3 Marks Question

Question 13 Marks

For any sets $A$ and $B$ show that
$i. (A \cap B) \cup(A-B)=A$
$ii. A \cup(B-A)=A \cup B$

Answer

$\text { i. }(A \cap B) \cup(A-B)=A$
$\text { L.H.S. }=(A \cap B) \cup(A-B)$
$=( A \cap B ) \cup\left(A \cap B^{\prime}\right)\left[\therefore( A - B )= A \cap B^{\prime}\right]$
$= A \cap\left(B \cup B^{\prime}\right)\ [$By distributive law$]$
$= A \cap U \ [\left(B \cup B^{\prime}\right) = U = $Universal set $]$
$= A$
$=\text { R.H.S. }$
$\text { ii. } A \cup(B-A)=A \cup B$
$\text { L.H.S. }=A \cup(B-A)$
$= A \cup\left( B \cap A^{\prime}\right)\left[\therefore( B - A )= B \cap A^{\prime}\right]$
$=( A \cup B ) \cap\left(A \cap A^{\prime}\right) \ [$By distributive law$]$
$=( A \cup B ) \cap u \ [\therefore A \cup A^{\prime} = u =$ Universal set $]$
$= A \cup B$
$=\text { R.H.S. }$

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Question 23 Marks

If the sum of an infinite decreasing $G.P.$ is $3$ and the sum of the squares of its term is $\frac{9}{2}$, then write its first term and common difference.

Answer

Let us take a $G.P.$ whose first is a and common difference is $r.$
$\therefore S_{\infty}=\frac{a}{1-r}$
$\Rightarrow \frac{a}{1-r}=3 \ldots (i)$
And, sum of the terms of the $G.P. a^2,(a r)^2,\left(a r^2\right)^2, \ldots \infty$
$S_{\infty}=\frac{a^2}{1-r^2}$
$\Rightarrow \frac{a^2}{1-r^2}=\frac{9}{2} \ldots(ii)$
$\Rightarrow 2 a^2=9\left(1-r^2\right)$
$\Rightarrow 2[3(1-r)]^2=9-9 r^2[$ From $(i)]$
$\Rightarrow 18\left(1+r^2-2 r\right)=9-9 r^2$
$\Rightarrow 18-9+18 r^2+9 r^2-36 r=0$
$\Rightarrow 27 r^2-36 r+9=0$
$\Rightarrow 3\left(9 r^2-12 r+3\right)=0$
$\Rightarrow 9 r^2-12 r+3=0$
$\Rightarrow 9 r^2-9 r-3 r+3=0$
$\Rightarrow 9 r(r-1)-3(r-1)=0$
$\Rightarrow(9 r-3)(r-1)=0$
$\Rightarrow r=\frac{1}{3}$ and $r=1$
But, $r = 1$ is not possible
$\therefore r=\frac{1}{3}$
Now, substituting $r=\frac{1}{3}$ in $\frac{a}{1-r}=3$
$ a =3\left(1-\frac{1}{3}\right)$
$\Rightarrow a=3 \times \frac{2}{3}=2$
Therefore the first term is $2$ and common difference is $\frac{1}{3}$

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Question 33 Marks

If $p^{th},$ qth and rth terms of an $A.P$. and $G.P$. are both $a, b,$ and $c$ respectively. Show that
$a^{b-c} \cdot b^{c-a} \cdot c^{a-b}=1$

Answer

Let $A$ and d be the first term and common difference respectively of an $A.P$. and $x$ and $R$ be the first term and common ratio respectively of the $G.P$
$\therefore A+(p-1) d=a \ldots \text { (i) }$
$A+(q-1) d=b \ldots \text { (ii) }$
And $A + (r – 1)d = c …..(iii)$
For $G.P.,$ we have
$x R^{p-1}=a \ldots \text { (iv) }$
$x R^{q-1}=b \ldots . .(v)$
and $x R ^{ r -1}= c \ldots \ldots (vi)$
Subracting eq. $(ii)$ from eq. $(i)$ we get
$(p – q)d = a – b ….(vii)$
Similarly $, (q – r)d = b – c ….(viii)$
and $(r – p)d = c – a ….(ix)$
Now we have to prove that
$a^{b-c} \cdot b^{c-a}-c^{a-b}=1$
$\text{L.H.S}. a ^{ b - c } \cdot b ^{ c - a } \cdot c ^{ a - b }$
$\text { L.H.S. } a^{b-c} \cdot b^{c-a} \cdot c^{a-b}$
$=\left[x R^{p-1}\right]^{(q-r) d} \cdot$
$\left[x R^{q-1}\right]^{(r-p) d}\left[x R^{r-1}\right]^{(p-q)} d$
$[$from $(i), (ii), (iii), (iv), (v), (vi), (vii), (viii), (ix), (x) ]$
$=x^{(q-r) d} \cdot R^{(p-1)(q-r) d} \cdot $
$x^{(r-p) d} \cdot R^{(q-1)(r-p) d} \cdot x^{(p-q) d} \cdot R^{(r-1)(p-q) d}$
$=x^{(q-r) d+(r-p) d+(p-q) d} $
$R^{(p-1)(q-r) d+(q-1)(r-p) d+(r-1)(p-q) d}$
$=x^{(q-r+r-p+p-q) d} \cdot $
$R^{(p d-p-q+r+q r-p-r+p+p r-q r-p+q) d}$
$=x^{(0) d} \cdot R^{(0) d}=x^0 \cdot R^0 \text { R.H.S. }$
$\text{L.H.S. = R.H.S}$. Hence proved.

