Question 15

Question: The term independent of $x$ in the expansion of $\begin{array} { r } { \bigg ( \frac { ( x + 1 ) } { \left( x ^ { 2 / 3 } + 1 - x ^ { 1 / 3 } \right) } - \frac { ( x - 1 ) } { \left( x - x ^ { 1 / 2 } \right) } \bigg ) ^ { 1 0 } , x > 1 , \mathrm { i s } : } \end{array}$

Options:

A. 240

B. 120

C. 150

D. 210

Correct Answer: D

Year: 2025 (Online) 2nd April Morning Shift

Solution: The internal terms are simplified using algebraic identities to reach $(x^{1/3} - x^{-1/2})^{10}$. By setting the exponent of $x$ in the general term formula to zero, $r$ is determined to be 4. The constant term is then calculated as $T_5 = \binom{10}{4} = 210$.

Step Solution:

1.  Simplify the first term: Use $a^3+b^3=(a+b)(a^2-ab+b^2)$ to rewrite $(x+1)$ as $((x^{1/3})^3+1^3)$, which simplifies to $(x^{1/3}+1)$.

2.  Simplify the second term: Factor the numerator and denominator: $\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}-1)} = \frac{\sqrt{x}+1}{\sqrt{x}} = 1 + x^{-1/2}$.

3.  Combine terms: The expression inside the brackets becomes $(x^{1/3}+1) - (1+x^{-1/2}) = x^{1/3} - x^{-1/2}$.

4.  Find $r$: For the expansion $(x^{1/3} - x^{-1/2})^{10}$, set the power of $x$ to zero: $\frac{10-r}{3} - \frac{r}{2} = 0$, giving $r = 4$.

5.  Calculate final value: $T_{4+1} = \binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$.

Difficulty level: Medium

The Concept Name: Simplification using Algebraic Identities and Binomial General Term.

Short cut solution: Use the formula $r = \frac{n\alpha - m}{\alpha + \beta}$ for an expansion $(x^{\alpha} \pm x^{-\beta})^n$ to find the coefficient of $x^m$. Here, $r = \frac{10(1/3) - 0}{1/3 + 1/2} = \frac{10/3}{5/6} = 4$, then calculate $\binom{10}{4} = 210$.

 Question 43

Question: The constant term in the expansion of $\begin{array} { r } { \left( 2 \mathbf { x } + \mathbf { \frac { 1 } { x ^ { 7 } } } + 3 \mathbf { x } ^ { 2 } \right) ^ { 5 } } \end{array}$ is

Options: (Options not provided in source)

Correct Answer: 1080

Year: 2023 (Online) 25-Jan Shift 1

Solution: The problem uses the Multinomial Theorem general term $\frac{n!}{n_1! n_2! n_3!} a^{n_1} b^{n_2} c^{n_3}$. By solving $n_1 + n_2 + n_3 = 5$ and the power of $x$ equation $n_1 - 7n_2 + 2n_3 = 0$, the indices are found to be $n_1=1, n_2=1, n_3=3$.

Step Solution:

1.  General Term: Write the multinomial general term as $\frac{5!}{n_1! n_2! n_3!} (2x)^{n_1} (x^{-7})^{n_2} (3x^2)^{n_3}$.

2.  Constraint 1: The sum of the exponents must be $n_1 + n_2 + n_3 = 5$.

3.  Constraint 2: For a constant term, the power of $x$ must be zero: $n_1 - 7n_2 + 2n_3 = 0$.

4.  Solve System: Substituting $n_1 = 5 - n_2 - n_3$ into the power equation yields $5 - 8n_2 + n_3 = 0$. For non-negative integers, $n_2 = 1$ forces $n_3 = 3$ and $n_1 = 1$.

5.  Compute: Coefficient $= \frac{5!}{1! 1! 3!} \cdot 2^1 \cdot 1^1 \cdot 3^3 = 20 \times 2 \times 27 = 1080$.

Difficulty level: Hard

The Concept Name: Multinomial Theorem.

Short cut solution: Inspect the indices by checking small integer values for $n_2$ that satisfy $n_1 + 2n_3 = 7n_2$ within the sum total of 5.

 Question 56

Question: If the constant term in the binomial expansion of $\left( \frac{x^{5/2}}{2} - \frac{4}{x^\ell} \right)^9$ is $-84$ and the coefficient of $x^{-3\ell}$ is $2^\alpha \beta$, where $\ell$ is an odd number, then $|\alpha \ell - \beta|$ is equal to

Options: (Options not provided in source)

Correct Answer: 98

Year: 2023 (Online) 31-Jan Shift 2

Solution: The general term $T_{r+1}$ is used to find $r$ by equating the constant term coefficient to $-84$. This reveals $r=3$ and $\ell=5$. The coefficient of $x^{-15}$ is then calculated to find values for $\alpha$ and $\beta$, leading to the final result.

Step Solution:

1.  General Term: $T_{r+1} = \binom{9}{r} (\frac{x^{5/2}}{2})^{9-r} (-\frac{4}{x^\ell})^r = (-1)^r \binom{9}{r} \frac{2^{2r}}{2^{9-r}} x^{\frac{45-5r}{2} - \ell r}$.

