5 Marks Questions

Question 15 Marks

If $A + B + C =\pi,$ prove that $\frac{\sin 2 A+\sin 2 B+\sin 2 C}{\sin A+\sin B+\sin C}=8 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$

Answer

Here it is given that, $A + B + C =\pi$
We need to prove that, $\frac{\sin 2 A+\sin 2 B+\sin 2 C}{\sin A+\sin B+\sin C}=8 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$
Proof: Taking $\text{LHS},$ we have,
$\text{L.H.S} =\frac{\sin 2 A+\sin 2 B+\sin 2 C}{\sin A+\sin B+\sin C}$
Where $, \sin 2A + \sin 2B + \sin 2C = 2\sin A \cos A + 2\sin(B + C)\cos(B - C)$
$[$ By using, $\sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$
and $\sin 2A = 2sin A \cos A]$
Since $ A+B+C=\pi$
$\Rightarrow B+C=180-A$
$\therefore \sin 2 A+\sin 2 B+\sin 2 C=2 \sin A \cos A+2 \sin (\pi-A) \cos (B-C)$
$=2 \sin A \cos A+2 \sin A \cos (B-C)$
$=2 \sin A\{\cos A+\cos (B-C)\}$
$($but $\cos A = \cos { 180 - ( B + C ) } = - \cos ( B + C )$
And now using
$\cos A -\cos B =2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{-A+B}{2}\right)$
So $, \sin 2A + \sin 2B + \sin 2C = 2\sin A{2\sin B \sin C}$
$= 4\sin A \sin B \sin C$
$=32 \sin \frac{A}{2} \cos \frac{A}{2} \sin \frac{B}{2} \cos \frac{B}{2} \sin \frac{C}{2} \cos \frac{C}{2}$
Now, take denominator we have
$\sin A +\sin B +\sin C =\sin A +\left\{2 \sin \left(\frac{B+C}{2}\right) \cos \left(\frac{B-C}{2}\right)\right\}$
$=\sin A +\left\{2 \sin \left(\frac{\pi-A}{2}\right) \cos \left(\frac{B-C}{2}\right)\right\}$
$=\sin A +\left\{2 \cos \left(\frac{A}{2}\right) \cos \left(\frac{B-C}{2}\right)\right\}$
$=2 \sin \frac{A}{2} \cos \frac{A}{2}+\left\{2 \cos \left(\frac{A}{2}\right) \cos \left(\frac{B-C}{2}\right)\right\}$
$=2 \cos \frac{A}{2}\left\{\sin \frac{A}{2}+\cos \left(\frac{B-C}{2}\right)\right\}$
$=2 \cos \frac{A}{2}\left\{\cos \left(\frac{B+C}{2}\right)+\cos \left(\frac{B-C}{2}\right)\right\}$
$=2 \cos \frac{A}{2}\left\{2 \cos \left(\frac{B}{2}\right) \cos \left(\frac{c}{2}\right)\right\}$
$=4 \cos \frac{A}{2} \cos \left(\frac{B}{2}\right) \cos \left(\frac{C}{2}\right)$
Therefore,
$\text { L. H. } S=\frac{\sin 2 A+\sin 2 B+\sin 2 C}{\sin A+\sin B+\sin C}=\frac{32 \sin \frac{A}{2} \cos \frac{A}{2} \sin \frac{B}{2} \cos \frac{B}{2} \sin \frac{C}{2} \cos \frac{C}{2}}{4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}}$
$=8 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$
$=\text { R.H.S }$

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Question 25 Marks

Prove that: $\cot x+\cot \left(\frac{\pi}{3}+x\right)+\cot \left(\frac{2 \pi}{3}+x\right)=3 \cot 3 x$

