JEE Advanced 2025 · Paper 2 Mathematics

Looking at this problem, I need to solve the differential equation and evaluate a specific expression. ## Setting Up the Substitution Starting with $x^2 \frac{dy}{dx} + xy = x^2 + y^2$, I use the substitution **$y = vx$**, so $\frac{dy}{dx} = v + x\frac{dv}{dx}$: $$x^2\left(v + x\frac{dv}{dx}\right) + x(vx) = x^2 + v^2x^2$$ $$2vx^2 + x^3\frac{dv}{dx} = x^2(1 + v^2)$$ $$x\frac{dv}{dx} = (1 - v)^2$$ ## Solving the Separable Equation Separating variables and integrating: $$\int \frac{dv}{(1-v)^2} = \int \frac{dx}{x} \implies \frac{1}{1-v} = \ln x + C$$ Substituting back $v = y/x$: $$\frac{x}{x - y} = \ln x + C$$ **Applying $y(1) = 0$:** $\frac{1}{1} = 0 + C \implies C = 1$ ## Finding the Explicit Solution From $\frac{x}{x-y} = \ln x + 1$, solving for $y$: $$y = x - \frac{x}{\ln x + 1} = x\cdot\frac{\ln x}{\ln x + 1}$$ **Verification:** Substituting back confirms $x^2y' + xy = x^2 + y^2$ ✓ ## Computing the Required Values **At $x = e$:** $$y(e) = e \cdot \frac{1}{1+1} = \frac{e}{2}$$ **At $x = e^2$:** $$y(e^2) = e^2 \cdot \frac{2}{2+1} = \frac{2e^2}{3}$$ ## Final Calculation $$2\cdot\frac{(y(e))^2}{y(e^2)} = 2\cdot\frac{\left(\frac{e}{2}\right)^2}{\frac{2e^2}{3}} = 2\cdot\frac{\frac{e^2}{4}}{\frac{2e^2}{3}} = 2\cdot\frac{e^2}{4}\cdot\frac{3}{2e^2} = 2\cdot\frac{3}{8} = \boxed{\dfrac{3}{4}}$$

Looking at this problem, I need to find which coefficient $a_r$ in the expansion of $\left(1 + \frac{2}{5}x\right)^{23}$ is the largest. ## Setting Up the Coefficients By the **Binomial Theorem**: $$a_i = \binom{23}{i}\left(\frac{2}{5}\right)^i, \qquad i = 0, 1, \ldots, 23$$ Since all coefficients are **positive**, I can find the maximum by analyzing consecutive ratios. ## Ratio Test for Consecutive Terms $$\frac{a_{i+1}}{a_i} = \frac{\binom{23}{i+1}}{\binom{23}{i}} \cdot \frac{2}{5} = \frac{23-i}{i+1} \cdot \frac{2}{5}$$ **The sequence increases** when this ratio $> 1$, and **decreases** when it $< 1$. Setting the ratio equal to 1: $$\frac{2(23-i)}{5(i+1)} = 1 \implies 46 - 2i = 5i + 5 \implies i = \frac{41}{7} \approx 5.857$$ ## Verifying the Transition Since $41/7$ is **not an integer**, I check the integers on either side: | $i$ | Ratio $\frac{a_{i+1}}{a_i}$ | Comparison | |---|---|---| | $i=4$ | $\frac{19}{5} \cdot \frac{2}{5} = \frac{38}{25} = 1.52$ | $a_5 > a_4$ | | $i=5$ | $\frac{18}{6} \cdot \frac{2}{5} = \frac{6}{5} = 1.20$ | $a_6 > a_5$ ✓ | | $i=6$ | $\frac{17}{7} \cdot \frac{2}{5} = \frac{34}{35} \approx 0.971$ | $a_7 < a_6$ ✓ | | $i=7$ | $\frac{16}{8} \cdot \frac{2}{5} = \frac{4}{5} = 0.80$ | $a_8 < a_7$ | ## Conclusion The coefficients satisfy: $$a_0 < a_1 < a_2 < a_3 < a_4 < a_5 < \boxed{a_6} > a_7 > a_8 > \cdots > a_{23}$$ The maximum occurs at $a_6$, so: $$r = \boxed{6}$$

