Ncert Solutions Maths class 12th

$CD=\sqrt[]{{\left(10+x^2\right)}^2-{\left(10-x^2\right)}^2}=2\sqrt[]{10}\left|x\right|$

Area 

$=\frac{1}{2}*CD*AB=\frac{1}{2}*2\sqrt[]{10}\left|x\right|\left(20-2x^2\right)$

$\Rightarrow 10-x^2=2x$

3x² = 10

 x = k

3k² = 10

By truth table

So F₁ (A, B, C) is not a tautology

Now again by truth table

So      F₂ (A, B) be a tautology.

From option let it be isosceles where AB = AC then

$x=\sqrt[]{r^2-{\left(h-r\right)}^2}$            

=  $\sqrt[]{r^2-h^2-r^2+2rh}$

$x=\sqrt[]{2hr-h^2}........\left(i\right)$

 Now ar $\left(\Delta ABC\right)=\Delta =\frac{1}{2}BC*AL$

$\Delta =\frac{1}{2}*2\sqrt[]{2hr-a^2}*h$        

then  $x=\sqrt[]{2*\frac{3r}{2}*r-\frac{9r^2}{4}}=\frac{\sqrt[]{3}}{2}rfrom\left(i\right)$

$\Rightarrow BC=\sqrt[]{3}r$    

So $AB=\sqrt[]{h^2+x^2}=\sqrt[]{\frac{9r^2}{4}+\frac{3r^2}{4}}=\sqrt[]{3}r$ .

Hence $\Delta$ be equilateral having each side of length $\sqrt[]{3}r.$

$A_1+A_2={\displaystyle {\int }_0^{\pi /2}cosxdx}$

$={\left(Sinx\right)}_0^{\pi /2}=1$
$A_1={\displaystyle {\int }_0^{\pi /4}\left(cosx-sinx\right)dx={\left(sinx+cosx\right)}_0^{\pi /4}}$

$=\frac{2}{\sqrt[]{2}}-1=\sqrt[]{2}-1$

$So{A}_2=1-\left(\sqrt[]{2}-1\right)=2-\sqrt[]{2}=\sqrt[]{2}\left(\sqrt[]{2}-1\right)$

$Now\frac{A_1}{A_2}=\frac{1}{\sqrt[]{2}}$             

Given f(X) = ${\displaystyle {\int }_1^x\frac{lo{g}_{e}t}{\left(1+t\right)}dt............\left(i\right)}$  

So  $f\left(\frac{1}{x}\right)={\displaystyle {\int }_1^{1/x}\frac{\lambda lo{g}_{e}t}{1+t}dt}............\left(ii\right)$

put $t=\frac{1}{z}thendt=-\frac{1}{z^2}dz$

$f\left(\frac{1}{x}\right)={\displaystyle {\int }_1^x\frac{lo{g}_{e}t}{t\left(1+t\right)}}dt............\left(iii\right)$    

(i) + (iii), f(x) + $f\left(\frac{1}{x}\right)={\displaystyle {\int }_1^x\left(\frac{lo{g}_{e}t}{1+t}+\frac{lo{g}_{e}t}{t\left(1+t\right)}\right)dt}$

$={\left[\frac{{\left(lo{g}_{e}t\right)}^2}{2}\right]}_1^x=\frac{{\left(lo{g}_{x}x\right)}^2}{2}$  

Hence f(e) + $f\left(\frac{1}{e}\right)=\frac{1}{2}$

For line of intersection of two planes

put z = $\lambda$ then

$x+2y=6-\lambda \&y=4-2\lambda \Rightarrow x+2\left(4-2\lambda \right)=6-\lambda$           

=> $x=3\lambda -2$

Now $\overrightarrow{a}.\overrightarrow{PQ}=0gives9\lambda -15+4\lambda -4+\lambda -1=0$

$So,P\left(\frac{16}{7},\frac{8}{7},\frac{10}{7}\right)=P\left(\alpha ,\beta ,\gamma \right)$

$\Rightarrow 21\left(\alpha +\beta +\gamma \right)=21*\frac{34}{7}=102$

Given f (k) = $\{\begin{array}{l}k+1,kisodd\\ k,kiseven\end{array}$  

$?g:A\to A$ such that g (f (x) = f (x)

Case I : If x is even then g (x) = x . (i)

Case II : If x is odd then g (x + 1) = x + 1 . (ii)

From (i) & (ii), g (x) = x, when x is even

So total no. of functions = 10⁵ * 1 = 10⁵

Now equation of line OA be

$\frac{x-1}{4}=\frac{y-3}{-5}=\frac{z-5}{2}=\lambda$

direction cosines of plane are 4, -5, 2

Equation of any point on OA be

$O\left(4\lambda +1,-5\lambda +3,2\lambda +5\right)$

Since O lies on given plane so

$4\left(4\lambda +1\right)-5\left(-5\lambda +3\right)+2\left(2\lambda +5\right)=8$

So, O (9/5,2,27/5). Hence by mid-point formula

B $\left(\frac{13}{5},1,\frac{29}{5}\right)\Rightarrow 5\left(\alpha +\beta +\gamma \right)=47$

$\underset{x\to a}{lim}\frac{xf\left(a\right)-af\left(x\right)}{x-a}\left(\frac{0}{0}\right)$

By L'hospital Rule

$=\underset{x\to a}{lim}\frac{f\left(a\right)-af\text{'}\left(x\right)}{1}=f\left(a\right)-af\text{'}\left(a\right)$

$=4-2a$

Now equation of line OA be