Students can Download Chapter 8 Binomial Theorem Questions and Answers, Plus One Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

## Kerala Plus One Maths Chapter Wise Questions and Answers Chapter 8 Binomial Theorem

### Plus One Maths Binomial Theorem Three Mark Questions and Answers

Question 1.

Write the middle term in the expansion of the following; (3 score each)

Answer:

i) Here 7 is an odd number. Therefore there are two middle terms \(\left(\frac{7+1}{2}=4\right)^{t h}\) and \(\left(\frac{7+1}{2}+1=5\right)^{t h}\), ie; 4^{th} and 5^{th} terms in the above expansion.

ii) Here 10 is an even number. Therefore middle terms \(\left(\frac{10}{2}+1=6\right)^{t h}\) term in the above expansion.

iii) Here 17 is an odd number. Therefore there are two middle terms \(\left(\frac{17+1}{2}=9\right)^{t h}\), ie; 9^{th} and 10^{th} terms in the above expansion. \(\left(x+\frac{2}{\sqrt{x}}\right)^{17}\).

Question 2.

Find the term independent of x in the following expansion. (3 score each)

Answer:

i) General term = t_{r+1} = (-1)^{r}12C_{r}(x)^{12-r} \(\left(\frac{1}{x}\right)^{r}\)

=(-1)^{r}12C_{r}(x)^{12-r-r} = (-1)^{r}12C_{r}(x)^{12-2r}

Term independent of x in the expansion will be the term in which the power of x is zero, ie; 12 – 2r = 0 ⇒ 12 = 2r

⇒ r = 6

t_{7} = (-1)^{6}12C^{6}x^{12-2(6)}

= \(\frac{12 \times 11 \times 10 \times 9 \times 8 \times 7}{1 \times 2 \times 3 \times 4 \times 5 \times 6}\) = 924.

ii) General term = t_{r+1} =(-1)^{r}9C_{r}(x^{2})^{9-r}\(\left(\frac{1}{x}\right)^{r}\)

= (-1)^{r}9C_{r}(x)^{18-2r-r} = (-1)^{r}9C_{r}(x)^{18-3r}

Term independent of x in the expansion will be the term in which the power of x is zero.

ie; 18 – 3r = 0 ⇒ 18 = 3r ⇒ r = 6

t_{7} = (-1)^{6}9C_{6}x^{18-3(6)}

= \(\frac{9 \times 8 \times 7}{1 \times 2 \times 3}\) = 84.

iii) General term = t_{r+1}

Term independent of x in the expansion will be the term in which the power of x is zero.

iv) General term = t_{r+1}

Term independent of x in the expansion will be the term in which the power of x is zero.

Question 3.

Find the coefficient of x^{10} in the expansion of \(\left(2 x^{2}-\frac{3}{x}\right)^{11}\).

Answer:

General term = t_{r+1}

Given; 22 – 3r = 10 ⇒ 12 = 3r ⇒ r = 4

t_{5} = (-1)^{4}11C_{4}2^{11-4 }x^{22-3(4) }3^{4}

= 11C_{4}2^{7}3^{4}x^{10}

Therefore the coefficient of x^{10} is 11C_{4}2^{7}3^{4}.

Question 4.

Find the coefficient of a^{5}b^{7} in the expansion of (a – 2b)^{12}.

Answer:

General term = t_{r+1} = (-1)^{r}12C_{r}(a)^{12-r}(2b)^{r}

= (-1)^{r}12C_{r}(a)^{12-r}2^{r}b^{r}

The term containing a^{5}b^{7} is obtained by putting r = 7

⇒ t_{8} = (-1)^{7}12C_{7}(a)^{12-7}2^{7}b^{7}

Therefore the coefficient of a^{5}b^{7} is

(-1)^{7}12C_{7}2^{7} = -12C_{7}2^{7}.

Question 5.

