Term independent of x MCQ Quiz in বাংলা - Objective Question with Answer for Term independent of x - বিনামূল্যে ডাউনলোড করুন [PDF]

Calculation:

\(\rm \left(2x^2-3x-\frac{9}{x}\right)\left(-x+\frac{7}{x}\right) \)

⇒ - 2x³ + 3x² + 9 + 14x - 21 - 63/x²

⇒ - 2x³ +  3x2 + 9 + 14x - 21 - 63x^-2

Here, coefficients of x2 and x  are 3 and 14 respectively,

Their product = 3 × 14 = 42

∴ The correct answer is 42

Concept:

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

Calculation:

The general term of \(\rm ( x -\frac{k}{x^2})^9\) is given by, 

\(\rm T_{r+1}=(-1)^r\times^9C_r\times(x)^{9-r}\times(\frac{k}{x^2})^r\)

\(\rm =(-1)^r\times ^9C_r\times(x)^{9-r}\times x^{-2r}\times k^r\)

\(\rm =(-1)^r\times ^9C_r\times(x)^{9-r}\times x^{-2r}\times k^r\)

\(\rm =(-1)^r\times ^9C_r\times(x)^{9-3r}\times k^r\)

Now, the term is free from x (independent term), then power of x is zero

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

9 - 3r = 0 

⇒r = 3

⇒r + 1 = 4

∴ \(\rm T_{3+1}=(-1)^3\times 9C_3\times(x)^{9-9}\times k^3\)

\(\rm =-\frac{9\times8\times7}{3\times2\times1}\times k^3\)

=  -84k³

∴ -84 = -84k³ 

⇒ k = 1

Hence, option (3) is correct. 

Concept:

The general term in a binomial expansion of (a + b)ⁿ is given by:

\(T_{r+1}={}^nC_ra^{n-r}b^r\)

Calculation:

Given binomial is \(\rm\Big( x^2+\frac{2}{x}\Big)^{15}\)

∴ General term = Tr+1 = \({}^{15}C_r(x^2)^{15-r}\left(\frac{2}{x}\right)^r\) = \({}^{15}C_r2^rx^{30-3r}\)...(i)

Now, for the coefficient of term containing x¹⁵, 30 – 3r = 15 ⇒ r = 5

∴ coefficient of x¹⁵  = ¹⁵C₅ 2⁵  [ using (i) ] 

For the term independent of x, put 30 – 3r = 0 ⇒ r = 10

∴ Constant term = ¹⁵C₁₀ 2¹⁰ [ using (i) ] 

∴ ratio of the coefficient of x15 to the term independent of x

= \(\frac{{}^{15}C_52^{5}}{{}^{15}C_{10}2^{10}}\)

= \(\frac{{}^{15}C_52^{5}}{{}^{15}C_{5}2^{10}}\) [∵ \({}^nC_r={}^nC_{n-r}\)]

= \(\frac{1}{2^5}\)

∴ Required ratio is 1 : 32.

Concept:

(1 + x)ⁿ = \(\rm C_0 + C_1x+C_2x^2+C_3x^3+...+C_nx^n\)

Calculation:

Given, 

\(\rm (1+x)^n\left[1+\left(\frac{1}{x}\right)\right]^n\)

⇒ \(\rm (C_0 + C_1x+C_2x^2+C_3x^3+...+C_nx^n)(C_0 + C_1{1 \over x}+C_2 {1 \over x^2}+C_3{1 \over x^3}+...+C_n{1 \over x^n})\)

The term independent of x in this expression is obtained when

x^k is multiplied by 1/x^k .

⇒ The term independent of x = \(\rm C_0^2+C_1^2+C_2^2+...+C_n^2\)

∴ The correct answer is option (3).

