JEE Main 2021 (25 July Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Binomial Theorem: The term independent of $x$ in the expansion of $(\frac{x + 1}{x ^{2/3} - x ^{1/3} + 1} - \frac{x - 1}{x - x ^{1/2}})^{10}$ , where $x \neq = 0, 1$ is equal to

Q: How do I know which algebraic identity to use?

Look at the exponents! The first fraction has $x^{2/3}$ and $x^{1/3}$, which are squares and first powers of $x^{1/3}$. This is a massive clue for the sum of cubes identity. The second fraction has $x$ and $x^{1/2}$, which are squares and first powers of $x^{1/2}$. This points directly to the difference of squares. Pattern recognition is key in JEE.

Q: What happens to the negative sign in the binomial $(x^{1/3} - x^{-1/2})^{10}$?

Great catch! When you use the general term formula $T_{r+1} = \binom{n}{r} a^{n-r} b^r$, you must treat the second term as $b = -x^{-1/2}$. This means your term will include $(-1)^r$. When you solve for $r$, if $r$ is even, the negative sign disappears; if $r$ is odd, it remains. In this problem, $r=4$, so $(-1)^4 = 1$, and the sign becomes positive.

Enter Numerical Value:

Visualized Solution

Analyzing the Complex Expression

Given expression: $(\frac{x + 1}{x ^{2/3} - x ^{1/3} + 1} - \frac{x - 1}{x - x ^{1/2}})^{10}$
Goal: Find the term independent of $x$ (where the exponent of $x$ is $0$ ).
Constraint: $x \neq = 0, 1$ .

Simplifying the First Fraction: Logic Bridge

Focus on the first term: $\frac{x + 1}{x ^{2/3} - x ^{1/3} + 1}$
Recall the algebraic identity: $a^{3} + b^{3} = (a + b) (a^{2} - ab + b^{2})$
Let $a = x^{1/3}$ and $b = 1$ .

Applying the Identity to the Numerator

Express $x$ as $(x^{1/3})^{3}$ and $1$ as $1^{3}$ .
Numerator becomes: $(x^{1/3})^{3} + 1^{3}$
Expand using the identity: $(x^{1/3} + 1) ((x^{1/3})^{2} - x^{1/3} \cdot 1 + 1^{2})$

Simplifying the First Fraction: Execution

Substitute the expanded numerator back into the fraction.
$\frac{( x ^{1/3} + 1 ) ( x ^{2/3} - x ^{1/3} + 1 )}{x ^{2/3} - x ^{1/3} + 1}$
Cancel the common factor $(x^{2/3} - x^{1/3} + 1)$ .
Result: $x^{1/3} + 1$

Simplifying the Second Fraction: Logic Bridge

Focus on the second term: $\frac{x - 1}{x - x ^{1/2}}$
Recall the difference of squares identity: $a^{2} - b^{2} = (a - b) (a + b)$
Let $a = x^{1/2}$ (or $x$ ) and $b = 1$ .

Factoring the Second Fraction

Numerator: $x - 1 = (x^{1/2})^{2} - 1^{2} = (x^{1/2} - 1) (x^{1/2} + 1)$
Denominator: $x - x^{1/2} = x^{1/2} (x^{1/2} - 1)$
Substitute back: $\frac{( x ^{1/2} - 1 ) ( x ^{1/2} + 1 )}{x ^{1/2} ( x ^{1/2} - 1 )}$

Simplifying the Second Fraction: Execution

Cancel the common factor $(x^{1/2} - 1)$ .
Result: $\frac{x ^{1/2} + 1}{x ^{1/2}}$
Split the fraction: $\frac{x ^{1/2}}{x ^{1/2}} + \frac{1}{x ^{1/2}}$
Final simplified second term: $1 + x^{- 1/2}$

Constructing the Simplified Binomial

Original Expression: $(First Term - Second Term)^{10}$
Substitute simplified forms: $((x^{1/3} + 1) - (1 + x^{- 1/2}))^{10}$
Distribute the negative sign: $(x^{1/3} + 1 - 1 - x^{- 1/2})^{10}$
Final simplified binomial: $(x^{1/3} - x^{- 1/2})^{10}$

Defining the General Term

General term formula for $(a + b)^{n}$ : $T_{r + 1} = (r n) a^{n - r} b^{r}$
For our binomial $(x^{1/3} - x^{- 1/2})^{10}$ :
$n = 10$ , $a = x^{1/3}$ , $b = - x^{- 1/2}$

