MATHS M.C.Q (1 Marks)

2. 7.

Solution:

$(1+\text{x})^\text{n}=\ ^\text{n}\text{C}_0+\ ^\text{n}\text{C}_1\text{x}+\ ^\text{n}\text{C}_2\text{x}^2+\ ^\text{n}\text{C}_3\text{x}^3+...+\ ^\text{n}\text{C}_\text{n}\text{x}^\text{n}$

So, coefficients of 2^nd, 3^rd and 4th terms are ⁿC₁, ⁿC₂ and ⁿC₃, respectively.

Given that, ⁿC₁, ⁿC₂ and ⁿC₃, are in A.P.

$\therefore2\ ^\text{n}\text{C}_2=\ ^\text{n}\text{C}_1+\ ^\text{n}\text{C}_3$

$\Rightarrow2\Big[\frac{\text{n}!}{(\text{n}-2)!2!}\Big]=\text{n}+\frac{\text{n}!}{3!(\text{n}-3)!}$

$\Rightarrow2\Big[\frac{\text{n}(\text{n}-1)}{2!}\Big]=\text{n}+\frac{\text{n}(\text{n}-1)(\text{n}-2)}{3!}\Rightarrow(\text{n}-1)=1+\frac{(\text{n}-1)(\text{n}-2)}{6}$

$\Rightarrow6\text{n}-6=6+\text{n}^2-3\text{n}+2\Rightarrow\text{n}^2-9\text{n}+14=0$ $\Rightarrow(\text{n}-7)(\text{n}-2)=0$

$\therefore\text{n}=2\ \text{or}\ \text{n}=7$

Since n = 2 is not possible, so n = 7.

4. 2 : 1.

Solution:

General Term $\text{T}_{\text{r}+1}=\ ^\text{n}\text{C}_\text{r}\text{x}^{\text{n}-\text{r}}\text{y}^\text{r}$

In the expansion of (1 + x)²ⁿ, we get $\text{T}_{\text{r}+1}=\ ^{2\text{n}}\text{C}_\text{r}\text{x}^\text{r}$

To get the coefficient of xⁿ, put r = n

$\therefore$ Coefficient of $\text{x}^\text{n}=\ ^{\text{2n}}\text{C}_\text{n}$

In the expansion of $(1+\text{x})^{2\text{n}-1},$ we get $\text{T}_{\text{r}+1}=\ ^{2\text{n}-1}\text{C}_\text{r}\text{x}^\text{r}$

$\therefore$ Coefficient of $\text{x}^\text{n}\ \ \text{is}=\ ^{\text{2n}-1}\text{C}_{\text{n}-1}$

The required ratio is $\frac{^{\text{2n}}\text{C}_{\text{n}-1}}{^{\text{2n}-1}\text{C}_{\text{n}-1}}$

$=\frac{\frac{2\text{n}!}{\text{n}!(\text{n}!)}}{\frac{(2\text{n}-1)!}{(\text{n}-1)!(2\text{n}-1-\text{n}+1)!}}=\frac{\frac{2\text{n}!}{\text{n}!.\text{n}!}}{\frac{(2\text{n}-1)!}{(\text{n}-1)!(\text{n}!)}}$

$=\frac{2\text{n}!}{\text{n}!\text{n}!}\times\frac{(\text{n}-\text{n})!\cdot\text{n}!}{(2\text{n}-1)!}=\frac{2\text{n}(2\text{n}-1)!}{\text{n}!\text{n}(\text{n}-1)!}\times\frac{(\text{n}-1)!\cdot\text{n}!}{(2\text{n}-1)!}$

$=\frac{2}{1}=2:1$

Hence, the correct option is (d).

1. n = 2r.

Solution:

The given expression is $(1 + \text{x})^{2\text{n}}$

$\therefore\text{T}_{3\text{r}}=\text{T}_{(3\text{r}-1)+1}=\ ^{2\text{n}}\text{C}_{3\text{r}-1}\ \text{x}^{3\text{r}-1}$

and $\text{T}_{\text{r}+2}=\text{T}_{(\text{r}+1)+1}=\ ^{2\text{n}}\text{C}_{\text{r}+1}\text{x}^{\text{r}+1}$

Given, $^{2\text{n}}\text{C}_{3\text{r}-1}=\ ^{2\text{n}}\text{C}_{\text{r}+1}$

$\Rightarrow3\text{r}-1+\text{r}+1=2\text{n}\ \ [\because\ ^\text{n}\text{C}_\text{x}=\ ^\text{n}\text{C}_\text{y}\Rightarrow\text{x}+\text{y}=\text{n}]$

$\therefore\text{n}=2\text{r}$

3. 5^th and 6^th.

Solution:

Let the two successive terms in the expansion of (1 + x)²⁴ be (r + 1)(r + 2)^th terms.

Now, $\text{T}_{\text{r}+1}=\ ^{24}\text{C}_\text{r}\text{x}^\text{r}\ \text{and}\ \text{T}_{\text{r}+2}=\ ^{24}\text{C}_{\text{r}+1}\text{x}^{\text{r}+1}$

Given that, $\frac{^{24}\text{C}_\text{r}}{^{24}\text{C}_{\text{r}+1}}=\frac{1}{4}$

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

$\Rightarrow4\text{r}+4=24-\text{r}\Rightarrow\text{r}=4$

$\therefore\text{T}_{4+1}=\text{T}_5\ \text{and}\ \text{T}_{4+2}=\text{T}_6$

Hence. 5^th and 6^th terms.

2. $2.$

Solution:

Given expression is (1 + x)²ⁿ

$\text{T}_{\text{r}+1}=\ ^{2\text{n}}\text{C}_\text{r}\text{x}^\text{r}$

$\therefore$ Coefficient of xn = ²ⁿC_n = A (Given)

In the expression of (1 + x)^{2n - 1}

$\text{T}_{\text{r}+1}=2^{\text{n}-1}\text{C}_\text{x}=\text{B}$ Given

So, $\frac{\text{A}}{\text{B}}=\frac{^{2\text{n}}\text{C}_\text{n}}{^{2\text{n}-1}\text{C}_\text{n}}=\frac{^{2\text{n}}\text{C}_\text{n}}{^{2\text{n}-1}\text{C}_\text{n}}=\frac{2}{1}$

Hence, the correct option is (b).

3. $\text{n}\pi+(-1)^\text{n}\frac{\pi}{6}.$

Solution:

Given expression is $\Big(\frac{1}{\text{x}}+\text{x}\sin\text{x}\Big)^{10}$

Since, n = 10 (even), so there is only one middle term which is, 6^th term.

$\therefore\text{T}_6=\text{T}_{5+1}=\ ^{10}\text{C}_5\Big(\frac{1}{\text{x}}\Big)^{10-5}(\text{x}\sin\text{x})^5$

$\Rightarrow\frac{63}{8}=\ ^{10}\text{C}_5\sin^5\text{x}$ (given)

$\Rightarrow\frac{63}{8 }=252\times\sin^5\text{x}\Rightarrow\sin^5\text{x}=\frac{1}{32}$ $\Rightarrow\sin\text{x}=\frac{1}{2}\Rightarrow\sin\text{x}=\sin\frac{\pi}{6}$

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