Math Ncert Exemplar 2019 Solutions for Class 11 Science Maths Chapter 8 - Binomial Theorem

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Page No 142:

Question 1:

Find the term independent of x, x ≠ 0, in the expansion of ${(\frac{3 x^{2}}{2} - \frac{1}{3 x})}^{15}$ .

Answer:

General term in expansion of (x – a)ⁿ is T_r+1 = ⁿCr (x)^n–r (–a)^r
The term in dependent of x is obtained by substituting n – r = 0

T^r+1 = ¹⁵C_r ${(\frac{3 x^{2}}{2})}^{15 - r} {(\frac{- 1}{3 x})}^{r}$

         $= 15 C_{r} 3^{15 - r} x^{30 - 3 r} 2^{r - 15} {(- 1)}^{r} 3^{- r} x^{- r} = 15 C_{r} 3^{15 - r} x^{30 - 3 r} 2^{r - 15} {(- 1)}^{r} 3^{- r} x^{- r}$
So, for independent term of x,
30 – 3r = 0
⇒ 3r = 30
⇒ r = 10
So, T^r+1= ¹⁵C₁₀(–1)¹⁰ 3^15–20 2^10–15
              = ¹⁵C₁₀3^–5 2^–5
              = ¹⁵C₁₀6^–5
              = ¹⁵C_{10 ${(\frac{1}{6})}^{5}$}â€‹

Page No 142:

Question 2:

If the term free from x in the expansion of ${(\sqrt{x} - \frac{k}{x^{2}})}^{10}$ is 405, find the value of k.

Answer:

Let T^r+1 be the general term,
Then,
T^r+1 = ¹⁰C_r ${(\sqrt{x})}^{10 - r} {(\frac{- k}{x^{2}})}^{r}$
$=^{10} C_{r} {(x)}^{\frac{1}{2} (10 - r)} {(- k)}^{r} . x^{- 2 r} =^{10} C_{r} x^{5 - \frac{r}{2}} {(- k)}^{r} . x^{- 2 r} =^{10} C_{r} x^{5 - \frac{r}{2} - 20} {(- k)}^{r} =^{10} C_{r} x^{\frac{10 - 5 r}{2}} {(- k)}^{r}$

For the term from x,

$\frac{10 - 5 r}{2} = 0$

⇒ 10 – 5r = 0
⇒ r = 2
Since, T²⁺¹ = T³ is free foam x,
∴ T²⁺¹ = ¹⁰C_r (–k)²
= 405

$\Rightarrow \frac{10 \times 9 \times 8!}{2! \times 8!} {(- k)}^{2} = 405$ â€‹

⇒ 45k² = 405
⇒ k = ± 3

Page No 142:

Question 3:

Find the coefficient of x in the expansion of (1 – 3x + 7x²) (1 – x)¹⁶.

Answer:

(1 – 3x + 7x²) (1 – x)¹⁶
= (1 – 3x + 7x²) (¹⁶C₀ 1¹⁶ – ¹⁶C₁ 1¹⁵x¹ + ... + ¹⁶C₁₆x¹⁶ )
= (1 – 3x + 7x²) (1 – 16x + 120 x² + ...)
∴ Coefficient of x = –3 – 16

= –19

Page No 142:

Question 4:

Find the term independent of x in the expansion of, ${(3 x - \frac{2}{x^{2}})}^{15}$ .

Answer:

Let T_{r + 1} be the general term

Therefore,

$T_{r + 1} = {}^{15}C_{r} {(3 x)}^{15 - r} {(\frac{- 2}{x^{2}})}^{r} = {}^{15}C_{r} {(3 x)}^{15 - r} {(- 2)}^{r} x^{- 2 r} = {}^{15}C_{r} 3^{15 - r} x^{15 - 3 r} {(- 2)}^{r}$

For the term in dependent of x,

15 – 3r = 0

⇒ r = 5

So, T_{5 + 1} is independent of x

⇒ T_{5 + 1} = ¹⁵C₅ 3^{15 – 5} (–2)⁵

$= \frac{- 15 \times 14 \times 13 \times 12 \times 11 \times 10!}{5 \times 4 \times 3 \times 2 \times 1 \times 10!} \times 3^{10} \times 2^{5} = - 3003 \times 3^{10} \times 2^{5}$

Page No 142:

Question 5:

Find the middle term (terms) in the expansion of

(i) ${(\frac{x}{a} - \frac{a}{x})}^{10}$

(ii) ${(3 x - \frac{x^{3}}{6})}^{9}$

Answer:

(i) The power of binomial = 10 (even)

middle term =

\frac{10}{2} + 1

= 6^th term

\begin{array}{rcl} \Rightarrow & T_{6} = T_{5 + 1} \\ = & {}^{10}C_{5} {(\frac{x}{a})}^{10 - 5} {(\frac{- a}{x})}^{5} \\ = & - {}^{10}C_{5} {(\frac{x}{a})}^{5} {(\frac{a}{x})}^{5} \\ = & \frac{- 10 \times 9 \times 8 \times 7 \times 6 \times 5!}{5! \times 5 \times 4 \times 3 \times 2 \times 1} \\ = & - 9 \times 4 \times 7 \\ = & - 252 \end{array}

