TS EAMCET 2018 :: Mathematics :: General questions

• Mathematics - General questions

$\lim_{x \rightarrow -\infty}\frac{3|x|-x}{|x|-2x}-\lim_{x \rightarrow 0}\frac{\log (1+x^{3})}{\sin^{3}x}=$

Answer:

Explanation:

$\lim_{x \rightarrow -\infty}\frac{3|x|-x}{|x|-2x}-\lim_{x \rightarrow 0}\frac{\log (1+x^{3})}{\sin^{3}x}$

 =  $\lim_{x \rightarrow -\infty}\frac{-3x-x}{-x-2x}-\lim_{x \rightarrow 0}\frac{\frac{\log (1+x^{3})}{x^{3}}\times x^{3}}{\left(\frac{\sin x}{x}\right)^{3}\times x^{3}}$

                                          [ $\because$    |x|= -x, x<0]

 = $\frac{3+1}{1+2}-1= \frac{4}{3}-1=\frac{1}{3}$

Assertion (A) if  (-1,3,2) and (5,3,2) are respectively the  orthocentre and circumcentre of a triangle , then (3,3,2) is its centroid.

 Reason  (R)  centroid of the triangle  divides  the line segment joining the orthocentrer and the circumcentre in the ratio 1:2

 Which  one of the following is true?

Answer:

Explanation:

 Except the equilateral  triangle,  the centroid , orthocentre  and circumcentre are collinear and centroid  divides the line segment joining  the orthocentre  and circumcentre in the ratio 2:1. So, if (-1,3,2) and (5,3,2)  are respectively the orthocentre  and circumcentre of triangle  , then coordinate  of centroid is 

$\left(\begin{array}{c}\frac{(-1\times1)+(5\times2)}{1+2},\frac{(3\times1)+(3\times2)}{1+2}\\ \frac{(2\times1)+(2\times2)}{1+2}\end{array}\right)=(3,3,2)$

 So, (A) is true and (R) is false

The equation of the straight lin ein the normal form which is parallel to the lines x+2y+3=0 and x+2y+8=0 and dividing  the distance between these two lines in the ratio 1:2 internally is 

Answer:

Explanation:

 Let  the equation  of required line is

    $x+2y+c=0  $   .................(i)

 According to the question,

    $\frac{2|C-3|}{\sqrt{5}}=\frac{|8-C|}{\sqrt{5}}$

  $\Rightarrow$      $2( C-3)=8-C  \Rightarrow   3C=14$

$\Rightarrow$       $C=\frac{14}{3}$

 So, equation will be

       $x+2y+\frac{14}{3}=0$.............(ii)

 Let the normal form is

   $x \cos \alpha+y \sin \alpha=p$  .....(iii)

 From Eqs.(ii) and (iii) ,

        $\frac{\cos \alpha}{-1}=\frac{\sin \alpha}{-2}=\frac{p}{\frac{14}{3}}$

 $\Rightarrow$    $\alpha=\pi+\tan^{-1}2 $  and   $p=\frac{14}{3\sqrt{5}}\Rightarrow p=\frac{14}{\sqrt{45}}$

 So, required equation is

 $ x \cos \alpha+y \sin \alpha$= $\frac{14}{\sqrt{45}}, \alpha= \pi+ \tan^{-1}2$

An executive in a company makes on an average 5 telephone calls per hour at a cost of rupees 2 per cell. The probability that in any hour the cost of the calls  exceeds  a sum of  rupees 4 is 

Answer:

Explanation:

 The number of telephone calls per hour is a random variable mean =5

 The required  probability is given by

  $P(r>2)=1-P(r\leq2)=1-\sum_{r=0}^2\frac{e^{-5}.5^{r}}{r!}$

  $=1-\left[\frac{5^{0}}{0!}+\frac{5^{1}}{1!}+\frac{5^{2}}{2!}\right]\frac{1}{e^{5}}=1-\frac{37}{2e^{5}}=\frac{2e^{5}-37}{2e^{5}}$

A random variable X takes  the values 1,2,3 and 4 such that 2P(X=1)=3P (X=2)=P(X=3)=5P(X=4) . If $\sigma^{2}$  is the variance and $\mu$  is the mean of X . Then,

 $\sigma^{2}+\mu^{2}$=

Answer:

Explanation:

Given that   ,

$\frac{P(x=1)}{\frac{1}{2}}=\frac{P(x=2)}{\frac{1}{2}}=\frac{P(x=3)}{1}=\frac{P(x=4)}{\frac{1}{5}}=k$   (let)

 $\because$   P(x=1)+P(x=2)+P(x=3)+P(x=4)=1

 $\Rightarrow$      $\frac{k}{2}+\frac{k}{3}+k+\frac{k}{5}=1\Rightarrow k=\frac{30}{61}$

 So,  $P(x=1)=p_{1}=\frac{30}{2 \times 61}=\frac{15}{61}$

       $P(x=2)=p_{2}=\frac{30}{3 \times 61}=\frac{10}{61}$

   $P(x=3) =p_{3}= \frac{30}{61}$

 and $P(x=4)=p_{4}= \frac{30}{5 \times 61}= \frac{6}{61}$

 $\because$    $\sigma^{2}=\left(\sum_{i=1}^4 p_{i}x_{i}^{2}\right)-\mu^{2}$

 $\Rightarrow$       $\sigma^{2}+\mu^{2}=\left(\frac{15}{61}\times 1^{2}\right)+\left(\frac{10}{61} \times2^{2}\right)+\left(\frac{30}{61} \times3^{2}\right)+\left(\frac{6}{61} \times4^{2}\right)$

 = $\frac{15+40+270+96}{61}= \frac{421}{61}$

• Mathematics - General questions