Binomial Theorem MCQ Quiz in मल्याळम - Objective Question with Answer for Binomial Theorem - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

General Term in the expansion (x - y)ⁿ = T_r+1 = (-1)^r × ⁿC_r × x^n-r ×  y^r

Calculation:

The general term in the expansion of \(\rm (\sqrt x-\frac3x)^ {12}\) is given by, 

\(\rm T_{r+1}=(-1)^r\times ^{12}C_r\times(\sqrt x)^{12-r}\times(\frac3x)^r\)

\(⇒ \rm(-1)^r\times^{12}C_r\times( x)^{6-\frac r 2}\times(x)^{-r}\times(3)^r\)) 

\(⇒ \rm(-1)^r\times ^{12}C_r\times( x)^{6-\frac r 2-r}\times(3)^r\)

For independent terms, the power of x is zero.

On equating power of x in (1) with zero, we get,

\(\rm 6-\frac r2-r=0 \)

\(⇒ 6-\frac{3r}{2}=0\)

⇒ 3r = 12

⇒ r = 4

⇒ r + 1 = 5

So, the 5th term in the given expansion is independent of x.

Hence, option (3) is correct.

Formula used:

\(^nC_r=\frac{n!}{r!(n-r)!}\)

n! = n × (n - 1) × (n - 2).......3 × 2 × 1 

Calculation:

Given that,

4 × nC5 = 9 × n-1C5

Using the above formula

\(4×\frac{n!}{5!(n-5)!}=9×\frac{(n-1)!}{5![(n-1)-5]!}\)

⇒ \(4×\frac{n× (n-1)!}{5!(n-5)(n-6)!}=9×\frac{(n-1)!}{5![(n-6]!}\)

⇒ \(\frac{4n}{(n-5)} = 9\)

⇒ 9n - 45 = 49

⇒ 5n = 4n

⇒ n = 9

Concept:

• ⁿC_n = 1
• ⁿC₁ = n

Calculation:

We have to find the value of \(\mathop \sum \limits_{{\rm{r\;}} = {\rm{\;}}0}^1 {{\rm{\;}}^{{\rm{n}} + {\rm{r}}}}{{\rm{C}}_{\rm{n}}}\)

\(⇒ \mathop \sum \limits_{{\rm{r\;}} = {\rm{\;}}0}^1 {{\rm{\;}}^{{\rm{n}} + {\rm{r}}}}{{\rm{C}}_{\rm{n}}}\)

⇒ ⁿ⁺⁰C_n + ⁿ⁺¹C_n

⇒ nCn + n+1Cn 

⇒ 1 + (n + 1)

⇒ (n + 2)

∴  The value of \(\mathop \sum \limits_{{\rm{r\;}} = {\rm{\;}}0}^1 {{\rm{\;}}^{{\rm{n}} + {\rm{r}}}}{{\rm{C}}_{\rm{ n}}}\) is ^{n + 2}C_1.

Concept:

We have (x + y) n = nC0 xn + nC1 xn-1 . y + nC2 xn-2. y2 + …. + nCn yn

• General term: General term in the expansion of (x + y) n is given by
• \({{\rm{T}}_{\left( {{\rm{r\;}} + {\rm{\;}}1} \right)}} = {\rm{\;}}{{\rm{\;}}^{\rm{n}}}{{\rm{C}}_{\rm{r}}} \times {{\rm{x}}^{{\rm{n}} - {\rm{r}}}} \times {{\rm{y}}^{\rm{r}}}\)
• In the binomial expansion of (x + y)n , the rth term from end is (n – r + 2)th term.

Note: 

• In the binomial expansion of (x + y)n, the rth term from end = In the binomial expansion of (y + x)n, the rth term from the start.
• If we interchange the term x → y, it will give rth term from the beginning.

Calculation:

We have to find 9th term from the end in (x – 1/x) 12

We know that rth term from end means (n – r + 2)th term from start.

