3 Marks Question

Question 13 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
$\begin{array}{l}\lim _{x \rightarrow 2}\left(\frac{x^2-4}{\sqrt{x+2}-\sqrt{3 x-2}}\right)=\lim _{x \rightarrow 2} \frac{\frac{d}{d x}\left(x^2-4\right)}{\frac{d}{d x}(\sqrt{x+2}-\sqrt{3 x-2})} \\ \lim _{x \rightarrow 2}\left(\frac{x^2-4}{\sqrt{x+2}-\sqrt{3 x-2}}\right)=\lim _{x \rightarrow 2} \frac{2 x}{\frac{1}{2 \sqrt{x+2}}-\frac{3}{2 \sqrt{x-2}}} \\ \lim _{x \rightarrow 2}\left(\frac{x^2-4}{\sqrt{x+2}-\sqrt{3 x-2}}\right)=\frac{4}{\frac{1}{2 \sqrt{2+2}}-\frac{3}{2 \sqrt{5-2}}} \\ \lim _{x \rightarrow 2}\left(\frac{x^2-4}{\sqrt{x+2}-\sqrt{3 x-2}}\right)=\frac{8}{\frac{1}{2}-\frac{3}{2}} \\ \lim _{x \rightarrow 2}\left(\frac{x^2-4}{\sqrt{x+2}-\sqrt{3 x-2}}\right)=-8\end{array}$
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 23 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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Question 33 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

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

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Question 43 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. $a^2,(a r)^2,\left(a r^2\right)^2, \ldots \infty$
$S_{\infty}=\frac{a^2}{1-r^2}$
$\begin{array}{l}
\Rightarrow \frac{a^2}{1-r^2}=\frac{9}{2} \ldots(ii) \\
\Rightarrow 2 a^2=9\left(1-r^2\right) \\
\Rightarrow 2[3(1-r)]^2=9-9 r^2[\text { From }(i)] \\
\Rightarrow 18\left(1+r^2-2 r\right)=9-9 r^2 \\
\Rightarrow 18-9+18 r^2+9 r^2-36 r=0 \\
\Rightarrow 27 r^2-36 r+9=0 \\
\Rightarrow 3\left(9 r^2-12 r+3\right)=0 \\
\Rightarrow 9 r^2-12 r+3=0 \\
\Rightarrow 9 r^2-9 r-3 r+3=0 \\
\Rightarrow 9 r(r-1)-3(r-1)=0 \\
\Rightarrow(9 r-3)(r-1)=0 \\
\Rightarrow r=\frac{1}{3} \text { and } r=1\end{array}$
But, r = 1 is not possible
$\therefore r=\frac{1}{3}$
Now, substituting $r=\frac{1}{3}$ in $\frac{a}{1-r}=3$
$\begin{array}{l} a =3\left(1-\frac{1}{3}\right) \\ \Rightarrow a=3 \times \frac{2}{3}=2\end{array}$
Therefore the first term is 2 and common difference is $\frac{1}{3}$

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Question 53 Marks

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

Answer

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

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Question 63 Marks

If pth, 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
$\begin{array}{l}\therefore A+(p-1) d=a \ldots \text { (i) } \\ A+(q-1) d=b \ldots \text { (ii) }\end{array}$
And A + (r – 1)d = c …..(iii)
For G.P., we have
$\begin{array}{l}x R^{p-1}=a \ldots \text { (iv) } \\ x R^{q-1}=b \ldots . .(v)\end{array}$
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
$a^{b-c} \cdot b^{c-a}-c^{a-b}=1$
L.H.S. $a ^{ b - c } \cdot b ^{ c - a } \cdot c ^{ a - b }$
$\begin{array}{l}\text { L.H.S. } a^{b-c} \cdot b^{c-a} \cdot c^{a-b} \\
=\left[x R^{p-1}\right]^{(q-r) d} \cdot\left[x R^{q-1}\right]^{(r-p) d}\left[x R^{r-1}\right]^{(p-q)} d \\ \text { [from (i), (ii), (iii), (iv), (v), (vi), (vii), (viii), (ix), (x) } \\
=x^{(q-r) d} \cdot R^{(p-1)(q-r) d} \cdot x^{(r-p) d} \cdot R^{(q-1)(r-p) d} \cdot x^{(p-q) d} \cdot R^{(r-1)(p-q) d} \\
=x^{(q-r) d+(r-p) d+(p-q) d} R^{(p-1)(q-r) d+(q-1)(r-p) d+(r-1)(p-q) d} \\
=x^{(q-r+r-p+p-q) d} \cdot R^{(p d-p-q+r+q r-p-r+p+p r-q r-p+q) d} \\
=x^{(0) d} \cdot R^{(0) d}=x^0 \cdot R^0 \text { R.H.S. }\end{array}$
L.H.S. = R.H.S. Hence proved.

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Question 73 Marks

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

Answer

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

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Question 83 Marks

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

Answer

$\begin{array}{l}\text { Here }(3+a x)^9={ }^9 C_0(3)^9+{ }^9 C_1(3)^8(a x)+{ }^9 C_2(3)^7(a x)^2+{ }^9 C_3(3)^6(a x)^3+\ldots \\ ={ }^9 C_0(3)^9+{ }^9 C_1(3)^8 \cdot a \cdot x+{ }^9 C_2(3)^7(a)^2 \cdot x^2+{ }^9 C_3(3)^6 \cdot a^3 x^3+\ldots\end{array}$
$\therefore$ 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
$\begin{array}{l}{ }^9 C_2(3)^7 a^2={ }^9 C_3(3)^6 a^3 \Rightarrow 36 \cdot 3^7 a^2=84 \cdot 3^6 \cdot a^3 \\ \Rightarrow a=\frac{36 \cdot 3^7}{84 \cdot 3^6}=\frac{108}{84}=\frac{9}{7} .\end{array}$

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Question 93 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 ABC. Then
$\begin{array}{l}A B=\sqrt{(-1-0)^2+(6-7)^2+(6-10)^2}=\sqrt{1+1+16}=\sqrt{18} \\ B C=\sqrt{(-4+1)^2+(9-6)^2+(6-6)^2}=\sqrt{9+9+0}=\sqrt{18} \\ A C=\sqrt{(-4-0)^2+(9-7)^2+(6-10)^2}=\sqrt{16+4+16}=\sqrt{36} \\ N o w,(A B)^2=18,(B C)^2=18,(A C)^2=36 \\ \therefore(A C)^2=(A B)^2+(B C)^2\end{array}$
Hence, $\triangle ABC$ is a right-angled triangle.

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