2 Marks Questions

Question 12 Marks

Find the domain of the relation, $R=\{(x, y): x, y \in z, y=4\}$

Answer

$\begin{array}{l}\text { Given, } R=\{(x, y): x, y \in z, x y=4\} \\
=\{(-4,-1),(-2,-2),(-1,-4),(1,4),(2,2),(4,1)\} \\
\therefore \text { Domain of } R=\{-4,-2,-1,1,2,4\}\end{array}$

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Question 22 Marks

The intercept cuts-off by a line from y-axis is twice than that from x-axis and the line passes through the point (1, 2). Find the equation of the line.

Answer

The equation of a line intercept form is
$\frac{x}{a}+\frac{y}{b}=1$
Given, b = 2a
$\begin{array}{l}\therefore(1) \Rightarrow \frac{x}{a}+\frac{y}{2 a}=1 \\ \Rightarrow 2 x + y =2 a \end{array}$
since the line passes through the point (1, 2),
$\begin{array}{l}2 \cdot 1+2=2 a \\ \Rightarrow a=2\end{array}$
$\therefore$ Equation of the line is $2 x+y-4=0$.

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Question 32 Marks

Write E = (14, 21, 28, 35, 42, ..., 98) in set-builder form.

Answer

Now,
$\begin{array}{l}14=7 \times 2 \\ 21=7 \times 3 \\ 28=7 \times 4 \\ 35=7 \times 5 \\ 42=7 \times 6 \\ 98=7 \times 14\end{array}$
Therefore,the given set can be write as
$E=\{x: x=7 n, n \in N$ and $1 \leq n \leq 14\}$

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Question 42 Marks

Find the vertex, focus, axis, directrix and latus-rectum of the following parabolas $y^2-4 y+4 x=0$

Answer

We are given:
$\begin{array}{l}\Rightarrow(y-2)^2-4+4 x=0 \\ \Rightarrow(y-2)^2=-4(x-1)\end{array}$
Let Y = y - 2
X = x - 1
Then, we have
$Y ^2=-4 X$
On comparing the given equation with $Y^2=-4 a X$
$\begin{array}{l}4 a=4 \Rightarrow a=1 \\
\therefore \text { Vertex }=(X=0, Y=0)=(x=1, y=2) \\
\text { Focus }=(X=-a, Y=0)=(x-1=-1, y-2=0)=(x=0, y=2)\end{array}$
Equation of the directrix:
x = a
ip $x=1=1 \Rightarrow x=2$
Axis = Y = 0
i.e. $y -2=0 \Rightarrow y =2$
Therefore, length of the latus rectum = 4a = 4 units

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Question 52 Marks

Find the equation of the hyperbola, referred to its principal axes as axes of coordinates, in the following cases: Vertices at $( \pm 5,0)$, Foci at $( \pm 7,0)$.

Answer

Since the vertices lie on the $x$-axis, so let the equation of the required hyperbola be $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \ldots$
The coordinates of its vertices and foci are ( $\pm a , 0$ ) and ( $\pm ae , 0$ ) respectively.
But, the coordinates of vertices and foci are given as $( \pm 5,0)$ and $( \pm 7,0)$ respectively. $\therefore a=5$ and $a e=7$ then $5 e=7 \Rightarrow e=\frac{7}{5}$
Now, $b^2=a^2\left(e^2-1\right) \Rightarrow b^2=25\left(\frac{49}{25}-1\right)=24$
Substituting the values of $a ^2$ and $b ^2$ in (i), we obtain $\frac{x^2}{25}-\frac{y^2}{24}=1$
Required equation of hyperbola is $\frac{x^2}{25}-\frac{y^2}{24}=1$

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Question 62 Marks

Differentiate $\sin ^3 x \cos ^3 x$ w.r.t x

Answer

We have, $\frac{d}{d x}\left(\sin ^3 x \cos ^3 x\right)=\sin ^3 x \cdot \frac{d}{d x} \cos ^3 x+\cos ^3 x \cdot \frac{d}{d x}\left(\sin ^3 x\right)$ [Using Product Rule of differentiation]
$\begin{array}{l}=\sin ^3 x \cdot 3 \cos ^2 x(-\sin x)+\cos ^3 x \cdot 3 \sin ^2 x \cdot \cos x \\ =-3 \sin ^4 x \cos ^2 x+3 \cos ^4 x \sin ^2 x \\ =3 \sin ^2 x \cos ^2 x\left(-\sin ^2 x+\cos ^2 x\right) \\ =3 \sin ^2 x \cos ^2 x \cdot \cos 2 x \\ =\frac{3}{4} \cdot 4 \sin ^2 x \cos ^2 x \cdot \cos 2 x=\frac{3}{4}(2 \sin x \cos x)^2 \cos 2 x \\ =\frac{3}{4} \sin ^2 2 x \cdot \cos 2 x\end{array}$

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Question 72 Marks

Find the domain and the range of the real function: $f(x)=\frac{x^2-16}{x-4}$

Answer

Here we are given that, $f(x)=\frac{x^2-16}{x-4}$
Need to find: where the function is defined.
Let, $f(x)=\frac{x^2-16}{x-4}=y$
To find the domain of the function f(x) we need to equate the denominator of the function to 0
Therefore,
x - 4 = 0 or x = 4
It means that the denominator is zero when $x=4$
So, the domain of the function is the set of all the real numbers except 4
The domain of the function, $D _{\{ f ( x )\}}=(-\infty, 4) \cup(4, \infty)$
Now if we put any value of $x$ from the domain set the output value will be either (-ve) or (+ve), but the value will never be 8
So, the range of the function is the set of all the real numbers except 8
The range of the function, $R _{ f ( x )}=(-\infty, 8) \cup(8, \infty)$

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MATHS 2 Marks Questions

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