(ii) Power of binomial = 9 (odd)

Two middle terms =

\frac{9 + 1}{2}, \frac{9 + 1}{2} + 1

\begin{array}{rcl} \Rightarrow & T_{5} = T_{4 + 1} 5^{th}, 6^{th} \\ = & {}^{9}C_{4} {(3 x)}^{9 - 4} {(- \frac{x^{3}}{6})}^{4} \\ = & \frac{9 \times 8 \times 7 \times 6 \times 5!}{4 \times 3 \times 2 \times 1 \times 5!} 3^{5} x^{5} x^{12} 6^{- 4} \\ = & \frac{7 \times 6 \times 3 \times 3}{2^{4}} x^{17} \\ = & \frac{189}{8} x^{17} \end{array} \begin{array}{rcl} \Rightarrow & T_{6} = T_{5 + 1} \\ = & {}^{9}C_{5} {(3 x)}^{9 - 5} {(\frac{- x^{3}}{6})}^{5} \\ = & \frac{- 9 \times 8 \times 7 \times 6 \times 5!}{5! \times 4 \times 3 \times 2 \times 1} 3^{4} x^{4} x^{15} 6^{- 5} \\ = & \frac{- 21 \times 6}{3 \times 2^{5}} x^{19} \\ = & \frac{- 21}{16} x^{15} \end{array}

Page No 143:

Question 6:

Find the coefficient of x¹⁵ in the expansion of (x – x²)¹⁰.

Answer:

Let the term T_{r + 1} be the general term

$\begin{array}{rcl} \Rightarrow & T_{r + 1} = {}^{10}C_{r} x^{10 - r} {(- x^{2})}^{r} \\ = & {(- 1)}^{r} {}^{10}C_{r} x^{10 - r} x^{2 r} \\ = & {(- 1)}^{r} {}^{10}C_{r} x^{10 + r} \end{array}$

For the coefficient of x¹⁵,

10 + r = 15

⇒ r = 5

$\begin{array}{rcl} So, T_{5 + 1} & = & {(- 1)}^{5} {}^{10}C_{5} x^{15} \\ = & - 1 \times \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5!}{5 \times 4 \times 3 \times 2 \times 1 \times 5!} \\ = & - 3 \times 2 \times 7 \times 6 \\ = & - 252 \end{array}$

Page No 143:

Question 7:

Find the coefficient of $\frac{1}{x^{17}}$ in the expansion of ${(x^{4} - \frac{1}{x^{3}})}^{15}$ .

Answer:

Let the term T_{r + 1} contain the coefficient of $\frac{1}{x^{17}}$

$Then, T_{r + 1} = {}^{15}C_{r} {(x^{4})}^{15 - r} {(\frac{- 1}{x^{3}})}^{r} = {}^{15}C_{r} x^{60 - 4 r} {(- 1)}^{r} x^{- 3 r} = {}^{15}C_{r} x^{60 - 7 r} {(- 1)}^{r}$

For the coefficient x^–17,

60 – 7r = –17

⇒ 7r = 77

⇒ r = 11

$\begin{array}{rcl} \Rightarrow & T_{11 + 1} = {}^{15}C_{11} x^{60 - 77} {(- 1)}^{11} \\ = & \frac{- 15 \times 14 \times 13 \times 12 \times 11!}{11! \times 4 \times 3 \times 2 \times 1} \\ = & - 15 \times 7 \times 13 \\ = & - 1365 \end{array}$

Page No 143:

Question 8:

Find the sixth term of the expansion ${(y^{\frac{1}{2}} + x^{\frac{1}{3}})}^{n}$ , if the binomial coefficient of the third term from the end is 45.

Answer:

6^th term in the expansion is,
T₆ = T₅₊₁
     = ⁿC₅ ${(y^{\frac{1}{2}})}^{n - 5} {(x^{\frac{1}{3}})}^{5}$

Binomial coefficient of the third term from the end = Binomial coefficient of third term from the beginning
= ⁿC₂
⇒ ⁿC₂ = 45

$\Rightarrow \frac{n (n - 1) (n - 2)!}{2! (n - 2)!} = 45$

⇒ n(n – 1) = 90
⇒ n² – n – 90 = 0
⇒  n² – 10n + 9n – 90 = 0
⇒ n(n – 10) + 9(n – 10) = 0
⇒ (n + 9) (n – 10) = 0

⇒ T₆ = ¹⁰C₅ $y^{\frac{5}{2}} x^{\frac{5}{3}}$ â€‹

          = 252 $y^{\frac{5}{2}} x^{\frac{5}{3}}$ â€‹

Page No 143:

Question 9:

Find the value of r, if the coefficients of (2r + 4)^th and (r – 2)^thterms in the expansion of (1 + x)¹⁸ are equal.

