Ncert Solutions Maths class 12th

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

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

$=p\vee \sim q$

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

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

x + y + z = 48

x, y, z  $\in$ {2, 4, 8, 16, 32, 32}

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

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

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

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

$2sin12°-sin72°$

$=sin12°-\left(sin72°-sin12°\right)$

$=sin12°-2cos42°sin30°$

$=2cos30°\left(-sin18°\right)$

$=-\sqrt[]{3}sin18°$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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)

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

$\downarrow$                                                       $\downarrow$                           &nb

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]

For $x^2+\alpha x=\beta >0\forall x\in R$  to hold, we should have   ${\alpha }^2-4\beta <0$

If $\alpha =1,\beta$  can be 1, 2, 3, 4, 5, 6 i.e., 6 choices

If a  = 4, b can be 5 or 6 i.e., 2 choices

hence total favourable outcomes

= 6 + 5 + 4 + 2 + 0 + 0 = 17

Required probability =  $\frac{17}{36}$

If n is number of trials, p is probability of success and q is probability of unsuccess then,

Mean = np and variance = npq.

Here np + npq = 24         …. (i)

np. npq = 128