2 Marks Questions

Question 12 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, $b = 2a$
$\therefore(1) \Rightarrow \frac{x}{a}+\frac{y}{2 a}=1$
$\Rightarrow 2 x + y =2 a $
since the line passes through the point $(1, 2),$
$2 \cdot 1+2=2 a$
$\Rightarrow a=2$
$\therefore$ Equation of the line is $2 x+y-4=0$.

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Question 22 Marks

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

Answer

Now,
$14=7 \times 2$
$21=7 \times 3$
$28=7 \times 4$
$35=7 \times 5$
$42=7 \times 6$
$98=7 \times 14$
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 32 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:
$\Rightarrow(y-2)^2-4+4 x=0$
$\Rightarrow(y-2)^2=-4(x-1)$
Let $Y = y - 2$
$X = x - 1$
Then, we have
$Y ^2=-4 X$
On comparing the given equation with $Y^2=-4 a X$
$4 a=4$
$\Rightarrow a=1$
$\therefore$ Vertex $=(X=0, Y=0)$
$=(x=1, y=2)$
Focus $=(X=-a, Y=0)$
$=(x-1=-1, y-2=0)$
$=(x=0, y=2)$
Equation of the directrix:
$x = a$
$ 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 42 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 52 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$]$
$=\sin ^3 x \cdot 3 \cos ^2 x(-\sin x)+\cos ^3 x \cdot 3 \sin ^2 x \cdot \cos x$
$=-3 \sin ^4 x \cos ^2 x+3 \cos ^4 x \sin ^2 x$
$=3 \sin ^2 x \cos ^2 x\left(-\sin ^2 x+\cos ^2 x\right)$
$=3 \sin ^2 x \cos ^2 x \cdot \cos 2 x$
$=\frac{3}{4} \cdot 4 \sin ^2 x \cos ^2 x \cdot \cos 2 x$
$=\frac{3}{4}(2 \sin x \cos x)^2 \cos 2 x$
$=\frac{3}{4} \sin ^2 2 x \cdot \cos 2 x$

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Question 62 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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Question 72 Marks

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

Answer

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

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Maths 2 Marks Questions

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