TS EAMCET 2019 :: Mathematics :: General questions

• Mathematics - General questions

If   $\sqrt{\frac{y}{x}}+4\sqrt{\frac{x}{y}}=4$ then  $\frac{dy}{dx}$=

Answer:

Explanation:

 Given,

 $\sqrt{\frac{y}{x}}+4\sqrt{\frac{x}{y}}=4$

 On squaring both sides , we get

   $\frac{y}{x}+\frac{16x}{y}+8=16\Rightarrow\frac{y}{x}+16\frac{x}{y}=8$

$\Rightarrow$   $y^{2}+16x^{2}=8xy$

 On differentiating both sides w.r.t  x, we get

    $2y\frac{dy}{dx}+32x=8y+8x \frac{dy}{dx}$

 $\Rightarrow$     $(4x-y) \frac{dy}{dx}=4(4x-y)\Rightarrow \frac{dy}{dx}=4$

 Hence, option (d) is correct

$\lim_{x \rightarrow 0}\frac{\sqrt{x^{2}+100}-10}{x^{2}}=$

Answer:

Explanation:

$\lim_{x \rightarrow 0}\frac{\sqrt{x^{2}+100}-10}{x^{2}}$

   =$\lim_{x \rightarrow 0}\frac{(\sqrt{x^{2}+100}-10)(\sqrt{x^{2}+100}+10)}{x^{2}(\sqrt{x^{2}+100}+10)}$

   =$\lim_{x \rightarrow 0}\frac{x^{2}+100-100}{x^{2}(\sqrt{x^{2}+100}+10)}$

   =  $\lim_{x \rightarrow 0}\frac{1}{\sqrt{x^{2}+100}+10}$

  = $\frac{1}{10+10}= \frac{1}{20}=0.05$

 Hence, option (c) is correct

Let P(1,-2,5) be the foot of the perpendicular drawn from the origin to the plane $\pi_{1}$  and  the same P be the foot of the perpendicular from (1,2,-1) to the plane $\pi_{2}$ . then the acute   angle between the planes $\pi_{1}$ and  $\pi_{2}$ is 

Answer:

Explanation:

According to given informations, the direction ratios of normal to the plane $\pi_{1}$ are 1,-0, -2-0, 5-0 ie  1,-2,5

 And similarly direction ratios of normal to the plane $\pi_{2} $ are

     1-1, 2-(-2), -1-5 i.e, 0,4,-6

 so, the acute angle between planes $\pi_{1} $  and $\pi_{2}$  is 

 $cos^{-1}|\left(\frac{(1)(0)+(-2)(4)+(5)(-6)}{\sqrt{1+4+25}\sqrt{0+16+36}}\right)|$

  =$cos^{-1}|\frac{-8-30}{\sqrt{30}\sqrt{52}}|=cos^{-1}\left(\frac{38}{2\sqrt{30\times 13}}\right)$

   = $\cos ^{-1}\left(\frac{19}{\sqrt{390}}\right)$

 hence, option (a) is correct

The volume of the tetrahedron (in cubix units)  formed by the plane 2x+y+z=K

 and the coordinate planes is $\frac {2V^{3}}{3}$ , then K:V=

Answer:

Explanation:

  Equation of given plane is 

      $2x+y+z=K  $           ........(i)

Point of intersection of plane  (i) with the coordinate axes is A ($\frac{K}{2},0,0$), B(0,K,0)  and C(0,0,K).

 Now , volume of the tetrahedron 0ABC

  = $\frac{1}{6}$  [OA OB  OC]=   $\frac{1}{6}\begin{bmatrix}\frac{K}{2} & 0& 0\\0 & K&0\\0&0&K \end{bmatrix}= \frac{2V^{3}}{3}$        (given)

  $\Rightarrow$     $\frac{1}{6} \frac {K^{3}}{2}= \frac{2V^{3}}{3}$

 $\Rightarrow$   $\left(\frac{K}{V}\right)^{3}=2^{3}\Rightarrow K:V=2:1$

If a variable circle S=0 touches  the line y=x and passes through  the point (0,0)  , then the fixed point that lies on the common chord of the circles  $x^{2}+y^{2}+6x+8y-7=0$ and S=0 is 

Answer:

Explanation:

 Let the equation of circle S=0, passes through the point (0,0) is

    S=$x^{2}+y^{2}+2gx+2fy=0$  ..........(i)

 Since , circle S=0  touches the line y=x , so

        $\frac{|-g+f|}{\sqrt{2}}=\sqrt{g^{2}+f^{2}}$

   $\Rightarrow$         $g^{2}+f^{2}-2gf=2(g^{2}+f^{2})$

$\Rightarrow$        $g^{2}+f^{2}+2gf$=0

 $\Rightarrow$       g+f=0 

 So, the  equation of circle S=0, becomes

  $x^{2}+y^{2}+2gx-2gy=0$.....(ii)

 Now, the equation of common chord of circles (ii) and $x^{2}+y^{2}+6x+8y-7=0$ is 

                 $(2g-6)x-(2g+8)y+7=0$

$\Rightarrow$   $2g(x-y)-(6x+8y-7)=0$   ........(iii)

 eq.(iii)  represents the family of lines , passes through the intersection of lines

               x-y=0  and 6x+8y-7=0

 And the point of intersection is $\left(\frac{1}{2},\frac{1}{2}\right)$

 Hence, option (a) is correct

• Mathematics - General questions