Class 12th

$y^{1/4}+\frac{1}{y^{1/4}}=2x$

${\left(y^{1/4}\right)}^2-2x{y}^{1/4}+1=0$    

=> ${\left(y^{1/4}\right)}^2-2x{y}^{1/4}+1=0$

$\Rightarrow \frac{dy}{dx}=\frac{4y}{\sqrt[]{x^2-1}}..........\left(i\right)$

$\Rightarrow \left(x^2-1\right)\frac{d^{2}y}{d{x}^2}+\frac{xdy}{dx}-16y=0from\left(i\right)$      

=>α = 1, β = -16

|α - β| = 17

Let $l={\displaystyle \int \frac{dx}{{\left(x^2+x+1\right)}^2}.............\left(i\right)}$

Now ${\displaystyle \int \frac{dx}{x^2+x+1}={\displaystyle \int \left(\frac{1}{x^2+x+1}\right)dx}}$

by integration by parts

=> ${\displaystyle \int \frac{dx}{{\left(x^2+x+1\right)}^2}=\frac{4}{3\sqrt[]{2}}ta{n}^{-1}\frac{2x+1}{\sqrt[]{3}}+\frac{1}{3}\frac{\left(2x+1\right)}{\left(x^2+x+1\right)}+c}$

$\Rightarrow a=\frac{4}{3\sqrt[]{3}},b=\frac{1}{3}$

Now, $9\left(\sqrt[]{3}a+b\right)=9\left(\frac{4}{3}+\frac{1}{3}\right)=15$

Equation of the plane will be $\left\{\overrightarrow{r}.\left(\widehat{i}+\widehat{j}+\widehat{k}\right)-1\right\}+\lambda \left\{\overrightarrow{r}.\left(2\widehat{i}+3\widehat{j}-\widehat{k}\right)+4\right\}=0$   

$\overrightarrow{r}.\left\{\left(1+2\lambda \right)i+\left(1+3\lambda \right)\widehat{j}+\left(1-\lambda \right)\widehat{k}\right\}+\left(4\lambda -1\right)=0$               

-> $\left(1+2\lambda \right)x+\left(1+3\lambda \right)y+\left(1-\lambda \right)z+\left(4\lambda -1\right)=0.........\left(i\right)$

(i) is parallel to x-axis so its d.r.s will be (1, 0, 0)

-> $1+2\lambda =0so\lambda =\frac{-1}{2}$

Hence required equation will be

$\overrightarrow{r}\left\{-\frac{1}{2}\widehat{j}.+\frac{3}{2}\widehat{k}\right\}+\left(-3\right)=0$

$\overrightarrow{r}.\left(\widehat{j}-3\widehat{k}\right)+6=0$              

One water molecule is associated with hydrogen bond.

Ans. = 1

$L.R.=\frac{\left|3*2+4x-3-5\right|}{\sqrt[]{3^2+4^2}}=\frac{11}{5}$              

Equation of family of parabolas

${\left(x-h\right)}^2=\frac{11}{5}\left(y-k\right)$              

Differentiate 2 (x – h) = $\frac{11}{5}\frac{dy}{dx}$  

Again differentiate 2 = $\frac{11}{5}\frac{d^{2}y}{d{x}^2}$  

$11\frac{d^{2}y}{d{x}^2}=10$              

Given 2x + y – z = 3         . (i)

x – y – z = α        . (ii)

3x + 3y + βz = 3                . (iii)

(i) x 2 – (ii) – (iii) – (1 + β) z = 3 - α

For infinite solution 1 + β = 0 = 3 - α

=> α = 3, β = -1

So, α + β - αβ = 5

Please find the below image 

Cl, Br & I form halic (V) acid i.e. HClO₃, HBrO₃ & HlO₃.

If S4 is ranked 6th and S2 is ranked 5th, we have

O > O > O > O > S2 > S4 > O

In this case, S3 will rank the highest and thus S5 will rank the least. We have, S3 > O > O > O > S2 > S4 > S5

Also, the order S7 > S1 > S2 will be maintained The possible arrangements are

S3 > S6 > S7 > S1 > S2 > S4 > S5

or   S3 > S7 > S6 > S1 > S2 > S4 > S5 or      S3 > S7 > S1 > S6 > S2 > S4 > S5

Thus, option (d), i.e., S6 can be ranked third or fourth is true.

If S4 ranks 2nd. S3 ranks 1st and S5 ranks the lowest. The order S7 > S1 > S2 will be followed out of all the given options, only the arrangement given as per option (d) i.e., S3> S4 > S6 > S7 follow. So, option (d) is correct.