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 The  Integral $\int_{2}^{4} \frac{\log x^{2}}{\log x^{2}+\log (36-12x+x^{2})}dx$    is equal to

Answer:

Explanation:

Central Idea.

   Apply  the property    $\int_{a}^{b} f(x) dx=\int_{a}^{b} f(a+b-x) dx$  and then add.

  Let          $I=\int_{2}^{4} \frac{\log x^{2}}{\log x^{2}+\log (36-12x+x^{2})}dx$

                    $I= \int_{2}^{4} \frac{2\log x}{2\log x+\log(6-x)^{2}}dx$

                        $= \int_{2}^{4} \frac{2\log x dx}{2[\log x+\log(6-x)^{}]}$

                       $\Rightarrow I=\int_{2}^{4} \frac{\log x dx}{[\log x+\log(6-x)^{}]}$      ......(i)

        $\Rightarrow I=\int_{2}^{4} \frac{\log (6-x) }{\log (6-x)+\log x}dx$    .......(ii)

                                              $\left[\therefore  \int_{a}^{b}  f(x) dx=\int_{a}^{b} f(a+b-x)dx\right]$

On adding Eqs .(i) and (ii) , we get

                 $2I= \int_{2}^{4}\frac{\log x +\log (6-x)}{\log x+\log(6-x)}dx$

                 $2I= \int_{2}^{4}dx=\left[ x\right]_2^4$

                     $\Rightarrow  2I=2\Rightarrow I=1$

The integral  $\int_{}^{} \frac{dx}{x^{2}(x^{4}+1)^{\frac{3}{4}}}$  equals

Answer:

Explanation:

$\int_{}^{} \frac{dx}{x^{2}(x^{4}+1)^{\frac{3}{4}}}$   = $\int_{}^{} \frac{dx}{x^{5}(1+\frac{1}{x^{4}})^{\frac{3}{4}}}$

 Put            $1+\frac{1}{x^{4}}=t^{4}$

    $\Rightarrow$        $-\frac{4}{x^{5}}dx=4t^{3}dt$

    $\Rightarrow$     $\frac{dx}{x^{5}}=-t^{3}dt=\int_{}^{} \frac{-t^{3}dt}{t^{3}}$

  =   $-\int_{}^{} dt=-t+C=-\left(1+\frac{1}{x^{4}}\right)^{\frac{1}{4}}+c$

Let f(x) be a polynomial of degree four having extreme values at x=1, amd x=2 . If   $\lim_{x \rightarrow 0}\left[1+\frac{f(x)}{x^{2}}\right]=3$   , then f(2) is equal to

Answer:

Explanation:

Central Idea. Any function have extreme values (maximum or minimum) at its critical points , where f " (x)=0

   Since , the function have extreme values at x=1 and x=2

 $\therefore$     $f'(x)=0$  = 0 at x=1 and x=2

$\Rightarrow$    $f'(1)=0$  and $f'(2)=0$

 Also it is given that

             $\lim_{x \rightarrow 0}\left[ 1+\frac{f(x)}{x^{2}}\right]=3$

  $\Rightarrow$                         $1+\lim_{x\rightarrow 0}\frac{f(x)}{x^{2}}=3$

    $\Rightarrow$   $\lim_{x\rightarrow 0}\frac{f(x)}{x^{2}}=2$

   $\Rightarrow$  f(x) will be of the form

                               $ax^{4}+bx^{2}+2x^{2}$

                  [  $\therefore$ f(x)  is four degree polynomial]

   Let     $f(x)=ax^{4}+bx^{2}+2x^{2}$

  $\Rightarrow$     $f'(x)=4ax^{3}+3bx^{2}+4x$

  $\Rightarrow$     $f'(1)=4a+3b+4=0$            ........(i)

            and       $f'(2)=32a+12b+8=0$

$\Rightarrow$    8a+3b+2=0           ......(ii)

   On solving Eqs(i) and (ii) , we get

            $a=\frac{1}{2},b=-2$

 $\therefore$                  $f(x)=\frac{x^{4}}{2}-2x^{3}+2x^{2}$

  $\Rightarrow$       f(2)= 8-16+8=0

 The normal to the curve  $x^{2}+2xy-3y^{2}=0 $    at (1,1)

Answer:

Explanation:

Given the equation of curve is   $x^{2}+2xy-3y^{2}=0 $       .......(i)

 On differentiating w.r.t x, we get

    $2x+2xy'+2y-6yy'=0$

   $\Rightarrow$                                  $y'=\frac{x+y}{3y-x}$       

 At x=1, y=1, $y'=1$

  i.e,          $\left(\frac{\text{d}y}{\text{d}x}\right)_{(1,1)}=1$

   Equation of normal at (1,1)

    $y-1=-\frac{1}{1}(x-1)$

  $\Rightarrow$      y-1=-(x-1)

$\Rightarrow$      x+y=2     ......(ii)

On solving Eqs .(i) and (ii) simultaneously , we get

          $x^{2}+2x(2-x)-3(2-x)^{2}=0$

$\Rightarrow$            $x^{2}+4x-2x^{2}-3(4+x^{2}-4x)=0$

$\Rightarrow$            $-x^{2}+4x-12-3x^{2}+12x=0$

$\Rightarrow$          $-4x^{2}+16x-12=0$

$\Rightarrow$        $4x^{2}-16x+12=0$

$\Rightarrow$      $x^{2}-4x+3=0$

$\Rightarrow$          (x-1) (x-3)

$\Rightarrow$    x=1,3

 Now, whern x=1 , then y=1

   and when x=3, then y=-1

      P =(1,1)    and Q =(3,-1)

 Hence, normal  meets the curve again at (3,-1) in fourth quadrant.

Alter:

Given,      $x^{2}+2xy-3y^{2}=0 $ 

$\Rightarrow$        (x-y)(x+3y)=0

$\Rightarrow$     x-y=0  or x+3y=0

 Equation of normal  at (1,1) is

                y-1=-1(x-1)

$\Rightarrow$      x+y-2=0

  It intersects  x+3y=0 at   (3,-1)

and hence normal meet the curve in fourth quadrant.

If  the function   $g(x)=\begin{cases}k\sqrt{x+1}, & 0 \leq x \leq3\\mx+2, &3<x \leq 5\end{cases}$ differentiable , then value of k+m is

Answer:

Explanation:

 Since , g(x) is differentiable  $\Rightarrow$ g(x)  must be continuous.

$\therefore$    $g(x)=\begin{cases}k\sqrt{x+1}, & 0 \leq x \leq3\\mx+2, &3<x \leq 5\end{cases}$

 At x=3,   RHL=3m+2

   and at x=3,   LHL=2k

   $\therefore$               2k=m+2     .........(i)

 Also, 

              $g'(x)=\begin{cases}\frac{k}{2\sqrt{k+1}}, & 0\leq x < 3\\m, & 3<x \leq5 \end{cases}$

   $\therefore$   $L\left\{g'(3)\right\}=\frac{k}{4}$   and   $R\left\{g'(3)\right\}=m$

   $\Rightarrow$  $\frac{k}{4}=m $    i.e, k=4m

 On , solving Eqs(i)   and (ii) , we get

   $k=\frac{8}{5},m=\frac{2}{5}$

  $\Rightarrow$             k+m=2

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