Ncert Solutions Maths class 12th

3, 4, 5, 5

In remaining six places you have to arrange

3, 4, 5,5

So no. of ways $=\frac{6!}{2!2!2!}$

Total no. of seven digits nos. = $\frac{7!}{2!3!2!}*1$

Hence Req. prob. $\frac{\frac{6!}{2!2!2!}}{\frac{7!}{2!3!2!}}=\frac{6!3!}{2!7!}=\frac{3}{7}$

${\overrightarrow{a}}_1=x\widehat{i}-\widehat{j}+\widehat{k}\&{\overrightarrow{a}}_2=\widehat{i}+y\widehat{j}+z\widehat{k}$

given ${\overrightarrow{a}}_1\&{\overrightarrow{a}}_2$ are collinear then ${\overrightarrow{a}}_1=\lambda {\overrightarrow{a}}_2$

$\Rightarrow \left(x\widehat{i}-\widehat{j}+\widehat{k}\right)=\lambda \left(\widehat{i}+y\widehat{j}+z\widehat{k}\right)$       

Since $\widehat{i},\widehat{j}\&\widehat{k}$ are not collinear so

$Sox\widehat{i}+y\widehat{j}+z\widehat{k}=\lambda \widehat{i}-\frac{1}{\lambda }\widehat{j}+\frac{1}{\lambda }\widehat{k}$     

Hence possible unit vector parallel to it be $\frac{1}{\sqrt[]{3}}\left(\widehat{i}-\widehat{j}+\widehat{k}\right)$ for $\lambda$ =

Given f(x) =   _{${\displaystyle {\int }_e^x{e}^{t}f\left(t\right)dt+e^x..........\left(i\right)}$}

using Leibniz rule then

f'(x) = e^xf(x) + e^x

_{$\Rightarrow \frac{dy}{dx}=e^{x}y+e^{x}wherey=f\left(x\right)then\frac{dy}{dx}=f\text{'}\left(x\right)$}            

P = -e^x, Q = e^x

Solution be y. (I.F.) =  ${\displaystyle \int Q\left(I.F.\right)dx+c}$

I. f. =  $e^{{\displaystyle \int -e^{x}dx}=e^{-e^x}}$

$\Rightarrow y.\left(e^{-e^x}\right)={\displaystyle \int e^x.e^{-e^x}dx+c}$   

$y.e^{-e^x}=-{\displaystyle \int dt+c=-t+c=-e^{-e^x}+c..........\left(ii\right)}$

Put x = 0 , in (i) f (0) = 1

$From\left(ii\right),\frac{1}{e}=-\frac{1}{e}+cgivenc=\frac{2}{e}$   

Hence f(x) = 2. $e^{\left(e^x-1\right)}-1$

Given $l_{m,n}={\displaystyle {\int }_0^1{x}^{m-1}{\left(1-x\right)}^{n-1}dx............\left(i\right)}$  

put 1 - x = $t\{\begin{array}{l}x=0,t=1\\ x=1,t=0\end{array}$

dx = -dt

From (i) $l_{m,n}={\displaystyle {\int }_1^0{\left(1-t\right)}^{m-1}.t^{n-1}\left(-dt\right)}$

$l_{m,n}={\displaystyle {\int }_0^1{t}^{n-1}{\left(1-t\right)}^{m-1}dt={\displaystyle {\int }_0^1{x}^{n-1}{\left(1-x\right)}^{m-1}dx..........\left(ii\right)}}$    

$Putx=\frac{1}{1+y}in\left(i\right)thendx=-\frac{1}{{\left(1+y\right)}^2}dy$                          

(i)  $l_{m,n}={\displaystyle {\int }_{\infty }^0\frac{1}{{\left(1+y\right)}^{m-1}}}{\left(1-\frac{1}{1+y}\right)}^{n-1}\left(\frac{-dy}{{\left(1+y\right)}^2}\right)={\displaystyle {\int }_0^{\infty }\frac{y^{n-1}}{{\left(1+y\right)}^{m+n}}dy}..........\left(iii\right)$

