CBSE Class 11 Mathematics Binomial Theorem Worksheet Set 03

CBSE Class 11 Mathematics Worksheet - Binomial Theorem (2). Students can download these worksheets and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

Question. Find the positive value of ‘m’ for which the coefficient of \(x^2\) in the expansion of \((1 + x)^m\) is 6?
(a) 6
(b) 9
(c) 4
(d) 1
Answer: (c) 4
Solution:
Given expansion: \( (1 + x)^m \), coefficient of \( x^2 = 6 \)
To find: \( m \)
General term: \( T_{r+1} = {}^mC_r (1)^{m-r}x^r = {}^mC_rx^r \)
For \( x^2 \), put \( r = 2 \)
\( \therefore T_3 = {}^mC_2x^2 \)
Here, coefficient of \( x^2 = 6 \)
\( \Rightarrow {}^mC_2 = 6 \)
\( \Rightarrow \frac{m(m - 1)}{2} = 6 \)
\( \Rightarrow m^2 - m - 12 = 0 \)
\( \Rightarrow (m - 4)(m + 3) = 0 \)
\( \Rightarrow m = 4 \text{ or } m = -3 \)
but \( m \) cannot be negative (-) \( \therefore m \neq -3 \)
\( \therefore m = 4 \) ans.

Question. If the coefficient of \((r - 5)^{\text{th}}\) and \((2r - 1)^{\text{th}}\) terms in the expansion of \((1 + x)^{34}\) are equal, find the value of ‘r’?
(a) 14
(b) 10
(c) 12
(d) 20
Answer: (a) 14
Solution:
Given expansion: \( (1 + x)^{34} \)
Coefficient of \( T_{r-5} = T_{2r-1} \)
To find: \( r \)
General term: \( T_{r+1} = {}^{34}C_r x^r \)
For \( T_{r-5} \), put \( r = r - 6 \)
\( \therefore T_{r-5} = {}^{34}C_{r-6} x^{r-6} \)
Here coefficient of \( T_{r-5} = {}^{34}C_{r-6} \)
For \( T_{2r-1} \), put \( r = 2r - 2 \)
\( \dots T_{2r-2} = {}^{34}C_{2r-2} x^{2r-2} \)
Here coefficient of \( T_{2r-1} = {}^{34}C_{2r-2} \)
We are given that, coefficients are equal
\( \Rightarrow {}^{34}C_{r-6} = {}^{34}C_{2r-2} \)
\( \Rightarrow r - 6 + 2r - 2 = 34 \dots\dots\dots\dots \) (if \( {}^nC_x = {}^nC_y \), then \( x + y = n \) or \( x = y \))
(Or) \( r - 6 = 2r - 2 \)
\( \Rightarrow 3r = 42 \Rightarrow r = 14 \)
or \( r = -4 \), but \( r \) cannot be negative (-)
\( \Rightarrow r = 14 \) ans.

Question. Find the term independent of \(x\) in the expansion of \( \left(\frac{3x^2}{2} - \frac{1}{3x}\right)^6 \)?
(a) \( \frac{-9}{8} \)
(b) \( \frac{5}{12} \)
(c) 10
(d) \( \frac{2}{3} \)
Answer: (b) \( \frac{5}{12} \)
Solution:
General term is given by \( T_{r+1} = (-1)^r \cdot {}^6C_r \cdot \frac{3^{6-2r}}{2^{6-r}} \cdot x^{12-3r} \)
For independent term of \(x\) i.e. \( x^0 \), put \( 12 - 3r = 0 \)
\( \Rightarrow r = 4 \)
\( \therefore T_5 = (-1)^4 \cdot {}^6C_4 \cdot \frac{3^{6-8}}{2^{6-4}} \cdot x^0 \)
\( = {}^6C_2 \cdot \frac{3^{-2}}{2^2} \dots\dots\dots\dots ({}^6C_4 = {}^6C_2) \)
\( = \frac{6 \times 5}{2} \times \frac{1}{9 \times 4} = \frac{5}{12} \)
\( \therefore 5^{\text{th}} \) term is the independent term of \(x\) and is given by \( \frac{5}{12} \) ans.

