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The integral  $\int\frac{dx}{x^{2}\left(x^{4}+1\right)^{\frac{3}{4}}}$  equals

Answer:

Explanation:

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If the standard deviation of the numbers 2,3,a and 11 is 3.5, then which of the following is true?

Answer:

Explanation:

We know that , if x₁ , x₂..........x_n   are n observations , then their standard deviation is given by 

$\sqrt{\frac{1}{n}\sum x_i^2}-(\frac{\sum x_{i}}{2})^{2}$

We have ,  $(3.5)^{2}=\frac{(2^{2}+3^{2}+11^{2})}{4}-(\frac{2+3+a+11}{4})$

  $\Rightarrow \frac{49}{4}=\frac{(4+9+a^{2}+121)}{4}-(\frac{16+a}{4})^{2}$

$\Rightarrow \frac{49}{4}=\frac{(134+a^{2})}{4}-(\frac{256+a^{2}+32a}{16})$

$\Rightarrow \frac{49}{4}=\frac{4a^{2}+536-256-a^{2}-32a}{16}$

$\Rightarrow 49\times4=3a^{2}-32a+280$

$\Rightarrow 3a^{2}-32a+84=0$

Let $\widetilde{a},\widetilde{b}$ and  $\widetilde{c}$ be three unit vectors such that  $\widetilde{a} \times (\widetilde{b}\times\widetilde{c})=\frac{\sqrt{3}}{2}(\widetilde{b}+\widetilde{c})$ . If  $\widetilde{b} $ is not parallel to $\widetilde{c} $ , then the angle between $\widetilde{a} $ and $\widetilde{b} $ is

Answer:

Explanation:

Given  $\mid\widetilde{a}\mid=\mid \widetilde{b}\mid=\mid \widetilde{c}\mid =1$

 and   $\widetilde{a}\times(\widetilde{b}\times \widetilde{c}) =\frac{\sqrt{3}}{2}(\widetilde{b}+\widetilde{c})$

Now consider $\widetilde{a}\times(\widetilde{b}\times \widetilde{c}) =\frac{\sqrt{3}}{2}(\widetilde{b}+\widetilde{c})$

$\Rightarrow (\widetilde{a}.\widetilde{c})\widetilde{b}-(\widetilde{a}.\widetilde{b})\widetilde{c}=\frac{\sqrt{3}}{2}\widetilde{b}+\frac{\sqrt{3}}{2}\widetilde{c}$

On comparing , we get

                   $\widetilde{a}.\widetilde{b}=-\frac{\sqrt{3}}{2}$

$\Rightarrow  \mid\widetilde{a}\mid \mid\widetilde{b}\mid  \cos\theta =-\frac{\sqrt{3}}{2}$

$\Rightarrow   \cos\theta =-\frac{\sqrt{3}}{2}[\because \mid\widetilde{a}\mid=\mid\widetilde{b}\mid=1]$

$\Rightarrow   \cos\theta =\cos(\pi-\frac{\pi}{6})$

$\Rightarrow   \theta =\frac{5\pi}{6}$

If the line, $\frac{x-3}{2}=\frac{y+2}{-1}=\frac{z+4}{3}$ lies in the plane, lx+my-z=9, then  $l^{2}+m^{2}$ is equal to

Answer:

Explanation:

Since, the line $\frac{x-3}{2}=\frac{y+2}{-1}=\frac{z+4}{3}$ lies in the plane, lx+my-z=9, therefore we have 2l-m-3=0

 [ $\because$ normal  will be perpendicular to the line]

   $\Rightarrow  2l-m=3$          .......(i)

   and  $ 3l-2m+4=9$

   $[\because point (3,-2,-4)$ lies  on the plane]

$\Rightarrow 3l-2m=5$     ........(ii)

On solvubg Eqs(i) and (ii) , we get

  l=1 and m=-1

$\therefore   l^{2}+m^{2}=2$

The distance of the point (1,-5,9) from the the plane x-y+z=5 measured along  the line x=y=z is 

Answer:

Explanation:

Equation of line passing through the point (1,-5,9) and parrallel to x=y=z is 

$\frac{x-1}{1}=\frac{y+5}{1}=\frac{z-9}{1}=\lambda$  (say)

Thus, any point on this  line is  of the  form $(\lambda+1,\lambda-5,\lambda+9)$

 Now, if $P(\lambda+1,\lambda-5,\lambda+9)$ is the point  of intersection of line and plane, then

 $(\lambda+1)-(\lambda-5)+\lambda+9=5$

$\Rightarrow \lambda+15=5$

   $\Rightarrow \lambda =-10$

$\therefore$  Cordinates of point P are

 (-9,-15,-1)

Hence, required distance

$=\sqrt{(1+9)^{2}+(-5+15)^{2}+(9+1)^{2}}$

$=\sqrt{(10)^{2}+(10)^{2}+(10)^{2}}=10\sqrt{3}$

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