TS EAMCET 2017 :: Mathematics :: General questions

• Mathematics - General questions

If A and B are events having probabilities P(A)=0.6 , P(B)=0.4  and  $P(A\cap B)=0$ , then probability that neither  A nor B occurs is 

Answer:

Explanation:

Given, P(A)=0.6 , P(B)=0.4 and $P(A\cap B)=0$  then

P(neither A nor B)= $P(\overline{A}\cap \overline{B})$

= $P(\overline{A\cap B})=1-P(A\cup B)$

    = $1-P(A)-P(B)+P(A\cap B)$

              $[\because P(A\cup B)=P(A)+P(B)-P(A\cap B)]$

  =1-0.6-0.4+0=1-1=0

 If the line x-y=-4K is a tangent  to the parabola $y^{2}=8x$ at P, then the perpendicular distance of normal at P from (K,2K) is 

Answer:

Explanation:

Since, x-y=-4k  or y=x+4k is a tangent to the parabola $y^{2}=8x$, therefore

    $4k=\frac{2}{1}$            $[\because  4a=8 \Rightarrow a=2]$

  $\Rightarrow$       $k=\frac{1}{2}$

 Also  points of concept  p is    $\left(\frac{a}{m^{2}},\frac{2a}{m}\right)=(2,4)$

  Now, equation of normal to the $y^{2}=8x$ at (2,4) is 

$(y-4)=\frac{-4}{2(2)}(x-2)$

 [$\because$   Equation of normal to the parabola , $y^{2}=4ax$ at

                                        $(x_{1},y_{1})is y-y_{1}=\frac{-b}{2a}(x-x_{1})$

$\Rightarrow$   $y-4=-1(x-2)$

$\Rightarrow$   $y-4=-x+2$

$\Rightarrow$   $x+y=6$

$\Rightarrow$   $x+y-6=0$

The perpendicular distance of normal from (k,2k) ie  $(\frac{1}{2},1)$

$=\frac{|\frac{1}{2}+1-6|}{\sqrt{1^{2}+1^{2}}}=\frac{|\frac{3}{2}-6|}{\sqrt{2}}=\frac{9}{2\sqrt{2}}$

The lengths of the sides  of a triangle are13,14 and 15. If R and r repectively  denote circumradius and inradius of that  triangle , then  8R+r=

Answer:

Explanation:

Let a=13, b=14 and c=15, Then

 $S= \frac{a+b+c}{2}=\frac{13+14+15}{2}=21$

And area of triangle

          $\triangle=\sqrt{s(s-a)(s-b)(s-c)}$

                              = $\sqrt{21(21-13)(21-14)(21-15)}$

            $\triangle=\sqrt{21\times8\times7\times6}=7\times3\times4=84$

Now, as we know   $R=\frac{abc}{4 \triangle}  $ and $r=\frac{\triangle}{s}$

 $\therefore$       $R= \frac{ 13 \times 14 \times 15}{4 \times 84}$ and

  $r=\frac{84}{21}$

$\Rightarrow$   $R=\frac{65}{8}$  and r=4

 So, 8R+r=65+4=69

 The number of solutions of $\cos 2 \theta=\sin \theta$ in $(0,2\pi)$ is 

Answer:

Explanation:

Consider the equations

  $\cos 2 \theta=\sin \theta$

$\Rightarrow$     $1-2\sin^{2} \theta=\sin \theta$

$\Rightarrow$   $2 \sin^{2} \theta+\sin \theta-1=0$

$\Rightarrow$   $2 \sin \theta( \sin \theta+1)-1(\sin \theta+1)=0$

$\Rightarrow$   $(\sin \theta+1)(2 \sin \theta-1)=0$

$\Rightarrow$                $\sin \theta=-1 $  or $\sin \theta=\frac{1}{2}$

$\Rightarrow$            $\theta=\frac{3 \pi}{2}$  or  $\theta= \frac{\pi}{6},\frac{5 \pi}{6}$

                                                $[\because \theta \in (0,2\pi)]$

Thus, number of solutions of the given equation is 3

If the pair of straight  lines xy-x-y+1=0  and the line x+ay-3=0 are concurrent, then the acute angle between  the pair of lines 

$ax^{2}-13xy-7y^{2}+x+23y-6=0$ is 

Answer:

Explanation:

Given equations of pair of straight lines, xy-x-y+1=0 can be rewritten as

          $x(y-1)-1(y-1)=0$

$\Rightarrow$  $(x-1) (y-1)=0$

$\Rightarrow$    $x-1$  and $y=1$

 Thus, the three concurrent  lines are

                    x=1 ........(i)

                     y=1 ......(ii)

 and x+ay-3=0......(iii)

 Since, Eqs.(i) and (ii)  , intersect  at only point , namely (1,1), therefore this point also satisfy  the Eq.(iii) , we get

           $1+a-3=0$

$\Rightarrow$       $a=2$

 Now, the pair  of lines

 $ax^{2}-13xy-7y^{2}+x+23y-6=0$ becomes

$2x^{2}-13xy-7y^{2}+x+23y-6=0$

 The acute angle $\theta$  between these lines is given by

$\tan \theta=\mid\frac{2\sqrt{h^{2}-ab}}{a+b}\mid=\mid\frac{2\sqrt{\left(-\frac{13}{2}\right)^{2}+14}}{-5}\mid$

  $=\mid\frac{2\sqrt{\left(\frac{169+56}{4}\right)^{}}}{-5}\mid=\mid\frac{2\sqrt{\frac{225}{4}}}{-5}\mid=$

                                         $=\mid\frac{2\times\frac{15}{2}}{-5}\mid=|-3|=3$

 $\theta= tan^{1}(3)= \cos^{-1}\left(\frac{1}{\sqrt{1+3^{2}}}\right)=\cos^{-1}\left(\frac{1}{\sqrt{10}}\right)$

• Mathematics - General questions