2 Marks Questions

Question 12 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 $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$\begin{array}{l}\therefore \text { foci }( \pm c, 0) \text { is }( \pm 3 \sqrt{5}, 0) \\ \Rightarrow c=3 \sqrt{5}\end{array}$
$\begin{array}{l}\therefore \text { foci }( \pm c, 0) \text { is }( \pm 3 \sqrt{5}, 0) \\ \Rightarrow c=3 \sqrt{5}\end{array}$
Length of latus rectum $\frac{2 b^2}{a}=8 \Rightarrow b^2=4 a$
We know that $c ^2= a ^2+ b ^2$
$\begin{array}{l}\therefore(3 \sqrt{5})^2=a^2+4 a \\ \Rightarrow a^2+4 a-45=0 \\ \Rightarrow(a+9)(a-5)=0 \\ \Rightarrow a=5(\because a=-9 \text { is not possible })\end{array}$
Also a = 5
$\Rightarrow b^2=4 \times 5=20$
Thus required equation of hyperbola is

$\frac{x^2}{25}-\frac{y^2}{20}=1$

View full question & answer→

Question 22 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}
$\therefore \quad A \times(B-C)=\{1,2,3\} \times\{4\}=\{(1,4),(2,4),(3,4)\}$
Now,
$\begin{array}{l}A \times B=\{1,2,3\} \times\{4\}=\{(1,4),(2,4),(3,4)\} \\
\text { and, } A \times C=\{1,2,3\} \times\{5\}=\{(1,5),(2,5),(3,5)\}\end{array}$
$\therefore \quad(A \times B)-(A \times C)=\{(1,4),(2,4),(3,4)\} \ldots \ldots . .( ii )$
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.

View full question & answer→

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,
$\begin{array}{l}3 x^2+2 y^2=6 \\ \Rightarrow \frac{x^2}{2}+\frac{y^2}{3}=1\end{array}$
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 $(0,-1)$.

View full question & answer→

Question 42 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 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.
$\begin{array}{l}\frac{-1+7}{2}=\frac{2+x}{2} \text { and } \frac{-6+2}{2}=\frac{-5+y}{2} \\ \Rightarrow x=4 \text { and } y=1\end{array}$
Therefore, the coordinates of the fourth vertex D are (4, 1).

View full question & answer→

Question 52 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
$\begin{array}{l} A =\{1,2,3\} \\ B =\{2,3,5\} \\ C =\{4,5,6\}\end{array}$
$\begin{array}{l}\text { Now }(A \cap B) \cup C=(\{1,2,3\} \cap\{2,3,5\}) \cup\{4,5,6\} \\ =\{2,3\} \cup\{4,5,6\} \\ =\{2,3,4,5,6\}\end{array}$
$\begin{array}{l}\text { And } A \cap(B \cup C)=\{1,2,3\} \cap[\{2,3,5\} \cup\{4,5,6\} \\ =\{1,2,3\} \cap\{2,3,4,5,6\}\end{array}$
$\begin{array}{l}=\{2,3\} \\ \text { Thus, }(A \cap B) \cup C \neq A \cap(B \cup C)\end{array}$

View full question & answer→

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, $f(x)=x+\frac{1}{x}$ $\qquad$ (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
$\begin{array}{l}\{f(x)\}^3=x^3+\frac{1}{x^3}+3 \times x \times \frac{1}{x} \times\left(x+\frac{1}{x}\right) \\ =\left(x^3+\frac{1}{x^3}\right)+3\left(\frac{1}{x}+x\right) \\ = f \left(x^3\right)+3 f\left(\frac{1}{x}\right)\end{array}$
$\begin{array}{l}\left|\because f\left(\frac{1}{x}\right)=\left\{\frac{1}{x}+\frac{1}{\frac{1}{x}}\right\}=\left(\frac{1}{x}+x\right)\right| \\
\text { Hence, }\{ f ( x )\}^3= f \left( x ^3\right)+3 f \left(\frac{1}{x}\right)\end{array}$

View full question & answer→

Question 72 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;
$\begin{array}{l}\lim _{x \rightarrow \frac{1}{2}} \frac{8 x^3-1}{16 x^4-1}\left(\frac{0}{0} \text { form }\right) \\ =\lim _{x \rightarrow \frac{1}{2}} \frac{(2 x)^3-1^3}{\left(4 x^2\right)^2-1^2} \\ =\lim _{x \rightarrow \frac{1}{2}} \frac{(2 x-1)\left(4 x^2+2 x+1\right)}{\left(4 x^2+1\right)\left(4 x^2-1\right)}\left(\frac{0}{0} \text { form }\right) \\ =\lim _{x \rightarrow \frac{1}{2}} \frac{(2 x-1)\left(4 x^2+2 x+1\right)}{\left(4 x^2+1\right)(2 x-1)(2 x+1)} \\ =\lim _{x \rightarrow \frac{1}{2}} \frac{4 x^2+2 x+1}{\left(4 x^2+1\right)(2 x+1)}=\frac{3}{4}\end{array}$

View full question & answer→

MATHS 2 Marks Questions

Take a timed test