Concept:

The (r + 1)th term of (a + b)n is given by Tr+1 = nCran-rbr 

Calculation:

Given, second, third and fourth terms in the expansion of (a + b)n are 135, 30 and \(\frac{10}{3}\) 

∴ \({ }^n C_1 a^{n-1} b=135 \quad ...(i)\)

and, \({ }^{\mathrm{n}} \mathrm{C}_2 a^{\mathrm{n}-2} \mathrm{b}^2=30\quad ...(ii)\)

and, \({ }^n C_3 a^{n-3} b^3=\frac{10}{3} \quad ...(iii)\)

By \(\frac{(i)}{(\text { ii) }}\)

\(\frac{{ }^n C_1}{{ }^n C_2} \frac{a}{b}=\frac{9}{2} \quad ...(iv)\)

By \(\frac{(ii)}{(\text { iii) }}\) 

\(\frac{{ }^n C_2}{{ }^n C_3} \frac{a}{b}=9\quad ...(v)\)

By \(\frac{(iv)}{(\text {v) }}\) 

\(\frac{{ }^{\mathrm{n}} C_1{ }^n C_3}{{ }^n C_2{ }^n C_2}=\frac{1}{2}\)

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

⇒ 4n – 8 = 3n – 3 

⇒ n = 5

∴ The value of n is 5.

Concept:

The (r + 1)^th term of (a + b)ⁿ is given T_r+1 = ⁿC_ra^n-rb^r 

Calculation:

Given, \(\left(1+2 x-3 x^3\right)\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9\)

The general term of \(\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9\)

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

Put r = 6 to get coeff. of \(x^0={ }^9 C_6 \cdot \frac{1}{6^3} \cdot x^0=\frac{7}{18} x^0\)

Put r = 7 to get coeff. of \(x^{-3}={ }^9 C_r \cdot \frac{3^{-5}}{2^2}(-1)^7 \cdot x^{-3}\)

= \(-{ }^9 C_7 \cdot \frac{1}{3^5 \cdot 2^2} \cdot x^{-3}=\frac{-1}{27} x^{-3}\)

∴ \(\left(1+2 x-3 x^3\right)\left(\frac{7}{18} x^0-\frac{1}{27} x^{-3}\right)\)

\(\frac{7}{18}+\frac{3}{27}=\frac{7}{18}+\frac{1}{9}\)

= \(\frac{7+2}{18}=\frac{9}{18}=\frac{1}{2}\)

⇒ p = \(\frac{1}{2}\)

∴ 108p = 180 × \(\frac{1}{2}\)

= 54

∴ The value of 108p is 54.

Concept: Binomial expansion.

The binomial (a + b) can be expanded to the power n (n is non-negative) in the following form:

(a + b)ⁿ = \( \sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^r\)

where \( T_{r+1}=\binom{n}{r}a^{n-r}b^r\), denotes the (r + 1)^th term.

Solution:

We have \( \Big(\frac{2x^3-1}{x}\Big)^{12}\)= \( \Big(2x^2 -\frac{1}{x}\Big)^{12}\)

⇒ a = 2x², b = \(\frac{-1}{x}\), n = 12

∴ \( T_{r+1}=\binom{12}{r}(2x^2)^{12-r}(\frac{-1}{x})^r\)

⇒ \( T_{r+1}=\binom{12}{r}2^{12-r}x^{(24-2r)}x^{-r}(-1)^{r}\)

⇒ \( T_{r+1}=(-1)^{r}\binom{12}{r}2^{12-r}x^{24-3r}\)

Now, for the term to be independent of x, the power of x should be zero

⇒ 24 - 3r = 0

⇒ 3r = 24

⇒ r = 8

∴ \(T_{9}\) is independent of x.

∴ The term independent of x is the 9^th term.

