Class 11th

x + 2y + z = 14

${\perp }^{r}linePQ:\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{1}=t$                

Q (1 + t, 2 + 2t, 3 + t)

 x + 2y + z = 14 ⇒ 1 + t + 4 + 4t + 3 + t = 14 ⇒ 6t = 6

t = 1

⇒ Q (2, 4, 4)

PQ = $\sqrt[]{1+4+1}=\sqrt[]{6}$  

$tan60°=\frac{PQ}{QR}\Rightarrow QR=\frac{PQ}{\sqrt[]{3}}=\sqrt[]{2}$

ar (PQR) = $\frac{1}{2}\sqrt[]{6}\cdot \sqrt[]{2}=\sqrt[]{3}$  

Circumcentre (D) $\equiv \left(5,\frac{\alpha }{4}\right)$

${\left(5-\alpha \right)}^2+{\left(\frac{\alpha }{4}+2\right)}^2={\left(5-\alpha \right)}^2+{\left(\frac{\alpha }{4}-6\right)}^2...............\left(i\right)$

${\left(5-\frac{\alpha }{4}\right)}^2+{\left(\frac{\alpha }{4}+2\right)}^2....................\left(ii\right)$

(i) $\frac{\alpha }{4}+2=\pm \left(\frac{\alpha }{4}-6\right)$

(ii) 9 + 16 = 9 + 16

$\oplus \to x$

$\left(-\right)\to \frac{\alpha }{2}=4\Rightarrow \alpha =8$

ar $\left(ABC\right)=24$

2S = 24

R = 5, r = $\frac{\Delta }{s}=\frac{24}{12}=2$

2cos $\left(\frac{x^2+x}{6}\right)=4^x+4^{-x}$

$-2\le LHS\le 2LHS=2\&RHS=2\Rightarrow x=0only\to thenLHS=2also$

RHS $\ge$ 2

$m_1{m}_2=-1,$ for square a,b,c,d let

$A\left(10\left(cos\alpha -sin\alpha \right),10\left(sin\alpha +cos\alpha \right)\right)$

Diagonal : (cosα - sinα)x + (sinα + cosα)y = 10

BD (diagonal)

Dist. Of BD from A is

$\frac{\left|10{\left(cos\alpha -sin\alpha \right)}^2+10{\left(sin\alpha +cos\alpha \right)}^2-10\right|}{\sqrt[]{2}}=\frac{a}{\sqrt[]{2}}$

$\frac{10}{\sqrt[]{2}}=\frac{a}{\sqrt[]{2}}\Rightarrow a=10$

Also, a² + 11a + 3 $\left(m_1^2+m_2^2\right)=220$

210 + 3 $\left(c{m}_1^2+m_2^2\right)=220$

$m_1^2+m_2^2=\frac{10}{3}$

Also, m₁ m₂ = -1

m² + $\frac{1}{m^2}=\frac{10}{3}$

or $-\sqrt[]{3},\frac{1}{\sqrt[]{3}}$

m = $\sqrt[]{3},\frac{-1}{\sqrt[]{3}}$

$m^4-\frac{10}{3}{m}^2+1=0\Rightarrow m^2=\frac{\frac{10}{3}\pm \sqrt[]{\frac{100}{9}-4}}{2}-\frac{\frac{10}{3}\pm \frac{8}{3}}{2}=3,\frac{1}{3}$

m = $\pm \sqrt[]{3},\pm \frac{1}{\sqrt[]{3}}$

Diagonal AC:

$\left(sin\alpha +cos\alpha \right)x-\left(cos\alpha -sin\alpha \right)y$

=10 cos2α - 10cos2α = 0

Slope of AC = $\frac{sin\alpha +cos\alpha }{cos\alpha -sin\alpha }=\frac{tan\alpha +1}{1-tan\alpha }=tan\left(\alpha +\frac{\pi }{4}\right)\alpha =30°$

FIGURE

? = $72\left(\frac{1}{16}+\frac{9}{16}\right)+100-30+13=\frac{720}{16}+83=128$

${\displaystyle \sum _{r=1}^{20}\left(r^2+1\right)r!}$  

t_r = (r² + 1)r!

= r²r! + r!

= r(r + 1 – 1)r! + r!

= r(r + 1)! – (r – 1)r!

