M.C.Q (1 Marks) - Page 4 - Maths STD 11 Science Questions - Vidyadip

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M.C.Q (1 Marks)

MCQ 1511 Mark

The number of irrational terms in the expansion of $\Big(4^{\frac{1}{5}}+7^{\frac{1}{10}}\Big)^{45}$ is:

A
$40$
B
$5$
✓
$41$
D
None of these.

Answer

Correct option: C.

$41$

The general term $T_{r+1}$ in the given expansion is given by ${^\text{45}}\text{C}_{\text{r}}\Big(4^{\frac{1}{5}}\Big)^{45-\text{r}}\Big(7^{\frac{1}{10}}\Big)^{\text{r}}$
For $T_{r+1}$ to be an integer, we must have $\frac{\text{r}}{5}$ and $\frac{\text{r}}{10}$ as integers i.e. $0\leq\text{r}\leq45$
$\therefore \text{r}=0,10,20,30,40$
Hence, there are $5$ rational and $41,$
i.e. $46 - 5,$ irrational terms.

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MCQ 1521 Mark

Sum of the coefficients in the expansion of $(5x - 4y)^n$ where $n$ is a positive integer is:

✓
$1$
B
$9^n$
C
$(-1)^n$
D
$5^n$

Answer

Correct option: A.

$1$

Substituting $x = 1$ and $y = 1$ in the above expression we get the sum of the binomial coefficients as
$(5x - 4y)^n$
$= (5 - 4)^n ($substituting $x=1$ and $y=1)$
$= (1)^n$
$= 1$

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MCQ 1531 Mark

Choose the correct answer.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:

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 1541 Mark

The coefficient of $x^5$ in the expansion of $(1 + x^2)^5(1 + x)^4$ is:

A
$20$
B
$30$
✓
$60$
D
$55$

Answer

Correct option: C.

$60$

Given equationnis $(1+x)^4\left(1+x^2\right)^5$
$=\left(1+4 x+6 x^2+4 x^3+x^4\right)\left(1+5 x^2+10 x^4+10 x^6+5 x^8+x^{10}\right)...(i)$
Hence the coefficient of $x^5$ from Eq $(i)$ will be.
$4(10) + 4(5)$
$= 40 + 20$
$= 60$

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MCQ 1551 Mark

If the fourth term of the binomial expansion of $\Big(\text{px}+\big(\frac{1}{\text{x}}\big)\Big)^\text{n}$ is $\frac{5}{2},$ then:

A
$\text{n}=6, \text{p}=6$
B
$\text{n}=8, \text{p}=6$
C
$\text{n}=8, \text{p}=\frac{1}{2}$
✓
$\text{n}=6, \text{p}=\frac{1}{2}$

Answer

Correct option: D.

$\text{n}=6, \text{p}=\frac{1}{2}$

Given: $\Big(\text{px}+\big(\frac{1}{\text{x}}\big)\Big)^\text{n}$
Hence, the fourth term, $ \text{T}_{\text{r}+1}={^{\text{n}}}{\text{C}}_{3}(\text{px})^{\text{n}-3}\big(\frac{1}{\text{x}}\big)3$
Given that the fourth term of the binomial expansion of $\Big(\text{px}+\big(\frac{1}{\text{x}}\big)\Big)^\text{n}$ is $\frac{5}{2},$ which is independent of $x.$
Hence, $\big(\frac{5}{2}\big)={^\text{n}}{\text{C}}_{3}(\text{px})^{\text{n}-3}\big(\frac{1}{\text{x}}\big)^3 …(1)$
On solving this, we get $n = 6.$
Now, substitute $n = 6$ in $\big(\frac{5}{2}\big)=6{ \ ^\text{n}}{\text{C}}_{3}(\text{p})3$
$20\text{p}^3=\frac{5}{2}$
$\text{p}^3=\frac{5}{2}$
$\text{p}=\frac{1}{2}$
Therefore, $n = 6$ and $\text{p}=\frac{1}{2}.$

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MCQ 1561 Mark

If tr is the $r^{th}$ term in the expansion of $(1+ x)101,$ then what is the ratio $\frac{\text{t}20}{\text{t}19}$ equal to?

A
$\frac{20\text{x}}{19}$
B
$83\text{x}$
C
$19\text{x}$
✓
$\frac{83\text{x}}{19}$

Answer

Correct option: D.

$\frac{83\text{x}}{19}$

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MCQ 1571 Mark

Number of irrational terms in the expansion of $\Big(5^{\frac{1}{6}}+2^{\frac{1}{8}}\Big)^{100}$ is:

A
$96$
✓
$97$
C
$98$
D
$99$

Answer

Correct option: B.

$97$

$\text{T}_{\text{r}+1}=^{100}\text{C}_\text{r}5^{\frac{(\text{r}-100)}{6}}2^{\frac{\text{r}}{8}}$
Hence we get rational terms when
$r = 8k$ where $k$ is an integer and $\frac{8\text{k}-100}{6}$ is an integer
$r = 16, 40, 64, 88$
Hence we get in total $4$ rational terms.
However, total number of terms will be $101$
Hence total number of irrational terms is $101 - 4$
$= 97$ terms.

