JEE 2017 :: Main 2017 :: Mathematics 2017

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Let a vertical tower AB have its end A on the level ground. Let C be the mid-pont of AB and P be a point on the ground such that AP=2AB. If $\angle$ BPC= β , then tanβ is equal to

Answer:

Explanation:

LET AB= h, then AD= 2h, and AC=BC= $\frac{h}{2}$

  Again , let $\angle$ CPA= α

 Now. in ΔABP

   $\tan(\alpha+\beta)=\frac{AB}{AP}=\frac{h}{2h}=\frac{1}{2}$

Also, in Δ ACP, $\tan\alpha=\frac{AC}{AP}=\frac{\frac{h}{2}}{2h}=\frac{1}{4}$

Now, $\tan\beta=\tan[(\alpha+\beta)-\alpha]$

                   $=\frac{\tan(\alpha+\beta)-\tan\alpha}{1+\tan(\alpha+\beta)\tan\alpha}$

$=\frac{\frac{1}{2}-\frac{1}{4}}{1+\frac{1}{2}\times\frac{1}{4}}=\frac{\frac{1}{4}}{\frac{9}{8}}=\frac{2}{9}$

If $(2+sin x)\frac{\text{d}y}{\text{d}x}+(y+1)\cos x=0$ and y(0)=1, then $y(\frac{\pi}{2})$ is equal to 

Answer:

Explanation:

We have 

   $(2+sin x)\frac{\text{d}y}{\text{d}x}+(y+1)\cos x=0$

$\Rightarrow  \frac{\text{d}y}{\text{d}x}+\frac{cosx}{2+sinx}y=\frac{-cosx}{2+sinx}$

which is a linear differential equation

$\therefore$     IF= $e^{\int_{}^{} \frac{cosx}{2+sin x}dx}=e^{\log(2+sin x)}$

                      = 2+ sin x

$\therefore$      Required solution is given by

               y+(2+sinx)  = $\int_{}^{} \frac{-cos x}{2+sin x}.(2+sin x)dx +C$

  $\Rightarrow    y(2+sin x)=-\sin x+C$

                            Also y(0)  =1

$\therefore$     $1(2+\sin 0)=-\sin 0 +C$

     $\Rightarrow$  C=2

$\therefore$                             $y= \frac{2-\sin x}{2+\sin x}$

$\Rightarrow$          $y(\frac{\pi}{2})= \frac{2-\sin \frac{\pi}{2}}{2+\sin \frac{\pi}{2}}=\frac{1}{3}$

For $x\epsilon(0,\frac{1}{4})$, if the derivative of  $\tan^{-1}(\frac{6x \sqrt{x}}{1-9x^{3}}) $ is √x .g(x), then g(x) equals 

Answer:

Explanation:

Let y= $tan^{-1}(\frac{6x\sqrt{x}}{1-9x^{3}})$

         = $tan^{-1}(\frac{2.(3x^{\frac{3}{2}})}{1-(3x^{\frac{3}{2}})^{2}}]$

= $2\tan^{-1}(3x^{\frac{3}{2}})$

                             [$\because 2 tan^{-1}x= \tan^{-1}\frac{2x}{1-x^{2}}$]

  $\therefore \frac{\text{d}y}{\text{d}x}=2.\frac{1}{1+(3x^{\frac{3}{2}})^{2}}.3\times \frac{3}{2}x^{\frac{1}{2}}$

   = $\frac{9}{1+9x^{3}}.\sqrt{x}$

  g(x) = $\frac{9}{1+9x^{3}}$

If the image of the point P(1,-2,3) in the plane 2x+3y-4z+22=0 measured parallel to the line  $\frac{x}{1}=\frac{y}{4}=\frac{z}{5}$ is Q, then PQ is equal to 

Answer:

Explanation:

Any line parallel to $\frac{x}{1}=\frac{y}{4}=\frac{z}{5}$ passing through P (1, -2,3) is

$\frac{x-1}{1}=\frac{y+2}{4}=\frac{z-3}{5} =\lambda (say)$

Any point on above line can be written as (λ +1, 4λ-2, 5λ+3)

 $\therefore$   Coordinates of R are

      (λ +1, 4λ-2, 5λ+3)

Since, point R lies on the above plane 

$\therefore$   2(λ+1)+3(4λ-2)-4(5λ +3) +22=0

    $\Rightarrow$        λ =1

 So, point R is (2,2,8)

  Now, PR=  $\sqrt{(2-1)^{2}+(2+2)^{2}+(8-3)^{2}}$

        =$\sqrt{42}$

  $\therefore$   PQ= 2PR= $2\sqrt{42}$

The area ( in sq units) of the region {(x,y):x ≥ 0, x +y ≤ 3, x² ≤ 4y and y ≤ 1+√x } is 

Answer:

Explanation:

Required area

        $=\int_{0}^{1} (1+\sqrt{x})dx +\int_{1}^{2}(3-x)dx-\int_{0}^{2}\frac{x^{2}}{4}dx  $

   = $=[ x+\frac{x^{\frac{3}{2}}}{3/2}]_{0}^{1}+[3x-\frac{x^{2}}{2}]_{1}^{2}- [\frac{x^{3}}{12}]_{0}^{2}$

   $=(1+\frac{2}{3})+(6-2-3+\frac{1}{2})-(\frac{8}{12})$

   $=\frac{5}{3}+\frac{3}{2}-\frac{2}{3}=1+\frac{3}{2}=\frac{5}{2} sq. units$

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