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JEE Main 2020
Binomial Theorem
Binomial Theorem
Hard

Question

Let [t][t] denote the greatest integer t\leq t. If the constant term in the expansion of (3x212x5)7\left(3 x^{2}-\frac{1}{2 x^{5}}\right)^{7} is α\alpha, then [α][\alpha] is equal to ___________.

Answer: 1

Solution

1. Introduction: The Binomial Theorem and General Term

To find the constant term in the expansion of a binomial expression, we first need to understand the general term of a binomial expansion. For any binomial expression of the form (a+b)n(a+b)^n, the general term, often denoted as Tr+1T_{r+1}, is given by the formula:

Tr+1=nCranrbrT_{r+1} = { }^n C_r a^{n-r} b^r

where:

  • nn is the power to which the binomial is raised.
  • rr is the index of the term (starting from r=0r=0 for the first term), and rr must be a non-negative integer (0rn0 \le r \le n).
  • nCr=n!r!(nr)!{ }^n C_r = \frac{n!}{r!(n-r)!} is the binomial coefficient.
  • aa is the first term of the binomial.
  • bb is the second term of the binomial.

In our given problem, the expression is (3x212x5)7\left(3 x^{2}-\frac{1}{2 x^{5}}\right)^{7}. Comparing this with (a+b)n(a+b)^n, we can identify:

  • a=3x2a = 3x^2
  • b=12x5b = -\frac{1}{2x^5}
  • n=7n = 7

2. Determining the General Term (Tr+1T_{r+1})

Now, we substitute these values into the general term formula:

Tr+1=7Cr(3x2)7r(12x5)rT_{r+1} = { }^7 C_r (3x^2)^{7-r} \left(-\frac{1}{2x^5}\right)^r

To simplify this, we separate the constant parts from the xx parts:

Tr+1=7Cr(3)7r(x2)7r(12)r(1x5)rT_{r+1} = { }^7 C_r (3)^{7-r} (x^2)^{7-r} \left(-\frac{1}{2}\right)^r \left(\frac{1}{x^5}\right)^r

Using the exponent rules (xp)q=xpq(x^p)^q = x^{pq} and 1xp=xp\frac{1}{x^p} = x^{-p}:

Tr+1=7Cr(3)7rx2(7r)(12)rx5rT_{r+1} = { }^7 C_r (3)^{7-r} x^{2(7-r)} \left(-\frac{1}{2}\right)^r x^{-5r}

Combining the powers of xx:

Tr+1=7Cr(3)7r(12)rx142r5rT_{r+1} = { }^7 C_r (3)^{7-r} \left(-\frac{1}{2}\right)^r x^{14-2r-5r}

Tr+1=7Cr(3)7r(12)rx147rT_{r+1} = { }^7 C_r (3)^{7-r} \left(-\frac{1}{2}\right)^r x^{14-7r}

This is the simplified general term of the expansion.

3. Finding the Term Independent of xx (Constant Term)

A term is considered "constant" or "independent of xx" if it does not contain the variable xx. This means that the power of xx in that term must be equal to zero.

From our general term Tr+1T_{r+1}, the power of xx is 147r14-7r. We set this exponent to zero to find the value of rr that corresponds to the constant term:

147r=014 - 7r = 0

Now, we solve for rr:

14=7r14 = 7r r=147r = \frac{14}{7} r=2r = 2

Since r=2r=2 is a non-negative integer and 0270 \le 2 \le 7, this is a valid value for rr.

Tip: Always ensure that the value of rr you find is a non-negative integer within the range of 00 to nn. If rr turns out to be a fraction or a negative number, it means there is no constant term in the expansion.

4. Calculating the Constant Term (α\alpha)

Now that we have found r=2r=2, we substitute this value back into the constant part of our simplified general term (the part without x147rx^{14-7r}) to find the constant term, which is given as α\alpha.

α=7C2(3)72(12)2\alpha = { }^7 C_2 (3)^{7-2} \left(-\frac{1}{2}\right)^2

Let's calculate each part:

  • Binomial Coefficient: 7C2=7!2!(72)!=7!2!5!=7×62×1=7×3=21{ }^7 C_2 = \frac{7!}{2!(7-2)!} = \frac{7!}{2!5!} = \frac{7 \times 6}{2 \times 1} = 7 \times 3 = 21
  • First part's coefficient: (3)72=(3)5=3×3×3×3×3=243(3)^{7-2} = (3)^5 = 3 \times 3 \times 3 \times 3 \times 3 = 243
  • Second part's coefficient: (12)2=(1×12)2=(1)2×(12)2=1×14=14\left(-\frac{1}{2}\right)^2 = \left(-1 \times \frac{1}{2}\right)^2 = (-1)^2 \times \left(\frac{1}{2}\right)^2 = 1 \times \frac{1}{4} = \frac{1}{4}

Common Mistake: Be careful with negative signs raised to powers. An even power of a negative number results in a positive number.

Now, multiply these values to find α\alpha:

α=21×243×14\alpha = 21 \times 243 \times \frac{1}{4} α=5103×14\alpha = 5103 \times \frac{1}{4} α=51034\alpha = \frac{5103}{4} α=1275.75\alpha = 1275.75

5. Evaluating the Greatest Integer Function ([α][\alpha])

The problem asks for the value of [α][\alpha], where [t][t] denotes the greatest integer less than or equal to tt. This is also known as the floor function.

For α=1275.75\alpha = 1275.75, we need to find the greatest integer that is less than or equal to 1275.751275.75.

[α]=[1275.75][\alpha] = [1275.75] [α]=1275[\alpha] = 1275

Therefore, the value of [α][\alpha] is 12751275.

6. Key Takeaways

This problem demonstrates a standard application of the Binomial Theorem. The key steps are:

  1. Identify a,b,na, b, n from the given binomial expansion.
  2. Write the general term (Tr+1T_{r+1}) and simplify it, carefully separating constant and variable parts.
  3. Set the power of the variable (xx) to zero to find the term independent of xx (constant term). Solve for rr.
  4. Substitute the value of rr back into the constant part of Tr+1T_{r+1} to calculate the desired term.
  5. Apply any additional functions (like the greatest integer function) as required by the problem. Paying close attention to algebraic manipulations, especially with exponents and signs, is crucial for accuracy.

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