The availability of step-by-step ISC Class 12 Maths Solutions OP Malhotra Chapter 13 Binomial Theorem Chapter Test can make challenging problems more manageable.
S Chand Class 11 ICSE Maths Solutions Chapter 13 Binomial Theorem Chapter Test
Question 1.
Expand \(\left(\frac{2}{x}-\frac{x}{2}\right)^5\), x ≠ 0.
Solution:
Comparing \(\left(\frac{2}{x}-\frac{x}{2}\right)^5\) with (x + a)n; we have ‘x’ = \(\frac { 2 }{ x }\); ‘a’ = –\(\frac { x }{ 2 }\); n = 5
We know that, Tr+1 = nCr xn – r ar = 5Cr \(\left(\frac{2}{x}\right)^{5-r}\) \(\left(\frac{-x}{2}\right)^r\)
Then by binomial theorem, we have
Question 2.
Using binomial theorem, write the value of (a + b)n + (a – b)n and hence find the value of (√3 + √2)6 – (√3 – √2)6.
Solution:
Using binomial theorem, we have
(a + b)n = nC0 an b0 + nC1 an-1 b1 + nC2 an-2 b2 + … + nCn ao bn …(1)
and (a – b)n = nCo an (-b)0 + nC1 an-1 (-b) + nC2 an-2 b2 + …+ nCn ao (-b)n …(2)
On adding (1) and (2); we have
(a + b)n + (a – b)n = 2 [nCo an b0 + nC2 an-2 b2 + ….]
Again (a + b)n – (a – b)n = 2[nC1 an-1 b + nC3 an-3 b3 + ….] …(3)
Putting a = √3 and b = √2 in eqn. (3); we have
(√3 + √2)6 – ((√3 – √2)6 = 2[6C1(√3)5√2 + 6C3(√3)3 (√2)3 + 6C5 (√3)1 (√2)5]
= 2[6 × 9√3 × √2 + \(\frac{6 \times 5 \times 4}{6}\) × 3√3 × 2√2 + 6 × √3 × 4√2]
= 2[54√6 + 120√6 + 24√6] = 2 × 198√6 = 396√6
Question 3.
Find the 9th term in the expansion of \(\left(3 x-\frac{1}{2 x}\right)^8\), x ≠ 0.
Solution:
On comparing \(\left(3 x-\frac{1}{2 x}\right)^8\) with (x + a)n; we have
‘x’ = 3x; ‘a’ = \(\frac{-1}{2 x}\) and n = 8
We know that Tr+1 = nCr xn-r ar = 8Cr (3x)8-r \(\left(\frac{-1}{2 x}\right)^r\)
⇒ Tr+1 = 8Cr 38-r \(\left(\frac{-1}{2}\right)^r\) x8-2r …(1)
For T9; we put r = 8 in eqn. (1); we have
T9 = 8C8 38-8\(\left(\frac{-1}{2}\right)^8\) x-8 = \(\frac{1}{256}\) \(\frac{1}{x^8}\)
Question 4.
Find the term independent of x in \(\left(2 x^2-\frac{1}{x}\right)^{12}\)
Solution:
On comparing \(\left(2 x^2-\frac{1}{x}\right)^{12}\) with (x + a)n; we have
‘x’ = 2x2 ‘a’ = \(\frac{-1}{x}\); n = 12
We know that, general term = Tr+1 = nCr xn-r ar
i.e. Tr+1 = 12Cr (2x2)12-r \(\left(\frac{-1}{x}\right)^r\)
⇒ Tr+1 = 12Cr 212-r (-1)r x24-3r …(1)
For term independent of x, we put exponent of x in Tr+1 be equal to 0 .
∴ 24 – 3r = 0 ⇒ r = 8
putting r = 8 in eqn. (1); we have
T9 = 12C8 212-8 (-1)8 = \(\frac{12 !}{8 ! 4 !}\) 24 = \(\frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2}\) × 24 = 7920
Question 5.
Find the middle terms in the expansion of \(\left(3-\frac{x^3}{6}\right)^7\).
