TS EAMCET 2017 :: Mathematics :: General questions

• Mathematics - General questions

 If the slope  of the tangent of the curve $y=ax^{3}+bx+4$ at (2,14)  is 21, then the values  of a and b respectively

Answer:

Explanation:

 The  curve $y=ax^{3}+bx+4$ is passes through (2,14)

$\therefore$     $14=a(2)^{3}+b(2)+4$

$\Rightarrow$   $14=8a+2b+4$

$\Rightarrow$    $5=4a+b$    ........(i)

 Slope of tangent to the curve $y=ax^{3}+bx+4$

 i.e, $\frac{dy}{dx}= 3ax^{2}+b$

$\Rightarrow$  $\left(\frac{dy}{dx}\right)_{(2,14)}=3a(2)^{2}+b$

$\Rightarrow$     $21=12a+b$      .........(ii)

                                        $\left[\because \frac{dy}{dx}=21\right]$

 Solving Eqs.(i)  and (ii) , we get

  a=2, b=-3

A point on the plane that  passes through the  points (1,-1,6),(0,0,7)  and perpendicular  to the plane x-2y+z=6 is 

Answer:

Explanation:

 Equation of plane passes through the points (1,-1,6) ,(0,0,7) and perpendicular to the plane x-2y+z=6 is 

    $\begin{bmatrix}x-1 & y+1&z-6 \\0-1 & 0+1&7-6\\1&-2&1 \end{bmatrix}=0$

 $\Rightarrow$      $\begin{bmatrix}x-1 & y+1&z-6 \\-1 & 1&1\\1&-2&1 \end{bmatrix}=0$

$\Rightarrow$     $(x-1)(1+2)-(y+1)(-1-1)+(z-6)(2-1)=0$

$\Rightarrow$   $3x-3+2y+2+z-6=0$

$\Rightarrow$    $3x+2y+z-7=0$

This plane is passes through (1,1,2)

If the coefficients of $(2r+1)^{th}$ term and $(r+1)^{th}$ term in the expansion  of $(1+x)^{42}$ are equal then r can be

Answer:

Explanation:

We have, $(1+x)^{42}$

        $T_{2r+1}=^{42}C_{2r} x^{2r}$

 $T_{r+1}= ^{42}C_{r}x^{r}$

Coefficient  of $(2r+1)^{th}$  and $(r+1)^{th}$ term are equal 

$\therefore$     $^{42}C_{2r}= ^{42}C_{r}$

$\therefore$   2r+r=42    $[ \because   ^{n}C_{x}=^{n}C_{y} \Rightarrow x+y=n]$

 $\Rightarrow$    r=14

The solution of $(y-3x^{2})dx+xdy=0$ is 

Answer:

Explanation:

We have,

   $(y-3x^{2})dx+xdy=0$

$\Rightarrow$    $ydx-3x^{2}dx+xdy=0$

$\Rightarrow$  $ydx+xdy=3x^{2}dx$

$\Rightarrow$      $dxy=3x^{2}dx$

On integrating both sides , we get 

     $xy=x^{3}+C \Rightarrow  y=x^{2}+\frac{C}{x}$

$\lim_{y \rightarrow1}\left(\frac{1}{y^{2}-1}-\frac{2}{y^{4}-1}\right)=$

Answer:

Explanation:

 We have,

$\lim_{y \rightarrow1}\left(\frac{1}{y^{2}-1}-\frac{2}{y^{4}-1}\right)=$

 $=\lim_{y \rightarrow 1}\left(\frac{y^{2}+1-2}{y^{4}-1}\right)$ 

$=\lim_{y \rightarrow 1}\left(\frac{y^{2}-1}{(y^{2}-1)(y^{2}+1)}\right)$

  $=\lim_{y \rightarrow 1}\frac{1}{y^{2}+1}=\frac{1}{1+1}=\frac{1}{2}$

• Mathematics - General questions