2 Marks Questions

Question 12 Marks

Find the of derivative of the function from the first principle: $\sin x^2$.

Answer

Let $y=\sin x^2$
Then, $y+\delta y=\sin (x+\delta x)^2$
$\Rightarrow \delta y=\sin (x+\delta x)^2-\sin x^2$
Using first principle,
$\begin{array}{l}\Rightarrow \frac{\delta y}{\delta x}=\frac{\sin (x+\delta x)^2-\sin x^2}{\delta x} \\ \Rightarrow \frac{d y}{d x}=\lim _{\delta x \rightarrow 0} \frac{\delta y}{\delta x}\end{array}$
$\begin{array}{l}=\lim _{\delta x \rightarrow 0} \frac{\sin (x+\delta x)^2-\delta \sin x^2}{\delta x} \\
=\lim _{\delta x \rightarrow 0} \frac{2 \cos \left[\frac{(x+\delta x)^2+x^2}{2}\right] \sin \left[\frac{(x+\delta x)^2-x^2}{2}\right]}{\delta x} \\
{\left[\text { using }(\sin C-\sin D)=2 \cos \left(\frac{C+D}{2}\right) \sin \left(\frac{C-D}{2}\right)\right]} \\
=\lim _{\delta x \rightarrow 0} 2 \cos \left[\frac{(x+\delta x)^2+x^2}{2}\right] \frac{\sin \left[\left(x+\frac{\delta x}{2}\right) \cdot \delta x\right]}{\left(x+\frac{\delta x}{2}\right) \cdot \delta x}\left(x+\frac{\delta x}{2}\right) \\
=2 \cdot \lim _{\delta x \rightarrow 0} \cos \left[\frac{(x+\delta x)^2+x^2}{2}\right] \cdot \lim _{\delta x \rightarrow 0} \frac{\sin \left[\left(x+\frac{\delta x}{2}\right) \cdot \delta x\right]}{\left(x+\frac{\delta x}{2}\right) \cdot \delta x} \\
\cdot \lim _{\delta x \rightarrow 0}\left(x+\frac{\delta x}{2}\right) \\
=\left[2 \times \cos x^2 \times 1 \times x\right]=2 x \cos x^2\end{array}$
Hence, $\frac{d}{d x}\left(\sin x^2\right)=2 x \cos x^2$

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Question 22 Marks

If $A=\{1,2,3), B=\{3,4\}$ and $C=\{4,5,6\}$, find
i. $A \times(B \cap C)$
ii. $(A \times B) \cap(A \times C)$

Answer

We have,
$\begin{array}{l} A =\{1,2,3), B =\{3,4\} \text { and } C =\{4,5,6\} \\ i . \therefore \quad B \cap C=\{3,4\} \cap\{4,5,6\}=\{4\} \\ \quad \therefore \quad A \times(B \cap C)=\{1,2,3,\} \times\{4\} \\ \quad=\{(1,4),(2,4),(3,4)\} \\ \quad \Rightarrow \quad A \times(B \cap C)=\{(1,4),(2,4),(3,4)\}\end{array}$
ii.
$\begin{array}{l}\therefore \quad A \times B=\{1,2,3,\} \times\{3,4\} \\ =\{(1,3),(1,4),(2,3),(2,4),(3,3),(3,4)\}\end{array}$
and
$\begin{array}{l}A \times C=\{1,2,3\} \times\{4,5,6\} \\ =\{(1,4),(1,5),(1,6),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6)\} \\ (A \times B) \cap(A \times C)=\{(1,4),(2,4),(3,4)\}\end{array}$

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Question 32 Marks

Let $A=\{x: x \in N), B=(x: x=2 n, n \in N\}, C=\{x: x=2 n-1, n \in N\}$ and, $D=\{x: x$ is a prime natural number}. Find: $A \cap B$.

Answer

According to the question ,
we can state, A = All natural numbers i.e. {1, 2, 3…..}
B = All even natural numbers i.e. {2, 4, 6, 8…}
C = All odd natural numbers i.e. {1, 3, 5, 7……}
D = All prime natural numbers i.e. {1, 2, 3, 5, 7, 11, …}
$A \cap B$
A contains all elements of $B$
$\begin{array}{l}\therefore B \subset A \\ \therefore A \cap B=B\end{array}$

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Question 42 Marks

Find the equation of the perpendicular bisector of the line joining the points (1, 3) and (3, 1).

Answer

Given points are, A(1, 3) and B(3, 1).
Let C be the mid point of AB.
$\therefore$ Coordinates of $C=\left(\frac{1+3}{2}, \frac{3+1}{2}\right)=(2,2)$
Slope of $AB =\frac{1-3}{3-1}=-1$
$\therefore$ Slope of the perpendicular bisector of $AB =1$
Hence, the equation of the perpendicular bisector of AB is
$\begin{array}{l}y-2=1(x-2) \\
\Rightarrow x-y=0\end{array}$
or, y = x

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Question 52 Marks

Find the equation of the circle which touches the lines $4 x-3 y+10=0$ and $4 x-3 y-30=0$ and whose centre lies on the line $2 x + y =0$.