Solution:
On comparing \(\left(3-\frac{x^3}{6}\right)^7\) with (x + a)n; we have
‘x’ = 3 ; ‘a’ = –\(\frac{x^3}{6}\) and n = 7
Here no. of terms = n + 1 = 8 (even)
So there are two middle terms given by
We know that, general term Tr+1 = nCr xn-r ar
∴ Tr+1 = 7Cr (3)7-r \(\left(-\frac{x^3}{6}\right)^r\) …(1)
For T4 : putting r = 3 in eqn. (1); we have
T4 = 7C3 37-3 \(\left(-\frac{x^3}{6}\right)^3\) = \(\frac{7 !}{4 ! 3 !}\) × 34 × \(\frac{(-1)^3}{6^3}\) x9
For T5 : putting r = 4 in eqn. (1); we have
T5 = 7C4 (3)7-4 \(\left(-\frac{x^3}{6}\right)^4\) = \(\frac{7 !}{4 ! 3 !}\) × 33 × (-1)4 \(\frac{x^{12}}{6^4}\) = \(\frac{7 \times 6 \times 5}{6}\) × 27 × \(\frac{x^{12}}{36 \times 36}\) = \(\frac{35}{48}\) x12
Question 6.
(i) In the binomial expansion of (1 + a)m+n, prove that the coefficients of am and an are equal.
(ii) Prove that the coefficients of xn in (1 + x)2n is twice the coefficient of xn in (1 + x)2n – 1.
Solution:
On comparing (1 + a)m+n with (x + a)n; we have
‘x’ = 1; ‘a’ = a; ‘n’ = m + n
We know that, general term Tr+1 = nCr xn-r ar
∴ Tr+1 = m+nCr 1m+n-r ar = m+nCr ar …(1)
Let Tr+1 containing term involving am.
Putting r = m in eqn. (1); we have
Tm+1 = m+nCm am
∴ coeff. of am = m+nCm
Let Tr+1 containing term involving an.
So putting r = n in eqn. (1); we have
Tn+1 = m+nCn an
∴ coeff. of an = m+nCn = m+nCm
[∵ nCr = nCn-r]
Thus coefficients of am and an in the expansion of (1 + a)m+n are equal.
(ii) On comparing (1 + 2x)2n with (x + a)n; we have
‘x’ = 1; ‘a’ = x; ‘n’ = 2n
We know that, general term Tr+1 = nCr xn-r ar ⇒ Tr+1 = 2nCr 12n-r xr = 2nCr xr …(1)
Let Tr+1 containing the term involving xn.
we put r = n in eqn. (1); we have Tn+1 = 2nCn xn
∴ Coeff. of xn in (1 + x)2n-1 = 2n-1Cn
Hence the coefficient of xn in (1 + x)2n is twice the coefficient of xn in (1 + x)2n-1.
Question 7.
Use binomial theorem to evaluate (10.1)5.
Solution:
(10.1)5 = (10 + 0.1)5
= 5C0 (0.1)0 + 5C1 104 (0.1) + 5C2 103 (0.1)3 + 5C4 101 (0.1)4 + 5C5 100 (0.1)5 [using binomial theorem]
= 100000 × 1 + 5 × 10000 × 0.1 + 10 × 100 × 0.01 + 10 × 100 × \(\frac{1}{1000}\) + 5 × 10 × \(\frac{1}{10000}\) + \(\frac{1}{100000}\)
= 100000 + 5000 + 100 + 1 + 0.005 + 0.00001 = 105101.00501
Question 8.
Examine whether or not there is any term containing x9 in the expansion of \(\left(2 x^2-\frac{1}{x}\right)^{20}\).
Solution:
On comparing \(\left(2 x^2-\frac{1}{x}\right)^{20}\) with (x + a)n; we have
‘x’ = 2x2; ‘a’ = –\(\frac{1}{x}\); n = 20
We know that, general term = Tr+1 = nCr xn-r ar
⇒ Tr+1 = 20Cr (2x2)20-r \(\left(-\frac{1}{x}\right)^r\) = 20Cr 220-r (-1)r x40-3r …(1)
Let Tr+1 containing term involving x9.
So we put, 40 – 3r = 9 ⇒ 3r = 31 ⇒ r = \(\frac{31}{3}\) ∉ N
Hence there is no term containing x9 in the expansion of \(\left(2 x^2-\frac{1}{x}\right)^{20}\).
Question 9.