2.  Constant Term: Set $(-1)^r \binom{9}{r} 2^{3r-9} = -84$. This is satisfied only when $r=3$ (since $\binom{9}{3}=84$ and $2^{9-9}=1$).

3.  Solve for $\ell$: The power of $x$ at $r=3$ must be zero: $\frac{5(6)}{2} - 3\ell = 0 \Rightarrow 15 - 3\ell = 0 \Rightarrow \ell = 5$.

4.  Find new $r$: For $x^{-3\ell} = x^{-15}$, set $\frac{5(9-r)}{2} - 5r = -15$, which gives $r=5$.

5.  Final Calculation: Coefficient at $r=5$ is $-\binom{9}{5} \frac{4^5}{2^4} = -126 \times 2^6 = -63 \times 2^7$. Thus $\alpha=7, \beta=-63$. $|\alpha \ell - \beta| = |(7)(5) - (-63)| = 98$.

Difficulty level: Hard

The Concept Name: General Term of Binomial Expansion and Power Comparison.

Short cut solution: Match the binomial coefficient $\binom{9}{r}$ to the given value 84 to immediately identify $r=3$, then solve for $\ell$.

 Question 60

Question: If the term without $x$ in the expansion of $\left(x^{\frac{2}{3}} + \frac{\alpha}{x^3}\right)^{22}$ is 7315, then $|\alpha|$ is equal to.

Options: (Options not provided in source)

Correct Answer: 1.

Year: 2023 (Online) 1-Feb Shift 2.

Solution: The general term $T_{r+1}$ is used to find the power of $x$ as $\frac{44-11r}{3}$. Setting this to zero gives $r=4$. The constant term coefficient $\binom{22}{4} \alpha^4$ is equated to 7315 to find $\alpha$.

Step Solution:

1.  General Term: $T_{r+1} = \binom{22}{r} (x^{2/3})^{22-r} (\alpha x^{-3})^r = \binom{22}{r} \alpha^r x^{\frac{44-2r}{3} - 3r}$.

2.  Find $r$: For the term independent of $x$, set the power $\frac{44-2r-9r}{3} = 0$, which simplifies to $11r = 44$, so $r = 4$.

3.  Coefficient Equivalence: The term is $\binom{22}{4} \alpha^4 = 7315$.

4.  Calculate Combination: $\binom{22}{4} = \frac{22 \times 21 \times 20 \times 19}{4 \times 3 \times 2 \times 1} = 7315$.

5.  Solve for $\alpha$: $7315 \alpha^4 = 7315 \Rightarrow \alpha^4 = 1 \Rightarrow \mathbf{|\alpha| = 1}$.

Difficulty level: Medium

The Concept Name: Term Independent of $x$ (Binomial General Term).

Short cut solution: Use $r = \frac{n\alpha}{\alpha + \beta} = \frac{22(2/3)}{2/3 + 3} = \frac{44/3}{11/3} = 4$. Then $|\alpha| = \left(\frac{7315}{\binom{22}{4}}\right)^{1/4} = 1$.

 Question 66

Question: Let $[t]$ denote the greatest integer $\le t$. If the constant term in the expansion of $\left(3x^2 - \frac{1}{2x^5}\right)^7$ is $a$, then $[a]$ is equal to.

Options: (Options not provided in source)

Correct Answer: 1275.

Year: 2023 (Online) 8-Apr Shift 1.

Solution: The general term $T_{r+1}$ is expressed, and the exponent of $x$ is set to zero, giving $r=2$. The value of $a$ is calculated as 1275.75, and its greatest integer value is 1275.

Step Solution:

1.  General Term: $T_{r+1} = \binom{7}{r} (3x^2)^{7-r} (-\frac{1}{2x^5})^r = \binom{7}{r} 3^{7-r} (-\frac{1}{2})^r x^{14-2r-5r}$.

2.  Find $r$: Set $14 - 7r = 0$, which gives $r = 2$.

3.  Calculate $a$: $a = T_3 = \binom{7}{2} \cdot 3^5 \cdot (-\frac{1}{2})^2 = 21 \cdot 243 \cdot \frac{1}{4}$.

4.  Decimal Value: $a = \frac{5103}{4} = 1275.75$.

5.  Greatest Integer: $[a] = [1275.75] = \mathbf{1275}$.

Difficulty level: Easy

The Concept Name: Binomial General Term and Greatest Integer Function.

Short cut solution: Quickly find $r=2$ using the power balance $2(7-r) = 5r$. Then calculate $21 \times 243 / 4 \approx 1275$.

 Question 82

Question: Let $u$ be the constant term in the binomial expansion of $(\sqrt{x} - \frac{6}{x^{3/2}})^n, n \le 15$. If the sum of the coefficients of the remaining terms in the expansion is 649 and the coefficient of $x^{-n}$ is $\lambda u$, then $\lambda$ is equal to.

Options: (Options not provided in source)

Correct Answer: 36.

Year: 2023 (Online) 13-Apr Shift 1.

