Binomial Theorem Questions and Answers

Q: Let \((1 + ax)(1 - 2x)^{3} = \sum_{n=0}^{4} a_{n}x^{n}\) , where \(a\) is a constant. If \(a_{2} = 0\) , then the value of \(a\) is

The value of a is 2.Expand \((1 - 2x)^{3} = 1 - 6x + 12x^{2} - 8x^{3}\). In the product with \((1 + ax)\), the \(x^{2}\) terms come from \(1 \times 12x^{2}\) and \(ax \times (-6x)\).\[ \begin{aligned} a_2 &= 12 - 6a = 0 \\ a &= 2 \end{aligned} \]Taking the \(x^{2}\) coefficient as 12 alone, or pairing ax with the \(x^{3}\) term, ignores how the product forms \(x^{2}\); only \(12 - 6a = 0\) uses both contributions.

Q: The constant term in the expansion of \(\left(2x^{2} - \frac{1}{x^{2}}\right)^{6}\) is

The constant term is −160.The general term of \(\left(2x^{2} - \frac{1}{x^{2}}\right)^{6}\) is\[ \begin{aligned} T_{r+1} &= {}^{6}C_{r}(2x^{2})^{6-r} \\ &\quad \times \left(-\frac{1}{x^{2}}\right)^{r} \end{aligned} \]with power of x \(12 - 4r\). Setting it to 0 gives \(r = 3\):\[ \begin{aligned} {}^{6}C_{3} \cdot 2^{3} \cdot (-1)^{3} &= 20 \times 8 \times (-1) \\ &= -160 \end{aligned} \]160 drops the sign from \((-1)^{3}\), and ±180 use the wrong binomial coefficient or power of 2.

21.

Let \((1 + ax)(1 - 2x)^{3} = \sum_{n=0}^{4} a_{n}x^{n}\) , where \(a\) is a constant. If \(a_{2} = 0\) , then the value of \(a\) is

Exam Year(s)

KEAM - 2026,

Answer: E

The value of a is 2.

Expand \((1 - 2x)^{3} = 1 - 6x + 12x^{2} - 8x^{3}\). In the product with \((1 + ax)\), the \(x^{2}\) terms come from \(1 \times 12x^{2}\) and \(ax \times (-6x)\).

\[ \begin{aligned} a_2 &= 12 - 6a = 0 \\ a &= 2 \end{aligned} \]

Taking the \(x^{2}\) coefficient as 12 alone, or pairing ax with the \(x^{3}\) term, ignores how the product forms \(x^{2}\); only \(12 - 6a = 0\) uses both contributions.

Discuss in Forum

22.

The constant term in the expansion of \(\left(2x^{2} - \frac{1}{x^{2}}\right)^{6}\) is

A.-160
B.160
C.-180
D.180
E.0

Exam Year(s)

KEAM - 2025,

Answer: A

The constant term is −160.

The general term of \(\left(2x^{2} - \frac{1}{x^{2}}\right)^{6}\) is

\[ \begin{aligned} T_{r+1} &= {}^{6}C_{r}(2x^{2})^{6-r} \\ &\quad \times \left(-\frac{1}{x^{2}}\right)^{r} \end{aligned} \]

with power of x \(12 - 4r\). Setting it to 0 gives \(r = 3\):

\[ \begin{aligned} {}^{6}C_{3} \cdot 2^{3} \cdot (-1)^{3} &= 20 \times 8 \times (-1) \\ &= -160 \end{aligned} \]

160 drops the sign from \((-1)^{3}\), and ±180 use the wrong binomial coefficient or power of 2.

Binomial Theorem Questions and Answers

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