Substituting into the General Term

Substitute the values into the formula:
$T_{r + 1} = (r 10) (x^{1/3})^{10 - r} (- x^{- 1/2})^{r}$
Separate the constant $- 1$ :
$T_{r + 1} = (r 10) (- 1)^{r} (x^{1/3})^{10 - r} (x^{- 1/2})^{r}$

Collecting the Powers of $x$

Multiply the exponents for $x$ :
First part: $(x^{1/3})^{10 - r} = x^{\frac{10 - r}{3}}$
Second part: $(x^{- 1/2})^{r} = x^{- \frac{r}{2}}$
Combine using $x^{m} \cdot x^{n} = x^{m + n}$ :
$T_{r + 1} = (r 10) (- 1)^{r} x^{\frac{10 - r}{3} - \frac{r}{2}}$

Condition for Term Independent of $x$

We need the term independent of $x$ .
This means the final exponent of $x$ must be exactly $0$ .
Set the combined exponent to zero:
$\frac{10 - r}{3} - \frac{r}{2} = 0$

Solving for $r$

Equation: $\frac{10 - r}{3} - \frac{r}{2} = 0$
Move $\frac{r}{2}$ to the right: $\frac{10 - r}{3} = \frac{r}{2}$
Cross-multiply: $2 (10 - r) = 3 r$
Expand: $20 - 2 r = 3 r$
Solve: $5 r = 20 \Rightarrow r = 4$

Calculating the Final Value

Substitute $r = 4$ back into the coefficient part of the general term.
Term $= (4 10) (- 1)^{4}$
$(- 1)^{4} = 1$
Calculate $(4 10) = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1}$
Final Answer: $210$

00:00 / 00:00

The Sigma Insight: General Term and Middle Term

The Illusion of Complexity

A JEE Masterclass

Welcome, fellow traveler of the JEE path. Today, we face a problem that looks like a mathematical fortress designed to intimidate you.

You see that expression,

(\frac{x + 1}{x ^{2/3} - x ^{1/3} + 1} - \frac{x - 1}{x - x ^{1/2}})^{10}

Your instinct might be to panic. But remember, in the world of JEE Advanced, complexity is often just a mask for elegance. Let's peel back the layers and see the beauty hidden underneath.

The Algebraic Surgeon

Simplifying the First Fraction

First, let's look at the first fraction:

\frac{x + 1}{x ^{2/3} - x ^{1/3} + 1}

. Does it ring a bell? It should! It is the classic sum of cubes identity in disguise.

Recall that

a^{3} + b^{3} = (a + b) (a^{2} - ab + b^{2})

. If we let

a = x^{1/3}

and

b = 1

, then

a^{3} = x

and

b^{3} = 1

The denominator is exactly that

a^{2} - ab + b^{2}

part. By rewriting

x + 1

(x^{1/3})^{3} + 1^{3}

, the numerator becomes

(x^{1/3} + 1) (x^{2/3} - x^{1/3} + 1)

When you divide this by the denominator, the complex part vanishes, leaving you with just

x^{1/3} + 1

. It is like magic, isn't it? The monster fraction has been tamed.

The Second Act

Difference of Squares

Now, look at the second fraction:

\frac{x - 1}{x - x ^{1/2}}

. This is a difference of squares.

We know that

x - 1

can be written as

(x^{1/2})^{2} - 1^{2}

, which factors into

(x^{1/2} - 1) (x^{1/2} + 1)

. The denominator

x - x^{1/2}

is just

x^{1/2} (x^{1/2} - 1)

The

(x^{1/2} - 1)

terms cancel out, leaving

\frac{x ^{1/2} + 1}{x ^{1/2}}

, which simplifies to

1 + x^{- 1/2}

. We have successfully broken down the fortress.

The Binomial Dance

Now, let's bring it all together. The expression becomes:

((x^{1/3} + 1) - (1 + x^{- 1/2}))^{10}

The ones cancel out perfectly, and we are left with

(x^{1/3} - x^{- 1/2})^{10}

. This is the beauty of the problem! We have transformed a nightmare into a simple binomial.

Now, we use the general term formula:

T_{r + 1} = (r 10) (x^{1/3})^{10 - r} (- x^{- 1/2})^{r}

We need the term independent of

x

, which means the total exponent of

x

must be zero. The exponent is:

\frac{10 - r}{3} - \frac{r}{2} = 0

Solving this linear equation gives

r = 4

. Finally, the coefficient is:

(4 10) = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210

You see? The complexity was just a test of your patience and your ability to see the underlying structure. Keep this clarity, and you will conquer any problem. The final answer is 210.

