JEE 2014 :: Advanced 2014 :: Mathematics 2014

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• Advanced 2014 - Mathematics 2014

A circle S passes through the point (0,1)  ans is othogonal to the circles  (x-1)²+y²=16  and x²+y² =1 . Then  , 

Answer:

Explanation:

Plan

    (i)  The general equation  of a circle is  

              $x^{2}+y^{2}+2gx+2fy+c=0$

  where, centre and radius are given by (-g, -f)  and   $\sqrt{g^{2}+f^{2}-c}$  respectively.

(ii)  If the two circles 

$x^{2}+y^{2}+2g_{1}x+2f_{1}y+c_{1}=0$  and

   $x^{2}+y^{2}+2g_{2}x+2f_{2}y+c_{2}=0$   are orthogonal  , then 

    $2g_{1}g_{2}+2f_{1}f_{2}=c_{1}+c_{2}$

 Let circle be

                 $x^{2}+y^{2}+2gx+2fy+c=0$

 It passes through (0,1)

 $\therefore$         1+2f+c=0         ........(i)

Orthogonal with

           $x^{2}+y^{2}-2x-15=0$

          2g(-1)=c-15

      $\Rightarrow$            c=15-2g          .............(ii)

 Orthogonal with

$x^{2}+y^{2}-1=0$

   c=1      ...........(iii)

$\Rightarrow$         g=7  and f=-1

 Centre is (-g, -f) =(-7,1)

 Radius= $\sqrt{g^{2}+f^{2}-c}=\sqrt{49+1-1}=7$

Let   $f: \left(-\frac{\pi}{2},\frac{\pi}{2}\right)\rightarrow R$  be given by  $f(x)=[log(\sec x+\tan x)]^{3}$  Then,

Answer:

Explanation:

Plan 

 (i)  f '(x)>0,   $\forall x \in(a,b)$    , then f(x) is an increasing function in (a,b)  and thus f(x) is one-one function in (a,b)

 (ii) if range of f(x)= codomain of f(x) , then f(x) is an  onto function.

 (iii) A function f(x)   is said to be odd function , if f(-x)= -f(x)   $\forall x \in R$, i.e, f(-x)+f(x)=0, $\forall x \in R$,

  $f(x)=[ln(\sec x+\tan x)]^{3}$

                 $f'(x)=\frac{3 [ln(\sec x+\tan x)]^{2}(\sec x \tan x+\sec^{2}x)}{(\sec x+\tan x)}$

                    $f'(x)=3\sec x[ln(\sec x+\tan x)]^{2}>0,$

                                                                         $\forall x \in(-\frac{\pi}{2},\frac{\pi}{2})$

 f(x)  is an increasing function.

  $\therefore$    f(x)  is an one-one function.

  $(\sec x+\tan x)=\tan (\frac{\pi}{4}+\frac{x}{2})$

       as   $x \epsilon (-\frac{\pi}{2},\frac{\pi}{2})$   then

     $0< \tan  (\frac{\pi}{4}+\frac{x}{2})< \infty$

  0  < sec x +tan x < $\infty$

$\Rightarrow$    -$\infty$  < (ln (sec x+tan x) <$\infty$

                        -$\infty  < [ln(sec x+tan x)]^{3} <  \infty$

$\Rightarrow$   - $\infty$ < f(x0<  $\infty$

 Range of f(x)  is R and thus f(x) is an onto function.

   $f(-x)= [ln(\sec x-\tan x)]^{3}$

                     =   $\left[ln\left(\frac{1}{\sec x+\tan x}\right)\right]^{3}$

   $f(-x)=-[ln(\sec x-\tan x)]^{3}$

                          f(x)+f(-x)=0

$\Rightarrow$        f(x)  is an odd function.

Let   $f:(0,\infty)\rightarrow R$   be given by $f(x)= \int_{1/x}^{x} e^{-(t+\frac{1}{t})}\frac{dt}{t}$  Then

Answer:

Explanation:

Plan  (i)  $f(x)= \int_{\phi(x)}^{\psi(x)} f(t)dt$    then

         $\frac{d}{dx}[I(x)]=f\left\{\psi(x)\right\}\left\{\frac{d}{dx}\psi(x)\right\}$ - $f\left\{\phi(x)\right\}\left\{\frac{d}{dx}\phi(x)\right\}$

(ii)     if f'(x) > o,  $\forall x \in[a,b]$  then f(x) is monotonically increasing on [a,b]

 (iii)   $\int_{a}^{b}  f(x) dx=-\int_{b}^{a} f(x) dx$

 (iv)     if f(-x) =-f(x) , $\forall x \in R$   , then f(x) is an odd function of x on R

Given,

           $f(x)= \int_{1/x}^{x}       \frac{e^{-(t+\frac{1}{t})}}{t} dt$

  $f'(x)=1.\frac{e^{-\left(x+\frac{1}{x}\right)}}{x}-\left(\frac{-1}{x^{2}}\right)\frac{e^{-(\frac{1}{x}+x)}}{1/x}$

   $=\frac{e^{-\left(x+\frac{1}{x}\right)}}{x}+\frac{e^{-\left(x+\frac{1}{x}\right)}}{x}$

                                 $=\frac{2e^{-\left(x+\frac{1}{x}\right)}}{x}$

 As, f'(x)>0,   $\forall x \in(0,\infty)$

 $\therefore$   f(x)  is monotonically increasing on (0, $\infty$)

 $\Rightarrow$     options (a) is correct and (b) is wrong.