Solution: Using the property that the sum of all coefficients is $(1-6)^n = (-5)^n$, and the relation $(-5)^n - u = 649$, $n$ is found to be 4. From this, $u$ and the coefficient of $x^{-n}$ are calculated to find $\lambda$.

Step Solution:

1.  Constraint Equation: Sum of all coefficients $= (-5)^n = u + 649$.

2.  Determine $n$: For a constant term, the power $\frac{n-k}{2} - \frac{3k}{2} = 0 \Rightarrow n = 4k$. For $n \le 15$, $n$ can be 4, 8, 12. Testing $n=4$: $(-5)^4 = 625$. $u = \binom{4}{1}(-6)^1 = -24$. Since $625 - (-24) = 649$, $n=4$ and $u=-24$.

3.  Coefficient of $x^{-n}$: For $x^{-4}$, set $\frac{4-k}{2} - \frac{3k}{2} = -4$, giving $k=3$. Coeff is $\binom{4}{3}(-6)^3 = 4(-216) = -864$.

4.  Find $\lambda$: $-864 = \lambda u \Rightarrow -864 = \lambda(-24)$.

5.  Final Calculation: $\lambda = \frac{864}{24} = \mathbf{36}$.

Difficulty level: Hard

The Concept Name: Sum of Coefficients and Binomial General Term.

Short cut solution: Recognize that $649$ is close to $5^4=625$. Test $n=4$ immediately to confirm the constant term $u$ matches the difference.

 Question 97

Question: The number of positive integers $k$ such that the constant term in the binomial expansion of $\left( 2x^3 + \frac{3}{x^k} \right)^{12}, x \neq 0$ is $2^8 \cdot I$, where $I$ is an odd integer, is:

Options: (Options not provided in source)

Correct Answer: 2

Year: 2022 (Online) 28th June Shift 1

Solution: The general term is used to find the condition for the constant term: $3r - k(12-r) = 0$, leading to $k = \frac{3r}{12-r}$. Testing possible $r$ values where $k$ is a positive integer, the binomial coefficients are checked for the exponent of 2. Only $k=3$ and $k=6$ satisfy the $2^8 \cdot I$ condition.

Step Solution:

1.  General Term: $T_{r+1} = \binom{12}{r} (2x^3)^r (3x^{-k})^{12-r} = \binom{12}{r} 2^r 3^{12-r} x^{3r - k(12-r)}$.

2.  Constraint: For the constant term, the power of $x$ is zero: $3r - k(12-r) = 0 \Rightarrow k = \frac{3r}{12-r}$.

3.  Identify $r$: Since $k$ is a positive integer, $(12-r)$ must be a factor of $3r$. Possible $r$ values are $3, 6, 8, 9, 10$.

4.  Test Coefficients for $2^8$:

    *   If $r=6, k=3$: Coefficient $= \binom{12}{6} 2^6 \cdot 3^6 = 924 \cdot 2^6 \cdot 3^6 = (2^2 \cdot 231) \cdot 2^6 \cdot 3^6 = 2^8 \cdot I$. (Works)

    *   If $r=8, k=6$: Coefficient $= \binom{12}{8} 2^8 \cdot 3^4 = 495 \cdot 2^8 \cdot 3^4 = 2^8 \cdot I$. (Works)

5.  Count: Other $r$ values ($3, 9, 10$) produce powers of 2 equal to $2^5$ or $2^{11}$, which do not match $2^8$. Thus, there are 2 such integers.

Difficulty level: Hard

The Concept Name: Binomial General Term and Prime Factorization of Coefficients.

Short cut solution: Quickly find $r$ such that $\binom{12}{r}$ contains $2^{8-r}$. $\binom{12}{8}$ is odd, so $r=8$ works immediately. $\binom{12}{6}$ is $4 \times 231$, so $r=6$ works.

 Question 98

Question: The term independent of $x$ in the expansion of $(1 - x^2 + 3x^3) \left( \frac{5}{2}x^3 - \frac{1}{5x^2} \right)^{11}, x \neq 0$ is:

Options:

A. 7/40

B. 33/200

C. 39/200

D. 11/50

Correct Answer: B

Year: 2022 (Online) 28th June Shift 2

Solution: The general term of the binomial part is expressed as $T_{r+1} = \binom{11}{r} (\frac{5}{2})^{11-r} (-\frac{1}{5})^r x^{33-5r}$. We search for $r$ such that $33-5r$ cancels the powers of $x$ in the trinomial ($x^0, x^{-2}, x^{-3}$). Only $r=7$ yields an integer for the $x^{-2}$ case.

Step Solution:

1.  Binomial General Term: $T_{r+1} = \binom{11}{r} (\frac{5}{2})^{11-r} (-\frac{1}{5})^r x^{33-5r}$.

2.  Trinomial cases: Constant term occurs when $33-5r = 0, -2,$ or $-3$.

3.  Solve for $r$: Only $33-5r = -2$ gives a valid integer $r = 7$.

4.  Calculate term: The constant term is $(-x^2) \cdot T_{7+1} = -1 \cdot \binom{11}{7} (\frac{5}{2})^4 (-\frac{1}{5})^7$.