Looking at this problem, I need to find the defective rate of M3 using the given information about production proportions, overall defective rate, and conditional probability. ## Setting Up the Problem **Production proportions** (2:2:1): $$P(M_1) = \frac{2}{5}, \quad P(M_2) = \frac{2}{5}, \quad P(M_3) = \frac{1}{5}$$ **Known values:** - $P(D) = 0.20$ (overall defective rate) - $P(D \mid M_1) = 0.15$ (defective rate from M1) - $P(M_2 \mid D) = \frac{2}{5}$ (given a defective bulb, probability it's from M2) **Unknowns:** $P(D \mid M_2) = p_2$ and $P(D \mid M_3) = p_3$ ## Step 1: Find $p_2$ Using Bayes' Theorem $$P(M_2 \mid D) = \frac{P(D \mid M_2) \cdot P(M_2)}{P(D)}$$ $$\frac{2}{5} = \frac{p_2 \cdot \frac{2}{5}}{\frac{1}{5}} = 2p_2$$ $$\boxed{p_2 = \frac{1}{5} = 0.20}$$ ## Step 2: Find $p_3$ Using the Law of Total Probability $$P(D) = P(D \mid M_1)\,P(M_1) + P(D \mid M_2)\,P(M_2) + P(D \mid M_3)\,P(M_3)$$ $$0.20 = (0.15)\!\left(\frac{2}{5}\right) + (0.20)\!\left(\frac{2}{5}\right) + p_3\!\left(\frac{1}{5}\right)$$ $$0.20 = 0.06 + 0.08 + \frac{p_3}{5}$$ $$0.06 = \frac{p_3}{5}$$ $$p_3 = 0.30$$ ## Verification $$P(D) = 0.06 + 0.08 + 0.06 = 0.20 \checkmark$$ $$P(M_2 \mid D) = \frac{0.08}{0.20} = \frac{2}{5} \checkmark$$ ## Answer The probability that a randomly chosen bulb from $M_3$ is defective is: $$\boxed{\dfrac{3}{10}}$$

Looking at this problem, I need to find when vectors $\vec{X}$, $\vec{Y}$, $\vec{Z}$ are coplanar, meaning their scalar triple product equals zero. ## Setting Up the Coefficient Matrix Expressing the vectors in terms of $\vec{x}$, $\vec{y}$, $\vec{z}$: | | $\vec{x}$ | $\vec{y}$ | $\vec{z}$ | |---|---|---|---| | $\vec{X}$ | $\alpha$ | $\beta$ | $-1$ | | $\vec{Y}$ | $-1$ | $\alpha$ | $\beta$ | | $\vec{Z}$ | $\beta$ | $-1$ | $\alpha$ | The coplanarity condition gives: $$[\vec{X}, \vec{Y}, \vec{Z}] = \det(M) \cdot [\vec{x}, \vec{y}, \vec{z}] = 0$$ ## Computing $[\vec{x}, \vec{y}, \vec{z}]$ $$[\vec{x}, \vec{y}, \vec{z}] = \begin{vmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2 \end{vmatrix} = 1(5) - 2(1) + 3(-7) = -18 \neq 0$$ Since $[\vec{x}, \vec{y}, \vec{z}] \neq 0$, I need **$\det(M) = 0$**. ## Computing $\det(M)$ $$\det(M) = \begin{vmatrix} \alpha & \beta & -1 \\ -1 & \alpha & \beta \\ \beta & -1 & \alpha \end{vmatrix}$$ Expanding along the first row: $$= \alpha(\alpha^2 + \beta) - \beta(-\alpha - \beta^2) + (-1)(1 - \alpha\beta)$$ $$= \alpha^3 + \alpha\beta + \alpha\beta + \beta^3 - 