Common Traps & FAQs

Why can't we just expand the binomial

(\frac{x + 1}{x ^{2/3} - x ^{1/3} + 1} - \frac{x - 1}{x - x ^{1/2}})^{10}

directly?

Expanding this directly would be a nightmare! You would have a binomial with two complex fractions, each raised to powers, and then you would have to deal with the expansion of the 10th power. In JEE problems, if an expression looks overly complicated, it is almost always a sign that there is a hidden simplification or identity waiting to be used. Always simplify the base expression first.

How do I know which algebraic identity to use?

Look at the exponents! The first fraction has

x^{2/3}

and

x^{1/3}

, which are squares and first powers of

x^{1/3}

. This is a massive clue for the sum of cubes identity. The second fraction has

x

and

x^{1/2}

, which are squares and first powers of

x^{1/2}

. This points directly to the difference of squares. Pattern recognition is key in JEE.

What happens to the negative sign in the binomial

(x^{1/3} - x^{- 1/2})^{10}

Great catch! When you use the general term formula

T_{r + 1} = (r n) a^{n - r} b^{r}

, you must treat the second term as

b = - x^{- 1/2}

. This means your term will include

(- 1)^{r}

. When you solve for

r

, if

r

is even, the negative sign disappears; if

r

is odd, it remains. In this problem,

r = 4

, so

(- 1)^{4} = 1

, and the sign becomes positive.

Animated Solution for Mathematics - Binomial Theorem: The term independent of $x$ in the expansion of $(\frac{x + 1}{x ^{2/3} - x ^{1/3} + 1} - \frac{x - 1}{x - x ^{1/2}})^{10}$ , where $x \neq = 0, 1$ is equal to

Enter Numerical Value:

Visualized Solution

Analyzing the Complex Expression

Simplifying the First Fraction: Logic Bridge

Applying the Identity to the Numerator

Simplifying the First Fraction: Execution

Simplifying the Second Fraction: Logic Bridge

Factoring the Second Fraction

Simplifying the Second Fraction: Execution

Constructing the Simplified Binomial

Defining the General Term

Substituting into the General Term

Collecting the Powers of $x$

Condition for Term Independent of $x$

Solving for $r$

Calculating the Final Value

The Sigma Insight: General Term and Middle Term

The Illusion of Complexity

The Algebraic Surgeon

The Second Act

The Binomial Dance

Similar Questions

The term independent of $x$ in the expansion of $[\frac{x + 1}{x ^{2/3} - x ^{1/3} + 1} - \frac{x - 1}{x - x ^{1/2}}]^{10}, x \neq = 1$ , is equal to ___

The term independent of $x$ in the expansion of $(\frac{( x + 1 )}{( x ^{2/3} + 1 - x ^{1/3} )} - \frac{( x + 1 )}{( x - x ^{1/2} )})^{10}, x > 1$ is:

The term independent of $x$ in expansion of $(\frac{x + 1}{x ^{2/3} - x ^{1/3} + 1} - \frac{x - 1}{x - x ^{1/2}})^{10}$ is

The term independent of $x$ in the expression of $(1 - x^{2} + 3 x^{3}) (\frac{5}{2} x^{3} - \frac{1}{5 x ^{2}})^{11}$ , $x \neq = 0$ is

If the term independent of $x$ in the expansion of $(a x^{2} + \frac{1}{2 x ^{3}})^{10}$ is 105 , then $a^{2}$ is equal to :

The term independent of x in the expansion of $(\frac{1}{60} - \frac{x ^{8}}{81}) (2 x^{2} - \frac{3}{x ^{2}})^{6}$ is equal to :

The maximum value of the term independent of 't' in the expansion of $(t x^{1/5} + \frac{( 1 - x ) ^{1/10}}{t})^{10}$ where $x \in (0, 1)$ is:

If the greatest value of the term independent of $x$ in the expansion of $(x sin α + a \frac{c o s α}{x})^{10}$ is $\frac{10 !}{( 5 ! ) ^{2}}$ , then the value of $a$ is equal to:

If the term independent of $x$ in the expansion of $(\frac{3}{2} x^{2} - \frac{1}{3 x})^{9}$ is $k$ , then $18 k$ is equal to :

Let $α > 0, β > 0$ be such that $α^{3} + β^{2} = 4$ . If the maximum value of the term independent of $x$ in the binomial expansion of $(α x^{1/9} + β x^{- 1/6})^{10}$ is $10 k$ , then $k$ is equal to :

