TS EAMCET 2018 :: Mathematics :: General questions

• Mathematics - General questions

 If  $f(x)=\begin{cases}\frac{\sqrt{1+ax}-\sqrt{1-ax}}{x} ,& -1\leq x <0\\ \frac{x^{2}+2}{x-2}, & 0\leq x \leq 1\end{cases} is $

 continuous  on [-1,1] , than a =

Answer:

Explanation:

We have ,

$f(x)=\begin{cases}\frac{\sqrt{1+ax}-\sqrt{1-ax}}{x} ,& -1\leq x <0\\ \frac{x^{2}+2}{x-2}, & 0\leq x \leq 1\end{cases} is $

 Since, f(x)  is continuous on [-1,1]

 $\therefore$  f(x) is continuous  at x=0

 $\therefore$  LHS  (at x=0)=RHL(at x=0)

 $\Rightarrow \lim_{x \rightarrow 0}\frac{\sqrt{1+ax}-\sqrt{1-ax}}{x}=\lim_{x \rightarrow 0}\frac{x^{2}+2}{x-2}$

 $\Rightarrow \lim_{x \rightarrow 0}\frac{(1+ax)-{(1-ax)}}{x[\sqrt{1+ax}+\sqrt{1-ax}]}=\lim_{x \rightarrow 0}\frac{x^{2}+2}{x-2}$

$\Rightarrow \lim_{x \rightarrow 0}\frac{2a}{\sqrt{1+ax}+\sqrt{1-ax}}=\lim_{x \rightarrow 0}\frac{x^{2}+2}{x-2}$

$\Rightarrow$      a=-1

Consider the following  differential equations.

$D_{1}:y=4\frac{dy}{dx}+3x\frac{dx}{dy};D_{2}:\frac{d^{2}y}{dx^{2}}=\left(3+\left(\frac{dy}{dx}\right)^{2}\right)^{4/3}$

$D_{3}:\left[1+\left(\frac{dy}{dx}\right)\right]^{2}=\left(\frac{dy}{dx}\right)^{2}$

 The ratio of the sum of the orders of $D_{1},D_{2} $ and $D_{3}$  to the sum of their degrees is 

Answer:

Explanation:

 We have

$D_{1}:y=4\frac{dy}{dx}+3x\frac{dx}{dy} \Rightarrow  y\frac{dy}{dx}=4\left(\frac{dy}{dx}\right)^{2}+3x$ 

 $\therefore$  Order=1, Degree=2

$D_{2}:\frac{d^{2}y}{dx^{2}}=\left(3+\left(\frac{dy}{dx}\right)^{2}\right)^{4/3}$

   =  $\left(\frac{d^{2}y}{dx^{2}}\right)^{3}=\left(3+\left(\frac{dy}{dx}\right)^{2}\right)^{4}$

 $\therefore$    Order=2, Degree=3

$D_{3}:\left[1+\left(\frac{dy}{x}\right)\right]^{2}=\left(\frac{dy}{dx}\right)^{2}$

$\Rightarrow$    $1+\left(\frac{d^{}y}{dx^{}}\right)^{2}+2\frac{dy}{dx}=\left(\frac{dy}{dx}\right)^{2}\Rightarrow1+2 \frac{dy}{dx}=0$ 

 $\therefore$    Order=1, Degree=1

 $\therefore$   Required ratio = $\frac{1+2+1}{2+3+1}=\frac{4}{6}=\frac{2}{3}$

If  $\int\frac{\sqrt{x}}{\sqrt{x}-\sqrt[3]{x}}dx=x+Ex^{5/6}+Dx^{2/3}+Cx^{1/2}+Bx^{1/3}+Ax^{1/6}+\log (\sqrt[6]{x}-1)^{6}+K,$ then A+B+C+D+E=

Answer:

Explanation:

