2 Marks Questions

Question 12 Marks

Three vertices of a parallelogram, taken in order, are $(-1, -6), (2, -5)$ and $(7, 2)$. Write the coordinates of its fourth vertex.

Answer

Let $A(-1, -6), B(2, -5)$ and $C(7, 2)$ be the vertices of the parallelogram $\text{ABCD}$ and $D$ be the fourth vertex of the parallelogram.
Let the coordinates of $D$ be $(x, y).$
Since, diagonals of a parallelogram bisect each other.
$\frac{-1+7}{2}=\frac{2+x}{2}$ and $ \frac{-6+2}{2}=\frac{-5+y}{2}$
$\Rightarrow x=4$ and $y=1$
Therefore, the coordinates of the fourth vertex $D$ are $(4, 1).$

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Question 22 Marks

For all sets $A, B$ and $C$ Is $(A \cap B) \cup C=A \cap(B \cup C)$ ? Justify your statement.

Answer

Let us consider the following sets $A, B$ and $C$ such that
$ A =\{1,2,3\}$
$B =\{2,3,5\}$
$C =\{4,5,6\}$
$\text { Now }(A \cap B) \cup C=(\{1,2,3\} \cap\{2,3,5\}) \cup\{4,5,6\}$
$=\{2,3\} \cup\{4,5,6\}$
$=\{2,3,4,5,6\}$
$\text { And } A \cap(B \cup C)=\{1,2,3\} \cap[\{2,3,5\} \cup\{4,5,6\}$
$=\{1,2,3\} \cap\{2,3,4,5,6\}$
$=\{2,3\}$
$\text { Thus, }(A \cap B) \cup C \neq A \cap(B \cup C)$

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Question 32 Marks

Find the lengths of major and minor axes, coordinates of foci, vertices and the eccentricity: $3 x^2+2 y^2=6$.

Answer

We have,
$3 x^2+2 y^2=6$
$\Rightarrow \frac{x^2}{2}+\frac{y^2}{3}=1$
This equation is of the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,
where $a ^2=2$ and $b ^2=3$
i.e. $a =\sqrt{2}$ and $b =\sqrt{3}$.
Clearly, $a < b,$
so the major and minor axes of the given ellipse are along $y$ and $x-$axes respectively.
$\therefore$ Length of the major axis $=2 b=2 \sqrt{3}$
and Length of the minor axis $=2 b=2 \sqrt{2}$
The coordinates of the vertices $=(0, b)$ and $(0,- b )=(0, \sqrt{3})$ and $(0,-\sqrt{3})$
The eccentricity e of the ellipse is $e=\sqrt{1-\frac{a^2}{b^2}}=\sqrt{1-\frac{2}{3}}=\frac{1}{\sqrt{3}}$
The coordinates of the foci $=(0$, be $)$ and $(0,-b e)=(0,1)$ and $(0,-1)$.

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Question 42 Marks

Find the equation of hyperbola having Foci $( \pm 3 \sqrt{5}, 0)$, the latus rectum is of length $8 .$

Answer

Here foci are $( \pm 3 \sqrt{5}, 0)$ which lie on $x-$axis.
So the equation of hyperbola in standard form is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$\therefore \text { foci }( \pm c, 0) \text { is }( \pm 3 \sqrt{5}, 0)$
$\Rightarrow c=3 \sqrt{5}$
$\therefore \text { foci }( \pm c, 0) \text { is }( \pm 3 \sqrt{5}, 0)$
$\Rightarrow c=3 \sqrt{5}$
Length of latus rectum $\frac{2 b^2}{a}=8 $
$\Rightarrow b^2=4 a$
We know that $c ^2= a ^2+ b ^2$
$\therefore(3 \sqrt{5})^2=a^2+4 a$
$\Rightarrow a^2+4 a-45=0$
$\Rightarrow(a+9)(a-5)=0$
$\Rightarrow a=5(\because a=-9)$ is not possible
Also $a = 5$
$\Rightarrow b^2=4 \times 5=20$
Thus required equation of hyperbola is
$\frac{x^2}{25}-\frac{y^2}{20}=1$

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Question 52 Marks

Evaluate: $\lim _{x \rightarrow \frac{1}{2}} \frac{8 x^3-1}{16 x^4-1}$.

