Ncert Solutions Maths 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

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

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

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

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

  $\overrightarrow{b}\times \overrightarrow{c}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 1\hfill & \hfill 3\hfill & \hfill \beta \hfill \\ \hfill -1\hfill & \hfill 2\hfill & \hfill -3\hfill \end{array}\right|=\widehat{i}\left(-9-2\beta \right)-\widehat{j}\left(-3+\beta \right)+\widehat{k}\left(5\right)$             

$\left|\overrightarrow{b}\times \overrightarrow{c}\right|=5\sqrt[]{3}gives\sqrt[]{{\left(9-2\beta \right)}^2+{\left(\beta -3\right)}^2+25}=5\sqrt[]{3}$         

$81+4{\beta }^2+36\beta +{\beta }^2+9-6\beta +25=75$

$\beta =-2,-4$    

also $\overrightarrow{a}\perp \overrightarrow{b}so\overrightarrow{a}.\overrightarrow{b}=0i.e.1+15+\alpha \beta =0$

So, ${\left|\overrightarrow{a}\right|}^2=1+25+{\alpha }^2=90$

Let f (x) = 3x⁴ + 4x³ – 12x² + 4

So, f’ (x) = 12x³ + 12x² – 24x = 12x (x² + x – 2)

=12x (x + 2) (x – 1)

So, number of distinct real roots = 4

$l={\displaystyle \underset{0}{\overset{1}{\int }}\frac{\sqrt[]{x}dx}{\left(1+x\right)\left(1+3x\right)\left(3+x\right)}}$

Put x = t² then dx = 2tdt

$l={\displaystyle \underset{0}{\overset{1}{\int }}\frac{dt}{\left(1+t^2\right)\left(3+t^2\right)}-{\displaystyle \underset{0}{\overset{1}{\int }}\frac{dt}{\left(1+3t^2\right)\left(3+t^2\right)}}}$

$=\frac{1}{2}{\left(ta{n}^{-1}t\right)}_0^1-\frac{3}{8\sqrt[]{3}}{\left(ta{n}^{-1}\frac{t}{\sqrt[]{3}}\right)}_0^1-\frac{8}{8\sqrt[]{3}}{\left(ta{n}^{-1}\sqrt[]{3}t\right)}_0^1$

$=\frac{\pi }{8}\left(1-\frac{\sqrt[]{3}}{2}\right)$      

Given $2\mathcal{l}+2m-n=0........\left(i\right)$

$mn+n\mathcal{l}+\mathcal{l}m=0...........\left(ii\right)$

$\&wehave{\mathcal{l}}^2+m^2+n^2=1..........\left(iii\right)$

$\left(i\right)\Rightarrow 2\left(\mathcal{l}+m\right)=n$  

  $\left(ii\right)\Rightarrow \mathcal{l}m+n\left(\mathcal{l}+m\right)=0$

$2{\mathcal{l}}^2+2m^2+5\mathcal{l}m=0$             

(a) $\frac{\mathcal{l}}{m}=-2$  

(i) $\Rightarrow \frac{2\mathcal{l}}{m}+2-\frac{n}{m}=0$  

$\frac{n}{m}=-2$  

$So,\left(\mathcal{l},m,n\right)=\left(-2m,m,-2m\right)$  

= (-2, 1, -2)

(b)   $\frac{\mathcal{l}}{m}=\frac{-1}{2}givesn=-2\mathcal{l}$

$\left(\mathcal{l},m,n\right)=\left(\mathcal{l},-2\mathcal{l},-2\mathcal{l}\right)=\left(1,-2,-2\right)$

$Now,cos\theta =\frac{-2-2+4}{3\times 3}=0$

$\Rightarrow \theta =\frac{\pi }{2}$  

$\left(p\wedge q\right)\Rightarrow \left(\left(r\wedge q\right)\wedge p\right)$

$\sim \left(p\wedge q\right)\vee \left(\left(r\wedge q\right)\wedge p\right)$

$\left[\sim \left(p\wedge q\right)\vee \left(r\wedge p\right)\right]$

$\sim \left(p\wedge q\right)\vee \left(r\wedge p\right)$

$\Rightarrow \left(p\wedge q\right)\Rightarrow \left(r\wedge p\right)$

$n\left(n+1\right)x^2+2\left(2n+1\right)x+4$

$=n\left(n+1\right)x^2+\left\{\left(2n+2\right)+2n\right\}x+4$

${\displaystyle \sum _{n=1}^9\frac{x}{\left(nx+2\right)\left(\left(n+1\right)x+2\right)}}$

$={\displaystyle \sum _{n=1}^9\left[\frac{1}{nx+2}-\frac{1}{\left(n+1\right)x+2}\right]}$

$\left[\left(\frac{1}{x+2}-\frac{1}{2x+2}\right)+\left(\frac{1}{2x+2}-\frac{1}{3x+2}\right)+....+\left(\frac{1}{9x+2}-\frac{1}{10x+2}\right)\right]$

$\therefore \underset{x\to 2}{Lt}\frac{9x}{\left(10x+2\right)\left(x+2\right)}=\frac{9}{44}$              

${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left[\frac{1+si{n}^{2}x}{1+{\pi }^{sinx}}+\frac{1+si{n}^{2}x}{1+{\pi }^{-sinx}}\right]dx}$

$={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(1+si{n}^{2}x\right)dx=\frac{\pi }{2}+\frac{1}{2}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(1-cos2x\right)dx}}$

$=\frac{\pi }{2}+\frac{1}{2}{\left[x-\frac{sin2x}{2}\right]}_0^{\frac{\pi }{2}}$
$=\frac{\pi }{2}+\frac{\pi }{4}=\frac{3\pi }{4}$