= V_r – V_r-1

  ${\displaystyle \sum _{r=1}^{20}\left(V_r-V_{r-1}\right)}$              

= V₁ – V₀

$+V_2-V_1$

$+V_3-V_2$

$+V_{20}-V_{19}$

$+V_{20}-V_{19}$

$=V_{20}-V_0=20\left(21!\right)-0$

$\left(22-2\right)\left(21!\right)=22!-2\left(21!\right)$        

a₀ = 0, a₁ = 0

a_n+2 = 3a_n+1 – 2a_n + 1

a₂₅ a₂₃ – 2a₂₅a₂₂ – a₂₃a₂₄ + 4a₂₂a₂₄ =?

a₂ = 3a₁ – 2a₀ + 1

a₃ = 3a₂ – 2a₁ + 1

a₄ = 3a₃ – 2a₂ + 1

a₅ = 3a₄ – 2a₃ + 1

a_n+2  = 3a_n+1 – 2a_n + 1

$\left(+\right)\Rightarrow \left(a_2+a_3+a_4+....+a_{n+1}+a_{n+2}\right)$                

⇒ a_n+2 = 2 (a₂ + a₃ + …. + a_n + a_n+1) –2 (a₁ + a₂ + ….+ a_n) + n + 1

a_n+2 = 2a_n+1 + n + 1

a₂₅ a₂₃ -2a₂₅ a₂₂ -a₂₃ a₂₄ + 4a₂₂ a₂₄

= a₂₅ (a₂₃ – 2a₂₂) -2a₂₄ (a₂₃ – 2a₂₂)

As a_n+2 = 2a_n+1 + n + 1

⇒ a_n+2 – 2a_n+1 = n + 1

⇒ a_n+1 -2a_n = n

⇒ 24 * 22 = 528

y = 2x $\left|3x^2-5x+2\right|,0\le x\le 1$

$=\{\begin{array}{cc}\hfill -3x^2+7x-2,\hfill & \hfill 3x^2-3x+2,\hfill \end{array}\begin{array}{cc}\hfill 0\le x\le \frac{2}{3}\hfill & \hfill \frac{2}{3}<x\le 1\hfill \end{array}$

3x² – 5x + 2 = 0

$x=\frac{+5\pm \sqrt[]{25-24}}{6}$

= $\frac{5+1}{6}=1,\frac{2}{3}$

3x² – 7x + 3 = 0

x = $\frac{7\pm \sqrt[]{49-39}}{6}=\frac{7\pm \sqrt[]{13}}{6}$

$-3x^2+7x-2$

$\frac{-7\pm \sqrt[]{49-24}}{-6}=\frac{7+5}{6}=2,\frac{1}{3}$

3x² + 7x – 2 = 1

3x² – 7x + 1 = 0

x = $\frac{7\pm \sqrt[]{49-12}}{6}=\frac{7\pm \sqrt[]{37}}{6}$

I = ${\displaystyle \underset{0}{\overset{6}{\int }}\left(\begin{array}{cc}\hfill \left(-2\right)\hfill & \hfill +1\hfill \end{array}\right)dx+{\displaystyle \underset{\frac{7-\sqrt[]{37}}{6}}{\overset{\frac{1}{3}}{\int }}\left(\begin{array}{cc}\hfill \left(-1\right)\hfill & \hfill +1\hfill \end{array}\right)dx}+{\displaystyle \underset{-\frac{1}{3}}{\overset{\frac{7-\sqrt[]{13}}{6}}{\int }}\left(\begin{array}{cc}\hfill 0\hfill & \hfill +1\hfill \end{array}\right)dx+{\displaystyle \underset{\frac{7-\sqrt[]{13}}{6}}{\overset{\frac{2}{3}}{\int }}\left(1+1\right)dx}}}+{\displaystyle \underset{\frac{2}{3}}{\overset{1}{\int }}\left(1+1\right)dx}$

$=\frac{\sqrt[]{37}+\sqrt[]{13}-4}{6}$

$\left|z-\frac{1}{z}\right|=2$

${\left|z\right|}_{max}=?$

$\left|z-\frac{1}{2}\right|\ge \left|\left|z\right|-\frac{1}{\left|z\right|}\right|$

$2\ge \left|r-\frac{1}{r}\right|$

0 $\le r^2+2r-1\&r^2-2r-1\le 0$

$r=\frac{-2\pm \sqrt[]{8}}{2}$ $r=\frac{2\pm \sqrt[]{8}}{2}$

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

r $\ge \sqrt[]{2}-1\&0\le r\le 1+\sqrt[]{2}$

$\sqrt[]{2}-1\le r\le \sqrt[]{2}+1$

In 1D kinematics, you use scalar equations for one direction. In 2D, position, velocity, and acceleration become vectors with x and y components. You apply the same kinematic equations independently to each dimension. Just remember to treat horizontal and vertical motions as separate 1D problems to be solved simultaneously.

In uniform circular motion, we know that the speed is constant. But the velocity vector's direction continuously changes as the object moves in a circle. This continuous change in direction leads to an acceleration. In physics, we call that centripetal acceleration. This is always directed towards the centre of the circle.