JEE Binomial Theorem PYQs with Video Solutions, Download PDF

The general term is

$$T_{r+1}=\binom{4}{r}x^{4-r}\left(\frac1x\right)^r$$

$$=\binom{4}{r}x^{4-2r}$$

For the constant term,

$$4-2r=0$$

$$r=2$$

Therefore, the constant term is

$$\binom42=6$$

Hence,

$$\boxed{6}$$

To find the degree of the polynomial, we expand the expression completely using the Binomial Theorem. This question contains a deliberate trap: the highest power ($$x^{12}$$) completely cancels out during simplification.

Step 1: Apply the Binomial Theorem expansion
Let $$a = x^2 - 3$$ and $$b = (1 - x^4)^{\frac{1}{2}}$$. The expression is in the form $$(a + b)^6 + (a - b)^6$$.

Expanding both parts with the Binomial Theorem and adding them together cancels out all terms with odd powers of $$b$$:
$$(a+b)^6 + (a-b)^6 = 2 \left[ \binom{6}{0}a^6 + \binom{6}{2}a^4b^2 + \binom{6}{4}a^2b^4 + \binom{6}{6}b^6 \right]$$

Substitute the combination values $$\binom{6}{0}=1$$, $$\binom{6}{2}=15$$, $$\binom{6}{4}=15$$, and $$\binom{6}{6}=1$$:
$$= 2 [ a^6 + 15a^4b^2 + 15a^2b^4 + b^6 ]$$

Step 2: Write out the full expansion of each component
Substitute $$a = x^2 - 3$$ and $$b^2 = 1 - x^4$$ back into the formula:
$$= 2 [ (x^2 - 3)^6 + 15(x^2 - 3)^4(1 - x^4) + 15(x^2 - 3)^2(1 - x^4)^2 + (1 - x^4)^3 ]$$

We fully expand each of the four components inside the brackets:

1. For $$(x^2 - 3)^6$$:
$$x^{12} - 18x^{10} + 135x^8 - 540x^6 + 1215x^4 - 1458x^2 + 729$$

2. For $$15(x^2 - 3)^4(1 - x^4)$$:
$$15(x^8 - 12x^6 + 54x^4 - 108x^2 + 81)(1 - x^4)$$
$$= -15x^{12} + 15x^8 + 180x^{10} - 180x^6 - 810x^8 + 810x^4 + 1620x^6 - 1620x^2 - 1215x^4 + 1215$$
$$= -15x^{12} + 180x^{10} - 795x^8 + 1440x^6 - 405x^4 - 1620x^2 + 1215$$

3. For $$15(x^2 - 3)^2(1 - x^4)^2$$:
$$15(x^4 - 6x^2 + 9)(x^8 - 2x^4 + 1)$$
$$= 15x^{12} - 30x^8 + 15x^4 - 90x^{10} + 180x^6 - 90x^2 + 135x^8 - 270x^4 + 135$$
$$= 15x^{12} - 90x^{10} + 105x^8 + 180x^6 - 255x^4 - 90x^2 + 135$$

4. For $$(1 - x^4)^3$$:
$$-x^{12} + 3x^8 - 3x^4 + 1$$

Step 3: Combine all terms to get the complete polynomial expression
Now, sum all four fully expanded parts together inside the brackets:
$$\text{Sum} = (x^{12} - 15x^{12} + 15x^{12} - x^{12}) + (-18x^{10} + 180x^{10} - 90x^{10}) + (135x^8 - 795x^8 + 105x^8 + 3x^8) + (-540x^6 + 1440x^6 + 180x^6) + (1215x^4 - 405x^4 - 255x^4 - 3x^4) + (-1458x^2 - 1620x^2 - 90x^2) + (729 + 1215 + 135 + 1)$$

Simplify each grouped power step-by-step:
* For $$x^{12}$$: $$1 - 15 + 15 - 1 = 0$$
* For $$x^{10}$$: $$-18 + 180 - 90 = 72$$
* For $$x^8$$: $$135 - 795 + 105 + 3 = -552$$
* For $$x^6$$: $$-540 + 1440 + 180 = 1080$$
* For $$x^4$$: $$1215 - 405 - 255 - 3 = 552$$
* For $$x^2$$: $$-1458 - 1620 - 90 = -3168$$
* For constants: $$729 + 1215 + 135 + 1 = 2080$$

Multiplying the entire simplified bracket by the front factor of $$2$$ gives the complete expanded polynomial:
$$\text{Complete Expression} = 2(72x^{10} - 552x^8 + 1080x^6 + 552x^4 - 3168x^2 + 2080)$$
$$\text{Complete Expression} = 144x^{10} - 1104x^8 + 2160x^6 + 1104x^4 - 6336x^2 + 4160$$

Since the highest remaining power of $$x$$ is 10, the true degree of the polynomial is 10.

To find the term independent of $$x$$ in the expansion, we first simplify the two fractions inside the parentheses.

Step 1: Simplify the First Fraction

Let the first term be $$A$$:

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

Using the sum of cubes factorization $$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$$, we can write $$x + 1$$ as:

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

Substituting this back into $$A$$:

$$A = \frac{(x^{1/3} + 1)(x^{2/3} - x^{1/3} + 1)}{x^{2/3} - x^{1/3} + 1} = x^{1/3} + 1$$

Step 2: Simplify the Second Fraction

Let the second term be $$B$$:

$$B = \frac{x-1}{x - x^{1/2}}$$

Using the difference of squares $$a^2 - b^2 = (a - b)(a + b)$$ and factoring the denominator:

• Numerator: $$x - 1 = (x^{1/2} - 1)(x^{1/2} + 1)$$
• Denominator: $$x - x^{1/2} = x^{1/2}(x^{1/2} - 1)$$

Substituting these back into $$B$$:

$$B = \frac{(x^{1/2} - 1)(x^{1/2} + 1)}{x^{1/2}(x^{1/2} - 1)} = \frac{x^{1/2} + 1}{x^{1/2}} = 1 + x^{-1/2}$$

Step 3: Set up the Binomial Expansion

Now, we subtract the simplified terms:

$$A - B = (x^{1/3} + 1) - (1 + x^{-1/2}) = x^{1/3} - x^{-1/2}$$

We need to find the term independent of $x$ in the expansion of:

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

The general term $$T_{r+1}$$ is given by:

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

Step 4: Find $$r$$ and Compute the Term

For the term to be independent of $$x$$, the exponent of $$x$$ must be $$0$$:

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

Multiply the entire equation by $$6$$ to clear the denominators:

$$2(10-r) - 3r = 0$$ $$20 - 2r - 3r = 0$$ $$5r = 20 \implies r = 4$$

Now substitute $$r = 4$$ back into the general term formula:

$$T_{5} = \binom{10}{4} (-1)^4 = \binom{10}{4}$$ $$\binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$$

Final Answer

The term independent of $$x$$ is $$210$$ .