5.  Simplify: $\frac{11 \cdot 10 \cdot 9 \cdot 8}{4 \cdot 3 \cdot 2 \cdot 1} \cdot \frac{5^4}{2^4 \cdot 5^7} = \frac{330}{16 \cdot 125} = \frac{330}{2000} = \frac{33}{200}$.

Difficulty level: Medium

The Concept Name: Binomial General Term and Product of Polynomials.

Short cut solution: Focus on $33-5r$. Since $x$ must vanish, check which $r \in \{0, \dots, 11\}$ makes $33-5r$ equal to $0, -2,$ or $-3$. Only $r=7$ works. Calculate coefficient for $r=7$.

 Question 99

Question: If the constant term in the expansion of $\left( 3x^3 - 2x^2 + \frac{5}{x^5} \right)^{10}$ is $2^k \cdot I$, where $I$ is an odd integer, then the value of $k$ is equal to:

Options:

A. 6

B. 7

C. 8

D. 9

Correct Answer: D

Year: 2022 (Online) 29th June Shift 1

Solution: Using the Multinomial Theorem, the general term is $\frac{10!}{n_1! n_2! n_3!} (3)^{n_1} (-2)^{n_2} (5)^{n_3} x^{3n_1 + 2n_2 - 5n_3}$. Solving the system $n_1+n_2+n_3=10$ and $3n_1+2n_2-5n_3=0$ gives indices $n_1=1, n_2=6, n_3=3$. The power of 2 in the final coefficient is then calculated.

Step Solution:

1.  Constraint 1: For constant term, power of $x$: $3n_1 + 2n_2 - 5n_3 = 0$.

2.  Constraint 2: Sum of indices: $n_1 + n_2 + n_3 = 10 \Rightarrow n_1 = 10 - n_2 - n_3$.

3.  Solve System: $3(10 - n_2 - n_3) + 2n_2 - 5n_3 = 0 \Rightarrow n_2 + 8n_3 = 30$. For non-negative integers, $n_3=3, n_2=6$ is the only solution (leaving $n_1=1$).

4.  Coefficient: $C = \frac{10!}{1! 6! 3!} \cdot 3^1 \cdot (-2)^6 \cdot 5^3 = \frac{10 \cdot 9 \cdot 8 \cdot 7}{6} \cdot 3 \cdot 2^6 \cdot 125$.

5.  Find power of 2: $C = 840 \cdot 3 \cdot 2^6 \cdot 125 = (2^3 \cdot 105) \cdot 3 \cdot 2^6 \cdot 125 = 2^9 \cdot I$. Thus $k=9$.

Difficulty level: Hard

The Concept Name: Multinomial Theorem.

Short cut solution: In $n_2 + 8n_3 = 30$, $n_3$ must be 3 because $n_3=0, 1, 2$ don't allow $n_i \le 10$ with the sum constraint. Calculate the power of 2 from $\frac{10!}{6!3!} \cdot 2^6$ directly.

 Question 102

Question: If the maximum value of the term independent of $t$ in the expansion of $\left( t^2 x^{1/5} + \frac{(1-x)^{1/10}}{t} \right)^{15}, x \ge 0$, is $K$, then $8K$ is equal to.

Options: (Options not provided in source)

Correct Answer: 6006

Year: 2022 (Online) 25th July Shift 1

Solution: The general term $T_{r+1}$ is expressed to find the condition where the power of $t$ is zero, which gives $r=10$. The resulting term is a function of $x$ in the form $x(1-x)$, which is maximized using calculus at $x=1/2$. The maximum value $K$ is then used to find $8K$.

Step Solution:

1.  General Term: $T_{r+1} = \binom{15}{r} (t^2 x^{1/5})^{15-r} \left( \frac{(1-x)^{1/10}}{t} \right)^r = \binom{15}{r} t^{30-2r-r} x^{\frac{15-r}{5}} (1-x)^{\frac{r}{10}}$.

2.  Independent of $t$: Set the power of $t$ to zero: $30 - 3r = 0 \Rightarrow \mathbf{r = 10}$.

3.  Define $K(x)$: The term is $T_{11} = \binom{15}{10} x^{\frac{15-10}{5}} (1-x)^{\frac{10}{10}} = \binom{15}{10} x(1-x)$.

4.  Maximize: Let $f(x) = x - x^2$. $f'(x) = 1 - 2x = 0 \Rightarrow x = 1/2$. Since $f''(x) = -2 < 0$, it is a maximum.

5.  Calculate $8K$: $K = \binom{15}{10} (\frac{1}{2})(1-\frac{1}{2}) = \frac{1}{4} \binom{15}{10}$. Thus, $8K = 2 \cdot \binom{15}{10} = 2 \times 3003 = \mathbf{6006}$.

Difficulty level: Medium

The Concept Name: Application of Derivatives (Maxima/Minima) and Binomial General Term.

Short cut solution: For any term in the form $Cx^a(1-x)^b$, the maximum value always occurs at $x = \frac{a}{a+b}$. Here $a=1$ and $b=1$, so $x=1/2$.

