5 Marks Questions

Question 15 Marks

Prove that $\cos 12^{\circ}+\cos 60^{\circ}+\cos 84^{\circ}=\cos 24^{\circ}+\cos 48^{\circ}$

Answer

$\ce{LHS} =\cos 12^{\circ}+\cos 60^{\circ}+\cos 84^{\circ}$
$=\cos 12^{\circ}+\left(\cos 84^{\circ}+\cos 60^{\circ}\right)$
$=\cos 12^{\circ}+\left[2 \cos \left(\frac{84^{\circ}+60^{\circ}}{2}\right) \times \cos \left(\frac{86^{\circ}-60^{\circ}}{2}\right)\right]$
$\left[\because \cos x+\cos y=2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)\right]$
$=\cos 12^{\circ}+\left[2 \cos \frac{144^{\circ}}{2} \times \cos \frac{24^{\circ}}{2}\right]$
$=\cos 12^{\circ}+\left[2 \cos 72^{\circ} \times \cos 12^{\circ}\right]=\cos 12^{\circ}\left[1+2 \cos 72^{\circ}\right]$
$=\cos 12^{\circ}\left[1+2 \cos \left(90^{\circ}-18^{\circ}\right)\right]$
$=\cos 12^{\circ}\left[1+2 \sin 18^{\circ}\right]\left[\because \cos \left(90^{\circ}-\theta\right)=\sin \theta\right]$
$=\cos 12^{\circ}\left[1+2\left(\frac{\sqrt{5}-1}{4}\right)\right]\left[\because \sin 18^{\circ}=\frac{\sqrt{5}-1}{4}\right]$
$=\left(1+\frac{\sqrt{5}-1}{2}\right) \cos 12^{\circ}=\left(\frac{\sqrt{5}+1}{2}\right) \cos 12^{\circ}$
$\ce{RHS} =\cos 24^{\circ}+\cos 48^{\circ}$
$=2 \cos \left(\frac{24^{\circ}+48^{\circ}}{2}\right) \cos \left(\frac{24^{\circ}-48^{\circ}}{2}\right)\left[\because \cos x+\cos y=2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)\right]$
$=2 \cos 36^{\circ} \cos \left(-12^{\circ}\right)$
$=2 \cos 36^{\circ} \times \cos 12^{\circ}[\because \cos (-\theta)=\cos \theta]$
$=2 \times \frac{\sqrt{5}+1}{4} \times \cos 12^{\circ}=\frac{\sqrt{5}+1}{2} \times \cos 12^{\circ}\left[\because \cos 36^{\circ}=\frac{\sqrt{5}+1}{4}\right]$
$\therefore \ce{LHS = RHS}$
Hence proved.

