JEE 2014 :: Main 2014 :: Mathematics 2014

• Main 2014 - Physics 2014
• Main 2014 - Chemistry 2014
• Main 2014 - Mathematics 2014

Let a, b, c and d be non-zero numbers. If the point of intersection of lines 4ax+2ay+c=0 and 5bx+2by+d=0 lies in the fourth quadrant and is equidistant from the two axes, then

Answer:

Explanation:

 Let coordinate of the intersction point in fourth quadrant be $(\alpha,-\alpha)$

 Since, $(\alpha,-\alpha)$   lies on both lines
        

                    4ax+2ay+c=0

 and 5bx+2by+d=0

$\therefore$      $4a\alpha-2a\alpha+c=0$

     $\Rightarrow$    $ \alpha=\frac{-c}{2a}$               ......(i)

  and                   $5b\alpha-2b\alpha+d=0$

   $\Rightarrow$  $\alpha=\frac{-d}{3b}$        .........(ii)

From eqs.(i) and (ii)  , we get

     $\frac{-c}{2a}=\frac{-d}{3b}$

$\Rightarrow $  3bc=2ad

$\Rightarrow  $   2ad-3bc=0

If PS  is the median of the triangle with vertices P(2,2), Q(6,-1). and R(7,3) then equation of the line passing through (1,-1) and parallel to PS is

Answer:

Explanation:

Coordinate of $S=\left(\frac{7+6}{2},\frac{3-1}{2}\right)$

= (13/2.1)

[  $\because$   S is mid-point of line QR]

slop of the line   PS is   $\frac{-2}{9}$
Required equation passes through (1,-1) and parallel to PS is 

  $y+1=\frac{-2}{9}(x-1)$

  $\Rightarrow $    $2x+9y+7=0$

Let  the population of rabbits surviving at a time t be governed by the differential equation  $\frac{dp(t)}{dt}=\frac{1}{2}p(t)-200$.If  p(0)=100, then  p(t) is equal to 

Answer:

Explanation:

Given, differential  equation $\frac{dp}{dt}-\frac{1}{2}p(t)=-200$ is linear ndifferential equation 

 Here,   $p(t)=-\frac{1}{2}, Q(t)= -200$

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

 Hence, solution is

          $p(t).IF=\int_{}^{} Q(t)IF dt$

   $p(t).e^{-t/2}=\int_{}^{} -200.e^{-\frac{t}{2}dt}$

                $p(t).e^{-t/2}= 400.e^{-\frac{t}{2}}+K$

           $\Rightarrow$   $ p(t).= 400+ke^{-\frac{1}{2}}$

if p(0)=100, then k=-300

$\Rightarrow$   $400-300e^{\frac{t}{2}}$

The area of the region described by $A= \left\{(X,Y):x^{2}+y^{2}\leq 1 \right\}    and   \left\{y^{2}\leq 1-x\right\}$ is

Answer:

Explanation:

Given,   $A= \left\{(X,Y):x^{2}+y^{2}\leq 1 \right\}    and   \left\{y^{2}\leq 1-x\right\}$ is

Required area =   $\frac{1}{2}\pi r^{2}+2\int_{0}^{1} (1-y^{2})dy$

                     =  $\frac{1}{2}\pi (1)^{2}+2\left(y-\frac{y^{3}}{3}\right)_0^1$

                           =$\frac{\pi}{2}+\frac{4}{3}$

 The integral   $\int_{0}^{\pi} \sqrt{1+4\sin^{2}\frac{x}{2}-4\sin\frac{x}{2}}dx$   is equal to 

Answer:

Explanation:

 Use the formula 

   $|x-a|=\begin{cases}x-a, & x \geq a\\-(x-a), & x < a\end{cases}$

 to break given integral in two parts and then integrate separetely

= $\int_{0}^{\pi} \sqrt{\left(1-2\sin\frac{x}{2}\right)^{2}}dx=\int_{0}^{\pi} |1-2\sin\frac{x}{2}| dx$

=$\int_{0}^{\pi/3} \left(1-2\sin\frac{x}{2}\right)^{}dx=\int_{\pi/3}^{\pi} \left(1-2\sin\frac{x}{2}\right) dx$

=$\left(x+4\cos\frac{x}{2}\right)_{0}^{\pi/3}-\left(x+4\cos\frac{x}{2}\right)_{\pi/3}^{\pi}$

   = $4\sqrt{3}-4-\frac{\pi}{3}$

• Main 2014 - Physics 2014
• Main 2014 - Chemistry 2014
• Main 2014 - Mathematics 2014