 Question 118

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

Options:

A. $\frac{10!}{\sqrt{3}(5!)^2}$

B. $\frac{2 \cdot 10!}{3\sqrt{3}(5!)^2}$

C. $\frac{2 \cdot 10!}{3(5!)^2}$

D. $\frac{10!}{3(5!)^2}$

Correct Answer: B

Year: 2021 (Online) 26th Feb Shift 1

Solution: The general term is used to find $r=5$ for the term independent of $t$. The term becomes a function $f(x) = \binom{10}{5} x(1-x)^{1/2}$. Differentiating this function shows it reaches a maximum at $x=2/3$, which is then substituted back to find the final value.

Step Solution:

1.  General Term: $T_{r+1} = \binom{10}{r} (tx^{1/5})^{10-r} (\frac{(1-x)^{1/10}}{t})^r = \binom{10}{r} t^{10-2r} x^{\frac{10-r}{5}} (1-x)^{\frac{r}{10}}$.

2.  Independent of $t$: $10 - 2r = 0 \Rightarrow \mathbf{r = 5}$. The term is $T_6 = \binom{10}{5} x(1-x)^{1/2}$.

3.  Maximize $f(x)$: Let $f(x) = x(1-x)^{1/2}$. $f'(x) = (1-x)^{1/2} + x \cdot \frac{1}{2}(1-x)^{-1/2}(-1) = 0$.

4.  Solve for $x$: $\sqrt{1-x} = \frac{x}{2\sqrt{1-x}} \Rightarrow 2(1-x) = x \Rightarrow \mathbf{x = 2/3}$.

5.  Final Value: Max value $= \binom{10}{5} (\frac{2}{3})(1-\frac{2}{3})^{1/2} = \binom{10}{5} \frac{2}{3\sqrt{3}} = \frac{10! \cdot 2}{(5!)^2 \cdot 3\sqrt{3}}$.

Difficulty level: Hard

The Concept Name: Application of Derivatives (Maxima/Minima) and Binomial General Term.

Short cut solution: Use the power rule for $x^a(1-x)^b$ maximization: $x = \frac{a}{a+b}$. Here $a=1, b=1/2$, so $x = \frac{1}{1.5} = 2/3$.

 Question 119

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

Options: (Options not provided in source)

Correct Answer: 210

Year: 2021 (Online) 18th March Shift 2

Solution: The expression inside the brackets is simplified using the sum of cubes identity and basic factoring to reach the binomial $(x^{1/3} - x^{-1/2})^{10}$. Setting the exponent of $x$ in the general term to zero yields $r=4$, and the final coefficient is $\binom{10}{4}$.

Step Solution:

1.  Simplify Term 1: Use $a^3+b^3 = (a+b)(a^2-ab+b^2)$ for $(x+1) = (x^{1/3})^3+1^3$, resulting in $(x^{1/3}+1)$.

2.  Simplify Term 2: Factor $\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}-1)} = \frac{\sqrt{x}+1}{\sqrt{x}} = 1 + x^{-1/2}$.

3.  Combined Expansion: $[(x^{1/3}+1) - (1+x^{-1/2})]^{10} = (x^{1/3} - x^{-1/2})^{10}$.

4.  Find $r$: General term $T_{r+1} = \binom{10}{r} (x^{1/3})^{10-r} (-x^{-1/2})^r$. Set power $\frac{10-r}{3} - \frac{r}{2} = 0$, which gives $\mathbf{r = 4}$.

5.  Final Calculation: $T_5 = \binom{10}{4} (-1)^4 = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = \mathbf{210}$.

Difficulty level: Medium

The Concept Name: Simplification using Algebraic Identities and Binomial General Term.

Short cut solution: After simplifying to $(x^{1/3} - x^{-1/2})^{10}$, use the formula $r = \frac{n\alpha}{\alpha+\beta} = \frac{10(1/3)}{1/3+1/2} = 4$. Then $\binom{10}{4} = 210$.

 Question 128

Question: The term independent of $x$ in the expansion of $\left( \frac{x+1}{x^{2/3} - x^{1/3} + 1} - \frac{x-1}{x-x^{1/2}} \right)^{10}$, where $x \neq 0, 1$ is equal to:

Options: (Options not provided in source for this specific question number).

Correct Answer: 210

Year: 2021, 2 July Shift I

Solution: The expression is simplified using algebraic identities to $(x^{1/3} - x^{-1/2})^{10}$. The general term formula is then applied, and the power of $x$ is set to zero to find $r = 4$. The coefficient is calculated as ${}^{10}C_4 = 210$.

Step Solution:

1.  Simplify Term 1: Use the identity $a^3+b^3 = (a+b)(a^2-ab+b^2)$ to rewrite $(x+1)$ as $((x^{1/3})^3+1^3)$, which simplifies the first fraction to $(x^{1/3}+1)$.

2.  Simplify Term 2: Factor the numerator and denominator: $\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}-1)} = \frac{\sqrt{x}+1}{\sqrt{x}} = \mathbf{1 + x^{-1/2}}$.