  Now,     $f(x)+f(\frac{1}{x})=\int_{1/x}^{x} \frac{e^{-\left(t+\frac{1}{t}\right)}}{t}dt+\int_{x}^{1/x} \frac{e^{-\left(t+\frac{1}{t}\right)}}{t}dt$

                                           = 0, $\forall x\in(0,\infty)$

Now, let

  g(x) =f(2^x)= $\int_{2^{-x}}^{2^{x}} \frac{e^{-(t+\frac{1}{t})}}{t}dt$

 g(-x)=f(2^-x)

                        =$\int_{2^{x}}^{2^{-x}} \frac{e^{-(t+\frac{1}{t})}}{t}dt$

                    = -g(x)

  $\therefore$     f(2^x)  is an odd function.

Let M and N  be two 3 x3  matrices such that MN=NM  . further , If   $M\neq N^{2}$  and $M^{2}= N^{4}$   then

Answer:

Explanation:

Plan

(i)_  If A and B  are two non-zero matrices and AB= BA  then

 (A-B) (A+B)= A²-B²

(ii)   The determinant of the product of the ,matrices is equal to product of their individual determinants

    i.e,  |AB|=|A||B|

 Given  M²=N⁴

   $\Rightarrow$           $ M^{2}-N^{4}=0$

 $\Rightarrow$       $(M-N^{2})(M+N^{2})=0$

                                                         (as MN=NM)

 Also,    $M\neq N^{2}$

  $\Rightarrow$       M+N²=0

$\Rightarrow$     Det (M+N² )=0

 Also,                 Det (M+MN² )

   =  (Det M) ((Det (M+N²)

    = (Det M)  (0)= 0

As,  Det (M²+MN²)=0

Thus, there exist non-zero matrix  U such that (M²+MN²)U=0

From  a point  $P(\lambda,\lambda,\lambda)$  perpendicular PQ and PR are drawn  respectively on the  lines y=x, z=1 and y=-x, z=-1. If P is such that  $\angle QPR$    is a right angle , then the possible value(s) of   $\lambda$  is(are)

Answer:

Explanation:

Plan (i) Direction ratios of a line joining two points  $(x_{1},y_{1},z_{1})$  and   $(x_{2},y_{2},z_{2})$  are 

 $x_{2}-x_{1},y_{2}-y_{1},z_{2}-z_{1}$

(ii) If the two lines with direction ratios a₁,b₁, c₁ ; a₂,b₂,c₂ are perpendicular , then 

a₁a₂+b₁b₂+c1c₂=0

Line L₁  given by  y=x ;z=1 can be expressed

     $L_{1}:\frac{x}{1}=\frac{y}{1}=\frac{z-1}{0}$

$\frac{x}{1}=\frac{y}{1}=\frac{z-1}{0}=\alpha$

   $\Rightarrow$             $x=\alpha,y=\alpha,z=1$

Let the coordinates od Q on L₁ be   $(\alpha,\alpha,1)$

Line L₂ Given by y=-x , z=-1 can be expressed as 

 $L_{2}:\frac{x}{1}=\frac{y}{-1}=\frac{z+1}{0}$

$\frac{x}{1}=\frac{y}{-1}=\frac{z+1}{0}=\beta$               (say)

 $\Rightarrow$       $x=\beta,y=-\beta,z=-1$

 Let the coordinates  of R on L₂  be ($ \beta$, -$\beta$, -1)

  Direction ratios of PQ are   $\lambda-\alpha,\lambda-\alpha,\lambda-1$

 Now PQ  $\bot$ L₁

$\therefore$          $1 (\lambda-\alpha)+1(\lambda-\alpha)+0.(\lambda-1)=0$

$\Rightarrow$        $ \lambda=\alpha$

$\therefore$     Q ( $\lambda$,  $\lambda$,1)

   direction ratio of PR are

             $\lambda-\beta,\lambda+\beta,\lambda+1$

  Now,    PR  $\bot$ L₂

           $\therefore$           $ 1(\lambda-\beta)+(-1)(\lambda+\beta)+0(\lambda+1)=0$

                                  $\lambda-\beta-\lambda-\beta=0$

          $\Rightarrow$            $ \beta=0$

$\therefore$     R(0,0,-1)

Now , as   $\angle QPR= 90^{0}$

   [ as a₁a₂+b₁b₂+c₁c₂ =0,

if two lines with DR'S a₁,b₁,c₁:a₂:b₂:c₂ are perpendicular)

$\therefore$    $(\lambda-\lambda)(\lambda-0)+(\lambda-\lambda)(\lambda-0)+(\lambda-1)((\lambda+1)=0$

  $\Rightarrow$     $ (\lambda-1)(\lambda+1)=0$

$\Rightarrow$    $\lambda=1 $    or $\lambda=-1 $

$\lambda=1 $   , rejected as P  and Q are different points

  $\Rightarrow$   $ \lambda$=-1

• Advanced 2014 - Physics 2014
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• Advanced 2014 - Mathematics 2014