In the binomial expansion of (a – b)n, n ≥ 5, the sum of 5th and 6th terms is zero, then \(\frac{a}{b}\) equals
(a) \(\frac{n-5}{6}\)
(b) \(\frac{n-4}{5}\)
(c) \(\frac{5}{n-4}\)
(d) \(\frac{6}{n-5}\)
Solution:
On comparing (a – b)n with (x + a)n; we have
‘x’ = a; ‘a’ = -b; n = n
We know that, general term = Tr+1 = nCr xn-r ar
⇒ Tr+1 = nCr an-r(- b)r …(1)
For T5; Putting r = 4 in eqn. (1); we have
T5 = nC4 an-4 (-b)4
For T6; Putting r = 5 in eqn. (1); we have
T6 = nC5 an-5 (-b)5
Since it is given that the sum of 5th and 6th terms is 0.
Question 10.
If m = nC4, then mC2 is equal to
(a) 3 . nC2
(b) n+1C4
(c) 3 . n+1C4
(d) 3 . n+1C3
Solution:
Question 11.
If the coeff. of rth and (r + 4)th terms are equal in the expansion of (1 + x)20, then the value of r will be
(a) 7
(b) 8
(c) 9
(d) 10
Solution:
On comparing (1 + x)20 with (x + a)n; we have
‘x’ = 1; ‘a’ = x; n = 20
We know that, general term = Tr+1 = nCr xn-r ar
∴ Tr+1 = 20Cr 120-r xr = 20Cr xr …(1)
For Tr; replacing r by r – 1 in eqn. (1); we have
Tr = 20Cr-1 xr-1
For Tr+4; replacing r by r + 3 in eqn. (1) ; we have
Tr+4 = 20Cr+3 xr+3
Given coefficient of rth and (r + 4) terms in the expansion of (1 + x)20 are equal.
Thus 20Cr-1 = 20Cr+3
⇒ r – 1 + r + 3 = 20 ⇒ 2r + 2 = 20
⇒ r = 9
[∵ nCr = nCs ⇒ r = s or r + s = n]
∴ Ans. (c)
Question 12.
If (1 + x – 2x2)6 = 1 + a1x + a2x2 + …. + a12x12, then find a2 + a4 + …. + a12.
Solution:
Given (1 + x – 2x2)6 = 1 + a1x + a2x2 + a3x3 + …. + a12x12 …(1)
put x = 1 in eqn. (1); we get
0 = (1 + 1 – 2)6 = 1 + a1 + a2 + a3 + a4 + …. + a12 …(2)
putting x = -1 in eqn. (1); we have
64 = (1 – 1 – 2)6 = 1 – a1 + a2 – a3 + …… + a12 …(3)
On adding eqn. (2) and eqn. (3) ; we have
64 = 2 + 2a2 + 2a4 + ….. + 2a12
⇒ 62 = 2 (a2 + a4 + ….. + a12)
⇒ a2 + a4 + ……+ a12 = \(\frac{62}{2}\) = 31
Question 13.
If the coefficients of x2 and x3 in the expansion of (3 + ax)9 be same, then the value of ‘a’ is
(a) \(\frac{3}{7}\)
(b) \(\frac{7}{3}\)
(c) \(\frac{7}{9}\)
(d) \(\frac{9}{7}\)
Solution:
On comparing (3 + ax)9 with (x + a)n; we have
‘x’ = 3; ‘a’ = ax and n = 9
We know that, General Term = Tr+1 = nCr xn-r ar
∴ Tr+1 = 9Cr 39-r (ax)r = 9Cr 39-r ar xr …(1)
For term containing x2, we put r = 2 in eqn. (1); T3 = 9C2 37 a2 x2
∴ Coeff. of x2 = 9C2 37 . a2
For term containing x3, we put r = 3 in eqn. (1) ; T4 = 9C3 36 a3 x3
∴ Coeff. of x3 = 9C3 36 a3
According to given condition, we have
When a = 0 the given expansion no longer binomial.
∴ a = \(\frac{9}{7}\)
∴ Ans. (d)
Question 14.
Using binomial theorem, the value of (0.999)3 correct to 3 decimal places is
(a) 0.999
(b) 0.998
(c) 0.997
(d) 0.995
Solution:
(0.999)3 = (1 – 0.001)3
using binomial theorem, we have
= [3C0 13(- 0.001)0 +3C1 12(- 0.001) + 3C2 11(- 0.001)2 + 3C3 10(- 0.001)3]
= [1 + 3 \times(- 0.001) + 3 \times(0.000001) – 0.000000001]
= 1 – 0.003 + 0.000003 – 0.000000001
= 0.997
∴ Ans. (c)