3 Marks Question

Question 13 Marks

Out of $25$ members in a family, $12$ like to take tea, $15$ like to take coffee and $7$ like to take coffee and tea both. How many like
$i.$ at least one of the two drinks
$ii.$ only tea but not coffee
$iii.$ only coffee but not tea
$iv.$ neither tea nor coffee

Answer

Given that, $n(T) = 12$
$n(C) = 15$
$n(T \cap C)=7$
$i. n(T \cup C)=n(T)+n(C)-n(T \cap C)$
$= 12 + 15 - 7$
$n(T \cup C)=20$
$20$ members like at least one of the two drinks.
$i.$ Only tea but not coffee
$=n(T)-n(T \cap C)$
$=12-7$
$=5$
$iii.$ Only coffee but not tea
$=n(C)-n(T \cap C)$
$=15-7$
$=8$
$v.$ Neither tea nor coffee
$=n(U)-n(T \cup C)$
$=25-20$
$=5$

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Question 23 Marks

The sum of first three terms of a $G.P.$ is $\frac{39}{10}$ and their product is $1.$ Find the common ratio and the terms.

Answer

Let $\frac{a}{r}, a, ar$ be first three terms of the given $G.P.$
$\frac{a}{r}+a+a r=\frac{39}{10} \ldots (i)$
$\left(\frac{a}{r}\right)( a )( ar )=1 \ldots( ii )$
From $(ii)$ we obtain $a ^3=1$
$\Rightarrow a =1 ($considering real roots only$)$
Substituting $a = 1$ in equation $(i),$ we obtain
$\frac{1}{r}+1+r=\frac{39}{10}$
$\Rightarrow 1+ r + r ^2=\frac{39}{10} r$
$\Rightarrow 10+10 r +10 r ^2-39 r =0$
$\Rightarrow 10 r ^2-29 r +10=0$
$\Rightarrow 10 r ^2-25 r -4 r +10=0$
$\Rightarrow 5 r (2 r -5)-2(2 r -5)=0$
$\Rightarrow(5 r -2)(2 r -5)=0$
$\Rightarrow r =\frac{2}{5} \text { or } \frac{5}{2}$
corresponding terms of the $G.P$
when $r=\frac{2}{5}$
$\Rightarrow \frac{5}{2}, 1, \frac{2}{5}$
when $r=\frac{5}{2}$
$ \Rightarrow \frac{2}{5}, 1, \frac{5}{2}$

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Question 33 Marks

The sum of three numbers in $G.P.$ is $14.$ If the first two terms are each increased by $1$ and the third term decreased by $1,$ the resulting numbers are in $A.P.$ Find the numbers.

Answer

Let three number in $G.P.$ are $\frac{a}{r}, a, ar$
Here,
$\frac{a}{r} \times a \times a r=729$
$\Rightarrow a^3=729$
$\Rightarrow a =9$
From the given conditions we can write ,
$\left(\frac{a}{r} \times a\right)+(a \times a r)+\left(\frac{a}{r} \times a r\right)=819$
$\Rightarrow \frac{81}{r}+81 r+81=819$
$\Rightarrow \frac{9}{r}+9 r+9=91$
$\Rightarrow 9+9 r^2+9 r=91 r$
$\Rightarrow 9 r^2-82 r+9=0$
$\Rightarrow 9 r^2-81 r-r+9=0$
$\Rightarrow 9 r(r-9)-1(r-9)=0$
$r=9, \frac{1}{9}$
So, required $G.P.$ are
$81, 9, 1, ....$
or, $1, 9, 81, ....$

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Question 43 Marks

Differentiate the function: $3^{ x -5}$

Answer

We have,
$\frac{d}{d x} x^n=n x^{n-1}$
Therefore,
$\frac{d}{d x} 3 x^{-5}=3(-5) x^{-5-1}$
$=-15 x^{-6}$

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Question 53 Marks

If f(x) = mx + c and f(0) = f'(0) = 1. What is value of f(2)?