Question. Find the value of ‘a’ so that the term independent of ‘x’ in \( \left(\sqrt{x} + \frac{a}{x^2}\right)^{10} \) is 405?
(a) \( a^2 = \frac{40 \times 9}{8 \times 11} \)
(b) \( a^2 = \frac{-405 \times 2}{9 \times 10} \)
(c) \( a^2 = \frac{405 \times 2}{9 \times 10} \)
(d) \( a^2 = \frac{205 \times 2}{3 \times 6} \)
Answer: (c) \( a^2 = \frac{405 \times 2}{9 \times 10} \)
Solution:
Given expansion: \( \left(x^{1/2} + \frac{a}{x^2}\right)^{10} \)
Independent term of \(x\) = 405
To find ‘a’
General term: \( T_{r+1} = {}^{10}C_r \left(x^{1/2}\right)^{10-r} \cdot \frac{a^r}{x^{2r}} \)
\( \Rightarrow T_{r+1} = {}^{10}C_r (x)^{\frac{10-r}{2} - 2r} \cdot a^r \)
\( \Rightarrow T_{r+1} = {}^{10}C_r (x)^{\frac{10-5r}{2}} \cdot a^r \)
Now, for independent term of \(x\) i.e. \( x^0 \), Put \( \frac{10-5r}{2} = 0 \)
\( \Rightarrow r = 2 \)
\( \therefore T_3 = {}^{10}C_2 (x)^0 \cdot a^2 \)
\( T_3 = {}^{10}C_2 a^2 \)
Also independent term of \(x = 405 \dots\dots\dots\dots \text{(given)} \)
\( \Rightarrow {}^{10}C_2 a^2 = 405 \)
\( \Rightarrow \frac{10 \times 9}{2} a^2 = 405 \)
\( \Rightarrow a^2 = \frac{405 \times 2}{9 \times 10} \) ans.

Question. Find the middle terms in the expansion of \( \left(3x - \frac{x^3}{6}\right)^7 \)?
(a) \( 42x^{13} \) and \( 35 \)
(b) \( -\frac{105}{8}x^{13} \) and \( \frac{35}{48}x^{15} \)
(c) \( \frac{25}{72}x^{13} \) and \( \frac{30}{48} \)
(d) \( -\frac{10}{1}x^1 \) and \( \frac{35}{8}x^5 \)
Answer: (b) \( -\frac{105}{8}x^{13} \) and \( \frac{35}{48}x^{15} \)
Solution:
Given expansion: \( \left(3x - \frac{x^3}{6}\right)^7 \)
To find ‘middle term’
Since, power is odd, \( \therefore \) there are two middle terms = \( \left(\frac{n+1}{2}\right)^{\text{th}} \) and \( \left(\frac{n+3}{2}\right)^{\text{th}} \)
i.e. \( \left(\frac{7+1}{2}\right)^{\text{th}} \) and \( \left(\frac{7+3}{2}\right)^{\text{th}} \)
\( \Rightarrow 4^{\text{th}} \) and \( 5^{\text{th}} \) terms
General term: \( T_{r+1} = (-1)^r \cdot {}^{7}C_r (3x)^{7-r} \left(\frac{x^3}{6}\right)^r \)
\( = (-1)^r \cdot {}^{7}C_r (3)^{7-r} \cdot x^{7-r} \cdot \frac{x^{3r}}{6^r} \)
\( = T_{r+1} = (-1)^r \cdot {}^{7}C_r \cdot \frac{3^{7-r}}{6^r} \cdot x^{7+2r} \)
For \( T_4 \), put \( r = 3 \)
\( \Rightarrow T_4 = (-1)^3 \cdot {}^{7}C_3 \cdot \frac{3^4}{6^3} \cdot x^{7+6} \)
\( = -\frac{7 \times 6 \times 5}{6} \times \frac{81}{216} \cdot x^{13} \)
\( \Rightarrow T_4 = -\frac{105}{8}x^{13} \)
For \( T_5 \), put \( r = 4 \)
\( \therefore T_5 = (-1)^4 \cdot {}^{7}C_4 \cdot \frac{3^3}{6^4} \cdot x^{7+8} \)
\( = T_5 = \frac{35}{48}x^{15} \)
\( \dots \) the middle terms are \( -\frac{105}{8}x^{13} \) and \( \frac{35}{48}x^{15} \) ans.