Find the coefficient of (3 score each)

- x
^{11}in the expansion of \(\left(x-\frac{2}{x^{2}}\right)^{17}\) - x
^{9}in the expansion of \(\left(3 x^{2}+\frac{5}{x^{3}}\right)^{12}\) - x
^{20}in the expansion of \(\left(3 x^{3}-\frac{2}{x^{2}}\right)^{40}\)

Answer:

1. General term

The term containing x^{11} is obtained by

17 – 3r = 11 ⇒ 6 = 3r ⇒ r = 2

⇒ t_{3} = (-1)^{2}17C_{2} (x)^{17-3(2) }2^{2} = 17C_{2}(x)^{11} × 4

Therefore the coefficient of x^{11} is 17C_{2} × 4

= \(\frac{17 \times 16}{1 \times 2}\) × 4 = 544

2. General term

The term containing x^{9} is obtained by

24 – 5r = 9 ⇒ 15 = 5r ⇒ r = 3

⇒ t_{4} = 12C_{3}(3)^{12-3}(x)^{24-5(3)}5^{3}

= 12C_{3}(3)^{9}(x)^{9}5^{3}

Therefore the coefficient of x^{9} is 12C_{3}(3)^{9}5^{3}.

3. General term = t_{r+1}

= (-1)^{r}40C_{r}(3x^{3})^{40-r} (\(\frac{2}{x^{2}}\))^{r}

= (-1)^{r}40C_{r}(3)^{40-r}x^{120-3r} (2)^{r} x^{-2r}

= (-1)^{r}40C_{r}(3)^{40-r}x^{120-5r}(2)^{r}

The term containing x^{20} is obtained by

120 – 5r = 20 ⇒ 100 = 5r ⇒ r = 20

⇒ t_{21} = (-1)^{20}40C_{20}(3)^{40-20}(x) ^{120-5(20)} 2^{20}

= 40C_{20}(3)^{20}(x)^{20}2^{20}

Therefore the coefficient of x^{20} is 40C_{20}(3) ^{20} 2^{20}.

Question 6.

- Find the term independent of x in the expansion of \(\left(x^{2}+\frac{2}{x}\right)^{6}\) (3)
- If the middle term in the expansion of \(\left(x^{m}+\frac{2}{x}\right)^{6}\) is independent of x, find the value of m.

Answer:

1. t_{r+1} = ^{n}C_{r}a^{n-r}b^{r} = ^{6}C_{r}(x^{2})^{6-r}\(\left(\frac{2}{x}\right)^{r}\)

= ^{6}C_{r}x^{12-2r}x^{-r}(2)^{r} = ^{6}C_{r}x^{12-3r}(2)^{r}

For term independent of x;

12 – 3r = 0 r = 4

t_{5} = ^{6}C_{4}(2)^{4} = ^{6}C_{2} × 16 = \(\frac{6 \times 5}{1 \times 2}\) × 16 = 240

2. m = 1

### Plus One Maths Binomial Theorem Four Mark Questions and Answers

Question 1.

- Write the general term in the expansion \(\left(\frac{3 x^{2}}{2}-\frac{1}{3 x}\right)^{6}\) (2)
- Find the term independent of x in the above expansion. (2)

Answer:

1. General term = t_{r+1}

2. Term independent of x in the expansion will be the term in which the power of x is zero.

ie; 12 – 3r = 0 ⇒ 12 = 3r ⇒ r = 4

### Plus One Maths Binomial Theorem Practice Problems Questions and Answers

Question 1.

Expand the following. (2 score each)

- (3a
^{2}– 2b)^{4} - (3 – 4x
^{2})^{5} - \(\left(\frac{x}{2}-2 y\right)^{6}\)
- \(\left(\frac{x}{2}-2 y\right)^{6}\)

Answer:

1.

2.

3.

4.

Question 2.

Write the general term in the expansion of the following; (2 score each)

Answer:

i) General term = t_{r+1}

= (-1)^{r}6C_{r}(x^{2})^{6-r}(y)^{r}

= (-1)^{r}6C_{r}x^{12-r}y^{r}.

ii) General term = t_{r+1}

iii) General term = t_{r+1}

iv) General term = t_{r+1}

Question 3.

If the coefficient of x^{2} in the expansion of (1 + x)^{n} is 6 then the positive value of n.

Answer:

t_{r+1} = nC_{r}x^{r}, the term containing x^{2} is obtained by putting r= 2.

nC_{2} = 6 ⇒ \(\frac{n(n-1)}{2}\) = 6 ⇒ n(n -1) = 12

⇒ n(n -1) = 4 × 3 ⇒ n = 3.

Question 4.

Find the 13^{th} term in the expansion of \(\left(9 x-\frac{1}{3 \sqrt{x}}\right)^{18}\).

Answer:

= 18C_{6}(3)^{12-12} = 18C_{6} = 18564