Answer

We have,
f(x) = mx + c ....(i)
Differentiating with respect to x, we get
f'(x) = m.1 + 0
$\Rightarrow f^{\prime}(x)=m \ldots$ (ii)
Put, x = 0 in (i) and (ii), we get
f(0) = c and f'(0) = m
$\Rightarrow 1=c$ and $1=m\left[\therefore f(0)=f^{\prime}(0)=1\right]$
Put the values of m and c in f(x) = mx + c, we get f(x) = x + 1.
$\therefore f(2)=2+1=3$. [Put $x=2$ in $f(x)=x+1]$

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Question 63 Marks

In the expansion of $(x+a)^n$, sums of odd and even terms are $P$ and $Q$ respectively, prove that
$i.\ 2\left(P^2+Q^2\right)=(x+a)^{2 n}+(x-a)^{2 n}$
$ii.\ P ^2- Q ^2=\left( x ^2- a ^2\right)^{ n }$

Answer

Here $(x+a)^n={ }^n C_0 x^n+{ }^n C_1 x^{n-1} a+{ }^n C_2 x^{n-2} a^2+\ldots+{ }^n C_n a^n$
$= P + Q . . . (i)$
where $P={ }^n C_0 x^n+{ }^n C_3 x^{n-3} a^3+\ldots$
$Q={ }^n C_1 x^{n-1} a+{ }^n C_3 x^{n-3} a^3+\ldots$
Also $(x-a)^n={ }^n C_0 x^n-{ }^n C x^{n-1} a+^n C_2 x^{n-2} a^2+\ldots+(-1)^{n^n} C_n a^n \ldots$
$= P - Q$
$(i)$ Squaring and adding $(i)$ and $(ii)$ we have
$(x+a)^{2 n}+(x-a)^{2 n}=(P+Q)^2+(P-Q)^2$
$=P^2+Q^2+2 P Q+P^2+Q^2-2 P Q$
$=2 P^2+2 Q^2=2\left(P^2+Q^2\right)$
$(ii)$ Multiplying $(i)$ and $(ii)$ we have
$(x+a)^n(x-a)^n=(P+Q)(P-Q)$
$\left(x^2-a^2\right)^n=P^2-Q^2$

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Question 73 Marks

Using binomial theorem, expand: $(\sqrt{x}+\sqrt{y})^8$

Answer

We hand to find value of $(\sqrt{x}+\sqrt{y})^8$
Formula used: ${ }^n C_r=\frac{n!}{(n-r)!(r)!}$
$( a + b )^{ n }={ }^n C_0 a^n+{ }^n C_1 a^{n-1} b+{ }^n C_2 a^{n-2} b^2+\ldots \ldots+{ }^n C_{n-1} a b^{n-1}+{ }^n C_n b^n$
We have, $(\sqrt{x}+\sqrt{y})^8$
We can write $\sqrt{x}$ as $x^{\frac{1}{2}}$ and $\sqrt{y}$ as $y^{\frac{1}{2}}$
Now, we have to solve for $\left(x^{\frac{1}{2}}+y^{\frac{1}{2}}\right)^8$
$=\left[8 C_0\left(x^{\frac{2}{2}}\right)^{8-0}\right]+\left[8 C_1\left(x^{\frac{1}{2}}\right)^{8-1}\left(y^{\frac{2}{2}}\right)^1\right]+\left[8 C_2\left(x^{\frac{1}{2}}\right)^{8-2}\left(y^{\frac{1}{2}}\right)^2\right]+\left[{ }^8 C_3\left(x^{\frac{1}{2}}\right)^{8-3}\left(y^{\frac{1}{2}}\right)^3\right]$
$+\left[{ }^8 C_4\left(x^{\frac{1}{2}}\right)^{8-4}\left(\frac{1}{y^2}\right)^4\right]+\left[{ }^8 C_5\left(x^{\frac{1}{2}}\right)^{8-5}\left(y^{\frac{2}{2}}\right)^5\right]+\left[{ }^8 C_6\left(x^{\frac{1}{2}}\right)^{8-6}\left(y^{\frac{1}{2}}\right)^6\right]$
$+\left[8 C_7\left(x^{\frac{1}{2}}\right)^{8-7}\left(y^{\frac{1}{2}}\right)^7\right]+\left[8 C_8\left(y^{\frac{1}{2}}\right)^8\right]$
$=\left[\frac{81}{0(8-0)!}\left(x^{\frac{5}{2}}\right)\right]+\left[\frac{81}{11(8-1)!}\left(x^{\frac{2}{2}}\right)\left(y^{\frac{1}{2}}\right)\right]+\left[\frac{81}{2(8-2)!}\left(x^{\frac{6}{2}}\right)\left(y^{\frac{2}{2}}\right)\right]$
$+\left[\frac{81}{3(8-3)!}\left(x^{\frac{5}{2}}\right)\left(y^{\frac{3}{2}}\right)\right]+\left[\frac{81}{4(8-4)!}\left(x^{\frac{4}{2}}\right)\left(y^{\frac{4}{2}}\right)\right]$
$+\left[\frac{81}{5(8-5)!}\left(x^{\frac{2}{2}}\right)\left(y^{\frac{5}{2}}\right)\right]+\left[\frac{81}{6(8-6)!}\left(x^{\frac{2}{2}}\right)\left(\frac{6}{y^2}\right)\right]+\left[\frac{81}{7(8-7)!}\left(x^{\frac{1}{2}}\right)\left(\frac{2}{y^2}\right)\right]+\left[\frac{81}{8(8-8)( }\left(y^{\frac{5}{2}}\right)\right]$
$=\left[1\left(x^4\right)\right]+\left[8\left(x^{\frac{7}{2}}\right)\left(y^{\frac{1}{2}}\right)\right]+\left[28\left(x^3\right)(y)\right]+\left[56\left(x^{\frac{5}{2}}\right)\left(\frac{2}{y^2}\right)\right]$
$+\left[70\left(x^2\right)\left(y^2\right)\right]+\left[56\left(x^{\frac{3}{2}}\right)\left(y^{\frac{5}{2}}\right)\right]+\left[28\left(x^2\right)\left(y^3\right)\right]+\left[8\left(x^{\frac{1}{2}}\right)\left(y^{\frac{2}{2}}\right)\right]+\left[1\left(y^4\right)\right]$

