TS EAMCET 2017 :: Mathematics :: General questions

• Mathematics - General questions

If   $a=x\hat{i}+y\hat{j}+z\hat{k}, $ then 

 $(a \times \hat{i}).(\hat{i}+\hat{j})+(a \times \hat{j}).(\hat{j} +\hat{k})+(a \times \hat{k}).(\hat{k}+\hat{i})$=

Answer:

Explanation:

We have,

$a=x\hat{i}+y\hat{j}+z\hat{k} $

    $(a \times \hat{i})(\hat{i}+\hat{j})+(\hat{a} \times \hat{j}) (\hat{j}+\hat{k})+(a \times \hat{k})(\hat{k} +\hat{i})$

  =$[a \hat{i}\hat{i}]+[a\hat{i}\hat{j}]+[a \hat{j}\hat{j}]+[a \hat{j} \hat{k}]+[a \hat{k} \hat{k}]+[a \hat{k}\hat{i}]$

             =$\left[a \hat{i}\hat{j}\right]+\left[a \hat{j} \hat{k}\right]+\left[a \hat{k} \hat{i}\right]$

                                                                          $[\because  [a \hat{i}\hat{i}]=[a \hat{j} \hat{j}]=[a \hat{k} \hat{k}]=0]$

=$a.(\hat{i} \times \hat{j})+a.(\hat{j} \times \hat{k})+a.(\hat{k} \times \hat{i})$

  =  $a.\hat{k}+a.\hat{i}+a.\hat{j}=a.(\hat{i}+\hat{j}+\hat{k})$

=$(x \hat{i}+y\hat{j}+z \hat{k}).(\hat{i}+\hat{j}+\hat{k})=x+y+z$

The area of the region bounded by the curves $x=y^{2}-2$ and x=y is 

Answer:

Explanation:

We have,

 $x=y^{2}-2$ and x=y

 Graph of $x=y^{2}-2$ and x=y are 

 Solving, $x=y^{2}-2$  and x=y , we get (-1,-1) and (2,2) 

Area of shaded region is 

$\int_{-1}^{2} ydy-\int_{-1}^{2} (y^{2}-2)dy$

$\left[\frac{y^{2}}{2}-\frac{y^{3}}{3}+2y\right]_{-1}^{2}$=

                           $\left(\frac{4}{2}-\frac{8}{3}+4\right)-\left(\frac{1}{2}+\frac{1}{3}-2\right)$

   = $\frac{10}{3}+\frac{7}{6}=\frac{27}{6}=\frac{9}{2}$

The sum of the complex roots of the equations $(x-1)^{3}+64$=0 is 

Answer:

Explanation:

We have,

  $(x-1)^{3}+64$=0 

 $\Rightarrow$           $(x-1)^{3}=-64$           

 $\Rightarrow$                $(x-1)^{3}=(-4)^{3}$

 $\Rightarrow$        x-1=-4,-4w,-4w²

 $\Rightarrow$                 x=-3,-4w+1,-4w²+1

 Complex roots of the equations are

  -4w+1,-4w²+1

Sum of complex roots are

   -4w+1-4w²+1=-4(w+w²)+2

  $=-4(-1)+2=4+2=6$    $[\because1+w+w^{2}=0]$

The sides of a triangle are in the ratio $1:\sqrt{3}:2$ . Then the  angles are in the ratio

Answer:

Explanation:

We have,

 Ratio of sides of triangle are $1: \sqrt{3}:2$

 Let the sides $k,\sqrt{3}k, 2k,$

 Since , this triangle is a right angle triangle

 $(2k)^{2}=(\sqrt{3}k)^{2}+(k)^{2}=(2k)^{2}$

 $\therefore$     $\sin A= \frac{\sqrt{3}k}{2k}=\frac{\sqrt{3}}{2}\Rightarrow A=60^{0}$

$\Rightarrow\sin C= \frac{k}{2k}=\frac{1}{2}\Rightarrow C=30^{0}$

 $\Rightarrow$  B=90⁰

 $\therefore$ Ratio of angles are $30^{0}:60^{0}:90^{0}$

$\Rightarrow$   1:2:3

$\int x^{4}e^{2x} dx=$

Answer:

Explanation:

Let I= $\int x^{4}e^{2x} dx$

$\Rightarrow I=x^{4}\int e^{2x} dx-\int \frac{dx^{4}}{dx}.\int e^{2x} .dx.dx+C$

    $=\frac{x^{4}.e^{2x}}{2}-\frac{1}{2}\int 4x^{3}e^{2x}dx+C$

 $\Rightarrow\frac{x^{4}.e^{2x}}{2}-2\left[\frac{x^{3}e^{2x}}{2}-\int\frac{3x^{2}e^{2x}}{2}dx\right]+C$

$\Rightarrow\frac{x^{4}.e^{2x}}{2}-x^{3}e^{2x}+3\left[\frac{x^{2}e^{2x}}{2}-\frac{1}{2}\int2xe^{2x} dx\right]+C$

 $\Rightarrow\frac{x^{4}.e^{2x}}{2}-x^{3}e^{2x}+\frac{3}{2} x^{2}e^{2x}-3$

                                                            $\left[\frac{xe^{2x}}{2}-\int\frac{e^{2x}}{2}dx\right]+C $

   $\Rightarrow\frac{x^{4}.e^{2x}}{2}-x^{3}e^{2x}+\frac{3}{2} x^{2}e^{2x}-\frac{3}{2}xe^{2x}+\frac{3e^{2x}}{4}+C$

$\Rightarrow\frac{e^{2x}}{4}\left[ 2x^{4}-4x^{3}+6x^{2}-6x+3\right]+C$

• Mathematics - General questions