Ncert Solutions Maths class 12th

P(W) = $\frac{1}{3};$  P(B) =   $\frac{2}{3}$

$\Rightarrow p=\frac{1}{3};q=\frac{2}{3}$           and r = 4 or 5 and n = 5

Use   $P\left(r\right){=}^n{C}_r{p}^r{q}^{n-r}$

 P(4) + P(5)

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

$=\frac{10}{3^5}+\frac{1}{3^5}=\frac{11}{3^5}=\frac{11}{243}$  

A: runner succeeded exactly 3 times out of 5.
B: runner succeeds on the first trial.
P (B/A) = P (B ∩ A) / P (A) = (p (? C? p² (1−p)²) / (? C? p³ (1-p)²) = 3/5 = 0.6

Volume = |a b c; b c a; c a b| = |3abc – a³ – b³ – c³|
= | (a + b + c) (a² + b² + c² – ab – bc – ca)| = | (a + b + c) (a + b + c)² – 3 (ab + bc + ca)|
= |9 (81-3 * 15)| = 324

  $\left(\lambda ,2\lambda ,3\lambda \right)$

$cos\theta =\frac{6}{\sqrt[]{42}}\Rightarrow sin\theta =\frac{\sqrt[]{6}}{\sqrt[]{42}}$ (where q is the angle between L₁ & L₂)

$Area\Delta OAB=\frac{1}{2}\left(OA\right)\left(OB\right)sin\theta$

$=\frac{1}{2}\left(\sqrt[]{3}\right)\left|\lambda \right|\left(\sqrt[]{14}\right)\frac{\sqrt[]{6}}{\sqrt[]{42}}=\sqrt[]{6}\Rightarrow \lambda =\pm 2$

Kindly consider the following figure

$k=\frac{\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]+2\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]+\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]-\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]}{\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]}$            

k = 3

$\underset{x\to 0}{lim}{\left[si{n}^2\left(\frac{\pi }{2-3x}\right)\right]}^{se{c}^2\left(\frac{\pi }{2-5x}\right)}$

$e^{\underset{x\to 0}{lim}\left[si{n}^2\left(\frac{\pi }{2-3x}\right)-1\right]se{c}^2\left(\frac{\pi }{2-5x}\right)}$

$=e^{\underset{x\to 0}{lim}\frac{-si{n}^2\left(\frac{-3\pi x}{2\left(2-3x\right)}\right)}{si{n}^2\left(\frac{-5\pi x}{2\left(2-5x\right)}\right)}}=e^{-\frac{9}{25}}$

For point of intersection
1 + 3λ = 3 + μ
2 + λ = 1 + 2μ
5λ = 5 ⇒ λ = 1, μ = 1
Point of intersection (4, 3, 5)
For the greatest distance from origin perpendicular from meet plane at point of intersection
Hence equation r . (4i + 3j + 5k) = 50

${\displaystyle \underset{-a}{\overset{a}{\int }}\left(\left|x\right|+\left|x-2\right|\right)dx=22,a>2}$

${\displaystyle \underset{-a}{\overset{0}{\int }}\left(-2x+2\right)dx+{\displaystyle \underset{0}{\overset{2}{\int }}\left(x-x+2\right)dx+{\displaystyle \underset{2}{\overset{a}{\int }}\left(2x-2\right)dx=22}}}$

$\Rightarrow 2a^2+2=20\Rightarrow a^2=9\Rightarrow a=3$

$\therefore {\displaystyle \underset{3}{\overset{-3}{\int }}\left(x+\left[x\right]\right)dx=-{\displaystyle \underset{-3}{\overset{3}{\int }}\left(2x-\left\{x\right\}\right)dx=}}-{\displaystyle \underset{-3}{\overset{3}{\int }}2xdx+6}{{\displaystyle \underset{0}{\overset{1}{\int }}xdx=6.\frac{x^2}{2}}|}_0^1=3$           

f (x) = λ (x-2)²
⇒ 12 = λ (2)² ⇒ λ = 3
f (x) = 3 (x-2)² f (6) = 3 * 4² = 48