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Question 43 Marks

Evaluate $\lim _{x \rightarrow \pi} \frac{\sin (\pi-x)}{\pi(\pi-x)}$

Answer

$\text { Let } y=\lim _{x \rightarrow \pi} \frac{\sin (\pi-x)}{\pi(\pi-x)}\left[\frac{0}{0} \text { from }\right]$
$\text { Put } x=\pi+y, \text { as } x \rightarrow \pi, y \rightarrow 0$
$\therefore y=\lim _{y \rightarrow 0} \frac{\sin [\pi-\pi-y]}{\pi[\pi-\pi-y]}=\lim _{y \rightarrow 0} \frac{\sin (-y)}{-\pi y}$
$=\lim _{y \rightarrow 0} \frac{-\sin y}{-\pi y}$
$=\frac{1}{\pi} \lim _{y \rightarrow 0} \frac{\sin y}{y}$
$=\frac{1}{\pi} \times 1$
$=\frac{1}{\pi}$

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Question 53 Marks

Evaluate $\lim _{x \rightarrow 2}\left(\frac{x^2-4}{\sqrt{x+2}-\sqrt{3 x-2}}\right)$

Answer

To evaluate: $\lim _{x \rightarrow 2}\left(\frac{x^2-4}{\sqrt{x+2}-\sqrt{3 x-2}}\right)$
Formula used:
L'Hospital's rule
Let $f(x)$ and $g(x)$ be two functions which are differentiable on an open interval I except at a point a where
$\lim _{x \rightarrow a} f(x)=\lim _{x \rightarrow a} g(x)=0$ or $\pm \infty$
then $\lim _{x \rightarrow a} \frac{f(x)}{g(x)}=\lim _{x \rightarrow a} \frac{f^{\prime}(x)}{g^{\prime}(x)}$
As $x \rightarrow 0$,
we have $\lim _{x \rightarrow 2}\left(\frac{x^2-4}{\sqrt{x+2}-\sqrt{3 x-2}}\right)=\frac{0}{0}$
This represents an indeterminate form. Thus applying L'Hospital's rule,
we get $\lim _{x \rightarrow 2}\left(\frac{x^2-4}{\sqrt{x+2}-\sqrt{3 x-2}}\right)=\lim _{x \rightarrow 2} \frac{\frac{d}{d x}\left(x^2-4\right)}{\frac{d}{d x}(\sqrt{x+2}-\sqrt{3 x-2})}$
$\lim _{x \rightarrow 2}\left(\frac{x^2-4}{\sqrt{x+2}-\sqrt{3 x-2}}\right)=\lim _{x \rightarrow 2} \frac{2 x}{\frac{1}{2 \sqrt{x+2}}-\frac{3}{2 \sqrt{x-2}}}$
$\lim _{x \rightarrow 2}\left(\frac{x^2-4}{\sqrt{x+2}-\sqrt{3 x-2}}\right)=\frac{4}{\frac{1}{2 \sqrt{2+2}}-\frac{3}{2 \sqrt{5-2}}}$
$\lim _{x \rightarrow 2}\left(\frac{x^2-4}{\sqrt{x+2}-\sqrt{3 x-2}}\right)=\frac{8}{\frac{1}{2}-\frac{3}{2}}$
$\lim _{x \rightarrow 2}\left(\frac{x^2-4}{\sqrt{x+2}-\sqrt{3 x-2}}\right)=-8$
Thus, the value of $\lim _{x \rightarrow 2}\left(\frac{x^2-4}{\sqrt{x+2}-\sqrt{3 x-2}}\right)$ is $-8$

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Question 63 Marks

Expand $\left(1+x+x^2\right)^3$ using binomial expansion.