Similarly by (ii) $l_{m,n}={\displaystyle {\int }_0^{\infty }\frac{y^{m-1}}{{\left(y+1\right)}^{m+n}}dy..........\left(iv\right)}$

Adding (iii) & (iv) $2l_{m,n}={\displaystyle {\int }_0^{\infty }\frac{y^{n-1}+y^{m+1}}{{\left(y+1\right)}^{m+n}}}$

Putting  $\frac{1}{z}\{\begin{array}{l}y=1,z=1\\ y=\infty ,z=0\end{array}$ $\Rightarrow dy=\frac{-1}{z^2}dz$  

Hence $l_{m,n}={\displaystyle {\int }_0^1\frac{x^{m-1}+x^{n-1}}{{\left(1+x\right)}^{m+n}}}$ dx = α lm, n

=> α = 1

Given $f\left(x\right)=2x^5+5x^4+10x^3+10x^2+10x+10$ ${}^{}{}^{}{}^{}{}^{}$

$f\left(-1\right)=3>0\&f\left(-2\right)=-34<0$

So at least one root will lie in (2, 1)

now $f\text{'}\left(x\right)=10x^4+20x^3+30x^2+20x+10$ ${}^{}{}^{}{}^{}$

$=10\left[x^4+2x^3+3x^2+2x+1\right]$

$=10x^2{\left(x+\frac{1}{x}+1\right)}^2>0\forall x\in R$

So, f (x) be purely increasing function so exactly one root of f (x) that will lie in (-2, 1). Hence |a| = 2

$Given\left|z+5\right|\le 4..........\left(i\right)$

->Represent a circle ${\left(x+5\right)}^2+y^2\le 16$

$=z\left(1+i\right)+\overline{z}\left(1-i\right)\ge -10..........\left(ii\right)$

->Represent a line X – y $\ge -5$

So max |z + 1|² = AQ²

$={\left(-4-2\sqrt[]{2}\right)}^2+8$

$\begin{array}{l}=32+16\sqrt[]{2}\\ =\alpha +\beta \sqrt[]{2}\end{array}$  

Hence α + β) = 48

18 = 3² * 2

For G.C.D to be 3. no. of four digits should be only multiple of 3, but not multiple of 9 & also should not be even.

As we know no. of the form                          9 k -> 1000

9 k + 1 -> 1000

9 k + 2 -> 1000

9 k + 3 -> 1000  -> Total no. = 2000

9 k + 4 -> 1000

9 k + 5 -> 1000

9 k + 6 ->1000

9 k + 7 -> 1000

9 k + 8 -> 1000

In which half will be even & half be odd so Required no. = 1000

Given curves $\frac{x^2}{9}+\frac{y^4}{4}=1.........\left(i\right)$

&     $x^2+y^2=\frac{31}{4}...........\left(ii\right)$      

Equation of any tangent to (i) be y = mx +  $\sqrt[]{9m^2+4}............\left(iii\right)$

For common tangent (iii) also should be tangent to (ii) so by condition of common tangency

$9m^2+4=\frac{31}{4}\left(1+m^2\right)$

OR 36m² + 16 = 31 + 31m²

=>m² = 3

${\displaystyle \underset{0}{\overset{\pi }{\int }}\left|sin2x\right|dx}$

$=2{\displaystyle \underset{0}{\overset{\pi /2}{\int }}sin2xdx}$ =2 ${\left[\frac{-cos2x}{2}\right]}_0^{\pi /2}$ = $2\left(\frac{1}{2}-\left(-\frac{1}{2}\right)\right)=2$

$30{}^{30}{C}_0+29{}^{30}{C}_1+.....+2{}^{30}{C}_{28}+1{.}^{30}{C}_{29}=n.2^m$

$\therefore n=15,m=0$

$\therefore n+m=15+30=45$