So. 9th term from the end = [12 – 9 + 2]th term from start = 5th term from start

General term: \({{\rm{T}}_{\left( {{\rm{r\;}} + {\rm{\;}}1} \right)}} = {\rm{\;}}{{\rm{\;}}^{\rm{n}}}{{\rm{C}}_{\rm{r}}} \times {{\rm{x}}^{{\rm{n}} - {\rm{r}}}} \times {{\rm{y}}^{\rm{r}}}\)

\(\Rightarrow {{\rm{T}}_5} = {\rm{\;}}{{\rm{T}}_{\left( {4 + {\rm{\;}}1} \right)}} = {\rm{\;}}{{\rm{\;}}^{12}}{{\rm{C}}_4} \times {{\rm{x}}^{12 - 4}} \times {\left( {\frac{{ - 1}}{{\rm{x}}}} \right)^4}\)

\( = {{\rm{\;}}^{12}}{{\rm{C}}_4} \times {{\rm{x}}^8} \times \frac{1}{{{{\rm{x}}^4}}}\)

= ¹²C₄ x⁴

Concept:

We have (x + y) ⁿ = ⁿC₀ xⁿ + ⁿC₁ x^n-1 . y + ⁿC₂ x^n-2. y² + …. + ⁿC_n yⁿ

General term: General term in the expansion of (x + y) ⁿ is given by\({{\rm{T}}_{\left( {{\rm{r\;}} + {\rm{\;}}1} \right)}} = {\rm{\;}}{{\rm{\;}}^{\rm{n}}}{{\rm{C}}_{\rm{r}}} \times {{\rm{x}}^{{\rm{n}} - {\rm{r}}}} \times {{\rm{y}}^{\rm{r}}}\)

In the binomial expansion of (x + y)ⁿ , the r^th term from end is (n – r + 2)^th term.

\({\;^n}{C_r} = \frac{{n!}}{{r!\left( {n\; - \;r} \right)!}}\)

Calculation:

We have to find 4th term in the expansion of \({\left( {\sqrt x + \;\frac{1}{x}} \right)^{12}}\)

We know that, \({{\rm{T}}_{\left( {{\rm{r\;}} + {\rm{\;}}1} \right)}} = {\rm{\;}}{{\rm{\;}}^{\rm{n}}}{{\rm{C}}_{\rm{r}}} \times {{\rm{x}}^{{\rm{n}} - {\rm{r}}}} \times {{\rm{y}}^{\rm{r}}}\)

\( \Rightarrow {T_4} = \;{{\rm{T}}_{\left( {3{\rm{\;}} + {\rm{\;}}1} \right)}} = {\rm{\;}}{{\rm{\;}}^{12}}{{\rm{C}}_3} \times {\left( {\sqrt {\rm{x}} } \right)^{12 - 3}} \times {\left( {\frac{1}{x}} \right)^3}\)

\( = {{\rm{\;}}^{12}}{{\rm{C}}_3} \times {\left( x \right)^{\frac{9}{2}}} \times {\rm{\;}}\frac{1}{{{x^3}}}\)

\(= \frac{{12!}}{{3!\left( {12\; - \;3} \right)!}} \times {\left( x \right)^{\frac{9}{2} - 3}}\)

= 220 x^3/2

∴ Option 3 is correct.

Concept:

General term: General term in the expansion of (x + y)n is given by

\(\rm {T_{\left( {r\; + \;1} \right)}} = \;{\;^n}{C_r} × {x^{n - r}} × {y^r}\)

Middle terms: The middle terms is the expansion of (x + y) n depends upon the value of n.

• If n is even, then total number of terms in the expansion of (x + y) n is n +1. So there is only one middle term i.e. \(\rm \left( {\frac{n}{2} + 1} \right){{\rm{\;}}^{th}}\) term is the middle term.
• If n is odd, then total number of terms in the expansion of (x + y) n is n + 1. So there are two middle terms i.e. \(\rm {\left( {\frac{{n\; + \;1}}{2}} \right)^{th}}\;\)and \(\rm {\left( {\frac{{n\; + \;3}}{2}} \right)^{th}}\) are two middle terms.

Calculation:

Here, we have to find the middle terms in the expansion of \(\rm \left(x - \frac 2 x \right)^{10}\)

Here n = 10 (n is even number)

∴ Middle term = \(\rm \left( {\frac{n}{2} + 1} \right) = \left( {\frac{10}{2} + 1} \right) = 6th\;term\)

T6 = T (5 + 1) = 10C5 × (x) (10 - 5) × \(\rm \left(\frac {-2}{x}\right)^5\)

T5 = -2⁵ × 10C5

Calculation:

For the given expression (x2 + x)10  ,  n = 10

(r + 1)th term

\(\rm T_{r+1}=^{10}\textrm{C}_{r}{(x^2)}^{10-r}x^r\)

\(\rm T_{r+1}=^{10}\textrm{C}_{r}x^{20-2r}x^r\)

\(\rm T_{r+1}=^{11}\textrm{C}_{r}x^{20-r}\)

If this term is independent of x then the index of x must be zero.

i.e.  20 - r = 0,   r = 20.