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Question 25 Marks

Prove that: $\sin 20^{\circ} \sin 40^{\circ} \sin 80^{\circ}=\frac{\sqrt{3}}{8}$

Answer

Given, $\ce{LHS} =\sin 20^{\circ} \sin 40^{\circ} \sin 80^{\circ}$
$=\frac{1}{2}\left[2 \sin 20^{\circ} \cdot \sin 40^{\circ}\right] \sin 80^{\circ}[\text { multiplying and dividing by } 2]$
$=\frac{1}{2}\left[\cos \left(20^{\circ}-40^{\circ}\right)-\cos \left(20^{\circ}+40^{\circ}\right)\right] \cdot \sin 80^{\circ}[\because 2 \sin x \cdot \sin y=\cos ( x - y )-\cos ( x + y )]$
$=\frac{1}{2}\left[\cos \left(-20^{\circ}\right)-\cos 60^{\circ}\right] \sin 80^{\circ}$
$=\frac{1}{2}\left[\cos 20^{\circ}-\frac{1}{2}\right] \cdot \sin 80^{\circ}\left[\because \cos (-\theta)=\cos \theta \text { and } \cos 60^{\circ}=\frac{1}{2}\right]$
$=\frac{1}{2} \times \frac{1}{2}\left[2\left(\cos 20^{\circ}-\frac{1}{2}\right) \cdot \sin 80^{\circ}\right] \text { [again multiplying and dividing by 2] }$
$=\frac{1}{4}\left[2 \cos 20^{\circ} \cdot \sin 80^{\circ}-\sin 80^{\circ}\right]$
$=\frac{1}{4}\left[\sin \left(20^{\circ}+80^{\circ}\right)-\sin \left(20^{\circ}-80^{\circ}\right)-\sin 80^{\circ}\right][\because 2 \cos x \cdot \sin y=\sin (x+y)-\sin (x-y)]$
$=\frac{1}{4}\left[\sin 100^{\circ}-\sin \left(-60^{\circ}\right)-\sin 80^{\circ}\right]$
$=\frac{1}{4}\left[\sin 100^{\circ}+\sin 60^{\circ}-\sin 80^{\circ}\right][\because \sin (-\theta)=-\sin \theta]$
$=\frac{1}{4}\left[\sin \left(180^{\circ}-80^{\circ}\right)+\sin 60^{\circ}-\sin 80^{\circ}\right]\left[\because \sin 100^{\circ}=\sin \left(180^{\circ}-80^{\circ}\right)\right]$
$=\frac{1}{4}\left[\sin 80^{\circ}+\sin 60^{\circ}-\sin 80^{\circ}\right][\because \sin (\pi-\theta)=\sin \theta]$
$=\frac{1}{4} \times \sin 60^{\circ}=\frac{1}{4} \times \frac{\sqrt{3}}{2}\left[\because \sin 60^{\circ}=\frac{\sqrt{3}}{2}\right]$
$=\frac{\sqrt{3}}{8}= \ce{RHS} $
Hence proved.

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Question 35 Marks

Solve the following system of linear inequalities $\frac{4 x}{3}-\frac{9}{4}< x+\frac{3}{4}$ and $\frac{7 x-1}{3}-\frac{7 x+2}{6} > x$.

Answer

We have, $\frac{4 x}{3}-\frac{9}{4} < x+\frac{3}{4} \ldots(i)$ and
$\frac{7 x-1}{3}-\frac{7 x+2}{6}>x \ldots (ii)$
From inequality $(i),$ we get
$\frac{4 x}{3}-\frac{9}{4} < x+\frac{3}{4}$
$ \Rightarrow \frac{16 x-27}{12}<\frac{4 x+3}{4}$
$\Rightarrow 16 x -27<12 x +9 \ [$multiplying both sides by $12 \text]$
$\Rightarrow 16 x -27+27<12 x +9+27 \ [$adding $27 $ on both sides$]$
$\Rightarrow 16 x <12 x +36$
$\Rightarrow 16 x -12 x <12 x +36-12 x \ [$subtracting $12 x$ from bot sides$]$
$\Rightarrow 4 x < 36 $
$\Rightarrow x < 9 \ [$dividing both sides by $4] $
Thus, any value of $x$ less than $9$ satisfies the inequality.
So, the solution of inequality $(i)$ is given by $x \in(-\infty, 9)$

From inequality $(ii)$ we get,
$\frac{7 x-1}{3}-\frac{7 x+2}{6}>x $
$\Rightarrow \frac{14 x-2-7 x-2}{6}>x$
$\Rightarrow 7 x-4>6 x \ [$multiplying by $6$ on both sides$]$
$\Rightarrow 7 x-4+4>6 x+4 \ [$adding $4$ on both sides$]$
$\Rightarrow 7 x>6 x+4$
$\Rightarrow 7 x-6 x>6 x+4-6 x \ [$subtracting $6 x$ from both sides$]$
$\therefore x>4$
Thus, any value of $x$ greater than $4$ satisfies the inequality.
So, the solution set is $x \in(4, \infty)$

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

Clearly, the common value of $x$ lie between $4$ and $9$.
Hence, the solution of the given system is, $4 < x < 9$ i.e., $x \in(4,9)$

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Question 45 Marks

Find the equation of the hyperbola, the length of whose latus rectum is $4$ and the eccentricity is $3$.

