JEE Advanced 1984

LEVELJEE Main

Animated Solution for Mathematics - Binomial Theorem: If in the expansion of $(1 + x)^{m} (1 - x)^{n}$ , the coefficients of $x$ and $x^{2}$ are 3 and $- 6$ respectively, then $m$ is

Q: How do I know which terms to multiply to get $x^2$?

Think of it as a matching game. To get $x^2$, you need pairs of terms whose powers add up to 2. You have three combinations: (constant $\times$ $x^2$), ($x^1 \times$ $x^1$), and ($x^2 \times$ constant). If you miss any of these, your coefficient will be wrong.

Q: Is there a faster way to simplify the $x^2$ coefficient?

Yes! After you write out the expression $\frac{m(m-1)}{2} + \frac{n(n-1)}{2} - mn$, don't just expand it blindly. Look for the structure of $(m-n)^2$. By grouping the terms, you can simplify the expression into $\frac{(m-n)^2 - (m+n)}{2}$. This is a classic JEE trick to save time.

Select Answer:

Visualized Solution

Problem Setup

Expression: $(1 + x)^{m} (1 - x)^{n}$
Coefficient of $x$ is $3$
Coefficient of $x^{2}$ is $- 6$
Goal: Find the value of $m$

Binomial Expansion Formula

Standard expansion: $(1 + x)^{k} = 1 + k x + \frac{k ( k - 1 )}{2} x^{2} + \dots$
For $(1 - x)^{n}$ , replace $x$ with $- x$ :
$(1 - x)^{n} = 1 - n x + \frac{n ( n - 1 )}{2} x^{2} - \dots$

Expanding the Product

$(1 + x)^{m} = 1 + m x + \frac{m ( m - 1 )}{2} x^{2} + \dots$
$(1 - x)^{n} = 1 - n x + \frac{n ( n - 1 )}{2} x^{2} - \dots$
Product: $(1 + m x + \frac{m ( m - 1 )}{2} x^{2}) \times (1 - n x + \frac{n ( n - 1 )}{2} x^{2})$

Extracting Coefficient of $x$

To get $x$ , multiply constant from first with $x$ term from second, and vice versa.
Term 1: $1 \times (- n x) = - n x$
Term 2: $(m x) \times 1 = m x$
Total $x$ term: $(m - n) x$

Forming Equation 1

Given: Coefficient of $x = 3$
Derived: Coefficient of $x = m - n$
Equation 1: $m - n = 3$

Extracting Coefficient of $x^{2}$

To get $x^{2}$ , we need three cross-multiplications:
1. Constant $\times$ $x^{2}$ term: $1 \times \frac{n ( n - 1 )}{2} x^{2}$
2. $x$ term $\times$ $x$ term: $(m x) \times (- n x) = - mn x^{2}$
3. $x^{2}$ term $\times$ Constant: \frac{m(m-1)}{2}x^2 \times 1

Simplifying Coefficient of $x^{2}$

Total coefficient: $\frac{m ( m - 1 )}{2} + \frac{n ( n - 1 )}{2} - mn$
Take common denominator $2$ : $\frac{m ^{2} - m + n ^{2} - n - 2 mn}{2}$
Group terms: $\frac{( m ^{2} + n ^{2} - 2 mn ) - ( m + n )}{2}$
Use identity $(m - n)^{2}$ : $\frac{( m - n ) ^{2} - ( m + n )}{2}$

Forming Equation 2

Given: Coefficient of $x^{2} = - 6$
$\frac{( m - n ) ^{2} - ( m + n )}{2} = - 6$
Substitute $m - n = 3$ : $\frac{3 ^{2} - ( m + n )}{2} = - 6$
$9 - (m + n) = - 12$
$m + n = 21$

Solving the System of Equations

Equation 1: $m - n = 3$
Equation 2: $m + n = 21$
Add both equations: $(m - n) + (m + n) = 3 + 21$
$2 m = 24$
$m = 12$

Final Answer & Key Takeaways

The value of $m$ is $12$ .
Key Takeaway 1: In product of series, only compute the cross-terms that yield the required power of $x$ .
Key Takeaway 2: Use algebraic identities like $(a - b)^{2}$ to simplify complex coefficient expressions.

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The Sigma Insight: Binomial Expansion for Positive Integral Index

Analyzing the Setup

We are examining the product of two binomial expansions:

(1 + x)^{m}

and

(1 - x)^{n}

. Our objective is to determine the value of

m

given the coefficients of

x

and

x^{2}

The standard binomial expansions are:

(1 + x)^{m} = 1 + m x + \frac{m ( m - 1 )}{2} x^{2} + \dots

(1 - x)^{n} = 1 - n x + \frac{n ( n - 1 )}{2} x^{2} + \dots

We only need to consider terms up to

x^{2}

to satisfy the problem constraints.

