VITEEE 2014 :: Mathematics :: General questions

• Mathematics - General questions

If $|z+4|  \leq 3$, then the greatest and the least value of |z+1|  are 

Answer:

Explanation:

$|z+4|\leq 3 \Rightarrow -3\leq z+4\leq3$

$\Rightarrow$   $-6\leq z+1\leq0$

$\Rightarrow$  $0\leq -(z+1)\leq 6$

$\Rightarrow$   $0\leq |z+1|\leq 6$

 Hence, the greatest  and least values are 6 and 0

If a=  $\cos \alpha +i \sin \alpha,b=\cos \beta+i \sin \beta$, $c=\cos \gamma+i \sin \gamma$ and $\frac{b}{c}+\frac{c}{a}+\frac{a}{b}=1$ then $\cos (\beta-\gamma)+\cos (\gamma-\alpha)+\cos (\alpha- \beta)$  is equal to 

Answer:

Explanation:

Given, a=  $\cos \alpha +i \sin \alpha,b=\cos \beta+i \sin \beta$,

and  $c=\cos \gamma+i \sin \gamma$

 Now, $\frac{b}{c} =\frac{ \cos \beta+i \sin \beta}{\cos \gamma+i \sin \gamma} \times \frac{\cos \gamma-i \sin \gamma}{\cos \gamma-i \sin \gamma}$

$\Rightarrow$    $\frac{b}{c}= \cos (\beta-\gamma)+i \sin (\beta-\gamma)$ ....(i)

 Similarly, $\frac{c}{a}= \cos (\gamma-\alpha)+i \sin (\gamma-\alpha)$......(ii)

 and $\frac{a}{b}=\cos (\alpha-\beta)+i \sin (\alpha- \beta)$.....(iii)

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

 $\cos (\beta-\gamma)+\cos (\gamma-\alpha)+\cos (\alpha+\beta)+i$

   $[(\sin (\beta-\gamma)+\sin(\gamma-\alpha)+ \sin (\alpha- \beta)]=1$

      $[ \because   \frac{b}{c}+\frac{c}{a}+\frac{a}{b}=1]$

On equating  real parts  , we get

 $\cos (\beta-\gamma)+\cos (\gamma-\alpha)+\cos (\alpha- \beta)$=1

If the line lx+my-n=0 will be a normal to the hyperbola , then $\frac{a^{2}}{l}-\frac{b^{2}}{m^{2}}=\frac{(a^{2}+b^{2})^{2}}{k}$ , where k  is equal to 

Answer:

Explanation:

 The equation of any normal

to $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is  $a x \cos \phi+by \cot \phi=a^{2}+b^{2}$

 $\Rightarrow$   $a x \cos \phi+by \cot \phi -(a^{2}+b^{2})=0....(i)$

The straight line lx+my-n=0 will be normal to the hyperbola 

 $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ , then eq.(i) and lx+my-n=0 represent  the same line, 

 $\therefore$   $\frac{a \cos \phi}{l}=\frac{b \cot \phi}{m}=\frac{a^{2}+b^{2}}{n}$

$\Rightarrow$  $\sec \phi= \frac{na}{l(a^{2}+b^{2})}$

  and $\tan \phi = \frac{nb}{m(a^{2}+b^{2})}$ 

$\frac{n^{2}a^{2}}{l^{2}(a^{2}+b^{2})^{2}}-\frac{n^{2}b^{2}}{m^{2}(a^{2}+b^{2})^{2}}=1$

                                                   $[ \because   \sec^{2} \phi-\tan^{2} \phi=1]$

$\frac{a^{2}}{l^{2}}-\frac{b^{2}}{m^{2}}=\frac{\left(a^{2}+b^{2}\right)^{2}}{n^{2}}$

But given equation  of normal is 

$\frac{a^{2}}{l^{2}}-\frac{b^{2}}{m^{2}}=\frac{\left(a^{2}+b^{2}\right)^{2}}{k}$

$\therefore$   $k=n^{2}$

The minimum area of the triangle formed by any tangent to the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ with the coordinate axes is  

Answer:

Explanation:

Equation of tangent at $(a \cos \theta,b \sin \theta)$ to the ellipse is 

$\frac{x}{a} \cos \theta+\frac{y}{b} \sin \theta=1$

Coordinates of P and Q are

   $\left(\frac{a}{\cos \theta},0\right)$  and   $\left(0,\frac{b}{\sin \theta}\right)$, respectively 

 Area of $\triangle OPQ$=$\frac{1}{2}|\frac{a}{\cos \theta} \times \frac{b}{\sin \theta}|=\frac{ab}{|\sin 2 \theta|}$

  $\therefore$   Minimum area =ab 

The normals at three points P, Q and R of the parabola $y^{2}=4ax$ meet at (h,k). The centroid of the $\triangle PQR$ lies on 

Answer:

Explanation:

 The sum of ordinates of feet of normals drawn from a point  to the parabola . $y^{2}=4ax$ is always zero

Now, as normals  at three points p,Q and R of parabola $y^{2}=4ax$  meet at (h,k)

$\Rightarrow$  The normals from (h,k) to $y^{2}=4ax$ meet the parabola at P,Q and R

$\Rightarrow$  y-coordinate $y_{1},y_{2},y_{3}$ of these points and R will be zero

$\Rightarrow$    y-coordinate of the centroid of $\triangle PQR$

i.e, $\frac{y_{1}+y_{2}+y_{3}}{3}=\frac{0}{3}=0$

$\therefore$   centroid lies on y=0

• Mathematics - General questions