Answer:

(2r + 4)^th term = T_{2r+3 + 1}
                        = 18_C2r+3 (1)^{18–2r –3}(x)^2r+3
= 18_C2r+3x^2r+3
Also, (r – 2)th term = Tr–3+1
                                = 18_C2r+3(1)
Now,
18_C2r+3= 18_Cr–3       $[{}^{n}C_{x} = {}^{n}C_{y} \Rightarrow x + y = n]$

⇒ 2r + 3 + r – 3 = 18
⇒ 3r = 18
⇒ r = 6

Page No 143:

Question 10:

If the coefficient of second, third and fourth terms in the expansion of (1 + x)²ⁿare in A.P. Show that 2n² – 9n + 7 = 0.

Answer:

Coefficient of 2nd, 3rd and 4th terms in the expansion of (1 + x)2n are in AP.
So, coefficient of 2nd term = ²ⁿC₁
Coefficient of 3rd term = ²ⁿC₂
Coefficient of 4th term = ²ⁿC₃
Then, 2 ²ⁿC₂ = ²ⁿC₁ + ²ⁿC₃

$\Rightarrow 2 [\frac{2 n (2 n - 1) (2 n - 2)!}{2 \times 1 \times (2 n - 2)!}] = \frac{2 n (2 n - 1)!}{(2 n - 1)!} + \frac{2 n (2 n - 1) (2 n - 2) (2 n - 3)!}{3! (2 n - 3)!}$

$\Rightarrow n (2 n - 1) = \frac{n + n (2 n - 1) (2 n - 2)}{6}$

⇒ n(12n – 6) = n(6 + 4n² – 4n – 2n + 2)
⇒ 12n – 6 = 4n² – 6n + 8
⇒ 6(2n – 1) = 2(2n² – 3n + 4)
⇒ 3(2n – 1) = 2n2 – 3n + 4
⇒ 2n² – 3n + 4 – 6n + 3 = 0
⇒ 2n² – 9n + 7 = 0

Page No 143:

Question 11:

Find the coefficient of x⁴ in the expansion of (1 + x + x² + x³)¹¹.

Answer:

(1 + x + x² + x³)¹¹
= [(1 + x) + x² (1 + x)]¹¹
= [(1 + x) (1 + x²)]¹¹
= ¹¹C₀ + ¹¹C₁x + ¹¹C₂x + . . . ) (¹¹C₀ + ¹¹C₁x² + ¹¹C₂x⁴ + . . . )
= (1 + 11x + 55x² + 165x³ + 330x⁴ + . . . ) (1 + 11x² + 55x⁴ + . . . )
So, coefficient of x⁴ = 55 + 605 + 33330
= 990

Page No 143:

Question 12:

If p is a real number and if the middle term in the expansion of ${(\frac{p}{2} + 2)}^{8}$ is 1120, find p.

Answer:

Here, n = 8, (even)
So, the binomial expansion has only one middle term i.e., ${(\frac{8}{2} + 1)}^{t h} = 5^{th} term$
⇒ T₅ = T₄₊₁
= ⁸C₄ ${(\frac{P}{2})}^{8 - 4} . 2^{4}$
⇒ 1120 = ⁸C₄ P⁴ 2^–4 2⁴

⇒ 1120 = $\frac{8 \times 7 \times 6 \times 5 \times 4!}{4! \times 4 \times 3 \times 2 \times 1} P^{3}$

⇒ P⁴ = $\frac{1120}{70}$
= 16
⇒ P = ± 2.

Page No 143:

Question 13:

Show that the middle term in the expansion of ${(x - \frac{1}{x})}^{2 n} is \frac{1 \times 3 \times 5 \times . . . (2 n - 1)}{n} \times {(- 2)}^{n}$ .

Answer:

Here, the binomial has even power. So, it will have one middle term.

i.e. ${(\frac{2 n}{2} + 1)}^{t h} term = {(n + 1)}^{th} term$

T_n+1 = ${}^{2 n}C_{n} {(x)}^{2 n - n} {(\frac{- 1}{x})}^{n}$
          = ²ⁿC_nxⁿ (–1)ⁿx^–n
          = ²ⁿC_n (–1)ⁿ
           $= {(- 1)}^{n} (\frac{2 n!}{n! n!}) = \frac{1 \times 2 \times 3 \times 4 \times 5 \times . . . \times (2 n - 1) (2 n)}{n! n!} {(- 1)}^{n} = \frac{1 \times 3 \times 5 \times . . . \times (2 n - 1) 2 \times 4 \times 6 \times . . . \times (2 n)}{1 \times 2 \times 3 \times . . . \times n \times n!} {(- 1)}^{n} = \frac{1 \times 3 \times 5 \times (2 n - 1) \times 2^{n} \times (1 \times 2 \times 3 . . . \times n)}{(1 \times 2 \times 3 \times . . . \times n) n!} {(- 1)}^{n} = \frac{[1 \times 3 \times 3 \times . . . \times [2 n - 1]]}{n!} {(- 2)}^{n}$
Hence, proved.