Here r > n

So no term is independent of x.

Concept:

In the binomial expansion of (a + b)ⁿ, the term which does not involve any variable is said to be an independent term.

The general term in the binomial expansion of (a + b)ⁿ is given by: \({T_{r + 1}} = {^n}{C_r} \times {a^{n - r}} \times {b^r}\)

Calculation:

Given: \({\left( {{x^2} - \frac{1}{x}} \right)^9}\)

Let (r + 1)^th be the independent term in the expansion of \({\left( {{x^2} - \frac{1}{x}} \right)^9}\).

We know that the general term in the binomial expansion of (a + b)ⁿ is given by:

\({T_{r + 1}} = {^n}{C_r} \times {a^{n - r}} \times {b^r}\)

Here, a = x², n = 9 and b = 1 / x.

\(\Rightarrow {T_{r + 1}} = {\;^9}{C_r} \times {x^{2\left( {9 - r} \right)}} \times {\left( {\frac{1}{x}} \right)^r} = \;{\;^9}{C_r} \times {x^{18 - 3r}}\;\)

∵ The (r + 1)^th term is the independent term in the expansion of \({\left( {{x^2} - \frac{1}{x}} \right)^9}\) ⇒ x^{18 – 3r} = x⁰

⇒ 18 – 3r = 0 ⇒ r = 6

⇒ 7^th term in the expansion of \({\left( {{x^2} - \frac{1}{x}} \right)^9}\) is the independent term.

We have to find the value of the 7^th term in the expansion of \({\left( {{x^2} - \frac{1}{x}} \right)^9}\)

Concept:

General term: General term in the expansion of (x + y)n is given by

\(\rm {T_{\left( {r\; + \;1} \right)}} = \;{\;^n}{C_r} × {x^{n - r}} × {y^r}\)

Middle terms: The middle terms is the expansion of (x + y) n depends upon the value of n.

• If n is even, then there is only one middle term i.e. \(\rm \left( {\frac{n}{2} + 1} \right){{\rm{\;}}^{th}}\) term is the middle term.
• If n is odd, then there are two middle terms i.e. \(\rm {\left( {\frac{{n\; + \;1}}{2}} \right)^{th}}\;\)and \(\rm {\left( {\frac{{n\; + \;3}}{2}} \right)^{th}}\) are two middle terms.

Calculation:

Here, we have to find the middle terms in the expansion of \(\rm \left(2x + \frac {1} {2x} \right)^{8}\)

Here n = 8 (n is even number)

∴ Middle term = \(\rm \left( {\frac{n}{2} + 1} \right) = \left( {\frac{8}{2} + 1} \right) =5th\;term\)

T5 = T (4 + 1) = 8C4 × (2x) (8 - 4) × \(\rm \left(\frac {1}{2x}\right)^4\)

T₅ =  8C₄

Concept:

Binomial Expansion:

• (a + b)n = C0 a0 bn + C1 a1 bn-1 + C2 a2 bn-2 + … + Cr ar bn-r + … + Cn-1 an-1 b1 + Cn an b0, where C0, C1, …, Cn are the Binomial Coefficients defined as Cr = nCr = \(\rm \frac{n!}{r!(n-r)!}\).
• The total number of terms in the expansion is n + 1.
• The (r + 1)^th term in the expansion is T_r+1 = C_r a^r b^n-r.

Calculation:

For the given expression \(\rm \left({x\over 3} - 3y\right)^7\), n = 7.

There will be 7 + 1 = 8 terms in the expansion of the above expression.

Expanding by reversing the order of the terms (i.e. a = -3y and b = \(\rm\frac{x}{3}\)), we have:

T₅ = T₄₊₁ = ⁷C₄ \(\rm\left(\frac{x}{3}\right)^{7-4}\) (-3y)⁴  = 35 \(\rm\left(\frac{x^3}{3^3}\right)\) (3⁴y⁴)  = 105x³y⁴.