Ncert Solutions Maths class 12th

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

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

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

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

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

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

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

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

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) -

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

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

For inconsistent, $\Delta =0,\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill \alpha \hfill & \hfill 2a\hfill & \hfill 3\hfill \\ \hfill 1\hfill & \hfill 3\alpha \hfill & \hfill 5\hfill \end{array}\right|=0$

a = 1

$\therefore {\Delta }_1=\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 4\hfill & \hfill 3\hfill & \hfill 5\hfill \end{array}\right|=1\left(1\right)+1\left(12+5\right)+1\left(-3-8\right)=7\ne 0$              

  $\frac{6}{11}$ = Required probability         

After solving, we get n = 4

Surface area, S = 4pr²

$?\frac{ds}{dt}=4?.2r\frac{dr}{dt}=8?\frac{dr}{dt}=constant=k\left(say\right)$                

$?\frac{ds}{dt}=k?s=kt+c$

$?4?r^2=kt+c$        

Initially t = 0, r = 3

c = 36 p

When t = 5, r = 7, k = 32p

When t = 9, r = r, r = 9

Since student guesses only two wrong. So there are three possibilities

(i) both wrong in section A

(ii) both wrong in section B

(iii) one wrong in each section A and B.

$\therefore$ Required possibilities =

${=}^4{C}_4{\times }^6{C}_4{\left(\frac{3}{4}\right)}^4\times {\left(\frac{1}{4}\right)}^4{\left(\frac{3}{4}\right)}^2{+}^4{C}_3{\times }^6{C}_5{\left(\frac{3}{4}\right)}^3{\left(\frac{1}{4}\right)}^5\times \frac{1}{4}\times \frac{3}{4}$ ${+}^4{C}_2{\times }^6{C}_6\times {\left(\frac{3}{4}\right)}^2{\left(\frac{1}{4}\right)}^2\times {\left(\frac{1}{4}\right)}^6$

$=\frac{27}{4^{10}}\left[15\times 27+24\times 3+2\right]=\frac{27\times 479}{4^{10}}$

Circle $\left|z-3\right|\le 1\Rightarrow {\left(x-3\right)}^2+y^2\le 1$

and line $z\left(4+3i\right)+\overline{z}\left(4-3i\right)\le 24$

$\Rightarrow 4x-3y\le 12\therefore slope=tan\theta =\frac{4}{3}$