1 + \alpha\beta = \alpha^3 + \beta^3 + 3\alpha\beta - 1$$ ## Factoring Using the Identity Applying $a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ with $a=\alpha$, $b=\beta$, $c=-1$: $$\alpha^3 + \beta^3 + 3\alpha\beta - 1 = (\alpha + \beta - 1)\underbrace{(\alpha^2 + \beta^2 + 1 - \alpha\beta + \alpha + \beta)}_{\text{second factor}}$$ **Checking the second factor** for positive $\alpha, \beta$: - $\alpha^2 + \beta^2 - \alpha\beta = \left(\alpha - \frac{\beta}{2}\right)^2 + \frac{3\beta^2}{4} > 0$ - $\alpha + \beta + 1 > 0$ So the second factor is **always positive** for positive real numbers. ## Solving the Equation The only solution is: $$\alpha + \beta - 1 = 0 \implies \alpha + \beta = 1$$ This admits infinitely many pairs of distinct positive reals (e.g., $\alpha = 0.3$, $\beta = 0.7$). ✓ ## Final Answer $$\alpha + \beta - 3 = 1 - 3 = \boxed{-2}$$

Looking at this problem, I need to find the cube root of unity ω with positive argument, compute the given sum, then determine the argument. ## Setting Up the Key Values **Finding ω:** The cube roots of unity are $1, e^{2\pi i/3}, e^{-2\pi i/3}$. Since $0 < \arg(\omega) < \pi$: $$\omega = e^{2\pi i/3} = -\tfrac{1}{2} + \tfrac{\sqrt{3}}{2}i$$ **Finding $-\omega$:** $$-\omega = \tfrac{1}{2} - \tfrac{\sqrt{3}}{2}i = e^{-i\pi/3}$$ **Key property:** $(-\omega)^6 = 1$, so $-\omega$ is a primitive 6th root of unity. ## Computing the Cyclic Pattern Since $(-\omega)^6 = 1$, I'll find the sum over one complete period: | $n$ | $(-\omega)^n$ | |-----|---------------| | 1 | $-\omega$ | | 2 | $\omega^2$ | | 3 | $-1$ | | 4 | $\omega$ | | 5 | $-\omega^2$ | | 6 | $1$ | **Sum over one cycle:** $(-\omega + \omega^2 - 1 + \omega - \omega^2 + 1) = 0$ ## Evaluating the Sum Since $2025 = 6 \times 337 + 3$, the first $2022$ terms form **337 complete cycles** (each summing to 0), leaving: $$S = (-\omega)^{2023} + (-\omega)^{2024} + (-\omega)^{2025} = (-\omega)^1 + (-\omega)^2 + (-\omega)^3$$ $$S = -\omega + \omega^2 - 1$$ **Substituting explicit values:** $$S = \underbrace{\left(\tfrac{1}{2} - \tfrac{\sqrt{3}}{2}i\right)}_{-\omega} + \underbrace{\left(-\tfrac{1}{2} - \tfrac{\sqrt{3}}{2}i\right)}_{\omega^2} + \underbrace{(-1)}_{-1} = -1 - \sqrt{3}\,i$$ ## Verification via Geometric Series $$S = \frac{e^{-i\pi/3}(1 - e^{-i\pi \cdot 2025/3})}{1 - e^{-i\pi/3}} = \frac{2e^{-i\pi/3}}{e^{i\pi/3}} = 2e^{-2i\pi/3} = -1 - \sqrt{3}\,i \checkmark$$ ## Finding the Argument The point $-1 - \sqrt{3}\,i$ lies in the **third quadrant** with: $$|S| = 2, \qquad \text{reference angle} = \arctan\!\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3}$$ $$\alpha = \arg(S) = -\pi + \frac{\pi}{3} = -\frac{2\pi}{3}$$ This satisfies $-\pi < \alpha \leq \pi$. ✓ ## Final Computation $$\frac{3\alpha}{\pi} = \frac{3\left(-\dfrac{2\pi}{3}\right)}{\pi} = \boxed{-2}$$

