TS EAMCET 2017 :: Mathematics :: General questions

• Mathematics - General questions

If A=[5,3], B=[3,-2] and a point P is such that the area of the triangle PAB is 9, then the locus of P  represents.

Answer:

Explanation:

Let P(x,y), A=(5,3),B=(3,-2)

Area of $\triangle PAB$= $\frac{1}{2}\begin{bmatrix}x & y&1 \\5 & 3&1\\3&-2&1 \end{bmatrix}$

$\Rightarrow$              $9=\frac{1}{2}|x(3+2)-y(5-3)+(-10-9)|$

$\Rightarrow$     $18=|5x-2y-19|$

$\Rightarrow$   $5x-2y-16=|18|$

$\Rightarrow$     $5x-2y-19=\pm18$

$\Rightarrow$   $5x-2y=19 \pm 18$

$\Rightarrow$     $5x-2y=1  $ or $5x-2y=37$

Which represents a pair of parallel lines

If   $\int e^{2x} f'(x) dx=g(x) , $  then 

$\int (e^{2x} f(x)+e^{2x} f'(x)) dx= $

Answer:

Explanation:

We have, 

$\int e^{2x} f'(x) dx= g(x)$

Let $I= \int (e^{2x} f(x)+e^{2x} f'(x)) dx$

 $=f(x)\int e^{2x} dx-\int f'(x)  \int e^{2x} dx) dx+\int e^{2x} f'(x) dx$

 $=\frac{f(x)e^{2x}}{2}-\frac{1}{2}\int e^{2x}f'(x) dx+\int e^{2x} f'(x)dx$

  $=\frac{e^{2x}}{2}f(x)-\frac{1}{2}\int e^{2x} f'(x) dx$

  $\frac{1}{2}[e^{2x}f(x)-\int e^{2x} f'(x) dx] $

  =  $\frac{1}{2}[e^{2x}f(x)-g(x) ]+C $

If  $ cosec \theta  - \cot \theta=2017$, then quadrant in which $\theta$ lies is 

Answer:

Explanation:

 We have 

$ cosec \theta  - \cot \theta=2017$    ........(i)

$\therefore$      $cosec \theta +\cot \theta=\frac{1}{2017}$ ......(ii)

  $\begin{bmatrix}\because & cosec^{2}\theta-\cot^{2}\theta=1 \\\Rightarrow & cosec \theta-\cot \theta=\frac{1}{cosec \theta+\cot \theta} \end{bmatrix}$

 Adding Eqs,(i) and (ii) , we get

     $2 cosec \theta=2017+ \frac{1}{2017}$

  $\Rightarrow$    $cosec \theta= \frac{1}{2} [2017+ \frac{1}{2017}]>0$

 $\theta$ lie in 1st or II nd quadrant

 Substracting  Eq.(i) from Eq.(ii)  , we get

 $2\cot \theta= \frac{1}{2017}-2017$

$\cot \theta= \frac{1}{2}\left(\frac{1}{2017}-2017\right)<0$

 $\theta$  lie in II nd and III rd quadrant

 Hence , $\theta$ lies in II nd quadrant 

The radical centre of the circle 

$x^{2}+y^{2}-4x-6y+5=0$, $  x^{2}+y^{2}-2x-4y-1=0$  and

$x^{2}+y^{2}-6x-2y=0=0 $ lies on the line 

Answer:

Explanation:

 $S_{1}= x^{2}+y^{2}-4x-6y+5=0$

$S_{2}= x^{2}+y^{2}-2x-4y-1=0$

$S_{3}= x^{2}+y^{2}-6x-2y=0$

 $S_{1}-S_{2}=0 \Rightarrow 2x+2y-6=0 \Rightarrow x+y-3=0 $ .....(i)

 $S_{2}-S_{3}=0 \Rightarrow 4x-2y-1=0$   ......(ii)

 Solving Eqs.(i) and (ii) , we get

 $x=\frac{7}{6}, y=\frac{11}{6}$

$\left(\frac{7}{6},\frac{11}{6}\right)$  satisfies the equation 18x-12y+1=0

The bad contains 5 red balls, 3 black balls and 4 white balls . Three balls are drawn at random . The probability that they are not of same colour is 

Answer:

Explanation:

We have,

 5 red balls, 3 black balls , 4 white balls

Total number of balls =(5+3+4)=12

Three balls are drawn= $ ^{12}C_{3}=220$

 Three balls are drawn of same colour

    $^{5}C_{3}+^{3}C_{3}+^{4}C_{3}=10+1+4=15$

 Probability of are not same colour 

   =1- Probability of same colour

  =$1-\frac{15}{220}=1- \frac{3}{44}=\frac{41}{44}$

• Mathematics - General questions