If =19((r + 3/2r))cdot9Cr = alpha ( rac32)9 - beta [cite: 596], where alpha, beta N [cite: 596], then (alpha + beta)2…

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Q jee_main_2025_02_april_morning Term Independent of x

The term independent of

x

in the expansion of

( (x + 1)(x ^ (2)/(3) + 1 - x ^ (1)/(3)) - (x - 1)(x - x ^ (1)/(2))) ^ 1 0, x > 1

is:

A. $210$ $210$
B. $150$ $150$
C. $240$ $240$
D. $120$ $120$

Solution

Related Formula

Algebraic factorizations:

x+1 = (x1/3)³ + 1³ = (x1/3+1)(x2/3-x1/3+1)

x-1 = (\sqrt(x))² - 1² = (\sqrt(x)-1)(\sqrt(x)+1)

Core Logic

Simplify inside the parentheses before using the Binomial general term formula.

Step 1: Simplify Bracket Terms

First fraction:

x+1x2/3-x1/3+1 = x1/3+1

Second fraction:

x-1x-x1/2 = (\sqrt(x)-1)(\sqrt(x)+1)\sqrt(x)(\sqrt(x)-1) = \sqrt(x)+1\sqrt(x) = 1 + x-1/2

Subtract the two simplified results:

(x1/3+1) - (1+x-1/2) = x1/3 - x-1/2

Step 2: Apply Binomial Theorem

The expression simplifies to $(x1/3 - x-1/2)¹⁰$ $\left(x^{1/3} - x^{-1/2}\right)^{10}$. Write the general term $Tᵣ₊₁$ $T_{r+1}$:

Tᵣ₊₁ = 10r (x1/3)10-r (-x-1/2)^r = 10r (-1)^r x(10-r)/(3) - (r)/(2)

Step 3: Solve for Independent Term

Set the net power of $x$ $x$ to zero:

(10-r)/(3) - (r)/(2) = 0 20 - 2r - 3r = 0 5r = 20 r = 4

Substitute $r=4$ $r=4$ into the general term expression:

Coefficient = 104 (-1)⁴ = (10 \times 9 \times 8 \times 7)/(4 \times 3 \times 2 \times 1) = 210

Pattern Recognition

The initial fractions look complex but contain simple hidden identities ( $a³+b³$ $a^3+b^3$ and $a²-b²$ $a^2-b^2$). Identifying these transforms a daunting fraction expansion into a classic two-term independent coefficient puzzle.

Chapter Mix

Class 11 Mathematics: Binomial Theorem

Q72 jee_main_2025_03_april_evening Trinomial Expansion

Let

(1 + x + x²)¹⁰ = a₀ + a₁x + a₂x² + + a₂₀x²⁰

. If

(a₁ + a₃ + a₅ + + a₁₉) - 11a₂ = 121k

, then

k

is equal to

Numerical Answer. Answer: 239 to 239

Solution

Related Formula

For trinomial expansion:

(1+x+x²)ⁿ = Σr=0²ⁿ aᵣ x^r

Symmetric coefficients can be extracted by substituting $x=1$ $x=1$ and $x=-1$ $x=-1$.

Core Logic

Let's perform the substitution:

For $x = 1$ $x = 1$:

3¹⁰ = a₀ + a₁ + a₂ + + a₂₀ --- (1)

For $x = -1$ $x = -1$:

1¹⁰ = a₀ - a₁ + a₂ - a₃ + + a₂₀ --- (2)

Step 1: Extracting odd term sum

Subtracting (2) from (1):

3¹⁰ - 1 = 2(a₁ + a₃ + a₅ + + a₁₉)

a₁ + a₃ + a₅ + + a₁₉ = 3¹⁰ - 12

Since $3⁵ = 243 3¹⁰ = 59049$ $3^5 = 243 \implies 3^{10} = 59049$:

a₁ + a₃ + a₅ + + a₁₉ = (59049 - 1)/(2) = 29524

Step 2: Finding $a₂$ $a_2$ and $k$ $k$

Using expansion formula to find $a₂$ $a_2$ (coefficient of $x²$ $x^2$):

(1 + x + x²)¹⁰ = Σ (10!)/(p! q! r!) xq + 2r (p+q+r = 10)

For $q + 2r = 2$ $q + 2r = 2$:

Case 1: $r=1, q=0 p=9$ $r=1, q=0 \implies p=9$. Coefficient = $(10!)/(9! 0! 1!) = 10$ $\frac{10!}{9! 0! 1!} = 10$
Case 2: $r=0, q=2 p=8$ $r=0, q=2 \implies p=8$. Coefficient = $(10!)/(8! 2! 0!) = 45$ $\frac{10!}{8! 2! 0!} = 45$

a₂ = 10 + 45 = 55

Now substitute into the target relation:

29524 - 11(55) = 29524 - 605 = 28919

121k = 28919 k = 239

Pattern Recognition

Substituting standard complex roots or positive units is the fastest way to resolve sum of subset coefficients in polynomial expansions. Always use partition calculations for the early index terms ( $a₁, a₂$ $a_1, a_2$) rather than standard multinomial permutations to avoid calculation mistakes.