Concept:

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

Calculation:

We have to find term independent of x in \(\rm (1+x)^3(1+\frac{1}{x})^4\)

⇒ \(\rm (1+x)^3(1+\frac{1}{x})^4 = \frac{(x+1)^3(x+1)^4}{x^4} = (x+1)^7x^{-4}\)

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

⇒  \({{\rm{T}}_{\left( {{\rm{r\;}} + {\rm{\;}}1} \right)}} = {\rm{\;}}{{\rm{\;}}^{\rm{7}}}{{\rm{C}}_{\rm{r}}} \times {{\rm{x}}^{{\rm{7}} - {\rm{r}}}} \rm \times 1^r \times {{\rm{x}}^{\rm{-4}}}\)

= ⁷C_r x^3-r

For the term independent of x, power of x should be zero

Therefore, 3 - r = 0

⇒ r = 3

Hence the term is T₃₊₁ and the coefficient is ⁷C₃

Concept:

• Binomial Coefficient: In the expansion of \((1+x)^{10}\), the coefficient of \(x^{10-r}\) is given by \(a_r = \binom{10}{r}\).
• We are required to evaluate the expression: \(\sum_{r=1}^{10} r^3 \left( \frac{a_r}{a_{r-1}} \right)^2\)
• Using property of binomial coefficients: \(\frac{a_r}{a_{r-1}} = \frac{\binom{10}{r}}{\binom{10}{r-1}} = \frac{10 - r + 1}{r} = \frac{11 - r}{r}\)
• So, \(\left( \frac{a_r}{a_{r-1}} \right)^2 = \left( \frac{11 - r}{r} \right)^2\)
• Final expression becomes: \(\sum_{r=1}^{10} r^3 \cdot \left( \frac{11 - r}{r} \right)^2 = \sum_{r=1}^{10} (11 - r)^2\cdot r\)

Calculation:

We simplify:

\(\sum_{r=1}^{10} r \cdot (11 - r)^2\)

⇒ For r=1: \(1 \cdot 10^2 = 100 \) 

⇒ For r=2: \(2 \cdot 9^2 = 162 \) 

⇒ For r=3: \(3 \cdot 8^2 = 192 \)

⇒ For r=4: \( 4 \cdot 7^2 = 196 \)

⇒ For r=5: \(5 \cdot 6^2 = 180 \) 

⇒ For r=6: \(6 \cdot 5^2 = 150 \) 

⇒ For r=7: \(7 \cdot 4^2 = 112 \) 

⇒ For r=8: \(8 \cdot 3^2 = 72 \) 

⇒ For r=9:  \(9 \cdot 2^2 = 36 \)

⇒ For r=10: \(10 \cdot 1^2 = 10 \) 

⇒ Sum = 100 + 162 + 192 + 196 + 180 + 150 + 112 + 72 + 36 + 10 = 1210

∴ The value of the given summation is 1210.

Calculation:

Given,

We are given the equation:

\( \left( \frac{x+1}{x^{2/3} + 1 - x^{1/3}} - \frac{x+1}{x - x^{1/2}} \right)^{10}, x > 1 \)

Simplify the terms inside the bracket:

\( = \left( \frac{x+1}{(x^{1/3})^2 - x^{1/3} + 1} - \frac{x+1}{\sqrt{x}(\sqrt{x} - 1)} \right)^{10} \)

\( = \left( (x^{1/3} + 1) - \frac{\sqrt{x} + 1}{\sqrt{x}} \right)^{10} \)

\( = \left( x^{1/3} + 1 - 1 - x^{-1/2} \right)^{10} = \left( x^{1/3} - x^{-1/2} \right)^{10} \)

The general term T{r+1}\) is given by:

\( T_{r+1} = {}^{10}C_r (x^{1/3})^{10-r} (-x^{-1/2})^r = {}^{10}C_r (-1)^r x^{\frac{10-r}{3} - \frac{r}{2}} \)

For the term independent of x, the exponent of x must be zero:

\( \frac{10-r}{3} - \frac{r}{2} = 0 \Rightarrow 2(10-r) - 3r = 0 \Rightarrow 20 - 5r = 0 \Rightarrow r = 4 \)

The required term is T₅:

\( T_5 = {}^{10}C_4 (-1)^4 x^0 = {}^{10}C_4 = \frac{10!}{4!6!} = 210 \)

∴ The term independent of x is 210.