If the term independent of x in the expansion of (x^(1/3) - x^(-1/2))^10 is a , then the number of proper divisors of a is

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q18 – AP-EAMCET 2026

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Home Maths Binomial TheoremQuestion #33

If the term independent of

x

in the expansion of

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

a

, then the number of proper divisors of

a

Clue 1 of 3: Binomial Expansion

The general term in the expansion of $(A + B)^{n}$ is given by $T_{r + 1} = (\binom{n}{r}) A^{n - r} B^{r}$ .

🧠 Full Solution Path

Step 1: Finding the General Term of the Expansion✦ Active

The given expression is ${(x^{1 / 3} - x^{- 1 / 2})}^{10}$ . We use the binomial expansion formula for $(A + B)^{n}$ , where $A = x^{1 / 3}$ , $B = - x^{- 1 / 2}$ , and $n = 10$ . The general term, $T_{r + 1}$ , is given by:

T_{r + 1} = (\binom{n}{r}) A^{n - r} B^{r}

Substituting the values:

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

💡 Teacher's Secret Hint

Remember to include the negative sign from the second term, $- x^{- 1 / 2}$ , when raising it to the power $r$ .

Step 2: Determining the Exponent of x and Solving for r○ Expand

Simplify the expression for $T_{r + 1}$ to combine the powers of $x$ :

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

Combine the $x$ terms:

T_{r + 1} = (\binom{10}{r}) (- 1)^{r} x^{\frac{10 - r}{3} - \frac{r}{2}}

For the term to be independent of $x$ , the exponent of $x$ must be 0. So, we set the exponent to 0 and solve for $r$ :

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

Multiply by the least common multiple of 3 and 2, which is 6:

2 (10 - r) - 3 r = 0

20 - 2 r - 3 r = 0

20 - 5 r = 0

5 r = 20

r = 4

💡 Teacher's Secret Hint

A common mistake is forgetting to find a common denominator when combining fractional exponents, or errors in solving the linear equation for $r$ .

Step 3: Calculating the Value of 'a'○ Expand

Now substitute $r = 4$ back into the expression for $T_{r + 1}$ (with the $x$ term disappearing since its exponent is 0) to find the value of $a$ :

a = (\binom{10}{4}) (- 1)^{4} x^{0}

Since $(- 1)^{4} = 1$ and $x^{0} = 1$ (for $x \neq 0$ ):

a = (\binom{10}{4})

Calculate the binomial coefficient:

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

a = 10 \times 3 \times 7 = 210

💡 Teacher's Secret Hint

Ensure you correctly calculate the binomial coefficient. A factorial definition $n! = n \times (n - 1) \times \dots \times 1$ is essential.

Step 4: Finding the Number of Proper Divisors of 'a'○ Expand

We need to find the number of proper divisors of $a = 210$ . First, find the prime factorization of 210:

210 = 2 \times 105 = 2 \times 3 \times 35 = 2 \times 3 \times 5 \times 7

The prime factorization is $2^{1} \times 3^{1} \times 5^{1} \times 7^{1}$ . The total number of positive divisors is given by the product of one more than each exponent:

Total divisors = (1 + 1) (1 + 1) (1 + 1) (1 + 1) = 2 \times 2 \times 2 \times 2 = 16

The term 'proper divisors' can have slightly different interpretations. If 'proper divisors' means all positive divisors excluding 1 and the number itself, then:

Number of proper divisors = Total divisors - 2

Using this common interpretation (which aligns with the provided correct option):

Number of proper divisors = 16 - 2 = 14

💡 Teacher's Secret Hint

The definition of 'proper divisors' can vary. Often, it excludes the number itself. In some contexts (like this problem, given the options), it excludes both 1 and the number itself. Always check context or common conventions for the specific exam.

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Maths Question 18 – AP-EAMCET 2026

Clue 1 of 3: Binomial Expansion

🧠 Full Solution Path