Page No 143:

Question 14:

Find n in the binomial ${(\sqrt[3]{2} + \frac{1}{\sqrt[3]{3}})}^{n}$ if the ratio of 7^th term from the beginning to the 7^th term from the end is $\frac{1}{6}$ .

Answer:

7th term from the beginning = T₁
= T₆₊₁
= ⁿC₆ ${(3 \sqrt{2})}^{n - 6} {(\frac{1}{3 \sqrt{3}})}^{6}$

7th term from the end = 7th term from the beginning of ${(\frac{1}{3 \sqrt{3}} + 3 \sqrt{2})}^{n}$
$\Rightarrow T_{7} = T_{6 +!} = {}^{n}C_{6} {(3 \sqrt{3})}^{n - 6} {(\frac{1}{3 \sqrt{3}})}^{6}$
Now,

$\frac{{}^{n}C_{6} {(3 \sqrt{2})}^{n - 6} {(\frac{1}{3 \sqrt{3}})}^{6}}{{}^{n}C_{6} {(\frac{1}{3 \sqrt{3}})}^{n - 6} {(3 \sqrt{2})}^{6}} = \frac{1}{6}$

$\Rightarrow \frac{2^{\frac{n - 6}{3}} . 3^{\frac{- 6}{3}}}{3^{- (\frac{n - 6}{3})} . 2^{\frac{6}{3}}} = \frac{1}{6} \Rightarrow (2^{\frac{n - 6}{3}} . 2^{\frac{- 6}{3}}) (3^{\frac{- 6}{3}} . 3^{\frac{(n - 6)}{3}}) = 6^{- 1} \Rightarrow \frac{n}{3} - 4 = - 1 \Rightarrow n = 9$

Page No 143:

Question 15:

In the expansion of (x + a)ⁿ if the sum of odd terms is denoted by O and the sum of even term by E.
Then prove that
(i) O²– E² = (x² – a²)ⁿ
(ii) 4OE = (x + a)²ⁿ – (x – a)²ⁿ

Answer:

(i) (x + a)ⁿ = ${}^{n}C_{0} x {}^{n}a^{0} + {}^{n}C_{1} x^{n - 1} a^{1} + {}^{n}C_{2} x^{n - 2} a^{2} + . . . + {}^{n}C_{n} a^{n}$

Sum of odd terms = O
                              = ${}^{n}C_{0} x^{n} + {}^{n}C_{2} x^{n - 2} a^{2} + . . .$

Sum of even terms = E
                               = ${}^{n}C_{1} x^{n - 1} a + {}^{n}C_{3} x^{n - 3} a^{3} + . . .$
Now, (x + a)ⁿ = O + E          . . .(i)
and, (x – a)ⁿ = O – E           . . . (ii)
∴ (O + E) (O – E) = (x + a)ⁿ . (x – a)ⁿ
⇒ O² – E² = (x + a)ⁿ (x – a)ⁿ
                   = (x² – a²)

(ii) 40 E = (O + E)² – (O – E)²
              = [(x + a)ⁿ]² – [(x – a)ⁿ]²
              = (x + a)²ⁿ – (x – a)²ⁿ
Hence proved.

Page No 144:

Question 16:

16. If x^p occurs in the expansion of ${(x^{2} + \frac{1}{x})}^{2 n}$ , prove that its coefficient is $\frac{2 n!}{(\frac{4 n - p}{3})! (\frac{2 n + p}{3})!}$ .

Answer:

Let xp occur in the expansion of ${(x^{2} + \frac{1}{x})}^{2 n}$
T^r+1 = ²ⁿC_r (x²)2^n–r ${(\frac{1}{x})}^{r}$
         = ²ⁿC_rx4^n–2r^x^–r
         = ²ⁿC_rx4^n–3r
Let 4n – 3r = p
⇒ 3r = 4n – p

⇒ r = $\frac{4 n - p}{3}$

∴ coefficient of xp = ²ⁿC_r

â€‹
                               $= \frac{(2 n)!}{r! (2 n - r)!} = \frac{(2 n)!}{(\frac{4 n - p}{3})! (2 n - \frac{4 n - p}{3})!} = \frac{(2 n)!}{(\frac{4 n - p}{3})! (\frac{6 n - 4 n + p}{3})!} = \frac{(2 n)!}{(\frac{4 n - p}{3})! (\frac{2 n + p}{3})!}$

Page No 144:

Question 17:

17. Find the term independent of x in the expansion of (1 + x + 2x³) ${(\frac{3}{2} x^{2} - \frac{1}{3 x})}^{9}$ .

Answer:

