Class 12th

$\left|\widehat{a}\right|=1,\left|\widehat{b}\right|=1$

${\left|\widehat{b}\right|}^2={\left|\overrightarrow{c}+2\left(\overrightarrow{c}*\widehat{a}\right)\right|}^2$

$\Rightarrow 1={\left|\overrightarrow{c}\right|}^2+4{\left|\overrightarrow{c}\right|}^2{\left(\frac{\sqrt[]{3}-1}{2\sqrt[]{2}}\right)}^2$

${\left|\overrightarrow{c}\right|}^2\left[1+4*\frac{{\left(\sqrt[]{3}-1\right)}^2}{8}\right]={\left|\overrightarrow{c}\right|}^2\left(3-\sqrt[]{3}\right)$

$\therefore {\left|\overrightarrow{c}\right|}^2=\frac{1}{3-\sqrt[]{3}}=\frac{3+\sqrt[]{3}}{6}$

$\Rightarrow {\left|6\overrightarrow{c}\right|}^2=6\left(3+\sqrt[]{3}\right)$

$\lambda x-2y=\mu ........(i)and\frac{x^2}{\left(\frac{b^2}{a^2}\right)}-\frac{y^2}{b^2}=1.........(ii)$

Let (x₁, y₁) be a point on the curve equation of tangent of eq (ii)

  $y=mx\pm \sqrt[]{\frac{b^2}{a^2}{m}^2-b^2}..........(iii)$

From (i) and (ii) are identical

$\therefore m=\frac{\lambda }{2}and\frac{b^2}{a^2}{m}^2-b^2=\frac{{\mu }^2}{4}$               

$\Rightarrow \frac{b^2}{a^2}*\frac{{\lambda }^2}{4}-b^2=\frac{{\mu }^2}{4}$

$\Rightarrow {\left(\frac{\lambda }{a}\right)}^2-{\left(\frac{\mu }{b}\right)}^2=4$

  $y\frac{dx}{dy}=2x+y^3\left(y+1\right)e^y$               

    $\frac{dx}{dy}-\frac{2}{y}x=y^3\left(1+y\right)e^y..........(i)$

Equation (i) is in linear form, so I.F. = y^-2

$\begin{array}{l}\therefore x{y}^{-2}=y{e}^y+c\Rightarrow 0=e+c\Rightarrow c=-e\\ \therefore x=e^3\left(e^e-1\right)\end{array}$                

f(x) =   $\left|2x^2+3x?2\right|+sinxcosx$

$=\left|\left(2x?1\right)\left(x+2\right)\right|+sinxcosx$

$f\left(x\right)=\{\begin{array}{cc}\hfill ?2x^2?3x+2+sinxcosx,\hfill & \hfill 0<x<\frac{1}{2}\hfill \\ \hfill 2x^2+3x?2+sinxcosx,\hfill & \hfill \frac{1}{2}?x<1\hfill \end{array}$

$f\text{'}\left(x\right)=\{\begin{array}{cc}\hfill ?4x?3+cos2x,\hfill & \hfill 0<x<\frac{1}{2}\hfill \\ \hfill 4x+3+cos2x,\hfill & \hfill \frac{1}{2}?x<1\hfill \end{array}$            

 f(1) = 3 + sin 1 cos 1 and   $f\left(\frac{1}{2}\right)=sin\frac{1}{2}cos\frac{1}{2}$

$?f\left(1\right)+f\left(\frac{1}{2}\right)=3+\frac{sin2}{2}+\frac{sin1}{2}=3+\frac{1}{2}\left(sin1+sin2\right)$

$=3+\frac{sin1}{2}\left(1+2cos1\right)$

f(x) =   $\left|2x^2+3x-2\right|+sinxcosx$

$=\left|\left(2x-1\right)\left(x+2\right)\right|+sinxcosx$

$f\left(x\right)=\{\begin{array}{cc}\hfill -2x^2-3x+2+sinxcosx,\hfill & \hfill 0<x<\frac{1}{2}\hfill \\ \hfill 2x^2+3x-2+sinxcosx,\hfill & \hfill \frac{1}{2}\le x<1\hfill \end{array}$

$f\text{'}\left(x\right)=\{\begin{array}{cc}\hfill -4x-3+cos2x,\hfill & \hfill 0<x<\frac{1}{2}\hfill \\ \hfill 4x+3+cos2x,\hfill & \hfill \frac{1}{2}\le x<1\hfill \end{array}$            

 f(1) = 3 + sin 1 cos 1 and   $f\left(\frac{1}{2}\right)=sin\frac{1}{2}cos\frac{1}{2}$

$\therefore f\left(1\right)+f\left(\frac{1}{2}\right)=3+\frac{sin2}{2}+\frac{sin1}{2}=3+\frac{1}{2}\left(sin1+sin2\right)$

$=3+\frac{sin1}{2}\left(1+2cos1\right)$

Ala – Gly – Leu-Ala- Val

Two amino acid units are connected by peptide linkage. So number of peptide linkage = 4.

f (x) = 4log_e (x – 1) -2x² + 4x + 5, x > 1

    $(A)f\text{'}(x)=\frac{4}{x-1}-4x+4$

$\Rightarrow f\text{'}(x)>0\Rightarrow \frac{x\left(-x+2\right)}{x-1}>0$

option (A) is correct.

(B) f (x) = -1, has two solution

option (B) is correct

$\left(C\right)f"(x)=-\frac{4}{{\left(x-1\right)}^2}-4$                

$f\text{'}(e)-f"(2)=\frac{4e\left(2-e\right)}{e-1}+8>0$              

option (C) is not correct

(D) f (e) = 4log_e (e – 1) -2 (e² – 2e + 1) + 7 > 0

f (e + 1) = 4 – 2 (e + 1)² + 4 (e + 1) +5+ < 0

option (D) is correct   

In acidic solution

$3\stackrel{+6}{Mn}{O}_4^{2-}+4H^{+}\to 2\stackrel{+7}{Mn}{O}_4^{-}+\stackrel{+4}{Mn}{O}_2+2H_{2}O$

Difference in oxidation state of Mn is product

= 7 – 4 = 3

Baryte – BaSO₄ [Sulphate based ore]

Galena – PbS

Zinc blende – ZnS

Copper pyrite – CuFeS₂

S = $\left\{\sqrt[]{n}:1\le n\le 50\&n=odd\right\}$  

= $\left\{\sqrt[]{1},\sqrt[]{3},\sqrt[]{5},......,\sqrt[]{49}\left(25terms\right)\right\}$  

$\left|A\right|=\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 4\hfill \\ \hfill -1\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill -a\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right|$ = 1 + a²

^{$\therefore {\displaystyle \sum _{a\in S}^{}det\left(adjA\right)}=100\lambda \Rightarrow \sum {\left(1+a^2\right)}^2=100\lambda \Rightarrow \lambda =221$}