\alpha z+\beta(1+i)$, i=sqrt-1$i=\sqrt{-1}$ and alpha,betain R$\alpha,\beta\in R$, then alpha+beta$\alpha+\beta$ is equal to
- A. -4$-4$
- B. 3$3$
- C. 2$2$
- D. -1$-1$
Solution
### Related Formula
|x + iy| = sqrtx^2 + y^2$$|x + iy| = \sqrt{x^2 + y^2}$$
Two complex numbers are equal if and only if their real and imaginary parts are respectively equal.
### Core Logic
Given z = frac12 - 2i$z = \frac{1}{2} - 2i$.
Calculate |z + 1|$|z + 1|$:
|z + 1| = left| left(frac12 - 2iright) + 1 right|$$|z + 1| = \left| \left(\frac{1}{2} - 2i\right) + 1 \right|$$
= left| frac32 - 2i right|$$= \left| \frac{3}{2} - 2i \right|$$
= sqrtleft(frac32right)^2 + (-2)^2 = sqrtfrac94 + 4 = sqrtfrac254 = frac52$$= \sqrt{\left(\frac{3}{2}\right)^2 + (-2)^2} = \sqrt{\frac{9}{4} + 4} = \sqrt{\frac{25}{4}} = \frac{5}{2}$$
Now substitute z$z$ and |z + 1|$|z + 1|$ into the original equation:
frac52 = alphaleft(frac12 - 2iright) + beta(1 + i)$$\frac{5}{2} = \alpha\left(\frac{1}{2} - 2i\right) + \beta(1 + i)$$
Expand and group real and imaginary components on the RHS:
frac52 = left(fracalpha2 - 2alpha iright) + (beta + beta i)$$\frac{5}{2} = \left(\frac{\alpha}{2} - 2\alpha i\right) + (\beta + \beta i)$$
frac52 = left(fracalpha2 + betaright) + i(beta - 2alpha)$$\frac{5}{2} = \left(\frac{\alpha}{2} + \beta\right) + i(\beta - 2\alpha)$$
### Step 1: Equate Parts
By equating the real and imaginary parts from both sides, we get a system of linear equations:
Imaginary part:
0 = beta - 2alpha Rightarrow beta = 2alpha$$0 = \beta - 2\alpha \Rightarrow \beta = 2\alpha$$
Real part:
frac52 = fracalpha2 + beta$$\frac{5}{2} = \frac{\alpha}{2} + \beta$$
Substitute beta = 2alpha$\beta = 2\alpha$ into the real part equation:
frac52 = fracalpha2 + 2alpha$$\frac{5}{2} = \frac{\alpha}{2} + 2\alpha$$
frac52 = frac5alpha2$$\frac{5}{2} = \frac{5\alpha}{2}$$
alpha = 1$\alpha = 1$
Using alpha = 1$\alpha = 1$, find beta$\beta$:
beta = 2(1) = 2$$\beta = 2(1) = 2$$
### Step 2: Final Calculation
Calculate the final requested value:
alpha + beta = 1 + 2 = 3$$\alpha + \beta = 1 + 2 = 3$$
### Pattern Recognition
Equating complex parts reduces single complex equations into two simultaneous linear equations. Treat |z+1|$|z+1|$ strictly as a scalar magnitude and parse directly into algebraic components.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Mathematics: Complex Numbers and Quadratic Equations
