Let a be the sum of all coefficients in the expansion of (1 - 2x + 2x2)2023 (3 - 4x2 + 2x3)2024 and b = 0 ( 0x ((1 + t)/t2024) + 1 dtx2 ). If the equations cx2 + dx + e = 0 and 2bx2 + ax + 4 = 0 have

To find the sum of all coefficients in a polynomial expansion, substitute x = 1. Evaluate b = 0 0x ((1 + t)/1 + t2024) dtx2 Using L'Hôpital's Rule (differentiating numerator via Newton-Leibniz): The given second equation is 2bx2 + ax + 4 = 0. Substitute a = 1 and b = (1/2): The discriminant of x2 + x + 4 = 0 is D = 1 - 16 < 0. Roots are non-real complex conjugates. Since c, d, e R and one root is common with a quadratic having non-real roots, both roots must be common. Thus, the coefficients mus

Let a be the sum of all coefficients in the expansion of (1 - 2x + 2x2)2023 (3 - 4x2 + 2x3)2024 and b = 0 ( 0x ((1 + t)/t2024) + 1 dtx2 ). If the equations cx2 + dx + e = 0 and 2bx2 + ax + 4 = 0 have

To find the sum of all coefficients in a polynomial expansion, substitute x = 1. Evaluate b = 0 0x ((1 + t)/1 + t2024) dtx2 Using L'Hôpital's Rule (differentiating numerator via Newton-Leibniz): The given second equation is 2bx2 + ax + 4 = 0. Substitute a = 1 and b = (1/2): The discriminant of x2 + x + 4 = 0 is D = 1 - 16 < 0. Roots are non-real complex conjugates. Since c, d, e R and one root is common with a quadratic having non-real roots, both roots must be common. Thus, the coefficients mus

Let a be the sum of all coefficients in the expansion of (1 - 2x + 2x2)2023 (3

Year	2026	2025	2024	Total
Questions	9	17	11	37

Year

2026

2025

2024

Total

Questions

Solution & Explanation

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)²⁰²³ (1)²⁰²⁴ = 1

Step 1: Evaluate Limit for b

Evaluate $b = x \to 0 \int₀x ln(1 + t)1 + t²⁰²⁴ dtx²$ Using L'Hôpital's Rule (differentiating numerator via Newton-Leibniz):

b = x \to 0 ln(1 + x)1 + x²⁰²⁴2x = x \to 0 (ln(1 + x))/(x) \times 12(1 + x²⁰²⁴)

b = 1 \times (1)/(2) = (1)/(2)

Step 2: Analyze Common Roots

The given second equation is $2bx² + ax + 4 = 0$ $2bx^2 + ax + 4 = 0$. Substitute $a = 1$ $a = 1$ and $b = (1)/(2)$ $b = \frac{1}{2}$:

2((1)/(2))x² + 1(x) + 4 = 0 x² + x + 4 = 0

The discriminant of $x² + x + 4 = 0$ $x^2 + x + 4 = 0$ is $D = 1 - 16 < 0$ $D = 1 - 16 < 0$. Roots are non-real complex conjugates.

Step 3: Final Ratio

Since $c, d, e in R$ $c, d, e \in \mathbb{R}$ and one root is common with a quadratic having non-real roots, both roots must be common. Thus, the coefficients must be proportional:

(c)/(1) = (d)/(1) = (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.

Chapter Mix

Class 11 Maths: Binomial Theorem Class 12 Maths: Limits and Derivatives Class 11 Maths: Quadratic Equations

More Binomial Theorem Previous-Year Questions — Page 8

Q25 jee_main_2024_31_jan_morning Coefficients in Expansion

In the expansion of

(1 + x)(1 - x²)(1 + (3)/(x) + (3)/(x²) + (1)/(x³))⁵, x \neq 0

, the sum of the coefficient of

x³

and

x⁻¹³

is equal to

Numerical Answer. Answer: 118 to 118

Solution

Core Logic

(1+x)(1-x²) ( (1 + (1)/(x))³ )⁵

= (1+x)(1-x)(1+x) (x+1)¹⁵x¹⁵

= (1-x)(1+x)¹⁷x¹⁵

= (1+x)¹⁷ - x(1+x)¹⁷x¹⁵

Step 1: Find Coefficient of x^3

To find coeff of $x³$ $x^3$ in $(1+x)¹⁷ - x(1+x)¹⁷x¹⁵$ , we need the coeff of $x¹⁸$ $x^{18}$ 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 $x¹⁸$ $x^{18}$ in $(1+x)¹⁷$ $(1+x)^{17}$ is 0. Coeff of $x¹⁸$ $x^{18}$ in $x(1+x)¹⁷$ $x(1+x)^{17}$ is the coeff of $x¹⁷$ $x^{17}$ in $(1+x)¹⁷$ $(1+x)^{17}$, which is $1717 = 1$ $\binom{17}{17} = 1$. Thus, coeff of $x¹⁸$ $x^{18}$ 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 $x²$ $x^2$ in the numerator $(1+x)¹⁷ - x(1+x)¹⁷$ $(1+x)^{17} - x(1+x)^{17}$. Coeff of $x²$ $x^2$ in $(1+x)¹⁷$ $(1+x)^{17}$ is $172$ $\binom{17}{2}$. Coeff of $x²$ $x^2$ in $x(1+x)¹⁷$ $x(1+x)^{17}$ is coeff of $x¹$ $x^1$ in $(1+x)¹⁷$ $(1+x)^{17}$, which is $171$ $\binom{17}{1}$. Value = $172 - 171 = (17 \times 16)/(2) - 17 = 136 - 17 = 119$ .

Step 3: Final Sum

Sum of coefficients = $-1 + 119 = 118$ $-1 + 119 = 118$.

Chapter Mix

Class 11 Maths: Binomial Theorem

Solution & Explanation

Core Logic

Step 1: Evaluate Limit for b

Step 2: Analyze Common Roots

Step 3: Final Ratio

Pattern Recognition

Chapter Mix

Reference Study Guides

More Binomial Theorem Previous-Year Questions — Page 8

Solution

Core Logic

Step 1: Find Coefficient of x^3

Step 2: Find Coefficient of x^{-13}

Step 3: Final Sum

Chapter Mix

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Solution & Explanation

Core Logic

Step 1: Evaluate Limit for b

Step 2: Analyze Common Roots

Step 3: Final Ratio

Pattern Recognition

Chapter Mix

Reference Study Guides

More Binomial Theorem Previous-Year Questions — Page 8

Solution

Core Logic

Step 1: Find Coefficient of x^3

Step 2: Find Coefficient of x^{-13}

Step 3: Final Sum

Chapter Mix

More Binomial Theorem Questions — jee_main_2024_31_jan_morning

Select Your Target Exam

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