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MATHS 3 Marks Question

Model Paper 3 · Rajasthan Board (RBSE) STD 11 Science (Rajasthan - English Medium). 9 questions.

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3 Marks Question

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9 questions · timed · auto-graded

Question 13 Marks
Solve inequation and represent the solution set on the number line: $\frac{2 x-1}{12}-\frac{x-1}{3}<\frac{3 x+1}{4}$ where $x \in R$
Answer
Given:
$\frac{2 x-1}{12}-\frac{x-1}{3}<\frac{3 x+1}{4}$, where $x \in R$.
Multiply by 12 on both sides in the above equation
$\begin{array}{l}\Rightarrow 12\left(\frac{2 x-1}{12}\right)-12\left(\frac{x-1}{3}\right)<12\left(\frac{3 x+1}{4}\right) \\ \Rightarrow(2 x-1)-4(x-1)<3(3 x+1) \\ \Rightarrow 2 x-1-4 x+4<9 x+3 \\ \Rightarrow 3-2 x<9 x+3\end{array}$
Now, subtracting 3 on both sides in the above equation
$\begin{array}{l}\Rightarrow 3-2 x-3<9 x+3-3 \\ \Rightarrow-2 x<9 x\end{array}$
Now, subtracting 9x from both the sides in the above equation
$\begin{array}{l}\Rightarrow-2 x-9 x<9 x-9 x \\ \Rightarrow-11 x<0\end{array}$
Multiplying -1 on both the sides in above equation
$\begin{array}{l}\Rightarrow(-11 x)(-1)>(0)(-1) \\ \Rightarrow 11 x>0\end{array}$
Dividing both sides by 11 in above equation
$\Rightarrow \frac{11 x}{11}>\frac{0}{11}$
Therefore,
$\Rightarrow>x>0$
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Question 33 Marks
Evaluate $\left[\frac{1}{1-4 i}-\frac{2}{1+i}\right]\left[\frac{3-4 i}{5+i}\right]$ to the standard form.
Answer

$\begin{array}{l}{\left[\frac{1}{1-4 i}-\frac{2}{1+i}\right]\left[\frac{3-4 i}{5+i}\right]=\left[\frac{1+i-2+8 i}{(1-4 i)(1+i)}\right]\left[\frac{3-4 i}{5+i}\right]} \\ =\left[\frac{-1+9 i}{1+i-4 i-4 i^2}\right]\left[\frac{3-4 i}{5+i}\right]=\left[\frac{-1+9 i}{5-3 i}\right]\left[\frac{3-4 i}{5+i}\right] \\ =\frac{-3+4 i+27 i-36 i^2}{25+5 i-15 i-3 i^2}=\frac{33+31 i}{28-10 i} \times \frac{28+10 i}{28+10 i} \\ =\frac{924+330 i+868 i+3102^2}{(28)^2-(10 i)^2}=\frac{614+1198 i}{784+100}\left(\because i^2=-1\right) \\ =\frac{2(307+599 i)}{884}=\frac{307+599 i}{442}\end{array}$
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Question 43 Marks
Let $A=\{a, e, i, o, u\}, B=\{a, d, e, o, v)$ and $C=\{e, o, t, m]$. Using Venn diagrams, verify that: $A \cap(B \cup C)=$ $(A \cap B) \cup(A \cap C)$
Answer
Here, it is given: $A=\{a, e, i, o, u\}, B=\{a, d, e, o, v\}$ and $C=\{e, o, t, m\}$ 
$B \cup C=\{a, d, v, e, o, t, m\}$ and $A \cap(B \cup C)=\{a, e, o\}$ 
LHS
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R.H.S: $A \cap B=\{a, e, o\}$ and $A \cap C=\{ e , o \}$
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$(A \cap B) \cup(A \cap C)=\{a, e, o\}$ 
L.H.S = R.H.S. [Verified]

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Question 53 Marks
If $( a + ib )=\frac{c+i}{c-i}$, where c is real, prove that $a ^2+ b ^2=1$ and $\frac{b}{a}=\frac{2 c}{c^2-1}$.
Answer

$\begin{array}{l}\text { Here } a+i b=\frac{c+i}{c-i} \\ =\frac{c+i}{c-i} \times \frac{c+i}{c+i}=\frac{(c+i)^2}{c^2-i^2} \\ =\frac{c^2+2 c i+i^2}{c^2+1} \\ =\frac{c^2-1}{c^2+1}+\frac{2 c}{c^2+1} i\end{array}$
Comparing real and imaginary parts on both sides, we have 
$a=\frac{c^2-1}{c^2+1}$ and $b=\frac{2 c}{c^2+1}$
Now $a^2+b^2=\left(\frac{c^2-1}{c^2+1}\right)^2+\left(\frac{2 c}{c^2+1}\right)^2$
$=\frac{\left(c^2-1\right)^2+4 c^2}{\left(c^2+1\right)^2}=\frac{\left(c^2+1\right)^2}{\left(c^2+1\right)^2}=1$
Also $\frac{b}{a}=\frac{\frac{2 c}{c^2+1}}{\frac{c^2-1}{c^2+1}}=\frac{2 c}{c^2-1}$
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Question 63 Marks
Expand the given expression $\left(x+\frac{1}{x}\right)^6$
Answer
Using binomial theorem for the expansion of $\left(x+\frac{1}{x}\right)^6$ we have
$\left(x+\frac{1}{x}\right)^6=={ }^6 C_0(x)^6+{ }^6 C_1(x)^5\left(\frac{1}{x}\right)+{ }^6 C_2(x)^4\left(\frac{1}{x}\right)^2+{ }^6 C_3(x)^3\left(\frac{1}{x}\right)^3$
$+{ }^6 C_4(x)^2\left(\frac{1}{x}\right)^4+{ }^6 C_5(x)\left(\frac{1}{x}\right)^5+{ }^6 C_6\left(\frac{1}{6}\right)^6$
$\begin{array}{l}=x^6+6 \cdot x^5 \cdot \frac{1}{x}+15 \cdot 4 x^4 \cdot \frac{1}{x^2}+20 \cdot x^3 \cdot \frac{1}{x^3}+15 \cdot x^2 \cdot \frac{1}{x^4}+6 \cdot x \cdot \frac{1}{x^5}+\frac{1}{x^6} \\ =x^6+6 x^4+15 x^2+20+\frac{15}{x^2}+\frac{6}{x^4}+\frac{1}{x^6}\end{array}$