Answer

We have to prove $\cot x+\cot \left(\frac{\pi}{3}+x\right)+\cot \left(\frac{2 \pi}{3}+x\right)=3 \cot 3 x$.
$\text{LHS} =\cot x+\cot \left(\frac{\pi}{3}+x\right)+\cot \left(\frac{2 \pi}{3}+x\right)$
We know,
$\cot \left(\frac{2 \pi}{3}+x\right)=\cot \left(\pi-\left(\frac{\pi}{3}-x\right)\right)=-\cot \left(\frac{\pi}{3}-x\right) \ldots\left(\right.$ as $-\cot \theta=\cot \left(180^{\circ}-\theta\right)$
Hence the above LHS becomes
$=\cot x+\cot \left(\frac{\pi}{3}+x\right)-\cot \left(\frac{\pi}{3}-x\right)$
$=\frac{1}{\tan x}+\frac{1}{\tan \left(\frac{\pi}{3}+x\right)} \cdot \frac{1}{\tan \left(\frac{\pi}{3}-x\right)}$
$=\frac{1}{\tan x}+\left(\frac{1-\tan x \tan \frac{\pi}{3}}{\tan \frac{\pi}{3}+\tan x}\right)-\left(\frac{1+\tan x \tan \frac{\pi}{3}}{\tan \frac{x}{3}-\tan x}\right) \cdots\left[\because \tan (A+B)=\left(\frac{\tan A+\tan B}{1-\tan A \tan B}\right) \text { and } \tan (A-B)=\left(\frac{\tan A-\tan B}{1+\tan A \tan B}\right)\right]$
$=\frac{1}{\tan x}+\left(\frac{1-\sqrt{3} \tan x}{\sqrt{3}+\tan x}\right)-\left(\frac{1+\sqrt{3} \tan x}{\sqrt{3}-\tan x}\right)$
$=\frac{1}{\tan x}+\left(\frac{(1-\sqrt{3} \tan x)(\sqrt{3}-\tan x)-(1+\sqrt{3} \tan x)(\sqrt{3}+\tan x)}{(\sqrt{3}+\tan x)(\sqrt{3}-\tan x)}\right)$
$=\frac{1}{\tan x}+\left(\frac{\left(\sqrt{3}-\tan x-3 \tan x+\sqrt{3} \tan ^2 x\right)-\left(\sqrt{3}+3 \tan x+\tan x+\sqrt{3} \tan ^2 x\right)}{\left(3-\tan ^2 x\right)}\right)$
$=\frac{1}{\tan x}+\left(\frac{(0-4 \tan x-4 \tan x+0)}{\left(3-\tan ^2 x\right)}\right)$
$=\frac{1}{\tan x}-\left(\frac{8 \tan x}{\left(\left(3-\tan ^2 x\right)\right)}\right)$
$=\left(\frac{\left(3-\tan ^2 x\right)-8 \tan ^2 x}{\tan x\left(3-\tan ^2 x\right)}\right)=\left(\frac{\left(3-\tan ^2 x\right)-8 \tan ^2 x}{\tan x\left(3-\tan ^2 x\right)}\right)$
$=3\left(\frac{1-3 \tan ^2 x}{\left(3 \tan x-\tan ^3 x\right)}\right)$
$=3 \times \frac{1}{\tan 3 x} \ldots\left(\operatorname{as} \tan 3 x=\frac{3 \tan x-\tan ^3 x}{1-3 \tan ^2 x}\right)$
$= \cot 3x$
$\text{LHS = RHS}$
Hence proved.

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Question 35 Marks

Solve the following system of linear inequalities $\frac{4 x}{3}-\frac{9}{4}x$

Answer

We have, $\frac{4 x}{3}-\frac{9}{4} $ and $\frac{7 x-1}{3}-\frac{7 x+2}{6}>x \ldots(ii)$
From inequality $(i),$
we get $\frac{4 x}{3}-\frac{9}{4}$
$\Rightarrow 16 x-27<12 x+9 \ [$multiplying both sides by $12]$
$\Rightarrow 16 x-27+27<12 x+9+27 \ [$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, $4$

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Question 45 Marks

A visitor with sign board 'DO NOT LITTER' is moving on a circular path in an exhibition. During the movement he stops at points represented by (3, - 2) and (-2, 0). Also, centre of the circular path is on the line 2x - y = 3. What is the equation of the path? What message he wants to give to the public?

Answer

Let the equation of circle whose centre (- g, - f) be
$x^2+y^2+2 g x+2 f y+c=0 \ldots$ (i)
Since, is passes through points (3, - 2) and (- 2, 0)
$\therefore(3)^2+(-2)^2+2 g(3)+2 f(-2)+c=0$
and $(-2)^2+(0)^2+2 g(-2)+2 f(0)+c=0$
$\Rightarrow 9+4+6 g-4 f+c=0$
and 4 + 0 - 4g + 0 + c = 0
$\Rightarrow 6 g-4 f+c=-13$
and c = 4g - 4 ...(ii)
$\therefore 6 g-4 f+(4 g-4)=-13$
$\Rightarrow 10 g-4 f =-9 \ldots( iii )$
Also, centre (- g, - f) lies on the line 2x - y = 3
$\therefore-2 g+f=3 \ldots$ (iv)
On solving Eqs. (iii) and (iv), we get
$g=\frac{3}{2}$ and $f=6$
On putting the values of g and f in Eq. (ii), we get
$c=4\left(\frac{3}{2}\right)-4=6-4=2$
On putting the values of g, f and c in Eq. (i), we get
$x^2+y^2+2\left(\frac{3}{2}\right) x+2(6) x+2=0$
$\Rightarrow x^2+y^2+3 x+12 x+2=0$
which is the required equation of the path
The message which he wants to give to the public is 'Keep your place clean'.

