5 Marks Questions

Question 15 Marks

Prove that: $\cos ^3 x \sin 3 x+\sin ^3 x \cos 3 x=\frac{3}{4} \sin 4 x$.

Answer

We have to prove that $\cos ^3 x \sin 3 x+\sin ^3 x \cos 3 x=\frac{3}{4} \sin 4 x$.
We know that,
$\begin{array}{l}\cos 3 \theta=4 \cos ^3 \theta-3 \cos \theta \\ \Rightarrow 4 \cos ^3 \theta=\cos 3 \theta+3 \cos \theta \\ \Rightarrow \cos ^3 \theta=\frac{\cos 39+3 \cos \theta}{4} \ldots \text { (i) }\end{array}$
And similarly
$\begin{array}{l}\Rightarrow \sin 3 \theta=3 \sin \theta-4 \sin ^3 \theta \\
\Rightarrow 4 \sin ^3 \theta=3 \sin \theta-\sin 3 \theta \\
\Rightarrow \sin ^3 \theta=\frac{3 \sin \theta-\sin 3 \theta}{4} \ldots\end{array}$
Now,
LHS $=\cos ^3 x \sin 3 x+\sin ^3 x \cos 3 x$
Substituting the values from equation (i) and (ii), we get
$\begin{array}{l}\Rightarrow\left(\frac{\cos 3 x+3 \cos x}{4}\right) \sin 3 x+\left(\frac{\cos 3 x-3 \cos x}{4}\right) \cos 3 x \\
=\frac{1}{4}(\sin 3 x \cos 3 x+3 \sin 3 x \cos x+3 \sin x \cos 3 x-\sin 3 x \cos 3 x) \\ =\frac{1}{4}[3(\sin 3 x \cos x+\sin x \cos 3 x)+0] \\
=\frac{1}{4}(3 \sin (3 x+x))\end{array}$
$\begin{array}{l}\text { (as } \sin (x+y)=\sin x \cos y+\cos x \sin y) \\
\Rightarrow \frac{3}{4} \sin 4 x \\
\text { LHS }=\text { RHS }\end{array}$
Henve proved

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Question 25 Marks

Draw the shape of the ellipse $4 x^2+9 y^2=36$ and find its major axis, minor axis, value of $c$, vertices, directrices, foci, eccentricity and length of latusrectum.

Answer

We have, equation of ellipse is $4 x^2+9 y^2=36$
$\text { or } \frac{x^2}{9}+\frac{y^2}{4}=1$
Since, the denominator of $\frac{x^2}{9}$ is greater than denominator of $\frac{y^2}{4}$ So, the major axis lies along X -axis.

i. Shape is shown above.
ii. Major axis, $2 a =2 \times 3=6$
iii. Minor axis, $2 b=2 \times 2=4$
iv. Value of $c =\sqrt{a^2-b^2}=\sqrt{9-4}=\sqrt{5}$
v. Vertices $=(-a, 0)$ and $(a, 0)$ i.e., $(-3,0)$ and $(3,0)$
vi. Directrices, $x = \pm \frac{a^2}{c}= \pm \frac{9}{\sqrt{5}}$
vii. Foci $=(- c , 0)$ and $(c, 0)$ i.e., $(-\sqrt{5}, 0)$ and $(\sqrt{5}, 0)$
viii. Eccentricity, $e =\frac{c}{a}=\frac{\sqrt{5}}{3}$
ix. Length of latusrectum, $2 l=\frac{2 b^2}{a}=\frac{2 \times 4}{3}=\frac{8}{3}$

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Question 35 Marks

Solve the following system of linear inequalities
$-2-\frac{x}{4} \geq \frac{1+x}{3}$ and $3-x<4(x-3)$

Answer

The given system of linear inequalities is
$-2-\frac{x}{4} \geq \frac{1+x}{3} \ldots \text { (i) }$
and $3-x<4(x-3) \ldots$ (ii)
From inequality (i), we get
$-2-\frac{x}{4} \geq \frac{1+x}{3}$
$\begin{array}{l}\Rightarrow-24-3 x \geq 4+4 x \text { [multiplying both sides by } 12] \\
\Rightarrow-24-3 x-4 \geq 4+4 x-4 \text { [subtracting } 4 \text { from both sides] } \\ \Rightarrow-28-3 x \geq 4 x \\
\Rightarrow-28-3 x+3 x \geq 4 x+3 x \text { [adding } 3 x \text { on both sides] } \\ \Rightarrow-28 \geq 7 x \\
\Rightarrow-\frac{28}{7} \geq \frac{7 x}{7}[\text { dividing both sides by } 7 \text { ] } \\
\Rightarrow-4 \geq x \text { or } x \leq-4 \ldots \text { (iii) }\end{array}$
Thus, any value of x less than or equal to - 4 satisfied the inequality
So, the solution set is $x \in(-\infty,-4]$z