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MCQ 1581 Mark

What is the middle term in the expansion of $\Big(\frac{\text{x}\sqrt{\text{y}}}{3}-\frac{3}{\text{y}\sqrt{\text{x}}}\Big)^{12}?$

A
$ C(12,7) x^3 y^{-3} $
B
$ C(12,6) x^{-3} y^3 $
C
$ C(12,7) x^{-3} y^3 $
✓
$ C(12,6) x^3 y^{-3} $

Answer

Correct option: D.

$ C(12,6) x^3 y^{-3} $

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MCQ 1591 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:

A
$5$
✓
$7$
C
$11$
D
$14$

Answer

Correct option: B.

$7$

For the terms to be in $A.P,$ it must follow
$(n-2 r)^2=n+2$
In the above case $r=2$
Substituting in the equation, we get
$ (n-4)^2=n+2 $
$ n^2-8 n+16=n+2 $
$ n^2-9 n+14=0 $
$ (n-2)(n-7)=0 $
$ n=2$ and $n=7$
However for $n = 2$ we will have only $3$ terms.
Hence the required answer is $7.$

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MCQ 1601 Mark

If $r^{th}$ term in the expansion of $\Big(2\text{x}^{2}-\frac{1}{\text{x}}\Big)^{12}$ is without $x$, then $r$ is equal to:

A
$8$
B
$7$
✓
$9$
D
$10$

Answer

Correct option: C.

$9$

$r^{th}$ term in the given expansion is ${^\text{20}}\text{C}_{\text{r}-1}\Big(2\text{x}^{2}\Big)^{12-\text{r}+1}\Big(\frac{-1}{\text{x}}\Big)^{\text{r}-1}$
$=(-1)^{\text{r}-1}\ {^\text{20}}\text{C}_{\text{r}-1}\ 2^{13-\text{r}}\ \text{x}^{26-2\text{r}-\text{r}+1}$
For this term to be independent of $x,$ we must have,
$27-3\text{r}=0$
$\Rightarrow \text{r}=9$
Hence, the term in the expansion is independent.

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MCQ 1611 Mark

The middle term in the expansion of$\Big(1+\frac{1}{\text{x}^{2}}\Big)\big(1+\text{x}^{2}\big)^{\text{n}}$ is:

A
$ { }^{2 n} \mathrm{Cn}{~x}^{2 n} $
B
$ { }^{2 n} \mathrm{Cn} {~x}^{-2 n} $
✓
$ { }^{2 n} \mathrm{Cn} $
D
$ { }^{2 n} C_{n-1} $

Answer

Correct option: C.

$ { }^{2 n} \mathrm{Cn} $

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MCQ 1621 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:

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 1631 Mark

The number of rational terms in the expansion of $(\sqrt3+\sqrt[4]{5})^{124}$ is:

A
$31$
✓
$32$
C
$33$
D
$34$

Answer

Correct option: B.

$32$

$\text{T}_\text{r}=^{124}\text{C}_{\text{r}-1}(\sqrt3)^{125-\text{r}}(\sqrt[4]{5})^{\text{r}-1}$
When both the terms are rational, $T_r$ will be rational.
Hence, $\frac{125-\text{r}}{2}$ and $\frac{\text{r}-1}{4}$ both must be integers.
Therefore, $r$ must be of the form $4k + 1,$ where $k$ is an integer.
There are $125$ terms in the expansion.
Hence, $k$ can assume values from $0$ to $31.$
Hence, there are $32$ values of $k$ and $32$ rational terms in the expansion.

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MCQ 1641 Mark

If in the expansion of $(1+\text{y})^{\text{n}},$ the coefficients of $5^{th}, 6^{th}$ and $7^{th}$ terms are in $A.P.,$ then $n$ is equal to:

A
$7, 11$
✓
$7, 14$
C
$8, 16$
D
None of these.

Answer

Correct option: B.

$7, 14$

Coefficients of the $5^{th}, 6^{th}$ and $7^{th}$ terms in the given expansion are ${^\text{n}}\text{C}_{\text{4}},{^\text{n}}\text{C}_{\text{5}}$ and ${^\text{n}}\text{C}_{\text{6}}.$
These coefficients are in $AP.$
Thus, we have
$2\ {^\text{n}}\text{C}_{\text{5}}={^\text{n}}\text{C}_{\text{4}}+{^\text{n}}\text{C}_{\text{6}}$
On dividing both sides by ${^\text{n}}\text{C}_{\text{5}},$ we get
$2=\frac{{^\text{n}}\text{C}_{\text{4}}}{{^\text{n}}\text{C}_{\text{5}}}+\frac{{^\text{n}}\text{C}_{\text{6}}}{{^\text{n}}\text{C}_{\text{5}}}$
$\Rightarrow 2=\frac{5}{\text{n}-4}+\frac{\text{n}-5}{6}$
$\Rightarrow 12\text{n}-48=30+\text{n}^{2}-4\text{n}-5\text{n}+20$
$\Rightarrow \text{n}^{2}-21\text{n}+98=0$
$\Rightarrow (\text{n}-14)(\text{n}-7)=0$
$\Rightarrow \text{n}=7,14$

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MCQ 1651 Mark

The fourth term in the expansion of $(x - 2y)^{12}$ is:

✓
$ -1760 x^9 \times y^3 $
B
$ -1670 x^9 \times y^3 $
C
$-7160 x^9 \times y^3 $
D
$ -1607 x^9 \times y^3 $

Answer

Correct option: A.

