M.C.Q (1 Marks)

MCQ 11 Mark

The two successive terms in the expansion of $(1 + x)^{24}$ whose coefficients are in the ratio $1 : 4$ are: $[$Hint:$\frac{^{24}\text{C}_\text{r}}{^{24}\text{C}_{\text{r}+1}}=\frac{1}{4}\ \frac{\text{r}+1}{24-\text{r}}\ \frac{1}{4}\Rightarrow4\text{r}+4=24-4\Rightarrow\text{r}=4]$

A
$3^{rd} and 4^{th}.$
B
$4^{th} and 5^{th}.$
✓
$5^{th} and 6^{th}.$
D
$6^{th} and 7^{th}.$

Answer

Correct option: C.

$5^{th} and 6^{th}.$

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.

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MCQ 21 Mark

If the coefficients of $2^{nd}, 3^{rd}$ and the $4^{th}$ terms in the expansion of $(1 + x)^n$ are in $A.P.,$ then value of n is: $[$Hint: $2^nC_2 = ^nC_1 + ^nC_3 \Rightarrow n^2 - 9n + 14 = 0 \Rightarrow n = 2\ $or $7.]$

A
$2.$
✓
$7.$
C
$11.$
D
$14.$

Answer

Correct option: B.

$7.$

$(1+\text{x})^\text{n}=\ ^\text{n}\text{C}_0+\ ^\text{n}\text{C}_1\text{x}+\ ^\text{n}\text{C}_2\text{x}^2+\ ^\text{n}\text{C}_3\text{x}^3+...+\ ^\text{n}\text{C}_\text{n}\text{x}^\text{n}$
So$,$ coefficients of $2^{nd}, 3^{rd}$ and $4^{th}$ terms are $^nC_1, ^nC_2$ and $^nC_3,$ respectively.
Given that$, ^nC_1, ^nC_2$ and $^nC_3,$ 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 $n = 7.$

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MCQ 31 Mark

Choose the correct answer.
The total number of terms in the expansion of $(x + a)^{100} + (x - a)^{100}$ after simplification is:

A
$50.$
B
$202.$
✓
$51.$
D
None of these.

Answer

Correct option: C.

$51.$

Number of terms in the expansion of $(x + a)^{100} = 101$
Number of terms in the expansion of $(x - a)^{100} = 101$
Now 50 terms of expansion will cancel out with negative $50$ terms of $(x - a)^{100}$
So$,$ the remaining $51$ terms of first expansion will be added to $51$ terms of other.
Therefore, the number of terms $= 51$
Hence$,$ the correct option is $(c)$.

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MCQ 41 Mark

The coefficient of $x^n$ in the expansion of $(1 + x)^{2n} and (1 + x)^{2n - 1}$ are in the ratio. $[$Hint: $^{2\text{n}}\text{C}_\text{n} : \ ^{2\text{n} - 1}\text{C}_\text{n}]$

A
$1 : 2.$
B
$1 : 3.$
C
$3 : 1.$
✓
$2 : 1.$

Answer

Correct option: D.

$2 : 1.$

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 $x^n, put r = n$
$\therefore$ 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 $(d).$

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MCQ 51 Mark

If A and B are coefficient of $x^n$ in the expansions of $(1 + x)^{2n}$ and $(1 + x)^{2n – 1}$ respectively, then $\frac{\text{A}}{\text{B}}$ equals:
$[$Hint: $\frac{\text{A}}{\text{B}}=\frac{^{2\text{n}}\text{C}_\text{n}}{^{2\text{n}-1}\text{C}_\text{n}}=2]$

A
$1.$
✓
$2.$
C
$\frac{1}{2}.$
D
$\frac{1}{\text{n}}.$

Answer

Correct option: B.

$2.$

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 $xn = ^{2n}C_n = A ($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 $(b).$

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MCQ 61 Mark

Choose the correct answer.
If the middle term of $\Big(\frac{1}{\text{x}}+\text{x}\sin\text{x}\Big)^{10}$ is equal to $7\frac{7}{8},$ then value of $x$ is:
$[$Hint: $\text{T}_6=\ ^{10}\text{C}_5\frac{1}{\text{x}^5}.\text{x}^5\ \sin^5\text{x}=\frac{63}{8}\Rightarrow\sin^5\text{x}=\frac{1}{2^5}\sin\frac{1}{2}$
$\Rightarrow\text{x}=\text{n}\pi+(-1)^\text{n}\frac{\pi}{6}]$

A
$2\text{n}\pi+\frac{\pi}{6}.$
B
$\text{n}\pi+\frac{\pi}{6}.$
✓
$\text{n}\pi+(-1)^\text{n}\frac{\pi}{6}.$
D
$\text{n}\pi+(-1)^\text{n}\frac{\pi}{3}.$

Answer

Correct option: C.

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

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

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MCQ 71 Mark

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:

✓
$n = 2r.$
B
$n = 3r.$
C
$n = 2r + 1.$
D
None of these.

Answer

Correct option: A.

$n = 2r.$

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, $^{2\text{n}}\text{C}_{3\text{r}-1}=\ ^{2\text{n}}\text{C}_{\text{r}+1}$
$\Rightarrow3\text{r}-1+\text{r}+1=2\text{n}\ \ [\because\ ^\text{n}\text{C}_\text{x}=\ ^\text{n}\text{C}_\text{y}\Rightarrow\text{x}+\text{y}=\text{n}]$
$\therefore\text{n}=2\text{r}$

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Maths M.C.Q (1 Marks)

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