Consider ${(\frac{3}{2} x^{2} - \frac{1}{3 x})}^{9}$

T^r+1 = ⁹Cr ${(\frac{3}{2} x^{2})}^{9 - r} {(- \frac{1}{3 x})}^{r}$

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

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

S, the general term of the expansion

$= {}^{9}C_{r} {(\frac{3}{2})}^{9 - r} {(\frac{- 1}{3})}^{r} x^{18 - 3 r} + {}^{9}C_{r} {(\frac{3}{2})}^{9 - r} {(\frac{- 1}{3})}^{r} x^{19 - 3 r} + 2 = {}^{9}C_{r} {(\frac{3}{2})}^{9 - r} {(\frac{- 1}{3})}^{r} x^{21 - 3 r}$

For the term independent of x,
18 – 3r = 0, 19 – 3r = 0 & 21 – 3r = 0

⇒ r = 6, $\frac{19}{3}, 7$

Since, the possible values of r are 6 & 7, hence, the second term is not independent of x

$= {}^{9}C_{6} {\frac{3}{2}}^{(9 - 6)} {(\frac{- 1}{3})}^{6} + 2 . {}^{9}C_{7} {\frac{3}{2}}^{(9 - 7)} {(\frac{- 1}{3})}^{7} = \frac{9 \times 8 \times 7 \times 6!}{6! \times 3 \times 2} . \frac{3^{3}}{2^{3}} . \frac{1}{3^{6}} - 2 . \frac{9 \times 8 \times 7!}{7! \times 2 \times 1} . \frac{3^{2}}{2^{2}} . \frac{1}{3^{7}} = \frac{8^{4}}{8} . \frac{1}{3^{3}} - \frac{36}{4} . \frac{2}{3^{5}} = \frac{7}{18} - \frac{2}{27} = \frac{17}{54}$

Page No 144:

Question 18:

Choose the correct answer from the given options:
The total number of terms in the expansion of (x + a)¹⁰⁰+ (x – a)¹⁰⁰after simplification is
(A) 50
(B) 202
(C) 51
(D) none of these

Answer:

Total number of terms in the expansion of (x + a)100 – (x – a)100 are 101.
Out of 101 terms, so terms are cancelled out due to same magnitude and opposite sign.
Total terms = 101 – 50
= 51
Hence, the correct answer is option C.

Page No 144:

Question 19:

Choose the correct answer from the given options:
Given the integers r > 1, n > 2, and coefficients of (3r)^th and (r + 2)^nd terms in the binomial expansion of (1 + x)²ⁿ are equal, then
(A) n = 2r
(B) n = 3r
(C) n = 2r + 1
(D) none of these

Answer:

Given that r > 1, n > 2 and the coefficients of (3r)^th and (r + 2)^th term are equal in the expansion of (1 + x)²ⁿ
Then,
T_3r = T_{3r –1 + 1}
      = ²ⁿC_3r–1x^3r–1
And,
T_r+2 = T_{r+1 +1}
        = ²ⁿC_r+1
Given,
²ⁿC_3r–1 = ²ⁿC_r+1        [âˆµ ⁿC_x = ⁿC_y ⇒ x + y = n]
⇒ 3_{r + –1 + r + 1} = 2n
⇒ 4r = 2n
⇒ n = 2r
Hence, the correct answer is option A.

Page No 144:

Question 20:

Choose the correct answer from the given options:
The two successive terms in the expansion of (1 + x)²⁴ whose coefficients are in the ratio 1:4 are
(A) 3^rd and 4^th
(B) 4^th and 5^th
(C) 5^th and 6^th
(D) 6^th and 7^th

Answer:

Let two successive terms in the expansion of (1 + x)²⁴ be (r + 1)^th term and (r + 2)^th term.
∴ T_r+1 = ²⁴C_rx^r

and T_r+2= ²⁴C_{r + 1}x^r⁺¹
Now,

$\frac{{}^{24}C_{r}}{{}^{24}C_{r + 1}} = \frac{1}{4}$

$\Rightarrow \frac{\frac{24!}{r! (24 - r)!}}{\frac{(24)!}{(r + 1)! (24 - r - 1)!}} = \frac{1}{4}$

$\Rightarrow \frac{{(r + 1)}^{r}! (23 - r)!}{r! (24 - r) (23 - r)!} = \frac{1}{4} \Rightarrow \frac{r + 1}{24 - r} = \frac{1}{4}$

⇒ 4r + 4 = 24 – r
⇒ 5r =20
⇒ r =4
⇒ T₄₊₁ = T₅ and T₄₊₂ = T₆
Hence, the correct answer is option C.

Page No 144:

Question 21:

Choose the correct answer from the given options:
The coefficient of xⁿin the expansion of (1 + x)²ⁿ and (1 + x)^{2n – 1} are in the ratio.
(A) 1 : 2
(B) 1 : 3
(C) 3 : 1
(D) 2 : 1

Answer:

Since, coefficient of xn in the expansion of (1 + x)2n = 2nCn
And coefficient of x4 in the expansion of (1 + x)2n–1cn

$∴ \frac{{}^{2 n}C_{r}}{{}^{2 n - 1}C_{n}} = \frac{\frac{(2 n)!}{n! n!}}{\frac{(2 n - 1)!}{n! (n - 1)!}}$

$= \frac{(2 n)! n! (n - 1)!}{n! n! (2 n - 1)!} = \frac{2 n (2 n - 1)! n 1 (n - 1)!}{n! n (n - 1)! (2 n - 1)!} = \frac{2 n}{n} = 2 : 1$

Hence, the correct answer is option D.

