### Core Logic To find the sum of all coefficients in a polynomial expansion, substitute x = 1$x = 1$. a = (1 - 2(1) + 2(1)^2)^2023 (3 - 4(1)^2 + 2(1)^3)^2024$a = (1 - 2(1) + 2(1)^2)^{2023} (3 - 4(1)^2 + 2(1)^3)^{2024}$ a = (1)^2023 (1)^2024 = 1$a = (1)^{2023} (1)^{2024} = 1$ ### Step 1: Evaluate Limit for b Evaluate b = lim_x to 0 fracint_0^x fracln(1 + t)1 + t^2024 dtx^2$b = \lim_{x \to 0} \frac{\int_{0}^{x} \frac{\ln(1 + t)}{1 + t^{2024}} dt}{x^2}$ Using L'Hôpital's Rule (differentiating numerator via Newton-Leibniz): b = lim_x to 0 fracfracln(1 + x)1 + x^20242x = lim_x to 0 fracln(1 + x)x times frac12(1 + x^2024)$b = \lim_{x \to 0} \frac{\frac{\ln(1 + x)}{1 + x^{2024}}}{2x} = \lim_{x \to 0} \frac{\ln(1 + x)}{x} \times \frac{1}{2(1 + x^{2024})}$ b = 1 times frac12 = frac12$b = 1 \times \frac{1}{2} = \frac{1}{2}$ ### Step 2: Analyze Common Roots The given second equation is 2bx^2 + ax + 4 = 0$2bx^2 + ax + 4 = 0$. Substitute a = 1$a = 1$ and b = frac12$b = \frac{1}{2}$: 2left(frac12right)x^2 + 1(x) + 4 = 0 implies x^2 + x + 4 = 0$2\left(\frac{1}{2}\right)x^2 + 1(x) + 4 = 0 \implies x^2 + x + 4 = 0$ The discriminant of x^2 + x + 4 = 0$x^2 + x + 4 = 0$ is D = 1 - 16 < 0$D = 1 - 16 < 0$. Roots are non-real complex conjugates. ### Step 3: Final Ratio Since c, d, e in mathbbR$c, d, e \in \mathbb{R}$ and one root is common with a quadratic having non-real roots, both roots must be common. Thus, the coefficients must be proportional: fracc1 = fracd1 = frace4$\frac{c}{1} = \frac{d}{1} = \frac{e}{4}$ This implies d : c : e = 1 : 1 : 4$d : c : e = 1 : 1 : 4$. ### Pattern Recognition If a quadratic equation with real coefficients shares a common root with another quadratic having complex roots (D < 0$D < 0$), both roots must be shared, meaning their coefficients are directly proportional. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Binomial Theorem Class 12 Maths: Limits and Derivatives Class 11 Maths: Quadratic Equations

### Core Logic (1+x)(1-x^2) left( left(1 + frac1xright)^3 right)^5$(1+x)(1-x^2) \left( \left(1 + \frac{1}{x}\right)^3 \right)^5$ = (1+x)(1-x)(1+x) frac(x+1)^15x^15$= (1+x)(1-x)(1+x) \frac{(x+1)^{15}}{x^{15}}$ = frac(1-x)(1+x)^17x^15$= \frac{(1-x)(1+x)^{17}}{x^{15}}$ = frac(1+x)^17 - x(1+x)^17x^15$= \frac{(1+x)^{17} - x(1+x)^{17}}{x^{15}}$ ### Step 1: Find Coefficient of x^3 To find coeff of x^3$x^3$ in frac(1+x)^17 - x(1+x)^17x^15$\frac{(1+x)^{17} - x(1+x)^{17}}{x^{15}}$, we need the coeff of x^18$x^{18}$ in the numerator (1+x)^17 - x(1+x)^17$(1+x)^{17} - x(1+x)^{17}$. The maximum power of x$x$ in (1+x)^17$(1+x)^{17}$ is 17$17$, and in x(1+x)^17$x(1+x)^{17}$ is 18$18$. Coeff of x^18$x^{18}$ in (1+x)^17$(1+x)^{17}$ is 0. Coeff of x^18$x^{18}$ in x(1+x)^17$x(1+x)^{17}$ is the coeff of x^17$x^{17}$ in (1+x)^17$(1+x)^{17}$, which is binom1717 = 1$\binom{17}{17} = 1$. Thus, coeff of x^18$x^{18}$ in the numerator is 0 - 1 = -1$0 - 1 = -1$. ### Step 2: Find Coefficient of x^{-13} To find coeff of x^-13$x^{-13}$, we need the coeff of x^2$x^2$ in the numerator (1+x)^17 - x(1+x)^17$(1+x)^{17} - x(1+x)^{17}$. Coeff of x^2$x^2$ in (1+x)^17$(1+x)^{17}$ is binom172$\binom{17}{2}$. Coeff of x^2$x^2$ in x(1+x)^17$x(1+x)^{17}$ is coeff of x^1$x^1$ in (1+x)^17$(1+x)^{17}$, which is binom171$\binom{17}{1}$. Value = binom172 - binom171 = frac17 times 162 - 17 = 136 - 17 = 119$\binom{17}{2} - \binom{17}{1} = \frac{17 \times 16}{2} - 17 = 136 - 17 = 119$. ### Step 3: Final Sum Sum of coefficients = -1 + 119 = 118$-1 + 119 = 118$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Binomial Theorem