TS EAMCET 2019 :: Mathematics :: General questions

• Mathematics - General questions

If 'a' is the point of discontinuity  of the function 

$f(x)=\begin{cases}\cos 2x & for -\infty<x<0\\e^{3x} & for 0\leq x<3\\ x^{2}-4x+3& ,for 3\leq x\leq6\\\frac{\log(15x-89)}{x-6}&, for x>6\end{cases}$

Then ,   $\lim_{x \rightarrow a}\frac{x^{2}-9}{x^{3}-5x^{2}+9x-9}$=

Answer:

Explanation:

We have ,

$f(x)=\begin{cases}\cos 2x & for -\infty<x<0\\e^{3x} & for 0\leq x<3\\ x^{2}-4x+3& ,for 3\leq x\leq6\\\frac{\log(15x-89)}{x-6}&, for x>6\end{cases}$

$\because \lim_{x \rightarrow 3-}e^{3x}= e^{9}$  and   $\lim_{x \rightarrow 3+} x^{2}-4x+3=9-12+3=0$

 Clearly ,   $\lim_{x \rightarrow 3-}f(x)\neq \lim_{x \rightarrow 3+}f(x) $

 $\therefore$  f(x)   is discontinuous at x=3

$\therefore$ a=3,

Now,   $\lim_{x \rightarrow 3}\frac{x^{2}-9}{x^{3}-5x^{2}+9x-9}=\lim_{x \rightarrow 3}\frac{(x-3)(x+3)}{(x-3)(x^{2}-2x+3)}$

  =  $\frac{3+3}{(3)^{2}-2(3)+3}= \frac{2}{2}=1$

If $e_{1}$  is the eccentricity  of the ellipse   $\frac{x^{2}}{16}+\frac{y^{2}}{25}=1$  and $e_{2}$ is the eccentricity of a hyperbola  passing through  the foci of the given ellipse  and $e_{1}e_{2}=1$ , then the equation of such a hyperbola among the following is 

Answer:

Explanation:

Equation of ellipse    $\frac{x^{2}}{16}+\frac{y^{2}}{25}=1$ 

 Foci   =(0,$\pm 3$)

 $\Rightarrow$    $e_{1}=\sqrt{1-\frac{16}{25}}=\frac{3}{5}$ given $e_{1} e_{2}$ =1

             (where $e_{2}$ is eccentricity of hyperbola )

Let equation of hyperbola -$\frac{x^{2}}{a^{2}}+\frac{ y^{2}}{b^{2}}$=1

 its passes through (0, $\pm 3$)

$\therefore$     $b^{2}$=9

$\Rightarrow$     $ e_{2}=\sqrt{1+\frac{a^{2}}{b^{2}}}$

$\Rightarrow$    $ e_{2}^{2}=1+\frac{a^{2}}{b^{2}}$

$\Rightarrow$    $ \frac{1}{e_{2}^{2}}=1+\frac{a^{2}}{9}\Rightarrow \frac{25}{9}=1+\frac{a^{2}}{9}$

 $\Rightarrow$      $a^{2}$=16

Hence, the equation of hyperbola  $\frac{y^{2}}{9}-\frac{x^{2}}{16}=1$

A tangent is drawn  at  $(3\sqrt{3}\cos\theta, \sin \theta)\left(0< \theta < \frac{\pi}{2}\right)$  to the ellipse  $\frac{x^{2}}{27}+\frac{y^{2}}{1}=1$ . The value of $\theta$  for which the sum of the intercepts on the coordinate axes made by this tangent attains the minimum , is 

Answer:

Explanation:

  Equation of tangent at $(3\sqrt{3}\cos\theta, \sin \theta)$  on the ellipse  $\frac{x^{2}}{27}+\frac{y^{2}}{1}=1$ is

 $\frac{3\sqrt{3}x \cos \theta}{27}+\frac{y \sin \theta}{1}=1$

$\frac{x}{3\sqrt{3}} \cos \theta+\frac{y \sin \theta}{1}=1$

 Sum of intercepts of tangent

 i.e, $L=3\sqrt{3} \sec \theta+cosec \theta$

 $\because \frac{dL}{d \theta}=3\sqrt{3} \sec \theta \tan \theta- cosec \theta \cot \theta$

For  maxima or minima  $\frac{dL}{d \theta}$= 0

$3\sqrt{3} \sec \theta \tan \theta- cosec \theta \cot \theta=0$

$\tan ^{3} \theta = \frac{3}{3\sqrt{3}}\Rightarrow \tan \theta =1\sqrt{3}\Rightarrow \theta = \frac{\pi}{6}$

 Minimum   at $\theta$   = $\frac{\pi}{6}$

If  one of the diameters of the circle $x^{2}+y^{2}-2x-6y+6=0$   is a chord to the circle with centre (2,1), then the radius of the bigger circle is 

Answer:

Explanation:

Equation of circle 

$x^{2}+y^{2}-2x-6y+6=0$  

Centre (1,3)   ,r= $\sqrt{1+9-6}=2$

OA =   $\sqrt{(2-1)^{2}+(1-3)^{2}}\Rightarrow OA=\sqrt{1+4}=\sqrt{5}$

$\because$   AB= r=2    and $R^{2}=r^{2}+OA^{2}$

$\Rightarrow$      $R^{2}=4+5  \Rightarrow$    R=3

If getting a head on a coin when it is  tossed is considered as success, then the probability  of having  more number of failures when  ten fair coins are tossed simultaneously , is 

Answer:

Explanation:

n=10, $p= \frac{1}{2}$, q= $\frac {1}{2}$

$\therefore$  Required probability = $P(r\geq6)$

 $=P(r=6)+P(r=7)+P(r=8)+P(r=9)+P(r=10)$

$=^{10}C_{6}\left(\frac{1}{2}\right)^{10}+^{10}C_{7}\left(\frac{1}{2}\right)^{10}+^{10}C_{8}\left(\frac{1}{2}\right)^{10}+^{10}C_{9}\left(\frac{1}{2}\right)^{10}+\left(\frac{1}{2}\right)^{10}$

$=\frac{1}{2^{10}}\left(^{10}C_{6}+^{10}C_{7}+^{10}C_{8}+^{10}C_{9}+1\right)$

 $=\frac{1}{2^{10}}\left(210+120+45+10+1\right)=\frac{386}{2^{10}}=\frac{193}{2^{9}}$

• Mathematics - General questions