From inequality (ii), we get
3 - x < 4 (x - 3)
$\begin{array}{l}\Rightarrow 3- x +12<4 x -12+12 \text { [adding } 12 \text { on both sides] } \\ \Rightarrow 15- x <4 x \\ \Rightarrow 15- x + x <4 x + x \text { [adding } x \text { on both sides] } \\ \Rightarrow 15<5 x \\ \Rightarrow 3< x \text { [dividing both sides by } 3 \text { ] } \\ \text { or } x >3 \ldots \text { (iv) }\end{array}$
Thus, any value of x greater than 3 satisfies the inequality.
So, the solution set is $x \in(3, \infty)$

The solution set of inequalities (i) and (ii) are represented graphically on number line as given below:

As no region is common, hence the given system has no solution.

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Question 45 Marks

Find the equation of the parabola whose focus is (1, -1) and whose vertex is (2, 1). Also find its axis and latus rectum.

Answer

Here, we are given the coordinates of the focus and vertex.
So, we require the equation of the directrix.
Let $Z \left( x _1, y _1\right)$ be the coordinates of the point of intersection of the axis and the directrix.
Then, the vertex $A(2,1)$ is the mid-point of the line segment joining $Z\left(x_1, y_1\right)$ and the focus $S(1,-1)$

$\therefore \frac{x_1+1}{2}=2 \text { and } \frac{y_1+(-1)}{2}=1 \Rightarrow x_1=3, y_1=3$
Thus, the directrix meets the axis at Z(3, 3)
Let $m _1$ be the slope of the axis. Then,
$m _1=($ Slope of the line joining the focus S and the vertex A $)=\frac{1+1}{2-1}=2 \ldots$. (i)
$\therefore$ Slope of the directrix $=-\frac{1}{2}[\because$ Directrix is perpendicular to the axis $]$
Thus, the directrix passes through (3, 3) and has slope - 1/2.
So its equation is
$\begin{array}{l}y-3=-\frac{1}{2}(x-3) \\
\text { or, } x+2 y-9=0\end{array}$
Let P (x, y) be a point on the parabola.
Then,
Distance of P from the focus = Distance of P from the directrix
$\begin{array}{l}\Rightarrow \sqrt{(x-1)^2+(y+1)^2}=\left|\frac{x+2 y-9}{\sqrt{1^2+2^2}}\right| \\
\Rightarrow(x-1)^2+(y+1)^2=\frac{(x+2 y-9)^2}{5} \\
\Rightarrow 5 x^2+5 y^2-10 x+10 y+10=x^2+4 y^2+81+4 x y-18 x-36 y\end{array}$
The axis passes through the focus $(1,-1)$, and its slope is $m_1=2$
Therefore, equation of the axis is $y+1=2(x,-1)$ or, $2 x-y-3=0$
Now,
Latus-rectum = 2 (Length of the perpendicular from the focus on the directrix)
$=2$ Length of the perpendicular from $(1,-1)$ on $x+2 y-9=0$
$=2\left|\frac{1-2-9}{\sqrt{1+4}}\right|=2 \times \frac{10}{\sqrt{5}}=4 \sqrt{5}$

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Question 55 Marks

Prove that: $\cos 10^{\circ} \cos 30^{\circ} \cos 50^{\circ} \cos 70^{\circ}=\frac{3}{16}$.