Page No 144:

Question 22:

Choose the correct answer from the given options:
If the coefficients of 2^nd, 3^rd and the 4^th terms in the expansion of (1 + x)ⁿare in A.P., then value of n is
(A) 2
(B) 7
(c) 11
(D) 14

Answer:

The expansion of (1 + x)ⁿ is ⁿC₀ + ⁿC₁x + ⁿC₂x2 + ⁿC_{3 x3 + . . . +}ⁿC_nxⁿ.
∴ Coefficient of 2nd term = ⁿC₁
Coefficient of 3rd term = ⁿC₂
Coefficient of 2nd term = ⁿC₃
Given that, â€‹ⁿC₁,ⁿC₂and ⁿC₃are in AP.

$\Rightarrow 2 [\frac{n!}{(n - 2)! 2!}] = \frac{n!}{(n - 1)!} + \frac{n!}{! (n - 3)!} \Rightarrow \frac{2 . n (n - 1) (n - 2)!}{(n - 2)! 2!} = \frac{n (n - 1)!}{(n - 1)!} + \frac{n (n - 1) (n - 2) (n - 3!)}{3 \times 2 \times 1 \times (n - 3)} \Rightarrow n (n - 1) = n + \frac{n (n - 1) (n - 2)}{6}$

⇒ 6n – 6 = 6 + n² – 3n + 2
⇒ n² – 9 n + 14 = 0
⇒ n(n – 7) – 2(n –7) = 0
⇒ (n – 7) (n – 2) = 0
⇒ n = 2 or n = 7
Since, n = 2 is not possible, so n = 7.
Hence, the correct answer is option B.

Page No 145:

Question 23:

Choose the correct answer from the given options:
If A and B are coefficient of xⁿin the expansions of (1 + x)²ⁿand (1 + x)^{2n – 1} respectively, then $\frac{A}{B}$ equals
(A) 1

(B) 2

(C) $\frac{1}{2}$

(D) $\frac{1}{n}$

Answer:

Since, the coefficient of xn in the expansion of (1 + x)²ⁿ is ²ⁿC_n, therefore
A = ²ⁿC_n
The coefficient of xn in the expansion of (1 + x)2^n–1 is ^2n–1C_n.
⇒ B = ^2n–1C_n
So,
$\frac{A}{B} = \frac{{}^{2 n}C_{n}}{{}^{2 n - 1}C_{n}} = \frac{2}{1} = 2$

Hence, the correct answer is option B.

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Question 24:

Choose the correct answer from the given options:
If the middle term of ${(\frac{1}{x} + x \sin x)}^{10} is equal to 7 \frac{7}{8}$ , then value of x is
(A) $2 n π + \frac{π}{6}$

(B) $n π + \frac{π}{6}$

(C) $n π + {(- 1)}^{n} \frac{π}{6}$

(D) $n π + {(- 1)}^{n} \frac{π}{3}$

Answer:

In the given expansion, n = 10 (even).
So, it has only one middle term
i.e., 6th term.
$\Rightarrow T_{6} = T_{5 +!}$

         $= {}^{10}C_{5} {(\frac{1}{x})}^{10 - 5} {(x \sin x)}^{5} \Rightarrow \frac{63}{8} = {}^{10}C_{5} x^{- 5} x^{5} \sin^{5 x} \Rightarrow \frac{63}{8} = \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5!}{5 \times 4 \times 3 \times 2 \times 1 \times 5!} \sin^{5} x$

         $\Rightarrow \frac{63}{8} = 2 \times 9 \times 2 \times 7 \times \sin^{5} x$

         $\Rightarrow \sin^{5} x = \frac{1}{32} \Rightarrow \sin^{2} = \frac{1}{2}$

           $\Rightarrow x = n π + {(- 1)}^{3} \frac{π}{6}$

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Question 25:

The largest coefficient in the expansion of (1 + x)³⁰ is _____________ .

Answer:

The largest coefficient in the expansion of (1 + x)³⁰ = ³⁰C $\frac{30}{2}$
= ³⁰C₁₅

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Question 26:

The number of terms in the expansion of (x + y + z)ⁿ ______________ .

Answer:

(x + y + z)ⁿ = [x + (y + z)]ⁿ
= ⁿC₀xⁿ + ⁿC₁x^n–¹ (y + 2) + ⁿC₂x^n–² (y + z)² + . . . + ⁿC_n (y + z)ⁿ
∴ Number of terms = 1 + 2 + 3 + . . . + n + (n + 1)

= $\frac{(n + 1) (n + 2)}{2}$

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Question 27:

In the expansion of ${(x^{2} - \frac{1}{x^{2}})}^{16}$ , the value of constant term is _____________ .