The Hunt for the Coefficient of $x$

To obtain the coefficient of

x

, we identify pairs from the two expansions that multiply to

x^{1}

. These pairs are

(1) \times (- n x)

and

(m x) \times (1)

Summing these gives the coefficient:

- n + m = 3

This provides our first fundamental equation:

m - n = 3

The $x^{2}$ Trap

To obtain the coefficient of

x^{2}

, we must account for three distinct cross-multiplications:

1. The constant

1

from the first bracket times the

x^{2}

term of the second:

\frac{n ( n - 1 )}{2} x^{2}

. 2. The

x

term from the first bracket times the

x

term from the second:

(m x) \times (- n x) = - mn x^{2}

. 3. The

x^{2}

term from the first bracket times the constant

1

from the second:

\frac{m ( m - 1 )}{2} x^{2}

Summing these, the total coefficient of

x^{2}

is:

\frac{m ( m - 1 )}{2} + \frac{n ( n - 1 )}{2} - mn = - 6

The Algebraic Symphony

We simplify the expression by taking a common denominator of

2

\frac{m ^{2} - m + n ^{2} - n - 2 mn}{2} = - 6

Recognizing that

m^{2} + n^{2} - 2 mn = (m - n)^{2}

, we rewrite the numerator:

\frac{( m - n ) ^{2} - ( m + n )}{2} = - 6

Substituting the known value

m - n = 3

into the equation:

\frac{3 ^{2} - ( m + n )}{2} = - 6

9 - (m + n) = - 12

m + n = 21

The Final Victory

We now solve the system of linear equations:

m - n = 3

m + n = 21

Adding these two equations yields

2 m = 24

. Therefore, the final value is:

$m = 12$

Common Traps & FAQs

Why can't we just expand the whole thing?

Time is your most precious resource in JEE Advanced. Expanding

(1 + x)^{m}

fully is an infinite series. We only care about the coefficients of

x

and

x^{2}

. By focusing only on the relevant terms, we save time and reduce the chance of calculation errors.

How do I know which terms to multiply to get

x^{2}

Think of it as a matching game. To get

x^{2}

, you need pairs of terms whose powers add up to 2. You have three combinations: (constant

\times

x^{2}

), (

x^{1} \times

x^{1}

), and (

x^{2} \times

constant). If you miss any of these, your coefficient will be wrong.

Is there a faster way to simplify the

x^{2}

coefficient?

Yes! After you write out the expression

\frac{m ( m - 1 )}{2} + \frac{n ( n - 1 )}{2} - mn

, don't just expand it blindly. Look for the structure of

(m - n)^{2}

. By grouping the terms, you can simplify the expression into

\frac{( m - n ) ^{2} - ( m + n )}{2}

. This is a classic JEE trick to save time.

Animated Solution for Mathematics - Binomial Theorem: If in the expansion of $(1 + x)^{m} (1 - x)^{n}$ , the coefficients of $x$ and $x^{2}$ are 3 and $- 6$ respectively, then $m$ is

Select Answer:

Visualized Solution

Problem Setup

Binomial Expansion Formula

Expanding the Product

Extracting Coefficient of $x$

Forming Equation 1

Extracting Coefficient of $x^{2}$

Simplifying Coefficient of $x^{2}$

Forming Equation 2

Solving the System of Equations

Final Answer & Key Takeaways

The Sigma Insight: Binomial Expansion for Positive Integral Index

Analyzing the Setup

The Hunt for the Coefficient of $x$

The $x^{2}$ Trap

The Algebraic Symphony

The Final Victory

Similar Questions

The coefficient of $x^{n}$ in expansion of $(1 + x) (1 - x)^{n}$ is

The coefficient of $x^{7}$ in the expansion of $(1 - x - x^{2} + x^{3})^{6}$ is

The coefficient of $x^{2}$ in the expansion of the product $(2 - x^{2}) \cdot ((1 + 2 x + 3 x^{2})^{6} + (1 - 4 x^{2})^{6})$ is :

The sum of the co-efficients of all even degree terms in $x$ in the expansion of $(x + x^{3} - 1)^{6} + (x - x^{3} - 1)^{6}, (x > 1)$ is equal to :

The coefficient of $x^{2012}$ in the expansion of $(1 - x)^{2008} (1 + x + x^{2})^{2007}$ is equal to

If in the expansion of $(1 + x)^{p} (1 - x)^{q}$ the coefficients of $x$ and $x^{2}$ are 1 and -2, respectively, then $p^{2} + q^{2}$ is equal to:

If the coefficients of $x$ and $x^{2}$ in $(1 + x)^{p} (1 - x)^{q}$ are 4 and $- 5$ respectively, then $2 p + 3 q$ is equal to

In the expansion of $(1 + x) (1 - x^{2}) (1 + \frac{3}{x} + \frac{3}{x ^{2}} + \frac{1}{x ^{3}})^{5}, x \neq = 0$ , the sum of the coefficient of $x^{3}$ and $x^{- 13}$ is equal to _______.