3.  Combine inside brackets: $(x^{1/3}+1) - (1+x^{-1/2}) = \mathbf{x^{1/3} - x^{-1/2}}$.

4.  Power Equation: Set the exponent of $x$ in the general term to zero: $\frac{10-r}{3} - \frac{r}{2} = 0$, which gives $20 - 2r - 3r = 0$, so $r = 4$.

5.  Final Calculation: $T_5 = {}^{10}C_4 = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = \mathbf{210}$.

Difficulty level: Medium

The Concept Name: Binomial General Term and Algebraic Simplification.

Short cut solution: Use the shortcut $r = \frac{n\alpha - m}{\alpha + \beta}$ for $(x^{\alpha} \pm x^{-\beta})^n$. Here $r = \frac{10(1/3) - 0}{1/3 + 1/2} = \frac{10/3}{5/6} = 4$. Then, ${}^{10}C_4 = 210$.

 Question 135

Question: If the constant term, in Binomial expansion of $\left( 2x^r + \frac{1}{x^2} \right)^{10}$ is 180, then $r$ is equal to:

Options: (Options not provided in source).

Correct Answer: 8

Year: 2021, 22 July Shift-II

Solution: The general term $T_{k+1}$ is expressed, and the power of $x$ is set to zero to create a relation between $r$ and $k$. By equating the coefficient part ${}^{10}C_k \cdot 2^{10-k}$ to 180, $k$ is found to be 8. Substituting $k=8$ into the power relation yields $r=8$.

Step Solution:

1.  General Term: Write $T_{k+1} = {}^{10}C_k (2x^r)^{10-k} (x^{-2})^k = {}^{10}C_k \cdot 2^{10-k} \cdot x^{10r - rk - 2k}$.

2.  Constraint 1 (Power): For the constant term, $10r - rk - 2k = 0$, which simplifies to $k = \frac{10r}{r+2}$.

3.  Constraint 2 (Coefficient): Set the coefficient ${}^{10}C_k \cdot 2^{10-k} = 180$.

4.  Solve for $k$: Since 180 is a multiple of 45 (${}^{10}C_2$ or ${}^{10}C_8$), we test $k=8$, where $45 \times 2^2 = 180$.

5.  Solve for $r$: Substitute $k=8$ into $8 = \frac{10r}{r+2} \Rightarrow 8r + 16 = 10r \Rightarrow 2r = 16 \Rightarrow \mathbf{r = 8}$.

Difficulty level: Hard

The Concept Name: Term Independent of $x$ (Constant Term).

Short cut solution: Focus on the coefficient $180 = 45 \times 4$. Recognize $45 = {}^{10}C_8$, implying $k=8$. Then use the power balance $r(10-k) = 2k \Rightarrow 2r = 16 \Rightarrow r=8$.

 Question 145

Question: If $\left( \frac{3^6}{4^4} \right) k$ is the term, independent of $x$, in the binomial expansion of $\left( \frac{x}{4} - \frac{12}{x^2} \right)^{12}$, then $k$ is equal to:

Options: (Options not provided in source).

Correct Answer: 55

Year: 2021, 31 Aug. Shift-I

Solution: The general term $T_{r+1}$ is used to find the term independent of $x$ by setting the power of $x$ to zero, resulting in $r=4$. The resulting constant term is then simplified and compared to the given expression to find $k = 55$.

Step Solution:

1.  General Term: $T_{r+1} = (-1)^r {}^{12}C_r \left( \frac{x}{4} \right)^{12-r} \left( \frac{12}{x^2} \right)^r = (-1)^r {}^{12}C_r \frac{12^r}{4^{12-r}} x^{12-3r}$.

2.  Find $r$: Set the power $12 - 3r = 0$, which gives $r = 4$.

3.  Evaluate Term: $T_5 = (-1)^4 {}^{12}C_4 \frac{12^4}{4^8} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} \cdot \frac{12^4}{4^8}$.

4.  Simplify Math: $495 \cdot \frac{3^4 \cdot 4^4}{4^8} = 495 \cdot \frac{3^4}{4^4} = (55 \times 9) \cdot \frac{3^4}{4^4}$.

5.  Solve for $k$: $55 \cdot \frac{3^2 \cdot 3^4}{4^4} = 55 \cdot \frac{3^6}{4^4}$. Comparing with the given form, $k = 55$.

Difficulty level: Medium

The Concept Name: Term Independent of $x$.

Short cut solution: Use $r = \frac{n\alpha}{\alpha + \beta} = \frac{12(1)}{1+2} = 4$. Constant term is ${}^{12}C_4 \cdot (1/4)^8 \cdot (-12)^4$. Simplify to reach $55 \cdot \frac{3^6}{4^4}$.