Answer:

Let constant be T_r+1
T_r+1= ¹⁶C_r(x²)16^–^r ${(\frac{1}{- x^{2}})}^{r}$
= ¹⁶C_rx^{32 – 20}(–1)^rx^–2^r
= â€‹¹⁶C_rx32–4r (–1)^r
For constant term, 32–4r = 0
⇒ r = 8
∴ T₈₊₁= ¹⁶C₈â€‹

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Question 28:

If the seventh terms from the beginning and the end in the expansion of ${(\sqrt[3]{2} + \frac{1}{\sqrt[3]{3}})}^{n}$ are equal, then n equals _____________ .

Answer:

T₇ = T₆₊₁ = ⁿC₆ $(3 \sqrt{2}) n - 6 {(\frac{1}{3 \sqrt{3}})}^{6}$
T₇ form the end = T₇ from the beginning of ${(\frac{1}{3 \sqrt{3}} + 3 \sqrt{2})}^{n}$
Then, T₇ = ${}^{n}C_{6} {(\frac{1}{3 \sqrt{3}})}^{n - 6} {(3 \sqrt{2})}^{6}$
Now,
${}^{n}C_{6} {(2)}^{\frac{1 - 6}{3}} {(3)}^{\frac{- 6}{3}} = {}^{n}C_{6} {(3)}^{\frac{- (n - 6)}{3}} 2^{\frac{6}{3}}$

$\Rightarrow 2^{\frac{n - 12}{3}} = (\frac{1}{3^{\frac{1}{3}}})$

This is true when $\frac{n - 12}{3} = 0$

⇒ n – 12 = 0
⇒ n = 12

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Question 29:

The coefficient of a^{– 6}b⁴ in the expansion of ${(\frac{1}{a} - \frac{2 b}{3})}^{10}$ is _________.

Answer:

Let T^r+1have the coefficient of a^–6b⁴.

T^r+1 = 10Cr ${(\frac{1}{a})}^{10 - r} {(- \frac{2 b}{3})}^{r}$

For the coefficient of a–6b4,
10 – r = 6
⇒ r = 4
⇒ coefficient of a^{– 6}b⁴ = ¹⁰C₄ ${(- \frac{2}{3})}^{4}$

                                      = $\frac{10 \times 9 \times 8 \times 7 \times 6!}{6! \times 4 \times 3 \times 2 \times 1} . \frac{2^{4}}{3^{4}}$

                                       $= \frac{1120}{27} .$ â€‹

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Question 30:

Middle term in the expansion of (a³ + ba)²⁸is _________ .

Answer:

Here, n = 28 (even)

⇒ middle term = ${(\frac{28}{2} +!)}^{th} term$
                         = 15^th term
⇒ T15 = T14+1
            = ²⁸C₁₄ (a³)^28–14 (ba)¹⁴
            = ²⁸C₁₄ (a³)⁴² b¹⁴a¹⁴
            = ²⁸C₁₄ a⁵⁴ b¹⁴

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Question 31:

The ratio of the coefficients of x^pand x^qin the expansion of (1 + x)^{p + q} is __________

Answer:

Coefficient of x^p = ^p+qC_p
And, coefficient of x^q = ^p+qC_q

$\Rightarrow \frac{{}^{p + q}C_{p}}{{}^{p + q}C_{q}} = \frac{{}^{p + q}C_{p}}{{}^{p + q}C_{p}}$

                  $= \frac{1}{1}$

                  = 1 : 1

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Question 32:

The position of the term independent of x in the expansion of ${(\sqrt{\frac{x}{3}} + \frac{3}{2 x^{2}})}^{10}$ is ____________.

Answer:

Let the constant term be Tr+1.
Then,
T_r+1 = ¹⁰C_r ${(\sqrt{\frac{x}{3}})}^{10 - r} {(\frac{3}{2 x^{2}})}^{r}$

= ¹⁰C_r $x^{\frac{10 - r}{2}} . 3^{\frac{- 10 + r}{2}} . 3^{r} . 2^{- r} . x^{- 2 r}$

= ¹⁰C_r $x^{\frac{10 - 5 r}{2}} . 3^{\frac{- 10 + 3 r}{2}} . 2^{- r}$
For the constant term,
10 – 5r = 0
⇒ r = 2
So, the third term is independent of x.

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Question 33:

If 25¹⁵ is divided by 13, the reminder is _________ .

Answer:

25¹⁵ = (26 – 1)¹⁵

$= {}^{15}C_{0} 26^{15} - {}^{15}C_{1} 26^{14} + . . . - {}^{15}C_{15} = {}^{15}C_{0} 26^{15} - {}^{15}C_{1} 26^{14} + . . . - 1 - 13 + 13 = {}^{15}C_{0} 26^{15} - {}^{15}C_{1} 26^{14} + . . . - 13 + 12$
Hence, when 25¹⁵ is divided by 13, then remainder will be 12.