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Question 83 Marks

Find the point in $yz-$plane which is equidistant from the points $A(3, 2, -1), B(1, -1, 0)$ and $C(2, 1, 2).$

Answer

The general point on $yz$ plane is $D(0, y, z).$
Consider this point is equidistant to the points $A(3, 2, -1), B(1, -1, 0)$ and $C(2, 1, 2).$
$\therefore \text{AD = BD}$
$\sqrt{(0-3)^2+(y-2)^2+(z+1)^2}=\sqrt{(0-1)^2+(y+1)^2+(z-0)^2}$
Squaring both sides,
$(0-3)^2+(y-2)^2+(z+1)^2=(0-1)^2+(y+1)^2+(z-0)^2$
$9+y^2-4 y+4+z^2+2 z+1=1+y^2+2 y+1+z^2$
$-6 y+2 z+12=0$
Also, $\text{AD = CD}$
$\sqrt{(0-3)^2+(y-2)^2+(z+1)^2}=\sqrt{(0-2)^2+(y-1)^2+(z-2)^2}$
Squaring both sides,
$(0-3)^2+(y-2)^2+(z+1)^2=(0-2)^2+(y-1)^2+(z-2)^2$
$9+y^2-4 y+4+z^2+2 z+1=4+y^2-2 y+1+z^2-4 z+4$
$-2y + 6z + 5 = 0 ….(2)$
By solving equation $(1)$ and $(2)$ we get
$y=\frac{31}{16} z=\frac{-3}{16}$
The point which is equidistant to the points $A (3,2,-1), B (1,-1,0)$ and $C (2,1,2)$ is $\left(\frac{31}{16}, \frac{-3}{16}\right)$.

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Question 93 Marks

Prove that: ${ }^{2 n} C_n=\frac{2^n[1 \cdot 3 \cdot 5 \ldots \ldots(2 n-1)]}{n!}$.

Answer

${ }^{2 n} C_n=\frac{2^n[1 \cdot 35 \ldots .(2 n-1)]}{n!}$
$=\frac{(2 n)!}{n!n!}$
$(2 n)(2 n-1)(2 n-2)(2 n-3) \ldots . .$
$=\frac{4 \cdot 3 \cdot 2 \cdot 1}{n!n!}$
$=\frac{\left[2^n \ldots .4 \cdot 2\right][(2 n-1) \ldots \cdot 3 \cdot 1]}{n!n!}$
$=\frac{2^n[1 \cdot 2 \cdot \ldots . n][1 \cdot 3 \cdot 5 \ldots \ldots \cdot(2 n-1)]}{n!n!}$
$=\frac{2^n \times n![1 \cdot 3 \cdot 5 \ldots . \cdot(2 n-1)]}{n!n!}$
$=\frac{2^n[1 \cdot 3 \cdot 5 \ldots \cdot(2 n-1)]}{n!}$

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