Answer

Let $y=x+x^2$.
Then, $\left(1+x+x^2\right)^3=(1+y)^3$
$={ }^3 C_0+{ }^3 C_1 y+{ }^3 C_2 y^2+{ }^3 C_3 y^3$
$=1+3 y+3 y^2+y^3$
$=1+3\left(x+x^2\right)+3\left(x+x^2\right)^2+\left(x+x^2\right)^3$
$=1+3\left(x+x^2\right)+3\left(x^2+2 x^3+x^4\right)+\left\{{ }^3 C_0 x^3\left(x^2\right)^0+{ }^3 C_1 x^{3-1}\left(x^2\right)^1+{ }^3 C_2 x^{3-2}\left(x^2\right)^2+{ }^3 C_3 x^0\left(x^2\right)^3\right\}$
$=1+3\left(x+x^2\right)+3\left(x^2+2 x^3+x^4\right)+\left(x^3+3 x^4+3 x^5+x^6\right)$
$=x^6+3 x^5+6 x^4+7 x^3+6 x^2+3 x+1$

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Question 73 Marks

Find a if the coefficient of $x^2$ and $x^3$ in the expansion of $(3+a x)^9$ are equal.

Answer

$\text { Here }(3+a x)^9={ }^9 C_0(3)^9+{ }^9 C_1(3)^8(a x)+{ }^9 C_2(3)^7(a x)^2+{ }^9 C_3(3)^6(a x)^3+\ldots$
$={ }^9 C_0(3)^9+{ }^9 C_1(3)^8 \cdot a \cdot x+{ }^9 C_2(3)^7(a)^2 \cdot x^2+{ }^9 C_3(3)^6 \cdot a^3 x^3+\ldots$
$\therefore$ Coefficient of $x^2={ }^9 C_2(3)^7 a^2$
Coefficient of $x^3={ }^9 C_3(3)^6 a^3$
It is given that
${ }^9 C_2(3)^7 a^2={ }^9 C_3(3)^6 a^3$
$\Rightarrow 36 \cdot 3^7 a^2=84 \cdot 3^6 \cdot a^3$
$\Rightarrow a=\frac{36 \cdot 3^7}{84 \cdot 3^6}$
$=\frac{108}{84}$
$=\frac{9}{7} .$

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Question 83 Marks

Verify that $(0, 7, 10), (-1, 6, 6)$ and $(-4, 9, 6)$ are the vertices of a right$-$angled triangle.

Answer

Let $A(0, 7, 10), B(-1, 6, 6)$ and $C(-4, 9, 6)$ be three vertices of triangle $\text{ABC.}$ Then
$A B=\sqrt{(-1-0)^2+(6-7)^2+(6-10)^2}=\sqrt{1+1+16}=\sqrt{18}$
$B C=\sqrt{(-4+1)^2+(9-6)^2+(6-6)^2}=\sqrt{9+9+0}=\sqrt{18}$
$A C=\sqrt{(-4-0)^2+(9-7)^2+(6-10)^2}=\sqrt{16+4+16}=\sqrt{36}$
Now,$\ce{(AB)^2=18,(BC)^2=18,(AC)^2}=36$
$\therefore \ce{(AC)^2=(AB)^2+(BC)^2}$
Hence, $\triangle ABC$ is a right$-$angled triangle.

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Question 93 Marks

It is required to seat $5$ men and $3$ women in a row so that the women occupy the even places. How many such arrangements are possible?

Answer

To find: number of arrangements in which women sit in even places
Condition: women occupy even places
Here the total number of people is $8$.

__	$W$	__	$W$	__	$W$	__	$W$
$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$

In this question first, the arrangement of women is required.
The positions where women can be made to sit is $2^{\text {nd }}, 4^{\text {th }}, 6^{\text {th }}, 8^{\text {th }}$.
There are $4$ even places in which $3$ women are to be arranged.
Women can be placed in $P (4,3)$ ways.
The rest $5$ men can be arranged in $5!$ ways.
Therefore, the total number of arrangements is $P (4,3) \times 5 !$
Formula:
Number of permutations of $n$ distinct objects among $r$ different places, where repetition is not allowed, is
$P(n, r)=\frac{n!}{(n-r)!}$
Therefore, a permutation of $4$ different objects in $3$ places and the arrangement of $5$ men are
$\begin{array}{l}P(4,3) \times 5!=\frac{4!}{(4-3)!} \times 5! \\ =\frac{24}{1} \times 120=2880\end{array}$
Hence number of ways in which they can be seated is $2880$

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