Class 11th

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.

When we speak of 2D or 3D motion, the velocity and acceleration vectors need not align or be in the same direction. They can have any angle between 0° and 180° between them. This is because acceleration accounts for changes in both the magnitude and direction of velocity.

Vectors are necessary because motion in a plane (two dimensions) or space (three dimensions) involves physical quantities, including velocity and acceleration. Both of these have both magnitude and direction. That helps us know how objects move in the real-world and in any real space. In one-dimensional motion, we only can know two directions, and show it as signs (+/-), and not more than that.  Motion in a plane requires vectors to accurately represent these directional aspects.