Ncert Solutions Maths class 12th

  $f\left(x\right)=\left[2x^2+1\right]$

  $g\left(x\right)=\{\begin{array}{l}2x-3,x<0\\ 2x+3,x\ge 0\end{array}$

  $f\left(g\left(x\right)\right)=\left[2{\left(g\left(x\right)\right)}^2\right]+1$              

 for 0 < x < 1

-3 < 2x 3 < -1

2 < 2 (2x 3)² < 18

31 + 15 = 46

  $ta{n}^{-1}\left(\frac{\frac{1}{\sqrt[]{2}-1}}{\frac{1}{\sqrt[]{2}}}\right)$

$=ta{n}^{-1}\left(1-\sqrt[]{2}\right)$

$=ta{n}^{-1}\left(\sqrt[]{2}-1\right)$

$=-\theta$

$tan\theta =\sqrt[]{2}-1$

$tan2\theta =\frac{2\left(\sqrt[]{2}-1\right)}{1-{\left(\sqrt[]{2}-1\right)}^2}$

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

tan 2 θ = 1

$\theta =\frac{\pi }{8}$                

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

$=-\left(\left(q\vee \sim p\right)\vee \left(\sim p\vee q\right)\right)$

$=p\vee \sim q$

$\sim \left(p\to q\right)$

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

x + y + z = 48

x, y, z  $\in$ {2, 4, 8, 16, 32, 32}

16 + 16 + 16 -> P_{1 $={\left(\frac{1}{16}\right)}^3$}  

 $32+8+8\to +P_2=\frac{1}{32}.{\left(\frac{1}{8}\right)}^2.\frac{3!}{2!}$

$P=\frac{1}{2^{12}}+\frac{3}{2^{11}}$

$=\frac{1+6}{2^{12}}=\frac{7}{2^{12}}$  

$2sin12°-sin72°$

$=sin12°-\left(sin72°-sin12°\right)$

$=sin12°-2cos42°sin30°$

$=2cos30°\left(-sin18°\right)$

$=-\sqrt[]{3}sin18°$

$=-\sqrt[]{3}\left(\frac{\sqrt[]{5}-1}{4}\right)$  

$\frac{dx}{dt}=12\left(1+cos2t\right)$

$\frac{dy}{dt}=24\left(1+sint\right)cost$

$\frac{1+sint}{cost}=\sqrt[]{3}$

$\Rightarrow \frac{1}{2}=\frac{\sqrt[]{3}}{2}cost-\frac{1}{2}sint$

$y_0=12{\left(1+\frac{1}{2}\right)}^2=12*\frac{9}{4}=27$

$S=\pi r\mathcal{l}$

$=\pi x\sqrt[]{x^2+y^2}$

$=\pi x\sqrt[]{x^2+25x^2}$

$\frac{ds}{dt}=\pi \sqrt[]{26}.2x.\frac{dx}{dt}$  

${\left(\frac{ds}{dt}\right)}_{y=10}\to x=2$                

$5\pi .3x^2\frac{dx}{dt}=1$                

$\left(\frac{dx}{dt}\right)=\frac{1}{15\pi \left(4\right)}=\frac{1}{60\pi }$

  $\left(x+3y-z-5\right)+\lambda \left(2x-y+z-3\right)=0$

$\Rightarrow \left(1+2\lambda \right)x+\left(3-\lambda \right)y+\left(\lambda -1\right)z-5-3\lambda =0$                

Point $\left(2,1,-2\right)\Rightarrow \left(1+2\lambda \right)\left(2\right)+\left(3-\lambda \right)\left(1\right)+\left(\lambda -1\right)\left(-2\right)-5-3\lambda =0$  

$-2\lambda +2=0\Rightarrow \lambda =1$                

 P : 3x + 2y + 0.z = 8

X (1, -2, 4)

Y (5, -1, 2)

X + Y = (6, -3, 6)

Y – X = (4, 1, -2)

$\underset{x\to \frac{\pi }{2}}{lim}\frac{si{n}^{2}x\left(si{n}^{2}x-3sinx+2\right)}{co{s}^{2}x.\left(\sqrt[]{2si{n}^{2}x+3sinx+4}+\sqrt[]{si{n}^{2}x+6sinx+2}\right)}$  

$\downarrow$                                                       $\downarrow$                            

3                                                       3

$\underset{x\to \frac{\pi }{2}}{lim}\frac{1}{9}.\frac{\left(sinx-1\right)\left(sinx-2\right)}{\left(1-sinx\right)\left(1+sinx\right)}=\frac{1}{6}\left(-1\right).\frac{-1}{2}=\frac{1}{12}$

Number of solution of the equation |cos x| = sin x for $x\in \left[-4\pi ,4\pi \right]$ $$ will be equal to 4 times the number of solutions of the same equation for $x\in \left[0,2\pi \right]$ $$

Graphs of y = |cos x| and y = sin x are as shown below.

Hence, two solutions of given equation in [0, 2]

$$ $\Rightarrow$  total of 8 solutions in [4, 4]