3 Marks Question

Question 13 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,
$\begin{array}{l}\frac{a}{r} \times a \times a r=729 \\ \Rightarrow \quad a^3=729 \\ \Rightarrow a =9\end{array}$
From the given conditions we can write ,
$\left(\frac{a}{r} \times a\right)+(a \times a r)+\left(\frac{a}{r} \times a r\right)=819$
$\begin{array}{l}\Rightarrow \quad \frac{81}{r}+81 r+81=819 \\ \Rightarrow \quad \frac{9}{r}+9 r+9=91 \\ \Rightarrow 9+9 r^2+9 r=91 r \\ \Rightarrow 9 r^2-82 r+9=0 \\ \Rightarrow 9 r^2-81 r-r+9=0 \\ \Rightarrow 9 r(r-9)-1(r-9)=0\end{array}$
$r=9, \frac{1}{9}$
So, required G.P. are
81, 9, 1, ....
or, 1, 9, 81, ....

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Question 23 Marks

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

Answer

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

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Question 33 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
$\begin{array}{l}=n(T)-n(T \cap C) \\ =12-7 \\ =5\end{array}$
iii. Only coffee but not tea
$\begin{array}{l}=n(C)-n(T \cap C) \\ =15-7 \\ =8\end{array}$
v. Neither tea nor coffee
$\begin{array}{l}=n(U)-n(T \cup C) \\ =25-20 \\ =5\end{array}$

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Question 43 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
$\frac{1}{r}+1+r=\frac{39}{10}$
$\begin{array}{l}\Rightarrow 1+ r + r ^2=\frac{39}{10} r \\ \Rightarrow 10+10 r +10 r ^2-39 r =0 \\ \Rightarrow 10 r ^2-29 r +10=0 \\ \Rightarrow 10 r ^2-25 r -4 r +10=0 \\ \Rightarrow 5 r (2 r -5)-2(2 r -5)=0 \\ \Rightarrow(5 r -2)(2 r -5)=0 \\ \Rightarrow r =\frac{2}{5} \text { or } \frac{5}{2}\end{array}$
corresponding terms of the G.P
$\begin{array}{l}\text { i. } \text { when } r=\frac{2}{5} \\ \Rightarrow \frac{5}{2}, 1, \frac{2}{5} \\ \text { ii. when } r=\frac{5}{2} \\ \quad \Rightarrow \frac{2}{5}, 1, \frac{5}{2}\end{array}$

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Question 53 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 63 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 73 Marks

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

Answer

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Question 83 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
$\begin{array}{l}(x+a)^{2 n}+(x-a)^{2 n}=(P+Q)^2+(P-Q)^2 \\ =P^2+Q^2+2 P Q+P^2+Q^2-2 P Q \\ =2 P^2+2 Q^2=2\left(P^2+Q^2\right)\end{array}$
(ii) Multiplying(i) and (ii) we have
$\begin{array}{l}(x+a)^n(x-a)^n=(P+Q)(P-Q) \\ \left(x^2-a^2\right)^n=P^2-Q^2\end{array}$

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Question 93 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).
$\begin{array}{l}\therefore AD = BD \\ \sqrt{(0-3)^2+(y-2)^2+(z+1)^2}=\sqrt{(0-1)^2+(y+1)^2+(z-0)^2}\end{array}$
Squaring both sides,
$\begin{array}{l}(0-3)^2+(y-2)^2+(z+1)^2=(0-1)^2+(y+1)^2+(z-0)^2 \\ 9+y^2-4 y+4+z^2+2 z+1=1+y^2+2 y+1+z^2 \\ -6 y+2 z+12=0\end{array}$
Also, 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,
$\begin{array}{l}(0-3)^2+(y-2)^2+(z+1)^2=(0-2)^2+(y-1)^2+(z-2)^2 \\ 9+y^2-4 y+4+z^2+2 z+1=4+y^2-2 y+1+z^2-4 z+4\end{array}$
-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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