Answer

Clearly, the lines 4x - 3y + 10 = 0 and 4x - 3y - 30 = 0 are parallel and are touching the circle.
It is given that the centre of the circle lies on the line 2x + y = 0 which intersects the lines 4x - 3y + 10 = 0 and 4x - 3y - 30 = 0 at A (-1, 2) and B (3, - 6) respectively.
Therefore, the centre of the circle is the mid-point of AB.
So, the coordinates of the centre C are (1, - 2)+

Let d be the distance between parallel lines 4x -3y + 10 =0 and 4x -3y - 30 = 0 Then
$\begin{array}{l} PQ =d=\left|\frac{10-(-30)}{\sqrt{4^2+(-3)^2}}\right|=8 \\ \Rightarrow PQ = d =8 \\ \text { Radius }=\frac{1}{2}(P Q)=\frac{1}{2} \times 8=4 \\ \Rightarrow \text { Radius }=4\end{array}$
Thus, the required circle has its centre at C (1, - 2) and radius = 4
Hence, its equation is $(x-1)^2+(y+2)^2=4^2$

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Question 62 Marks

Find the equation of the parabola whose: focus is $(2,3)$ and the directrix $x-4 y+3=0$.

Answer

Let P(x, y) be any point on the parabola whose focus is S(2, 3) and the directrix is x - 4y + 3 = 0
Draw PM perpendicular to x - 4y + 3 = 0
Thus, we have:
SP = PM
$\begin{array}{l}\Rightarrow SP ^2= PM ^2 \\ \Rightarrow( x -2)^2+( y -3)^2=\left|\frac{x-4 y+3}{\sqrt{1+16}}\right|^2 \\ \Rightarrow( x -2)^2+( y -3)^2=\left(\frac{x-4 y+3}{\sqrt{17}}\right)^2 \\ \Rightarrow 17\left( x ^2+4-4 x + y ^2-6 y+9\right)= x ^2+16 y ^2+9-8 xy -24 y +6 x \\ \Rightarrow\left(17 x ^2-68 x -102 y +17 y ^2+13 \times 17\right)= x ^2+16 y ^2+9-8 xy -24 y +6 x \\ \Rightarrow 16 x ^2+ y ^2+8 xy -74 x -78 y +212=0\end{array}$
Which is the required equation of parabola.

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Question 72 Marks

Find the domain and range of the real valued function $f(x)=\frac{1}{\sqrt{16-x^2}}$.

Answer

We know the square of a real number is never negative.
Clearly, $f ( x )$ takes real values only when $16- x ^2 \geq 0$
$\begin{array}{l}=16 \geq x ^2 \\ = x ^2 \leq 16 \\ = x ^2-16 \leq 0 \\ = x ^2-4^2 \leq 0 \\ =( x +4)( x -4) \leq 0 \\ = x \geq-4 \text { and } x \leq 4 \\ \therefore x \in[-4,4]\end{array}$
In addition, $f ( x )$ is also undefined when $16- x ^2=0$ because denominator will be zero and the result will be indeterminate.
$\begin{array}{l}16-x^2=0 \Rightarrow x= \pm 4 \\
\text { Hence, } x \in[-4,4]-\{-4,4\} \\
\therefore x \in(-4,4)\end{array}$
Thus, domain of $f=(-4,4)$
$\begin{array}{l}\text { Let } f ( x )= y \\ \Rightarrow \frac{1}{\sqrt{16-x^2}}=y \\ \Rightarrow\left(\frac{1}{\sqrt{16-x^2}}\right)^2=y^2 \\ \Rightarrow \frac{1}{16-x^2}=y^2 \\ =1=\left(16-x^2\right) y ^2 \\ =1=16 y^2-x^2 y^2 \\ =x^2 y^2+1-16 y^2=0 \\ =\left(y^2\right) x^2+(0) x +\left(1-16 y^2\right)=0\end{array}$
As $x \in R$, the discriminant of this quadratic equation in $x$ must be non-negative.
$\begin{array}{l}=0^2-4\left(y^2\right)\left(1-16 y^2\right) \geq 0 \\ =-4 y^2\left(1-16 y^2\right) \geq 0 \\ =4 y^2\left(1-16 y^2\right) \leq 0 \\ =1-16 y^2 \leq 0\left[\because y^2 \geq 0\right] \\ =16 y^2-1 \geq 0 \\ \Rightarrow(4 y)^2-1^2 \geq 0 \\ =(4 y+1)(4 y-1) \geq 0 \\ =4 y \leq-1 \text { and } 4 y \geq \\ \Rightarrow y \leq-\frac{1}{4} \text { and } y \geq \frac{1}{4} \\ \Rightarrow y \in\left(-\infty,-\frac{1}{4}\right] \cup\left[\frac{1}{4}, \infty\right) \\ \Rightarrow f(x) \in\left(-\infty,-\frac{1}{4}\right] \cup\left[\frac{1}{4}, \infty\right)\end{array}$
However, y is always positive because it is the reciprocal of a non-zero square root.
$\therefore f(x) \in\left[\frac{1}{4}, \infty\right)$
Thus, range of $f =\left[\frac{1}{4}, \infty\right)$
Thus , is the required domain and range of the function

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