Chapter Mix

Class 11 Mathematics: Binomial Theorem

Q54 jee_main_2025_07_april_morning Remainder Problems

The remainder when

((64)(64))(64)

is divided by 7 is equal to

A. $4$ $4$
B. $1$ $1$
C. $3$ $3$
D. $6$ $6$

Solution

Related Formula

Binomial expansion for checking remainders:

(1 + kx)ⁿ = 1 + n(kx) + (n(n-1))/(2)(kx)² + \equiv 1 x

Core Logic

Let the target expression be $N = ((64)⁶⁴)⁶⁴$ $N = \left((64)^{64}\right)^{64}$. Using power rules, $N = 6464 \times 64 = 6464²$ $N = 64^{64 \times 64} = 64^{64^2}$.

Let the large exponent exponent be $n = 64²$ $n = 64^2$. We need to find the remainder of $64ⁿ$ $64^n$ when divided by $7$ $7$.

Step 1: Express Base in Terms of Modulus

Observe that $64 = 63 + 1 = 7 \times 9 + 1$ $64 = 63 + 1 = 7 \times 9 + 1$. Substituting this into the expression: $N = (1 + 63)ⁿ$ $N = (1 + 63)^n$

Expanding using the binomial theorem:

N = 1 + ⁿC₁(63) + ⁿC₂(63)² + + ⁿCₙ(63)ⁿ

N = 1 + 63 \cdot λ = 1 + 7(9λ)

Since $7(9λ)$ $7(9\lambda)$ is perfectly divisible by $7$ $7$, the remaining term is $1$ $1$.

Pattern Recognition

Whenever the base can be written as $km + 1$ $km + 1$, where $m$ $m$ is the divisor, the value of $(km + 1)ⁿ \equiv 1ⁿ \equiv 1 m$ instantly, regardless of the size or complexity of the exponent layout.

Chapter Mix

Class 11 Mathematics: Binomial Theorem

Q jee_main_2025_08_april_evening Integral Terms in Binomial Expansion

The number of integral terms in the expansion of

(5(1)/(2) + 7(1)/(8))¹⁰¹⁶

is

A. $127$ $127$
B. $130$ $130$
C. $129$ $129$
D. $128$ $128$

Solution

Related Formula

Tᵣ = nr an-r b^r

Core Logic

Formulate general progression steps. Isolate divisor common multiples to find non fraction power configurations tracking perfectly over boundaries.

Step 1: State the General Binomial Term

Tᵣ = 1016r 5(1016-r)/(2) 7(r)/(8)

Step 2: Apply Rational Power Constraints

For expressions to map purely to integer categories, indexing tracker index paths $r$ $r$ must explicitly scale as multiples of 8 over the domain range:

r in 0, 8, 16, 24, , 1016

Step 3: Enumerate Arithmetic Progression Size

Using standard progression length mapping tools:

1016 = 0 + (n - 1) \cdot 8

n - 1 = (1016)/(8) = 127 n = 128

Pattern Recognition

Finding pure integer steps matches finding values that fit LCM tracking parameters for base radical roots across whole block lengths.

Chapter Mix

Class 11 Mathematics: Binomial Theorem

Q74 jee_main_2025_08_april_evening Last Two Digits of a Number

The product of the last two digits of

(1919)¹⁹¹⁹

is

Numerical Answer. Answer: 63 to 63

Solution

Related Formula

(10k - 1)ⁿ = + nn-1(10k)(-1)ⁿ⁻¹ + (-1)ⁿ

Core Logic

Break the base parameter into a multiple of 10 form ( $1920 - 1$ $1920 - 1$) and use binomial expansions to decouple high factor blocks from final terminal digits.

Step 1: Set Up Binomial Expansion

(1919)¹⁹¹⁹ = (1920 - 1)¹⁹¹⁹

= 19190(1920)¹⁹¹⁹ - + 19191918(1920)¹ - 19191919

Step 2: Isolate Low Factor Coefficients

All higher structural tracks sitting above index points present factors multiple loops over 100. Isolate trailing values:

= 100λ + 1919 \times 1920 - 1

Step 3: Deduce Final Trailing Quotient Product

1919 \times 1920 - 1 = 3684480 - 1 = 3684479

Trailing structural indicators: $79$ $79$. Product calculation response: $7 \times 9 = 63$ $7 \times 9 = 63$

Pattern Recognition

Using expansions around tens bases shifts focus entirely to the last two expansion expressions to find digit answers rapidly.

Chapter Mix

Class 11 Mathematics: Binomial Theorem

Solution & Explanation

Related Formula

Core Logic

Step 1: Combining the components

Pattern Recognition

Chapter Mix

Reference Study Guides

More Binomial Theorem Previous-Year Questions — Page 3

Solution

Related Formula

Core Logic

Step 1: Simplify Bracket Terms

Step 2: Apply Binomial Theorem

Step 3: Solve for Independent Term

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Extracting odd term sum

Step 2: Finding a₂$a_2$ and k$k$

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Express Base in Terms of Modulus

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: State the General Binomial Term

Step 2: Apply Rational Power Constraints

Step 3: Enumerate Arithmetic Progression Size

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Set Up Binomial Expansion

Step 2: Isolate Low Factor Coefficients

Step 3: Deduce Final Trailing Quotient Product

Pattern Recognition

Chapter Mix

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Step 2: Finding $a₂$ $a_2$ and $k$ $k$