Let  I=$\int\frac{\sqrt{x}}{\sqrt{x}-\sqrt[3]{x}}dx$

 Put   $x=t^{6}\Rightarrow dx=6t^{5}dx$

 $\therefore$        $I=\int\frac{t^{3}.6t^{5}}{t^{3}-t^{2}}dt$

$=6\int\frac{t^{8}}{t^{2}(t-1)}dt=6\int\frac{t^{6}}{t-1}dt$

$=6\int\frac{(t^{6}-1)+1}{(t-1)}dt=6\int\frac{(t^{3}-1)(t^{3}+1)+1}{t-1}dt$

 $=6\int\frac{(t-1)(t^{2}+t+1)(t^{3}+1)+1}{t-1}dt$

 $=6\int\left[ t^{5}+t^{2}+t^{4}+t+t^{3}+1+\frac{1}{t-1}\right]dt$

$=6\int\left[ t^{5}+t^{4}+t^{3}+t^{2}+t+1+\frac{1}{t-1}\right]dt$

 $=6\left[\frac{t^{6}}{6}+\frac{t^{5}}{5}+\frac{t^{4}}{4}+\frac{t^{3}}{3}+\frac{t^{2}}{2}+t+\log(t-1)\right]+k$

 $= x+ \frac{6}{5}x^{5/6}+\frac{3}{2}x^{2/3}+2x^{1/2}+3x^{1/3}+6x^{1/6}+\log (\sqrt[6]{x-1})^{6}+k$

 $\therefore$    $A=6,B=3,C=2,D=\frac{3}{2},E=\frac{6}{5}$

 $therefore$     $A+B+C+D+E=6+3+2+\frac{3}{2}+\frac{6}{5}$

$=11+\frac{3}{2}+\frac{6}{5}= \frac{110+15+12}{10}=\frac{137}{10}$

 If the curves $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$ and $ \frac{x^{2}}{16}-\frac{y^{2}}{k}=1$ cut each other orthogonally , then k=

Answer:

Explanation:

 We have,

$\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$ and $\frac{x^{2}}{16}-\frac{y^{2}}{k}=1$ 

 On solving  these equation , we get

 $x^{2}=\frac{144+16k}{36+k}$ and $y^{2}= \frac{-27k}{36+k}$  ..........(i)

 Now,     $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$

 $\Rightarrow$    $\frac{2x}{4}+\frac{2yy'}{9}=0\Rightarrow y'=-\frac{9}{4}\frac{x}{y}$    ..........(ii)

 again ,       $\frac{x^{2}}{16}-\frac{y^{2}}{k}=1 \Rightarrow \frac{2x}{16}-\frac{2yy'}{k}=0$

$\Rightarrow$        $y'= \frac{k}{16} \frac{x}{y}$          ............(iii)

 Since both curves are orthogonal

 $\therefore$    $\frac{-9}{4}\frac{x}{y}\times\frac{k}{16}\frac{x}{y}=-1\Rightarrow 9kx^{2}=64y^{2}$

 From Eq.(i) , we have

$9k\left(\frac{144+16k}{36+k}\right)=64\left(\frac{-27k}{36+k}\right)\Rightarrow k=-21$

 The locus  of the mid-point of the line segment joining the focus to a moving point on the parabola, $y^{2}=4ax$  is a conic . The equation of the directrix of that conic is 

Answer:

Explanation:

Let Q(h,k)  be the mid-point of the line  joining  the  focus F(a,0)  and variable point $p(x_{0},y_{0})$.

$\therefore$       $(h,k)=\left(\frac{x_{0}+a}{2},\frac{y_{0}}{2}\right)$

$\Rightarrow$     $h=\frac{x_{0}+a}{2}$ and $k=\frac{y_{0}}{2}$

$\Rightarrow$      $x_{0}=2h-a  $   and $y_{0}=2k$

Since P(x₀,y₀) lies on parabola $y^{2}=4ax$

 $\therefore$         $y_{0}^{2}=4ax_{0} \Rightarrow (2k)^{2}=4a(2h-a)$

$\Rightarrow$     $4k^{2}=4a(2h-a)\Rightarrow  k^{2}= 2a \left(h- \frac{a}{2}\right)$

Which is equation  of parabola 

$\therefore$   $y^{2}=2a\left(x-\frac{a}{2}\right)$

$\therefore$    Equation of directrix is given by

$x-\frac{a}{2}=\frac{-a}{2}\Rightarrow x=0$

• Mathematics - General questions