Let the sum of the coefficients of the first three terms in the expansion of $(x - \frac{3}{x ^{2}})^{n}, x \neq = 0, n \in N$ , be 376. Then the coefficient of $x^{4}$ is ______.

If $α$ and $β$ be the coefficient of $x^{4}$ and $x^{2}$ respectively in the expansion of $(x + x^{2} - 1)^{6} + (x - x^{2} - 1)^{6}$ , then :

Animated Solution for Mathematics - Binomial Theorem: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If in the expansion of (1+x)m(1−x)n, the coefficients of x and x2 are 3 and −6 respectively, then m is

Select Answer:

Visualized Solution

Problem Setup

Binomial Expansion Formula

Expanding the Product

Extracting Coefficient of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Forming Equation 1

Extracting Coefficient of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2

Simplifying Coefficient of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2

Forming Equation 2

Solving the System of Equations

Final Answer & Key Takeaways

The Sigma Insight: Binomial Expansion for Positive Integral Index

Analyzing the Setup

The Hunt for the Coefficient of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

The .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2 Trap

The Algebraic Symphony

The Final Victory

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The coefficient of xn in expansion of (1+x)(1−x)n is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The coefficient of x7 in the expansion of (1−x−x2+x3)6 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The coefficient of x2 in the expansion of the product (2−x2)⋅((1+2x+3x2)6+(1−4x2)6) is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum of the co-efficients of all even degree terms in x in the expansion of (x+x3−1​)6+(x−x3−1​)6,(x>1) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The coefficient of x2012 in the expansion of (1−x)2008(1+x+x2)2007 is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If in the expansion of (1+x)p(1−x)q the coefficients of x and x2 are 1 and -2, respectively, then p2+q2 is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the coefficients of x and x2 in (1+x)p(1−x)q are 4 and −5 respectively, then 2p+3q is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In the expansion of (1+x)(1−x2)(1+x3​+x23​+x31​)5,x=0, the sum of the coefficient of x3 and x−13 is equal to _______.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the sum of the coefficients of the first three terms in the expansion of (x−x23​)n,x=0,n∈N, be 376. Then the coefficient of x4 is ______.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If α and β be the coefficient of x4 and x2 respectively in the expansion of (x+x2−1​)6+(x−x2−1​)6, then :

Animated Solution for Mathematics - Binomial Theorem: If in the expansion of $(1 + x)^{m} (1 - x)^{n}$ , the coefficients of $x$ and $x^{2}$ are 3 and $- 6$ respectively, then $m$ is

Extracting Coefficient of $x$

Extracting Coefficient of $x^{2}$

Simplifying Coefficient of $x^{2}$

The Hunt for the Coefficient of $x$

The $x^{2}$ Trap

The coefficient of $x^{n}$ in expansion of $(1 + x) (1 - x)^{n}$ is

The coefficient of $x^{7}$ in the expansion of $(1 - x - x^{2} + x^{3})^{6}$ is

The coefficient of $x^{2}$ in the expansion of the product $(2 - x^{2}) \cdot ((1 + 2 x + 3 x^{2})^{6} + (1 - 4 x^{2})^{6})$ is :

The sum of the co-efficients of all even degree terms in $x$ in the expansion of $(x + x^{3} - 1)^{6} + (x - x^{3} - 1)^{6}, (x > 1)$ is equal to :

The coefficient of $x^{2012}$ in the expansion of $(1 - x)^{2008} (1 + x + x^{2})^{2007}$ is equal to

If in the expansion of $(1 + x)^{p} (1 - x)^{q}$ the coefficients of $x$ and $x^{2}$ are 1 and -2, respectively, then $p^{2} + q^{2}$ is equal to:

If the coefficients of $x$ and $x^{2}$ in $(1 + x)^{p} (1 - x)^{q}$ are 4 and $- 5$ respectively, then $2 p + 3 q$ is equal to

In the expansion of $(1 + x) (1 - x^{2}) (1 + \frac{3}{x} + \frac{3}{x ^{2}} + \frac{1}{x ^{3}})^{5}, x \neq = 0$ , the sum of the coefficient of $x^{3}$ and $x^{- 13}$ is equal to _______.

Let the sum of the coefficients of the first three terms in the expansion of $(x - \frac{3}{x ^{2}})^{n}, x \neq = 0, n \in N$ , be 376. Then the coefficient of $x^{4}$ is ______.

If $α$ and $β$ be the coefficient of $x^{4}$ and $x^{2}$ respectively in the expansion of $(x + x^{2} - 1)^{6} + (x - x^{2} - 1)^{6}$ , then :