Answer

We have to find the value of
$\lim _{x \rightarrow \frac{1}{2}} \frac{8 x^3-1}{16 x^4-1}$
When $x =\frac{1}{2}$, the expression $\frac{8 x^3-1}{16 x^4-1}$ assumes
the for $n \frac{0}{0}$.
Therefore, $\left(x-\frac{1}{2}\right)$ or, $2 x -1$ is a factor common to
numerator and denominator.
Factorizing the numerator and denominator, we obtain;
$\lim _{x \rightarrow \frac{1}{2}} \frac{8 x^3-1}{16 x^4-1}\left(\frac{0}{0} \text { form }\right)$
$=\lim _{x \rightarrow \frac{1}{2}} \frac{(2 x)^3-1^3}{\left(4 x^2\right)^2-1^2}$
$=\lim _{x \rightarrow \frac{1}{2}} \frac{(2 x-1)\left(4 x^2+2 x+1\right)}{\left(4 x^2+1\right)\left(4 x^2-1\right)}\left(\frac{0}{0} \text { form }\right)$
$=\lim _{x \rightarrow \frac{1}{2}} \frac{(2 x-1)\left(4 x^2+2 x+1\right)}{\left(4 x^2+1\right)(2 x-1)(2 x+1)}$
$=\lim _{x \rightarrow \frac{1}{2}} \frac{4 x^2+2 x+1}{\left(4 x^2+1\right)(2 x+1)}=\frac{3}{4}$

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Question 62 Marks

If $f ( x )= x +\frac{1}{x}, $ show that $\{ f ( x )\}^3= f \left( x ^3\right)+3 f \left(\frac{1}{x}\right)$

Answer

We have, $f(x)=x+\frac{1}{x} ... (i)$
To prove: $\{ f ( x )\}^3= f \left( x ^3\right)+3 f \left(\frac{1}{x}\right)$
Proof : On cubing both sides of $(i),$ we get
$\{f(x)\}^3=x^3+\frac{1}{x^3}+3 \times x \times \frac{1}{x} \times\left(x+\frac{1}{x}\right)$
$=\left(x^3+\frac{1}{x^3}\right)+3\left(\frac{1}{x}+x\right)$
$= f \left(x^3\right)+3 f\left(\frac{1}{x}\right)$
$\left|\because f\left(\frac{1}{x}\right)=\left\{\frac{1}{x}+\frac{1}{\frac{1}{x}}\right\}=\left(\frac{1}{x}+x\right)\right|$
Hence $, \{ f ( x )\}^3= f \left( x ^3\right)+3 f \left(\frac{1}{x}\right)$

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Question 72 Marks

If $A =(1,2,3), B =\{4\}, C =\{5\}$, then verify that $A \times(B-C)=(A \times B)-(A \times C)$.

Answer

As given in the question we have, $A = \{1, 2, 3\}, B = \{4\}$ and $C = \{5\}$
From set theory, $(B - C) = \{4\}$
$\therefore A \times(B-C)=\{1,2,3\} \times\{4\}=\{(1,4),(2,4),(3,4)\}$
Now,
$A \times B=\{1,2,3\} \times\{4\}=\{(1,4),(2,4),(3,4)\}$
$\text { and, } A \times C=\{1,2,3\} \times\{5\}=\{(1,5),(2,5),(3,5)\}$
$\therefore (A \times B)-(A \times C)=\{(1,4),(2,4),(3,4)\} \ldots \ldots . .( ii )$
From equation $(i)$ and equation $(ii),$ we get
$A \times(B-C)=(A \times B)-(A \times C)$
We can see the equations $(i)$ and $(ii)$ have same ordered pairs.
Hence verified.

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