Answer

$\begin{array}{l}\cos 10^{\circ} \cos 30^{\circ} \cos 50^{\circ} \cos 70^{\circ}=\frac{3}{16} \\ \text { LHS }=\cos 10^{\circ} \cos 30^{\circ} \cos 50^{\circ} \cos 70^{\circ} \\
=\cos 30^{\circ} \cos 10^{\circ} \cos 50^{\circ} \cos 70^{\circ}\end{array}$
$\begin{array}{l}=\frac{\sqrt{3}}{2}\left(\cos 10^{\circ} \cos 50^{\circ} \cos 70^{\circ}\right) \\ =\frac{\sqrt{3}}{2}\left(\cos 10^{\circ} \cos 50^{\circ}\right) \cos 70^{\circ} \\ =\frac{\sqrt{3}}{4}\left(2 \cos 10^{\circ} \cos 50^{\circ}\right) \cos 70^{\circ}[\text { Multiplying and dividing by } 2] \\ =\frac{\sqrt{3}}{4} \cos 70^{\circ}\left(\cos \left(50^{\circ}+10^{\circ}\right)+\cos \left(10^{\circ}-50^{\circ}\right)\right\}[\text { Using } 2 \cos A \cos B=\cos (A+B)+\cos (A-B)] \\ =\frac{\sqrt{3}}{4} \cos 70^{\circ}\left\{\cos 60^{\circ}+\cos \left(-40^{\circ}\right)\right\} \\ =\frac{\sqrt{3}}{4} \cos 70^{\circ}\left[\frac{1}{2}+\cos 40^{\circ}\right]\left[\because \cos 60^{\circ}=\frac{1}{2} \text { and } \cos (-x)=\cos x \right] \\ =\frac{\sqrt{3}}{8} \cos 70^{\circ}+\frac{\sqrt{3}}{4} \cos 70^{\circ} \cos 40^{\circ} \\ =\frac{\sqrt{3}}{8} \cos 70^{\circ}+\frac{\sqrt{3}}{8}\left(2 \cos 70^{\circ} \cos 40^{\circ}\right) \\ =\frac{\sqrt{3}}{8}\left[\cos 70^{\circ}+\cos \left(70^{\circ}+40^{\circ}\right)+\cos \left(70^{\circ}-40^{\circ}\right)\right] \\ =\frac{\sqrt{3}}{8}\left[\cos 70^{\circ}+\cos 110^{\circ}+\cos 30^{\circ}\right] \\ =\frac{\sqrt{3}}{8}\left[\cos 70^{\circ}+\cos \left(180^{\circ}-70^{\circ}\right)+\frac{\sqrt{3}}{2}\right]\left[\because \cos 30^{\circ}=\frac{\sqrt{3}}{2}\right] \\ =\frac{\sqrt{3}}{8}\left[\cos 70^{\circ}-\cos 70^{\circ}+\frac{\sqrt{3}}{2}\right]\left[\because \cos \left(180^{\circ}-x\right)=-\cos x\right] \\ =\frac{\sqrt{3}}{8} \times \frac{\sqrt{3}}{2}=\frac{3}{16} \\ =\text { RHS }\end{array}$
Hence proved.

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Question 65 Marks

Calculate the mean, median and standard deviation of the following distribution:

Class-interval	31-35	36-40	41-45	46-50	51-55	56-60	61-65	66-70
Frequency	2	3	8	12	16	5	2	3

Answer

1st of all we will prepare the below table with the help of given information.

Class-interval	$f_i$	Mid point $x_i$	$u _{ i }=\frac{x_i-53}{4}$	$u_i^2$	$f _{ i } u _{ i }$	$f _{ i } u_{ i }^2$
31-35	2	33	-5	25	-10	50
36-40	3	38	-3.75	14.06	-11.25	42.18
41-45	8	43	-2.5	6.25	-20	50
46-50	12	48	-1.25	1.56	-15	18.72
51-55	16	53	0	0	0	0
56-60	5	58	1.25	1.56	6.25	7.8
61-65	2	63	2.5	6.25	5	12.5
66-70	3	68	3.75	14.06	11.25	42.18
	N = 51					$\sum_{i=1}^n f_i u_i^2=223.38$

$\begin{array}{l} X = a + h \left(\frac{\sum_{i=1}^n f_i u_i}{N}\right) \\ =53+4\left(\frac{-33.75}{51}\right) \\
=50.36\end{array}$
$\begin{array}{l}\sigma^2=h^2\left(\frac{\sum_{i-1}^n f_i u_i^2}{N}\left(\frac{\sum_{i=1}^n f_i u_i}{N}\right)^2\right) \\
=16\left(\frac{223.38}{51}-\frac{1139.06}{2601}\right) \\
=63.07 \\
\sigma=\sqrt{63.07} \\
=7.94\end{array}$

$f_i$	Cumulative frequency
2	2
3	5
8	13
12	25
16	41
5	46
2	48
3	51

$\begin{array}{l}\sum_N f_i=51=N \\
\frac{N}{2}=25.5\end{array}$
Median class interval is 51 - 55
L = 51
F = 25
f = 16
h = 4
$\begin{array}{l}\text { Median }= L +\frac{\frac{N}{2}-F}{f} \times h \\
=51+\frac{25.5-25}{16} \times 4 \\
=51+\frac{0.5}{4} \\
=51.125\end{array}$

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