Answer

Given: The length of latus rectum is $4,$ and the eccentricity is $3$
Let, the equation of the hyperbola be: $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

The length of the latus rectum is $4$ units.
$\Rightarrow$ length of the latus rectum $=\frac{2 b^2}{a}=4$
$\Rightarrow \frac{2 b^2}{a}=4 $
$\Rightarrow b^2=2 a \ldots \text { (i) }$
And also given, the eccentricity $, e = 3$
We know that, $e=\sqrt{1+\frac{b^2}{a^2}}$
$\Rightarrow \sqrt{1+\frac{b^2}{a^2}}=3$
$\Rightarrow 1+\frac{b^2}{a^2}=9\ [$Squaring both sides$]$
$\Rightarrow \frac{b^2}{a^2}=8$
$\Rightarrow b^2=8 a ^2$
$\Rightarrow 2 a =8 a ^2 \ [ $From $(i)]$
$\Rightarrow a=\frac{1}{4} $
$\Rightarrow a ^2=\frac{1}{16}$
From $(i)\ \Rightarrow b^2=2 a$
$=2 \times \frac{1}{4}=\frac{1}{2}$
$ \Rightarrow b^2=\frac{1}{2}$
So, the equation of the hyperbola is,
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 $
$\Rightarrow \frac{x^2}{1 / 16}-\frac{y^2}{1 / 2}=1$
$\Rightarrow 16 x^2-2 y^2=1$

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Question 55 Marks

Find the equation of a circle concentric with the circle $x^2+y^2+4 x+6 y+11=0$ and passing through the point $(5,4)$.

Answer

Here, the equation of circle is $x^2+y^2+4 x+6 y+11=0$
$\Rightarrow\left(x^2+4 x\right)+\left(y^2+6 y\right)=-11$
On adding 4 and 9 both sides to make perfect squares,
we get $\left(x^2+4 x+4\right)+\left(y^2+6 y+9\right)=-11+4+9$
$\Rightarrow( x +2)^2+( y +3)^2=(\sqrt{2})^2 \ldots( i )$
Its centre is $(- 2, - 3)$

The required circle is concentric with circle $1,$ therefore its centre is $(- 2 , - 3)$.
Since, it passes through $(5, 4),$ therefore radius is
$ r = CP =\sqrt{(5+2)^2+(4+3)^2}\left[\because \text { distance }=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\right]$
$=\sqrt{49+49}=7 \sqrt{2}$
Hence, the equation of required circle having centre $(-2,-3)$ and radius $7 \sqrt{2}$ is,
$(x+2)^2+(y+3)^2=(7 \sqrt{2})^2$
$\Rightarrow x^2+4 x+4+y^2+6 y+9=98$
$\Rightarrow x^2+4 x+y^2+6 y-85=0$

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Question 65 Marks

Find the mean deviation about the mean for the following data:

$x_i$	$3$	$5$	$7$	$9$	$11$	$13$
$f_i$	$6$	$8$	$15$	$3$	$8$	$4$

Answer

We have
$N=\sum_{i=1}^6 f_i=(6+8+15+3+8+4)=44$
$\bar{x}=\frac{\sum_{i=1}^6 f_i x_i}{N}=\frac{(6 \times 3)+(8 \times 5)+(15 \times 7)+(3 \times 9)+(8 \times 11)+(4 \times 13)}{44}$
$=\frac{(18+40+105+27+88+52)}{44}=\frac{330}{44}=\frac{15}{2}=7.5$

$x_i$	$3$	$5$	$7$	$9$	$11$	$13$
$f_i$	$6$	$8$	$15$	$3$	$8$	$4$
$cf$	$6$	$14$	$29$	$32$	$40$	$44$

Here we have$, N = 44,$ which is even.
Therefore$,$ median $=\frac{1}{2} \cdot\left\{\frac{N}{2}\right.$ th observation $+\left(\frac{N}{2}+1\right)$ th observation $\}$
$=\frac{1}{2}(22)$ nd observation $(23)$ rd observation
$=\frac{1}{2}(7+7)=7$
Thus$, M = 7.$
Now$,$ we have$:$

$\left\|x_i-M\right\|$	$4$	$2$	$0$	$2$	$4$	$6$
$f_i$	$6$	$8$	$15$	$3$	$8$	$4$
$f_i\left\|x_i-M\right\|$	$24$	$16$	$0$	$6$	$32$	$24$

$\therefore \sum_{i=1}^6 f_i=44 \text { and } \sum_{i=1}^6 f_i\left|x_i-M\right|=102$
$\therefore M D(\bar{x})=\frac{\sum_{i=1}^6 f_i\left|x_i-M\right|}{N}=\frac{102}{44}=2.32$

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