Animated Solution for Mathematics - Binomial Theorem: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The term independent of x in the expansion of (x2/3−x1/3+1x+1​−x−x1/2x−1​)10, where x=0,1 is equal to

Enter Numerical Value:

Visualized Solution

Analyzing the Complex Expression

Simplifying the First Fraction: Logic Bridge

Applying the Identity to the Numerator

Simplifying the First Fraction: Execution

Simplifying the Second Fraction: Logic Bridge

Factoring the Second Fraction

Simplifying the Second Fraction: Execution

Constructing the Simplified Binomial

Defining the General Term

Substituting into the General Term

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

Condition for Term Independent of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r

Calculating the Final Value

The Sigma Insight: General Term and Middle Term

The Illusion of Complexity

The Algebraic Surgeon

The Second Act

The Binomial Dance

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The term independent of x in the expansion of [x2/3−x1/3+1x+1​−x−x1/2x−1​]10,x=1, is equal to ___

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The term independent of x in the expansion of ((x2/3+1−x1/3)(x+1)​−(x−x1/2)(x+1)​)10,x>1 is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The term independent of x in expansion of (x2/3−x1/3+1x+1​−x−x1/2x−1​)10 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The term independent of x in the expression of (1−x2+3x3)(25​x3−5x21​)11, x=0 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the term independent of x in the expansion of (ax2​+2x31​)10 is 105 , then a2 is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The term independent of x in the expansion of (601​−81x8​)(2x2−x23​)6 is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The maximum value of the term independent of 't' in the expansion of (tx1/5+t(1−x)1/10​)10 where x∈(0,1) is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the greatest value of the term independent of x in the expansion of (xsinα+axcosα​)10 is (5!)210!​, then the value of a is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the term independent of x in the expansion of (23​x2−3x1​)9 is k, then 18k is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let α>0,β>0 be such that α3+β2=4. If the maximum value of the term independent of x in the binomial expansion of (αx1/9+βx−1/6)10 is 10k, then k is equal to :

Animated Solution for Mathematics - Binomial Theorem: The term independent of $x$ in the expansion of $(\frac{x + 1}{x ^{2/3} - x ^{1/3} + 1} - \frac{x - 1}{x - x ^{1/2}})^{10}$ , where $x \neq = 0, 1$ is equal to

Collecting the Powers of $x$

Condition for Term Independent of $x$

Solving for $r$

The term independent of $x$ in the expansion of $[\frac{x + 1}{x ^{2/3} - x ^{1/3} + 1} - \frac{x - 1}{x - x ^{1/2}}]^{10}, x \neq = 1$ , is equal to ___

The term independent of $x$ in the expansion of $(\frac{( x + 1 )}{( x ^{2/3} + 1 - x ^{1/3} )} - \frac{( x + 1 )}{( x - x ^{1/2} )})^{10}, x > 1$ is:

The term independent of $x$ in expansion of $(\frac{x + 1}{x ^{2/3} - x ^{1/3} + 1} - \frac{x - 1}{x - x ^{1/2}})^{10}$ is

The term independent of $x$ in the expression of $(1 - x^{2} + 3 x^{3}) (\frac{5}{2} x^{3} - \frac{1}{5 x ^{2}})^{11}$ , $x \neq = 0$ is

If the term independent of $x$ in the expansion of $(a x^{2} + \frac{1}{2 x ^{3}})^{10}$ is 105 , then $a^{2}$ is equal to :

The term independent of x in the expansion of $(\frac{1}{60} - \frac{x ^{8}}{81}) (2 x^{2} - \frac{3}{x ^{2}})^{6}$ is equal to :

The maximum value of the term independent of 't' in the expansion of $(t x^{1/5} + \frac{( 1 - x ) ^{1/10}}{t})^{10}$ where $x \in (0, 1)$ is:

If the greatest value of the term independent of $x$ in the expansion of $(x sin α + a \frac{c o s α}{x})^{10}$ is $\frac{10 !}{( 5 ! ) ^{2}}$ , then the value of $a$ is equal to:

If the term independent of $x$ in the expansion of $(\frac{3}{2} x^{2} - \frac{1}{3 x})^{9}$ is $k$ , then $18 k$ is equal to :

Let $α > 0, β > 0$ be such that $α^{3} + β^{2} = 4$ . If the maximum value of the term independent of $x$ in the binomial expansion of $(α x^{1/9} + β x^{- 1/6})^{10}$ is $10 k$ , then $k$ is equal to :