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Question 55 Marks

Fine the lengths major and minor axes, coordinates of the vertices, coordinates of the foci, eccentricity, and length of the latus rectum of the ellipse $25 x^2+4 y^2=100$.

Answer

Given that:
$25 x^2+4 y^2=100$
after divide by 100 to both the sides, we get
$\frac{25}{100} x^2+\frac{4}{100} y^2=1 \Rightarrow \frac{x^2}{4}+\frac{y^2}{25}=1 \ldots$
Now, above equation is of the form,
$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1 \ldots$ (ii)
Comparing eq. (i) and (ii), we get
$a^2=25$ and $b^2=4 \Rightarrow a=\sqrt{25}$ and $b=\sqrt{4} \Rightarrow a=5$ and $b=2$
i. Length of major axes
$\therefore$ Length of major axes $=2 a=2 \times 5=10$ units
ii. Length of minor axes
Length of minor axes $=2 b=2 \times 2=4$ units
iii. Coordinates of the vertices
$\therefore$ Coordinates of the vertices $=(0, a)$ and $(0,-a)=(0,5)$ and $(0,-5)$
iv. Coordinates of the foci
As we know that,
Coordinates of foci $=(0, \pm c)$ where $c ^2= a ^2- b ^2$
Now
$c^2=25-4 \Rightarrow c^2=21 \Rightarrow c=\sqrt{21} \ldots$ (iii)
$\therefore$ Coordinates of foci $=(0, \pm \sqrt{2} 1)$
v. Eccentricity
As we know that, Eccentricity $=\frac{c}{a} \Rightarrow e=\frac{\sqrt{21}}{5}$
vi. Length of the Latus Rectum
As we know that Length of Latus Rectum $=\frac{2 b^2}{a}=\frac{2 \times(2)^2}{5}=\frac{8}{5}$

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Question 65 Marks

The mean and standard deviation of $100$ observations were calculated as $40$ and $5.1,$ respectively by a student who took by mistake $50$ instead of $40$ for one observation. What are the correct mean and standard deviation?

Answer

We have, $n =100, \bar{x}=40$ and $\sigma=5.1$
$\therefore \bar{x}=\frac{1}{n} \Sigma x_i$
$\Rightarrow \Sigma x_i=n \bar{x}=100 \times 40=4000$
$\therefore$ Incorrect $\Sigma x _{ i }=4000$
and,
$\sigma=5.1$
$\Rightarrow \sigma^2=26.01$
$\Rightarrow \frac{1}{n} \Sigma x _{ i }^2-(\text { mean })^2=26.01$
$\Rightarrow \frac{1}{100} \Sigma x _{ i }^2-1600=26.01$
$\Rightarrow \Sigma x _{ i }{ }^2=1626.01 \times 100$
$\therefore$ Incorrect $\Sigma x _{ i }^2=162601$
To correct the $\sum x_i$, we need to subtract the incorrect observation $50$ and add correct observation is $40$ .
We have, incorrect $\Sigma x _{ i }=4000$
$\therefore$ Correct $\Sigma x _{ i }=4000-50+40=3990$
and,
Similarly, to obtain correct $\sum x_i^2$ we need to subtract $50^2$ and add $40^2$ to incorrect one.
Incorrect $\Sigma x _{ i }^2=162601$
$\therefore$ Correct $\Sigma x _{ i }{ }^2=162601-50^2+40^2=161701$
Now, Correct mean $=\frac{3990}{100}=39.90$
Correct variance $=\frac{1}{100}\left(\right.$ Correct $\left.\Sigma xi _{ i }{ }^2\right)-(\text { Correct mean })^2$
$\Rightarrow \text { Correct variance }=\frac{161701}{100}-\left(\frac{3990}{100}\right)^2$
$\Rightarrow \text { Correct variance }=\frac{161701 \times 100-(3990)^2}{(100)^2}$
$\Rightarrow \text { Correct variance }=\frac{16170100-15920100}{10000}=25$
$\therefore$ Correct standard deviation $=\sqrt{25}=5$

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