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Question 34:

The sum of the series $\sum_{r = 0}^{10} {}^{20}C_{r} is 2^{19} + \frac{{}^{20}C_{10}}{2}$

Answer:

$\sum_{r = 0}^{10} {}^{20}C_{r} = {}^{20}C_{0} + {}^{20}C_{1} + {}^{20}C_{2} + . . . + {}^{20}C_{10}$

$= {}^{20}C_{0} + {}^{20}C_{1} + {}^{20}C_{2} + . . . + {}^{20}C_{10} + {}^{20}C_{11} + . . . {}^{20}C_{20} - ({}^{20}C_{11} + . . . + {}^{20}C_{20})$

$= 2^{20} - ({}^{20}C_{11} + . . . + {}^{20}C_{20})$

Hence, the given statement is false.

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Question 35:

The expression 7⁹+ 9⁷is divisible by 64.

Answer:

7⁹ + 9⁷ = (1 + 8)⁷ – (1 – 8)⁹
             = (⁷C₀+ ⁷C₁8 +  ⁷C₂8 + . . . + ⁷C₇₈⁷) – (⁹C₀+ ⁹C₁8 +  ⁹C₂8^{2 –} . . . – ⁹C₉8⁹)
             = (1 + 7 × 8 + 21 × 82 + . . . ) – (1 + 9 × 8 + 36 × 8² + . . .–8⁹ )
             = (7 × 8 + 9 × 8) + (21 – 8²– 36 × 8²)
             = 2 × 64 + (21 – 36) – 64.
Which is divisible by 64.
Hence, the given statement is true.

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Question 36:

The number of terms in the expansion of [(2x + y³)⁴]⁷ is 8

Answer:

[(2x + y³)⁴]⁷ = (2x + y)²⁸
The expansion has 29 terms.
Hence, the given statement is false.

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Question 37:

The sum of coefficients of the two middle terms in the expansion of (1 + x)^{2n – 1}is equal to ^{2n – 1}C_n.

Answer:

The expansion (1 + x)^{2n – 1}has 2 middle terms i.e. ${(\frac{2 n + 1 - 1}{2})}^{th}$ term and ${(\frac{2 n - 1 + 1}{2} + 1)}^{th}$ term.
∴ The coefficients of n^th term = ^{2n – 1}C_{n – 1}

Coefficients of (n + 1)^th term = ^{2n – 1}C_n

Sum of coefficients as per conditions

= ^{2n – 1}C_{n – 1} + ^{2n – 1}C_n

= ^{2n – 1 + 1}C_n[Since ⁿC_r + ⁿC_{r – 1} = ^{n + 1}C_r]

= ²ⁿC_n

Hence the given statement is false.

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Question 38:

The last two digits of the numbers 3⁴⁰⁰ are 01.

Answer:

As per given condition

3⁴⁰⁰ = [3²]²⁰⁰ = [9]²⁰⁰ = [10 – 1]²⁰⁰

⇒ [10 – 1]²⁰⁰ = ²⁰⁰C₀10²⁰⁰ – ²⁰⁰C₁10¹⁹⁹ + ..... – ²⁰⁰C₁₉₉10¹ 1¹⁹⁹ + ²⁰⁰C₂₀₀10⁰ 1²⁰⁰[As per binomial expansion]

⇒ [10 – 1]²⁰⁰ = 10²⁰⁰ – 200 × 10¹⁹⁹ ..... –10 × 200 + 01

so it is clear from above expansion that the last two digits are 01.
Hence the given statement is true.

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Question 39:

If the expansion of ${(x - \frac{1}{x^{2}})}^{2 n}$ contains a term independent of x, then n is a multiple of 2.

Answer:

As per given conditions Binomial expansion is ${[x - \frac{1}{x^{2}}]}^{2 n}$

Let T_{r + 1} term be independent of x.

$\begin{array}{rcl} Then T_{r + 1} & = & ^{2 n} C_{r} {[x]}^{2 n - r} {[- \frac{1}{x^{2}}]}^{r} \\ = & ^{2 n} C_{r} {[x]}^{2 n - r} {[- 1]}^{r} {[x]}^{- 2 r} \\ = & ^{2 n} C_{r} {[x]}^{2 n - r - 2 r} {[- 1]}^{r} \\ =^{2 n} C_{r} x^{2 n - 3 r} {[- 1]}^{r} \end{array}$

for the term independent of x

x^{2n – 3r} = x⁰
⇒ 2n – 3r = 0

$\Rightarrow r = \frac{2 n}{3}$ which is not an integer

Hence the given statement is false.

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Question 40:

Number of terms in the expansion of (a + b)ⁿ where n ∈ N is one less than the power n.

Answer:

We know that the number of terms in the expansion of [a + b]ⁿ, n∈N is one more than the power n.
Hence the given statement is false.

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