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Question 73 Marks
Using binomial theorem, prove that $\left(2^{3 n }-7 n -1\right)$ is divisible by 49 , where $n \in N$
Answer
To prove: $\left(2^{3 n}-7 n-1\right)$ is divisible by 49 , where $n \in N$
$\begin{array}{l}\left(2^{3 n}-7 n-1\right)=\left(2^3\right) n-7 n-1 \\ =8^n-7 n-1 \\ =(1+7)^n-7 n-1\end{array}$
Now using binomial theorem.. 
$\begin{array}{l}\Rightarrow{ }^n C_0 1^n+{ }^n C_1 1^{n-1} 7+{ }^n C_2 1^{n-2} 7^2+\ldots \ldots+{ }^n C_{n-1} 7^{n-1}+{ }^n C_n 7^n-7 n-1 \\ ={ }^n C_0+{ }^n C_1 7+{ }^n C_2 7^2+\ldots \ldots+{ }^n C_{n-1} 7^{n-1}+{ }^n C_n 7^n-7 n-1 \\ =1+7 n+7^2\left[{ }^n C_2+{ }^n C_3 7+\ldots+{ }^n C_{n-1} 7^{n-3}+{ }^n C_n 7^{n-2}\right]-7 n-1 \\ =7^2\left[{ }^n C_2+{ }^n C_3 7+\ldots+{ }^n C_{n-1} 7^{n-3}+{ }^n C_n 7^{n-2}\right] \\ =49\left[{ }^n C_2+{ }^n C_3 7+\ldots+{ }^n C_{n-1} 7^{n-3}+{ }^n C_n 7^{n-2}\right] \\ =49 K, \text { where } K=\left({ }^n C_2+{ }^n C_3 7+\ldots++{ }^n C_{n-1} 7^{n-3}+{ }^n C_n 7^{n-2}\right)\end{array}$
Now, $\left(2^{3 n}-7 n-1\right)=49 K$
Therefore $\left(2^{3 n}-7 n-1\right)$ is divisible by 49 .
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Question 83 Marks
Show that the points (a, b, c), (b, c, a) and (c, a, b) are the vertices of an equilateral triangle.
Answer
Let $A(a, b, c) B(b, c, a)$, and $C(c, a, b)$ be the vertices of $\triangle A B C$. Then.
$\begin{array}{l}AB=\sqrt{(b-a)^2+(c-b)^2+(a-c)^2} \\
=\sqrt{b^2-2 a b+a^2+c^2-2 b c+b^2+a^2-2 c a+c^2} \\
=\sqrt{2 a^2+2 b^2+2 c^2-2 a b-2 b c-2 c a} \\
AB=\sqrt{2\left(a^2+b^2+c^2-a b-b c-c a\right)} \\
BC=\sqrt{(c-b)^2+(a-c)^2+(b-a)^2} \\
=\sqrt{c^2-2 b c+b^2+a^2-2 c a+c^2+b^2-2 a b+a^2} \\
=\sqrt{2 a^2+2 b^2+2 c^2-2 a b-2 b c-2 c a} \\
BC=\sqrt{2\left(a^2+b^2+c^2-a b-b c-c a\right)} \\
CA=\sqrt{(a-c)^2+(b-a)^2+(c-b)^2} \\
=\sqrt{a^2-2 c a+c^2+b^2-2 a b+a^2+c^2-2 b c+b^2} \\
=\sqrt{2 a^2+2 b^2+2 c^2-2 a b-2 b c-2 c a} \\
CA=\sqrt{2\left(a^2+b^2+c^2-a b-b c-c a\right)}\end{array}$
$\therefore A B=B C=C A$
Therefore, $\triangle ABC$ is an equilateral triangle.

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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). 
$\therefore A D=B D$
$\sqrt{(0-3)^2+(y-2)^2+(z+1)^2}=\sqrt{(0-1)^2+(y+1)^2+(z-0)^2}$
Squaring both sides, 
$\begin{array}{l}(0-3)^2+(y-2)^2+(z+1)^2=(0-1)^2+(y+1)^2+(z-0)^2 \\ 9+y^2-4 y+4+z^2+2 z+1=1+y^2+2 y+1+z^2 \\ -6 y+2 z+12=0 \ldots(1)\end{array}$
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, 
$\begin{array}{l}(0-3)^2+(y-2)^2+(z+1)^2=(0-2)^2+(y-1)^2+(z-2)^2 \\ 9+y^2-4 y+4+z^2+2 z+1=4+y^2-2 y+1+z^2-4 z+4 \\ -2 y+6 z+5=0 \ldots(2)\end{array}$
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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