Question. Show that the middle term in the expansion of \((1 + x)^{2n}\) is \( \frac{1 \cdot 3 \cdot 5 \dots (2n-1) \cdot 2^n \cdot x^n}{n!} \)?
(a) False
(b) True
(c) Not proved
(d) Negative
Answer: (b) True
Solution:
Given expansion: \( (1 + x)^{2n} \)
Since, power \( (2n) \) is even, only 1 middle term = \( \left(\frac{2n}{2} + 1\right)^{\text{th}} = (n + 1)^{\text{th}} \) term
General term: \( T_{r+1} = {}^{2n}C_r x^r \)
For \( T_{n+1} \), put \( r = n \)
\( = T_{n+1} = {}^{2n}C_n x^n \)
\( = \frac{(2n)!}{n!n!} x^n \)
\( = \frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \dots (2n-1) \cdot (2n)}{n!n!} x^n \)
\( = \frac{[1 \cdot 3 \cdot 5 \dots (2n-1)] [2 \cdot 4 \cdot 6 \dots (2n)]}{n!n!} x^n \)
\( = \frac{[1 \cdot 3 \cdot 5 \dots (2n-1)] \cdot 2^n \cdot (1 \cdot 2 \cdot 3 \dots n)}{n!n!} x^n \)
\( = \frac{1 \cdot 3 \cdot 5 \dots (2n-1) \cdot 2^n \cdot n!}{n!n!} x^n \)
\( = T_{n+1} = \frac{1 \cdot 3 \cdot 5 \dots (2n-1) \cdot 2^n \cdot x^n}{n!} \) ans.

Question. Show that the coefficient of the middle term in the expansion of \((1 + x)^{2n}\) is equal to the sum of the coefficients of two middle terms in the expansion of \((1 + x)^{2n-1}\)?
(a) True
(b) False
(c) Negative
(d) Positive
Answer: (a) True
Solution:
\( 1^{\text{st}} \) expansion: \( (1 + x)^{2n} \)
Since, power \( (2n) \) is even, only 1 middle term = \( \left(\frac{2n}{2} + 1\right)^{\text{th}} = (n + 1)^{\text{th}} \) term
General term: \( T_{r+1} = {}^{2n}C_r x^r \)
For \( T_{n+1} \), put \( r = n \)
\( = T_{n+1} = {}^{2n}C_n x^n \dots\dots\dots\dots (\text{coefficient} = {}^{2n}C_n) \)

\( 2^{\text{nd}} \) expansion: \( (1 + x)^{2n-1} \)
Since, power \( (2n - 1) \) is odd, only 2 middle terms = \( \left(\frac{2n-1+1}{2}\right)^{\text{th}} \) and \( \left(\frac{2n-1+3}{2}\right)^{\text{th}} \) terms
\( = n^{\text{th}} \) and \( (n + 1)^{\text{th}} \) terms
General term: \( T_{r+1} = {}^{2n-1}C_r x^r \)
For \( T_n \), put \( r = n - 1 \)
\( \therefore T_n = {}^{2n-1}C_{n-1} x^{n-1} \)
\( \text{Coefficient} = {}^{2n-1}C_{n-1} \)
For \( T_{n+1} \), put \( r = n \)
\( \therefore T_{n+1} = {}^{2n-1}C_n x^n \)
\( \text{Coefficient} = {}^{2n-1}C_n \)