Looking at this problem, I need to find the derivative of $f \circ g^{-1}$ at $x = 2$ using the chain rule. ## Setting Up the Inverse Function **Finding $g^{-1}(x)$:** Starting with $y = \frac{4}{1 + e^{-2x}}$, I solve for $x$: $$y(1 + e^{-2x}) = 4 \implies e^{-2x} = \frac{4-y}{y} \implies x = \frac{1}{2}\ln\left(\frac{y}{4-y}\right)$$ So: $g^{-1}(x) = \frac{1}{2}\ln\left(\frac{x}{4-x}\right)$ ## Evaluating at $x = 2$ **Finding $g^{-1}(2)$:** $$g^{-1}(2) = \frac{1}{2}\ln\left(\frac{2}{4-2}\right) = \frac{1}{2}\ln(1) = 0$$ ## Computing the Derivative via Chain Rule $$(f \circ g^{-1})'(x) = f'\!\big(g^{-1}(x)\big) \cdot (g^{-1})'(x)$$ **Computing $f'(t)$:** $$f'(t) = \frac{2t + 2}{t^2 + 2t + 4}$$ At $t = g^{-1}(2) = 0$: $$f'(0) = \frac{0 + 2}{0 + 0 + 4} = \frac{1}{2}$$ **Computing $(g^{-1})'(x)$:** $$g^{-1}(x) = \frac{1}{2}\big[\ln x - \ln(4-x)\big]$$ $$(g^{-1})'(x) = \frac{1}{2}\left[\frac{1}{x} + \frac{1}{4-x}\right] = \frac{1}{2}\cdot\frac{4}{x(4-x)} = \frac{2}{x(4-x)}$$ At $x = 2$: $$(g^{-1})'(2) = \frac{2}{2 \cdot 2} = \frac{1}{2}$$ ## Final Computation $$(f \circ g^{-1})'(2) = f'(0) \cdot (g^{-1})'(2) = \frac{1}{2} \cdot \frac{1}{2} = \boxed{\dfrac{1}{4}}$$

Looking at this problem, I need to evaluate a sum with a specific pattern and then compute a squared ratio. ## Setting Up the Sum The sum has terms of the form $\frac{1}{\sin(60+2k)°\sin(61+2k)°}$ for $k = 0, 1, \ldots, 29$ (that's **30 terms**). **Key identity:** Since each pair of angles differs by 1°: $$\frac{1}{\sin n° \sin(n+1)°} = \frac{1}{\sin 1°}\big(\cot n° - \cot(n+1)°\big)$$ Applying this to each term: $$\alpha = \frac{1}{\sin 1°}\Big[(\cot 60°-\cot 61°) + (\cot 62°-\cot 63°) + \cdots + (\cot 118°-\cot 119°)\Big]$$ ## Separating Even and Odd Cotangents Regrouping: $$\alpha = \frac{A - B}{\sin 1°}$$ where $A = \cot 60° + \cot 62° + \cdots + \cot 118°$ and $B = \cot 61° + \cot 63° + \cdots + \cot 119°$. **Evaluating $B$ (odd angles 61° to 119°):** Using $\cot(180°-x) = -\cot x$, I pair complementary angles: $$(61°,119°),\; (63°,117°),\; \ldots,\; (89°,91°) \quad \Longrightarrow \quad 15 \text{ pairs}$$ Each pair sums to zero, covering all 30 terms. So $B = 0$. **Evaluating $A$ (even angles 60° to 118°):** Similarly pairing: $$(62°,118°),\; (64°,116°),\; \ldots,\; (88°,92°) \quad \Longrightarrow \quad 14 \text{ pairs}$$ Each pair sums to zero, using 28 of the 30 terms. The **remaining