TS EAMCET 2017 :: Mathematics :: General questions

• Mathematics - General questions

The product of all the real roots  of  $x^{2}-8x+9-\frac{8}{x}+\frac{1}{x^{2}}=0$ is 

Answer:

Explanation:

We have 

$x^{2}-8x+9-\frac{8}{x}+\frac{1}{x^{2}}=0$

 $\Rightarrow$      $x^{4}-8x^{3}+9x^{2}-8x+1=0$

Product of roots =1

If the  line x+y+k=0 is a normal to the hyperbola  $\frac{x^{2}}{9}-\frac{y^{2}}{4}=1$, then k

Answer:

Explanation:

 We know that  equation of normal of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is 

   $\frac{a^{2}x}{x_{1}}+\frac{b^{2}y}{y_{1}}=a^{2}-b^{2}$

$\therefore$      Equations of normal to the hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{4}=1$

 is 

  $\frac{9x}{x_{1}}+\frac{4y}{y_{1}}=9+4$

 $\Rightarrow$   $\frac{9x}{x_{1}}+\frac{4y}{y_{1}}=13$

  Since,  line x+y=k is normal to the given hyperbola 

 $\therefore$    $\frac{\frac{9}{x_{1}}}{1}=\frac{\frac{4}{y_{1}}}{1}=\frac{13}{k}$

 $\Rightarrow$     $\frac{9}{x_{1}}=\frac{4}{y_{1}}$

   = $\frac{13}{k}$

    $\Rightarrow$      $x_{1}= \frac{9k}{13}$

                $y_{1}=\frac{4k}{13}$

 Since ($(x_{1},y_{1})$  lie on the hyperbola

$\therefore$    $\frac{\left(\frac{9k}{13}\right)^{2}}{9}-\frac{\left(\frac{4k}{13}\right)^{2}}{9}=1$

  $\Rightarrow$       $\frac{9k^{2}}{169}-\frac{4k^{2}}{169}=1$

 $\Rightarrow$    $5k^{2}=169$

   $\Rightarrow$   $k=\pm\frac{13}{\sqrt{5}}$

The equation of tangent to the curve 

$\left(\frac{x}{a}\right)^{n}+\left(\frac{y}{b}\right)^{n}=2$  at the point (a,b)  is 

Answer:

Explanation:

We have,

$\left(\frac{x}{a}\right)^{n}+\left(\frac{y}{b}\right)^{n}=2$

on differentiating w.r.t x, we get

       $\frac{nx^{n-1}}{a^{n}}+\frac{ny^{n-1}}{b^{n}}\frac{dy}{dx}=0$

$\Rightarrow$        $\frac{dy}{dx}=\frac{-b^{n}x^{n-1}}{a^{n}y^{n-1}}$

$\Rightarrow$             $\left(\frac{dy}{dx}\right)_{(a,b)}=\frac{-b^{n}a^{n-1}}{a^{n}b^{n-1}}=\frac{-b}{a}$

 Equations of tangent (a,b) is 

  $y-b=\frac{-b}{a}(x-a)$

$\Rightarrow$      ay-ab=-bx+ab

$\Rightarrow$    bx+ay=2ab

$\Rightarrow$    $\frac{bx}{ab}+\frac{ay}{ab}= \frac{2ab}{ab}$

$\Rightarrow$   $\frac{x}{a}+\frac{y}{b}=2$

Let f(x) be a quadratic expression such that f(0)+f(1)=0.If f(-2)=0 , then 

Answer:

Explanation:

We have ,

 f(x) is a quadratic  equation

$\therefore$     $f(x)=ax^{2}+bx+c$

   f(0)= c, f(1)=a+b+c

$\Rightarrow$      f(0)+f(1)=0

$\Rightarrow$   c+a+b+c=0

$\Rightarrow$  a+b+2c=0        ........(i)

  $f(-2)=a(-2)^{2}+b(-2)+c$

0=4a-2b+c

$\Rightarrow$      $4a-2b+c=0$        .........(ii)

 From Eqs.(i) and (ii) , we getr

     $\frac{a}{5}=\frac{b}{7}=\frac{c}{-6}$

$\Rightarrow$      a=5k,b=7k,c=-6k

 $\therefore$   $f(x)=k(5x^{2}+7x-6)$

$\Rightarrow$      $5x^{2}+7x-6=0$

  $\Rightarrow$     $5x^{2}+10x-3x-6=0$

$\Rightarrow$  $5x(x+2)-3(x+2)=0$

$\Rightarrow$     (x+2)(5x-3)=0

                         $x=-2,x=\frac{3}{5}$

Hence ,$f(\frac{3}{5})=0$

 The probability distribution of a  random variable  X is given below

If  mean of X is 4.2,  then a and b are respectively equal to 

Answer:

Explanation:

 We have, 

$\sum P_{i}=a+a+a+b+b+b+0.3$

$\Rightarrow$     $1=3a+2b+0.3$

$\Rightarrow$      $3a+2b=0.7$  .........(i)

$\sum P_{i}x_{i}=a+2a+3a+4b+5b+1.8$

$\Rightarrow$    $4.2=6a+9b+1.8$

$\Rightarrow$    $2a+3b=0.8$       .........(ii)

 Solving Eqs,(i) and (ii), we get

  a=0.1, b=0.2 

• Mathematics - General questions