Now, we have to prove that
\( {}^{2n}C_n = {}^{2n-1}C_{n-1} + {}^{2n-1}C_n \)
\( \text{R.H.S} = {}^{2n-1}C_{n-1} + {}^{2n-1}C_n \)
\( = {}^{2n-1+1}C_n \dots\dots\dots\dots ({}^nC_r + {}^nC_{r-1} = {}^{n+1}C_r) \)
\( = {}^{2n}C_n = \text{L.H.S} \) (proved)

Question. Prove that the coefficient of \(x^n\) in the expansion of \((1 + x)^{2n}\) is twice the coefficient of \(x^n\) in the expansion of \((1 + x)^{2n-1}\)?
(a) True
(b) False
(c) Negative
(d) Positive
Answer: (a) True
Solution:
\( 1^{\text{st}} \) expansion: \( (1 + x)^{2n} \)
General term: \( T_{r+1} = {}^{2n}C_r x^r \)
For \( x^n \), put \( r = n \)
\( = T_{n+1} = {}^{2n}C_n x^n \)
Coefficient of \( x^n = {}^{2n}C_n \)

\( 2^{\text{nd}} \) expansion: \( (1 + x)^{2n-1} \)
General term: \( T_{r+1} = {}^{2n-1}C_r x^r \)
For \( x^n \), put \( r = n \)
\( = T_{n+1} = {}^{2n-1}C_n x^n \)
Coefficient of \( x^n = {}^{2n-1}C_n \)

Now, we have to prove that
\( {}^{2n}C_n = 2({}^{2n-1}C_n) \)
\( \text{R.H.S} = 2 \cdot {}^{2n-1}C_n \)
\( = \frac{2 \cdot (2n-1)!}{n!(n-1)!} \dots\dots\dots\dots (1) \)
\( \text{L.H.S} = {}^{2n}C_n \)
\( = \frac{(2n)!}{n!n!} = \frac{(2n)(2n-1)!}{n \cdot n!(n-1)!} = \frac{2(2n-1)!}{n!(n-1)!} \dots\dots\dots\dots (2) \)
From (1) & (2) , R.H.S = L.H.S (proved)

Question. The sum of the coefficients of the \(1^{\text{st}}\) three terms in the expansion of \( \left(x - \frac{3}{x^2}\right)^m \) is 559. Find the term containing \(x^3\) in the expansion?
(a) \( 2582x^3 \)
(b) \( -5940x^3 \)
(c) \( 5900 \)
(d) \( 5940x^3 \)
Answer: (b) \( -5940x^3 \)
Solution:
Given expansion: \( \left(x - \frac{3}{x^2}\right)^m \)
To find ‘m’
General term: \( T_{r+1} = (-1)^r \cdot {}^mC_r (x)^{m-r} \left(\frac{3}{x^2}\right)^r \)
\( = (-1)^r \cdot {}^mC_r (x)^{m-r} \frac{3^r}{x^{2r}} \)
\( = T_{r+1} = (-1)^r \cdot {}^mC_r (3)^r (x)^{m-3r} \)
For \( T_1 \), put \( r = 0 \)
\( = T_1 = (-1)^0 \cdot {}^mC_0 (3)^0 (x)^m = x^m \)
\( \therefore \) coefficient of \( T_1 = 1 \)
For \( T_2 \), put \( r = 1 \)
\( = T_2 = (-1)^1 \cdot {}^mC_1 \cdot 3^1 \cdot x^{m-3} = -3mx^{m-3} \)
\( \dots \) coefficient of \( T_2 = -3m \)
For \( T_3 \), put \( r = 2 \)
\( = T_3 = (-1)^2 \cdot {}^mC_2 \cdot 3^2 \cdot x^{m-6} = 9 \cdot \frac{m(m-1)}{2} x^{m-6} \)
\( \therefore \) coefficient of \( T_3 = \frac{9m(m-1)}{2} \)
We are given that,
\( 1 - 3m + \frac{9m(m-1)}{2} = 559 \)
\( \Rightarrow 2 - 6m + 9m^2 - 9m = 1118 \)
\( \Rightarrow 9m^2 - 15m - 1116 = 0 \)
\( \Rightarrow 3m^2 - 5m - 372 = 0 \) (divide by 3)
\( a = 3, b = -5, c = -372 \)
By quadratic formula,
\( m = \frac{5 \pm \sqrt{25 + (4)(3)(372)}}{2 \times 3} \)
\( m = \frac{5 \pm \sqrt{4489}}{6} \)
\( m = \frac{5 \pm 67}{6} \)
\( m = \frac{5+67}{6}, m = \frac{5-67}{6} \)
\( m = \frac{72}{6}, m = \frac{-62}{6} \)
\( m = 12 \text{ or } m = -10.33 \)
\( \therefore m = 12 \)
\( \therefore \) general term becomes
\( T_{r+1} = (-1)^r \cdot {}^{12}C_r \cdot (3)^r \cdot (x)^{12-3r} \)
For \( x^3 \), put \( r = 3 \)
\( \therefore T_4 = (-1)^3 \cdot {}^{12}C_3 \cdot (3)^3 \cdot x^3 \)
\( = -\frac{12 \times 11 \times 10}{6} \times 27 \times x^3 = -5940x^3 \) ans.