terms** are: - $\cot 90° = 0$ - $\cot 60° = \dfrac{1}{\sqrt{3}}$ So $A = \dfrac{1}{\sqrt{3}}$. ## Computing $\alpha$ $$\alpha = \frac{A - B}{\sin 1°} = \frac{1/\sqrt{3}}{\sin 1°} = \frac{1}{\sqrt{3}\,\sin 1°}$$ ## Final Calculation $$\frac{\csc 1°}{\alpha} = \frac{1/\sin 1°}{1/(\sqrt{3}\,\sin 1°)} = \frac{1}{\sin 1°} \cdot \sqrt{3}\,\sin 1° = \sqrt{3}$$ $$\left(\frac{\csc 1°}{\alpha}\right)^2 = (\sqrt{3})^2 = \boxed{3}$$

# Evaluating $\alpha = \int_{1/2}^{2} \frac{\tan^{-1} x}{2x^2 - 3x + 2}\, dx$ ## Step 1: Exploit the Reciprocal Limits Since the limits $\frac{1}{2}$ and $2$ are reciprocals, I substitute $x = \frac{1}{t}$, so $dx = -\frac{1}{t^2}\,dt$ and $\tan^{-1}\!\left(\frac{1}{t}\right) = \frac{\pi}{2} - \tan^{-1} t$ for $t > 0$. The denominator transforms as: $$2x^2 - 3x + 2 \;\Big|_{x=1/t} = \frac{2t^2 - 3t + 2}{t^2}$$ After substitution (and flipping limits): $$\alpha = \int_{1/2}^{2} \frac{\frac{\pi}{2} - \tan^{-1} t}{2t^2 - 3t + 2}\, dt$$ ## Step 2: Add Original and Transformed Integrals Adding the two expressions for $\alpha$: $$2\alpha = \int_{1/2}^{2} \frac{\frac{\pi}{2}}{2x^2 - 3x + 2}\,dx \quad\Longrightarrow\quad \alpha = \frac{\pi}{4}\int_{1/2}^{2} \frac{dx}{2x^2 - 3x + 2}$$ ## Step 3: Evaluate the Remaining Integral Completing the square: $2x^2 - 3x + 2 = 2\!\left[\left(x - \tfrac{3}{4}\right)^2 + \tfrac{7}{16}\right]$ $$\int \frac{dx}{2\left[\left(x-\frac{3}{4}\right)^2 + \left(\frac{\sqrt{7}}{4}\right)^2\right]} = \frac{2}{\sqrt{7}}\tan^{-1}\!\left(\frac{4x-3}{\sqrt{7}}\right)$$ **Evaluating at the bounds:** $$\frac{2}{\sqrt{7}}\left[\tan^{-1}\!\left(\frac{5}{\sqrt{7}}\right) - \tan^{-1}\!\left(\frac{-1}{\sqrt{7}}\right)\right] = \frac{2}{\sqrt{7}}\left[\tan^{-1}\!\left(\frac{5}{\sqrt{7}}\right) + \tan^{-1}\!\left(\frac{1}{\sqrt{7}}\right)\right]$$ **Using the addition formula** ($ab = \frac{5}{7} < 1$, so no correction needed): $$\tan^{-1}\!\left(\frac{5}{\sqrt{7}}\right) + \tan^{-1}\!\left(\frac{1}{\sqrt{7}}\right) = \tan^{-1}\!\left(\frac{\frac{6}{\sqrt{7}}}{1 - \frac{5}{7}}\right) = \tan^{-1}\!\left(\frac{6/\sqrt{7}}{2/7}\right) = \tan^{-1}(3\sqrt{7})$$ ## Step 4: Assemble $\alpha$ and Compute the Answer $$\alpha = \frac{\pi}{4} \cdot \frac{2}{\sqrt{7}}\,\tan^{-1}(3\sqrt{7}) = \frac{\pi}{2\sqrt{7}}\,\tan^{-1}(3\sqrt{7})$$ Now compute the argument: $$\frac{2\alpha\sqrt{7}}{\pi} = \frac{2\sqrt{7}}{\pi}\cdot\frac{\pi}{2\sqrt{7}}\,\tan^{-1}(3\sqrt{7}) = \tan^{-1}(3\sqrt{7})$$ Therefore: $$\sqrt{7}\,\tan\!\left(\frac{2\alpha\sqrt{7}}{\pi}\right) = \sqrt{7}\,\tan\!\Big(\tan^{-1}(3\sqrt{7})\Big) = \sqrt{7}\cdot 3\sqrt{7} = 3\times 7 = \boxed{21}$$