 Question 156

Question: If the constant term in the binomial expansion of $\left( \sqrt { \mathbf { x } } - \frac { \mathbf { k } } { \mathbf { x } ^ { 2 } } \right) ^ { \mathbf { 1 } 0 }$ is 405, then $|k|$ equals:

Options: 

A. 9 

B. 1 

C. 3 

D. 2

Correct Answer: C

Year: Sep. 06, 2020 (II)

Solution: The general term is $T_{r+1} = {}^{10}C_r (\sqrt{x})^{10-r} \cdot \left( -\frac{k}{x^2} \right)^r = {}^{10}C_r (-k)^r x^{\frac{10-r}{2} - 2r} = {}^{10}C_r (-k)^r x^{\frac{10-5r}{2}}$. For the constant term, setting the exponent to zero gives $\frac{10-5r}{2} = 0 \Rightarrow r = 2$. The coefficient is ${}^{10}C_2 (-k)^2 = 405$, which simplifies to $45k^2 = 405$, resulting in $k^2 = 9$ and $|k| = 3$.

Step Solution:

1. Write the general term $T_{r+1} = {}^{10}C_r (x^{1/2})^{10-r} (-k \cdot x^{-2})^r$.

2. Simplify the powers of $x$: $x^{\frac{10-r}{2} - 2r} = x^{\frac{10-5r}{2}}$.

3. For the term independent of $x$, set the power $\frac{10-5r}{2} = 0$, giving $r = 2$.

4. Set the resulting coefficient to 405: ${}^{10}C_2 (-k)^2 = 405$.

5. Calculate the value: $45k^2 = 405 \Rightarrow k^2 = 9$, so $|k| = 3$.

Difficulty level: Easy

The Concept Name: Binomial General Term (Term independent of $x$)

Short cut solution: Use the formula $r = \frac{n\alpha}{\alpha + \beta} = \frac{10(1/2)}{1/2 + 2} = 2$. Solve $45k^2 = 405$ to get $|k|=3$.

 Question 159

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

Options: 

A. 5 

B. 9 

C. 7 

D. 11

Correct Answer: C

Year: Sep. 03, 2020 (II)

Solution: The general term is $T_{r+1} = {}^9C_r \left( \frac{3x^2}{2} \right)^{9-r} \left( -\frac{1}{3x} \right)^r = {}^9C_r \left( \frac{3}{2} \right)^{9-r} \left( -\frac{1}{3} \right)^r x^{18-3r}$. Setting $18 - 3r = 0$ gives $r = 6$. The term $k = T_7 = {}^9C_6 \left( \frac{3}{2} \right)^3 \left( -\frac{1}{3} \right)^6 = \frac{7}{18}$, so $18k = 7$.

Step Solution:

1. Determine the general term $T_{r+1} = {}^9C_r (\frac{3}{2}x^2)^{9-r} (-\frac{1}{3}x^{-1})^r$.

2. Isolate the exponent of $x$: $2(9-r) - r = 18 - 3r$.

3. For independence, set $18 - 3r = 0$, which gives $r = 6$.

4. Substitute $r=6$ to find $k$: $k = {}^9C_6 (\frac{3}{2})^3 (-\frac{1}{3})^6 = 84 \cdot \frac{27}{8} \cdot \frac{1}{729} = \frac{7}{18}$.

5. Calculate the final answer: $18k = 18 \times \frac{7}{18} = 7$.

Difficulty level: Medium

The Concept Name: Term Independent of $x$

Short cut solution: Use $r = \frac{n\alpha}{\alpha + \beta} = \frac{9(2)}{2+1} = 6$. Term is $k = {}^9C_3 \cdot \frac{3^3}{2^3 \cdot 3^6} = \frac{84}{8 \cdot 27} = \frac{7}{18}$.

 Question 178

Question: The term independent of $\mathbf { X }$ in the expansion of $\left( \frac { 1 } { 6 0 } - \frac { \mathrm { ~ x ~ } ^ { 8 } } { 8 1 } \right) \cdot \left( 2 \mathrm { ~ x ~ } ^ { 2 } - \frac { 3 } { \mathrm { ~ x ~ } ^ { 2 } } \right) ^ { 6 }$ is equal to:

Options: 

A. -72 

B. 36 

C. -36 

D. -108

Correct Answer: C

Year: April 12, 2019 (II)

Solution: The expression is split into two parts: $\frac{1}{60}(2x^2 - \frac{3}{x^2})^6 - \frac{x^8}{81}(2x^2 - \frac{3}{x^2})^6$. The term independent of $x$ is the sum of the $x^0$ coefficient from the first part and the $x^{-8}$ coefficient from the second part. These correspond to $r=3$ and $r=5$ in the binomial expansion respectively, yielding $-72 + 36 = -36$.

Step Solution:

1. Split into two products: $P_1 = \frac{1}{60}(2x^2 - 3x^{-2})^6$ and $P_2 = -\frac{1}{81}x^8(2x^2 - 3x^{-2})^6$.

2. For $P_1$, the constant term occurs when $12 - 4r = 0 \Rightarrow r = 3$. First value $= \frac{1}{60}{}^6C_3(2)^3(-3)^3 = -72$.