Question. The coefficients of three consecutive terms in the expansion of \((1 + a)^n\) are in ratio 1:7:42. Find the value of ‘n’?
(a) 33
(b) 26
(c) 55
(d) 78
Answer: (c) 55
Solution:
Given expansion: \( (1 + a)^n \)
General term: \( T_{r+1} = {}^nC_r a^r \)
Let the three consecutive terms are \( (r - 1)^{\text{th}} \), \( (r)^{\text{th}} \) and \( (r + 1)^{\text{th}} \) terms
For \( T_{r-1} \), put \( r = r - 2 \)
\( \therefore T_{r-1} = {}^nC_{r-2}a^{r-2} \)
Coefficient of \( T_{r-1} = {}^nC_{r-2} \)
For \( T_r \), put \( r = r - 1 \)
\( \therefore T_r = {}^nC_{r-1}a^{r-1} \)
Coefficient of \( T_r = {}^nC_{r-1} \)
\( T_{r+1} = {}^nC_r a^r \)
Coefficient of \( T_{r+1} = {}^nC_r \)
We are given that,
\( {}^nC_{r-2} : {}^nC_{r-1} : {}^nC_r = 1 : 7 : 42 \)
Consider, \( \frac{{}^nC_{r-2}}{{}^nC_{r-1}} = \frac{1}{7} \)
\( \Rightarrow \frac{\frac{n!}{(r-2)!(n-r+2)!}}{\frac{n!}{(r-1)!(n-r+1)!}} = \frac{1}{7} \)
\( \Rightarrow \frac{(r-1)!(n-r+1)!}{(r-2)!(n-r+2)!} = \frac{1}{7} \)
\( \Rightarrow \frac{r-1}{n-r+2} = \frac{1}{7} \)
\( \Rightarrow 7r - 7 = n - r + 2 \)
\( \Rightarrow 8r - 9 = n \) ………… (1)

Now, consider \( \frac{{}^nC_{r-1}}{{}^nC_r} = \frac{7}{42} = \frac{1}{6} \)
\( \Rightarrow \frac{\frac{n!}{(r-1)!(n-r+1)!}}{\frac{n!}{r!(n-r)!}} = \frac{1}{6} \)
\( \Rightarrow \frac{r!(n-r)!}{(r-1)!(n-r+1)!} = \frac{1}{6} \)
\( \Rightarrow \frac{r}{n-r+1} = \frac{1}{6} \)
\( \Rightarrow 6r = n - r + 1 \)
\( \Rightarrow 7r - 1 = n \) ………… (2)

From (1) and (2), \( 8r - 9 = 7r - 1 \)
\( \Rightarrow r = 8 \), put in eq. (1)
\( \Rightarrow n = 56 - 1 = 55 \) ans.