$ -1760 x^9 \times y^3 $

We know that the general term of an expansion $(a+b) n$ is $T_{r+1}={ }^n C_r a^{n-r}$ br.
Now, we have to find the fourth term in the expansion $(x-2 y)^{12}$
Hence, $r=3, a=x, b=-2 y, n=12$.
Now, substitute the values in the formula, we get
$T_{3+1}={ }^{12} C_3 x^{12-3}(-2 y)^3$
On solving this, we get
$T_4=-1760 x^9 y^3$

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MCQ 1661 Mark

${ }^5 \mathrm{C}_0+2 \cdot{ }^5 \mathrm{C}_1+2^2 \cdot{ }^5 \mathrm{C}_2+2^3 \cdot{ }^5 \mathrm{C}_3+2^4 \cdot{ }^5 \mathrm{C}_4+2^5 \cdot{ }^5 \mathrm{C}_5=$

A
$32$
✓
$243$
C
$64$
D
$729$

Answer

Correct option: B.

$243$

Just calculate it by substituting the values of $^5C_1,\ ^5C_2$ etc as $5, 10$ and just add them to get $243.$

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MCQ 1671 Mark

In the expansion of $\Big(\sqrt{2}+\sqrt[5]{3}\Big)^{120}$ the number of irrational terms is:

A
12
B
13
✓
108
D
54

Answer

Correct option: C.

108

Solution:

Total number of rational terms is

$\frac{120}{\text{L}.\text{C}.\text{M}(5,2)}+1$

$\frac{120}{10}+1$

$=13$

Hence total number of irrational terms are

= 121 - 13

= 108

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MCQ 1681 Mark

$(1-\sqrt{2})^{6}=$

A
$98-70\sqrt{2}$
✓
$99-70\sqrt{2}$
C
$99+70\sqrt{2}$
D
$98+70\sqrt{2}$

Answer

Correct option: B.

$99-70\sqrt{2}$

$(1-\sqrt{2})^{6}$
$=((1-\sqrt{2})^{2})^{3}$
$=(1+2-2\sqrt{2})^{3}$
$=(3-2\sqrt{2})^{3}$
$=27-16\sqrt{2}-54\sqrt{2}+72$
$=99-70\sqrt{2}.$

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MCQ 1691 Mark

If $\Big(\frac{\text{P}}{\text{q}}\Big)=0$ for $p < q p, \text{q}\in\text{W}$ then $\sum^\limits{\infty}_{\text{r}=0}\big(\frac{\text{n}}{2\text{r}})$

A
$ 2^n $
✓
$ 2^{n-1} $
C
$ 2^{2 n-1} $
D
$ 2^n C_n $

Answer

Correct option: B.

$ 2^{n-1} $

$\sum{^\text{n}}\text{C}_{2\text{r}}$
Is the sum of even odd term in the binomial expansion of $(1 + x)^n$
Hence
$\sum{^\text{n}}\text{C}_{2\text{r}}$will always be equal to $2^{n-1}$.

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MCQ 1701 Mark

The coefficient of $x - 7$ in the expansion of $\Big[\text{ax}2-\frac{1}{\text{bx}}^{2}\Big]11$ will be:

A
$\frac{462}{\text{b}^5}\times\text{a}^6$
✓
$\frac{462\text{a}^2}{\text{b}^6}$
C
$\frac{462\text{a}^5}{\text{b}^6}$
D
$\frac{-462\text{a}^6}{\text{b}^5}$

Answer

Correct option: B.

$\frac{462\text{a}^2}{\text{b}^6}$

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MCQ 1711 Mark

Choose the correct answer.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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MCQ 1721 Mark

The number of irrational terms in the expansion of $\Big(4^{\frac{1}{5}}+7^{\frac{1}{10}}\Big)^{45}$ is:

A
$40$
B
$5$
✓
$41$
D
None of these

Answer

Correct option: C.

$41$

Total number of terms in the expansion of $\Big(4^{\frac{1}{5}}+7^{\frac{1}{10}}\Big)^{45}$ is $45 + 1 = 46$
The general term in the expansion is $\text{T}{\text{r}+1}={^{45}}\text{C}_{\text{r}}\times4\frac{45-\text{r}}{5}\times7^{\frac{\text{r}}{10}}\text{T}_{\text{r}+1}$ is rational if $r = 0, 10, 20, 30, 40$
$\therefore$ Number of rational terms $= 5$
$\therefore$ Number of irrational terms $= 46 - 5 = 41$

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M.C.Q (1 Marks) - Page 4 - Maths STD 11 Science Questions - Vidyadip