3. For $P_2$, independence requires $x^8 \cdot x^{12-4r} = x^0 \Rightarrow r = 5$.

4. Calculate the second value: $-\frac{1}{81}{}^6C_5(2)^1(-3)^5 = -\frac{1}{81}(6 \cdot 2 \cdot -243) = 36$.

5. Combine the parts: $-72 + 36 = -36$.

Difficulty level: Hard

The Concept Name: Term Independent of $x$ in a Polynomial Product

Short cut solution: Mentally find $r$ for $x^0$ and $x^{-8}$ in $(2x^2 - 3/x^2)^6$ using the $r = \frac{n\alpha - m}{\alpha + \beta}$ shortcut. Sum the resulting coefficients adjusted by the outside constants.

 Question 193

Question: The term independent of $x$ in the binomial expansion of $\left( 1 - \frac { 1 } { x } + 3 x ^ { 5 } \right) \left( 2 x ^ { 2 } - \frac { 1 } { x } \right) ^ { 8 }$ is:

Options:

A. 496

B. -496

C. 400

D. -400

Correct Answer: C

Year: Online April 11, 2015

Solution: The general term of the binomial part $(2x^2 - x^{-1})^8$ is $T_{r+1} = {}^8C_r (2x^2)^{8-r} (-x^{-1})^r$. This is expanded and multiplied by the trinomial $(1 - x^{-1} + 3x^5)$. By analyzing the powers of $x$ for each product, we find that terms independent of $x$ occur when $r=5$ (for the second term) and $r=7$ (for the third term). The final sum of these coefficients yields 400.

Step Solution:

1.  General Term: Identify the general term for the binomial part $(2x^2 - x^{-1})^8$ as $T_{r+1} = {}^8C_r 2^{8-r} (-1)^r x^{16-3r}$.

2.  Product Distribution: The full expression is $1 \cdot T_{r+1} - x^{-1} \cdot T_{r+1} + 3x^5 \cdot T_{r+1}$.

3.  Find valid $r$: For a constant term, the power of $x$ must be 0:

    *   (i) $16-3r=0 \Rightarrow$ no integer $r$.

    *   (ii) $16-3r-1=0 \Rightarrow 15=3r \Rightarrow \mathbf{r=5}$.

    *   (iii) $16-3r+5=0 \Rightarrow 21=3r \Rightarrow \mathbf{r=7}$.

4.  Calculate Coefficients:

    *   For $r=5$: $-1 \cdot {}^8C_5 \cdot 2^3 \cdot (-1)^5 = -1 \cdot 56 \cdot 8 \cdot -1 = \mathbf{448}$.

    *   For $r=7$: $3 \cdot {}^8C_7 \cdot 2^1 \cdot (-1)^7 = 3 \cdot 8 \cdot 2 \cdot -1 = \mathbf{-48}$.

5.  Final Addition: Sum the individual constant contributions: $448 + (-48) = \mathbf{400}$.

Difficulty level: Hard

The Concept Name: Binomial General Term and Product of Polynomials.

Short cut solution: Quickly determine which $r$ values satisfy $16-3r = \{0, 1, -5\}$. Only $r=5$ and $r=7$ work. Calculate the coefficients for these two values only and sum them.

 Question 204

Question: The term independent of $x$ in expansion of $\left( \frac { x + 1 } { x ^ { 2 / 3 } - x ^ { 1 / 3 } + 1 } - \frac { x - 1 } { x - x ^ { 1 / 2 } } \right) ^ { 1 0 }$ is:

Options:

A. 4

B. 120

C. 210

D. 310

Correct Answer: C

Year: 2013

Solution: The internal algebraic fractions are simplified using the sum of cubes identity and factoring. This reduces the expression to $(x^{1/3} - x^{-1/2})^{10}$. The general term formula is applied, setting the exponent of $x$ to zero, which gives $r=4$. The coefficient is calculated as ${}^{10}C_4 = 210$.

Step Solution:

1.  Simplify First Fraction: Use $a^3+b^3$ identity to rewrite $(x+1)$ as $(x^{1/3}+1)(x^{2/3}-x^{1/3}+1)$, resulting in $(x^{1/3}+1)$.

2.  Simplify Second Fraction: Factor the numerator $(x-1)$ as $(\sqrt{x}-1)(\sqrt{x}+1)$ and the denominator as $\sqrt{x}(\sqrt{x}-1)$, resulting in $1 + x^{-1/2}$.

3.  Combine Terms: The simplified binomial is $(x^{1/3}+1) - (1+x^{-1/2}) = \mathbf{x^{1/3} - x^{-1/2}}$.

4.  Determine $r$: For the expansion $(x^{1/3} - x^{-1/2})^{10}$, set the general power $\frac{10-r}{3} - \frac{r}{2} = 0$, giving $20-2r-3r=0$, so $\mathbf{r=4}$.

5.  Calculate Value: $T_{4+1} = {}^{10}C_4 (-1)^4 = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = \mathbf{210}$.

Difficulty level: Medium

The Concept Name: Simplification using Algebraic Identities and Binomial General Term.

Short cut solution: After simplifying to $(x^{1/3} - x^{-1/2})^{10}$, use the index formula $r = \frac{n\alpha}{\alpha + \beta} = \frac{10(1/3)}{1/3 + 1/2} = \frac{10/3}{5/6} = 4$. Then calculate ${}^{10}C_4 = 210$.