5 Marks Questions

Question 15 Marks

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
$\frac{x^2}{100}+\frac{y^2}{400}=1$

Answer

The equation of given ellipse is $\frac{x^2}{100}+\frac{y^2}{400}=1$
Now $400>100 \Rightarrow a^2=400$ and $b^2=100$
So the equation of ellipse in standard form is $\frac{y^2}{a^2}+\frac{x^2}{b^2}=1$
$\therefore a^2=400 \Rightarrow a=20$ and $b^2=100 \Rightarrow b=10$
We know that $c=\sqrt{a^2-b^2}$
$\begin{array}{l}\therefore c=\sqrt{400-100}=\sqrt{300}=10 \sqrt{3} \\ \therefore \text { Coordinates of foci are }(0, \pm c) \text { i.e. }(0, \pm 10 \sqrt{3}) \\ \text { Coordinates of vertices are }(0, \pm a) \text { i.e. }(0, \pm 20)\end{array}$
$\begin{array}{l}\text { Length of major axis }=2 a =2 \times 20=40 \\ \text { Length of minor axis }=2 b=2 \times 10=20\end{array}$
Eccentricity $( e )=\frac{c}{a}=\frac{10 \sqrt{3}}{20}=\frac{\sqrt{3}}{2}$
Length of latus rectum $=\frac{2 b^2}{a}=\frac{2 \times 100}{20}=10$

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Question 25 Marks

Prove that: $\cos 36^{\circ} \cos 42^{\circ} \cos 60^{\circ} \cos 78^{\circ}=\frac{1}{16}$.

Answer

We have to prove that $\cos 36^{\circ} \cos 42^{\circ} \cos 60^{\circ} \cos 78^{\circ}=\frac{1}{16}$.
LHS $=\cos 36^{\circ} \cos 42^{\circ} \cos 60^{\circ} \cos 78^{\circ}$
By regrouping the LHS and multiplying and dividing by 2 we get,
$=\frac{1}{2} \cos 36^{\circ} \cos 60^{\circ}\left(2 \cos 78^{\circ} \cos 42^{\circ}\right)$
But 2 cos A cos B = cos (A + B) + cos (A - B)
Then the above equation becomes,
$\begin{array}{l}=\frac{1}{2} \cos 36^{\circ} \cos 60^{\circ}\left(\cos \left(78^{\circ}+42^{\circ}\right)+\cos \left(48^{\circ}-42^{\circ}\right)\right) \\ =\frac{1}{2} \cos 36^{\circ} \cos 60^{\circ}\left(\cos \left(120^{\circ}\right)+\cos \left(36^{\circ}\right)\right) \\ =\frac{1}{2} \cos 36^{\circ} \cos 60^{\circ}\left(\cos \left(180^{\circ}-60^{\circ}\right)+\cos \left(36^{\circ}\right)\right)\end{array}$
$=\frac{1}{2} \cos 36^{\circ} \cos 60^{\circ}\left(-\cos \left(60^{\circ}\right)+\cos \left(36^{\circ}\right)\right)$
Now, $\cos \left(36^{\circ}\right)=\frac{\sqrt{5}+1}{4}$
$\cos \left(60^{\circ}\right)=\frac{1}{2}$
Substituting the corresponding values, we get
$\begin{array}{l}=\frac{1}{2}\left(\frac{\sqrt{5}+1}{4}\right)\left(\frac{1}{2}\right)\left(-\frac{1}{2}+\frac{\sqrt{5}+1}{4}\right) \\ =\left(\frac{\sqrt{5}+1}{16}\right)\left(\frac{\sqrt{5}+1-2}{4}\right) \\ =\left(\frac{(\sqrt{5})^2-1^2}{16 \times 4}\right) \\ =\left(\frac{5-1}{64}\right)\end{array}$
$\frac{1}{16}$
LHS = RHS
Hence proved.

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Question 35 Marks

If $\sin x=\frac{\sqrt{5}}{3}$ and $x$ lies in the 2nd quadrant, find the values of $\cos \frac{x}{2}, \sin \frac{x}{2}$ and $\tan \frac{x}{2}$.

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Question 45 Marks

Find the mean and standard deviation for the following data:

Class interval	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80	80-90	90-100
Frequency	3	2	4	6	5	5	5	2	8	5

Answer

We make the table from the given data:

Class marks	Mid value $\left(x_i\right)$	$d _{ i }= x _{ i }- a$ $= x _{ i }=45$Z	$f _{ i }$	$f _{ i } d _{ i }$	$d_i^2$	$f _{ i } d_i^2$
0-10	5	-40	3	-120	1600	4800
10-20	15	-30	2	-60	900	1800
20-30	25	-20	4	-80	400	1600
30-40	35	-10	6	-60	100	800
40-50	45	0	5	0	0	0
50-60	55	10	5	50	100	500
60-70	65	20	5	100	400	2000
70-80	75	30	2	60	900	1800
80-90	85	40	8	320	1600	12800
90-100	95	50	5	250	2500	12500
			$\sum f _{ i }=45$	$\sum f _{ i } d _{ i }=460$		$\sum f _{ i } d_i^2=38400$

