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The function $f:R\rightarrow [-\frac{1}{2},\frac{1}{2}]$ defined as  $f(x)=\frac{x}{1+x^{2}} is$

Answer:

Explanation:

We have $f(x)=\frac{x}{1+x^{2}}$

$\therefore$   $f(\frac{1}{x})=\frac{\frac{1}{x}}{1+\frac{1}{x^{2}}}=\frac{x}{1+x^{2}}=f(x)$

$\therefore$  $f(\frac{1}{2})=f(2) $ or $f(\frac{1}{3})=f(3) $   and so on 

  So, f(x) is many -one function 

   Again , let y=f(x)

$\Rightarrow$    $y=\frac{x}{1+x^{2}}$

$\Rightarrow$     $y+x^{2}y=x$

$\Rightarrow$  $yx^{2}-x+y=0$

As,    $x\epsilon  R

$\therefore$   $(-1)^{2}-4(y)(y)\geq0$

$\Rightarrow$   $1-4y^{2}\geq 0$

$\Rightarrow$   $y\epsilon [\frac{-1}{2},\frac{1}{2}]$

$\therefore$ Range= Codomain= $ [\frac{-1}{2},\frac{1}{2}]$

    So, f(x) is surjective.

   Hence , f(x) is surjective but not injective.

If a hyperbola passes through the point $P(\sqrt{2},\sqrt{3})$ and has foci at (± 2,0), then the tangent to this hyperbola at P also passes through the point

Answer:

Explanation:

Let the equation of hyperbola be

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

$\therefore$       $ae=2\Rightarrow a^{2}e^{2}=4$

$\Rightarrow    a^{2}+b^{2}=4$

$\Rightarrow    b^{2}=4-a^{2}$

 $\therefore$   $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{4-a^{2}}=1$

  Since ($\sqrt{2},\sqrt{3}$) lie on hyperbola

$\therefore$    $\frac{2}{a^{2}}-\frac{3}{4-a^{2}}=1$

     $\Rightarrow$     $8-2a^{2}-3a^{2}=a^{2}(4-a^{2})$

$\Rightarrow$   $8-5a^{2}=4a^{2}-a^{4}$

$\Rightarrow$     $a^{4}-9a^{2}+8=0$

$\Rightarrow$     $(a^{4}-8)(a^{4}-1)=0$

$\Rightarrow$     $a^{4}=8,a^{4}=1$

$\therefore$     a=1

   Now equation of hyperbola is 

$\frac{x^{2}}{1}-\frac{y^{2}}{3}=1$

$\therefore$  Equation of tangent at ($\sqrt{2},\sqrt{3}$) is given by

          $\sqrt{2}x-\frac{\sqrt{3}y}{3}=1$

  $\Rightarrow$    $\sqrt{2}x-\frac{y}{\sqrt{3}}=1$

which passes through the point ($2\sqrt{2},3\sqrt{3}$) 

The eccentricity of an ellipse whose centre is at the orgin is 1/2 . If one of its directrices is x=-4, then the equation of the normal to it at (1, $\frac{3}{2}$) is 

Answer:

Explanation:

We have e =$\frac{1}{2}$   and $\frac{a}{e}$=4

$\therefore$    a=2

  Now, $b^{2}=a^{2}(1-e^{2})=(2)^{2}[1-(\frac{1}{2})^{2}]$

= $4(1-\frac{1}{4})=3\Rightarrow b=\sqrt{3}$

$\therefore$   Equation of the ellipse is

            $\frac{x^{2}}{(2)^{2}}+\frac{y^{2}}{(\sqrt{3})^{2}}=1$

$\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$

  Now, the equation of normal at (1,$\frac{3}{2}$) is

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

$\Rightarrow\frac{4x}{1}-\frac{3y}{(3/2)}=4-3$

  $\Rightarrow 4x-2y=1$

Let  $I_{n}=\int_{}^{} tan^{n}x dx (n>1), If $   $I_{4}+I_{6}=a\tan^{5}x+bx^{5}+C,$   , where C is a constant of integration, then  orderd pair (a,b) is equal to

Answer:

Explanation:

We have, $I_{n}=\int_{}^{} tan^{n}x dx$

$\therefore   I_{n}+I_{n+2}=\int_{}^{}\tan^{n}x dx+\int_{}^{}tan^{n}  x dx$

$=\int_{}^{}\tan^{n}x (1+tan^{2}  x) dx$

$=\int_{}^{}\tan^{n}x \sec ^{2}x dx$

$=\frac{\tan^{n+1}x}{n+1}+C$

    Put n=4, we get   $I_{4}+I_{6}=\frac{\tan^{5}x}{5}+C$

  $\therefore  a=\frac{1}{5}$ and $ b=0$

For any three positive real numbers a,b and c. If 9(25a²+b²)+25(c²-3ac)  =15b (3a+c), then

Answer:

Explanation:

We have,

 225a²+9b²+25c²-75ac-45ab-15bc=0

$\Rightarrow (15a)^{2}+(3b)^{2}+(5c)^{2}-(15a)(5c)-(15a)(3b)-(3b)(5c)=0$

$\Rightarrow \frac{1}{2}[(15a-3b)^{2}+(3b-5c)^{2}+(5c-15a)^{2}]=0$

$\Rightarrow 15a=3b,3b=5c $  and  $5c=15a$

$\therefore$         15a=3b=5c

$\Rightarrow  \frac{a}{1}=\frac{b}{5}=\frac{c}{3}=\lambda(say)$

$\Rightarrow  a=\lambda,b=5\lambda ,c=3\lambda$

   Hence,  a,b and c are in A.P

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