Let a = 45.
$\therefore$ Mean $=a+\frac{\sum f_i d_i}{\sum f_i}$
$\begin{array}{l}=45+\frac{460}{45} \\ =45+10.22=55.22\end{array}$
$\therefore$ Standard deviation $=\sqrt{\frac{\sum f_i d_i^2}{\sum f_i}-\left(\frac{\sum f_i d_i}{\sum f_i}\right)^2}$
$\begin{array}{l}=\sqrt{\frac{38400}{45}-(10.22)^2} \\ =\sqrt{853.33-104.45} \\ =\sqrt{748.88} \\ =27.36\end{array}$

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Question 55 Marks

Find the lengths major and minor axes, coordinates of the vertices, coordinates of the foci, eccentricity, and length of the latus rectum of the ellipse $9 x^2+y^2=36$

Answer

Given that:
$9 x^2+y^2=36$
After divide by 36 to both the sides, we get
$\frac{9}{36} x^2+\frac{1}{36} y^2=1 \Rightarrow \frac{x^2}{4}+\frac{y^2}{36}=1 \ldots$
Now, the above equation is of the form,
$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1 \ldots (1)$
$a ^2=36$ and $b ^2=4 \Rightarrow a=\sqrt{36}$ and $b=\sqrt{4} \Rightarrow$ and $a =6$ and $b =2$
i. Length of major axes
$\therefore$ Length of major axes $=2 a =2 \times 6=12$ units
ii. Length of major axes
$\therefore$ Length of major axes $=2 b=2 \times 2=4$ units
iii. Coordinates of the Vertices
$\therefore$ Coordinates of the Vertices $=(0, a)$ and $(0,-a)=(0,6)$ and $(0,-6)$
iv. Coordinates of the foci
As we know that, Coordinates of foci $=(0, \pm c)$ where $c ^2= a ^2- b ^2$
Now
$c^2=36-4 \Rightarrow c^2=32 \Rightarrow c=\sqrt{32} \ldots$
$\therefore$ Coordinates of foci $=(0, \pm \sqrt{32})$
v. Eccentricity
As we know that, Eccentricity $=\frac{c}{a} \Rightarrow e=\frac{\sqrt{32}}{6}$ [from (iii)]
vi. Length of the Latus Rectum
As we know that, Length of Latus Rectum
$=\frac{2 b^2}{a}=\frac{2 \times(2)^2}{6}=\frac{8}{6}$ $=\frac{4}{3}$

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Question 65 Marks

Solve the following system of linear inequalities.
$\begin{array}{l}2(2 x +3)-10<6( x -2) \\ \text { and } \frac{2 x-3}{4}+6 \geq 2+\frac{4 x}{3}\end{array}$

Answer

The given system of linear inequalities is
|$\begin{array}{l}2(2 x+3)-10<6(x-2) \ldots \text { (i) } \\ \text { and } \frac{2 x-3}{4}+6 \geq 2+\frac{4 x}{3} \ldots \text { (ii) }\end{array}$
From inequality (i), we get
2(2x + 3) - 10 < 6(x - 2)
$\begin{array}{l}\Rightarrow 4 x +6-10<6 x -12 \\ \Rightarrow 4 x -4<6 x -12 \\ \Rightarrow
4 x -4+4<6 x -12+4 \text { [adding } 4 \text { on both sides] } \\ \Rightarrow 4 x <6 x -8 \\ \Rightarrow 4 x -6 x <6 x -8-6 x \text { [subtracting } 6 x \text { from both sides] } \\ \Rightarrow-2 x <-8 \\ \Rightarrow 2 x >8 \text { [dividing both sides by }-1 \text { and then inequality sign will change] } \\ \Rightarrow \quad \frac{2 x}{2}>\frac{8}{2} \text { [dividing both sides by } 2 \text { ] }\end{array}$
$\therefore \quad x>4 \ldots$ (iii)
Thus, any value of x greater than 4 satisfies the inequality.
$\therefore$ Solution set is $x \in(4, \infty)$
The representation of solution of inequality (i) is

From inequality (ii), we get
$\frac{2 x-3}{4}+6 \geq 2+\frac{4 x}{3} \Rightarrow \frac{2 x-3+24}{4} \geq \frac{6+4 x}{3}$
$\begin{array}{l}\Rightarrow \quad \frac{2 x+21}{4} \geq \frac{6+4 x}{3} \Rightarrow 3(2 x+21) \geq 4(6+4 x) \\ \Rightarrow \quad 6 x+63 \geq 24+16 x \\ \Rightarrow \quad-10 x \geq-39 \Rightarrow 10 x \leq 39 \\ \Rightarrow \quad \frac{10 x}{10} \leq \frac{39}{10} \\ \Rightarrow \quad x<3.9\end{array}$
From Eqs. (iii) and (iv), it is clear, that there is no common value of x, which satisfies both inequalities (iii) and (iv). Hence, the given system of inequalities has no solution.

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