Ncert Solutions Maths class 12th

P (2W and 2B) = P (2B, 6W) × P (2W and 2B)

+ P (3B, 5W) × P (2W and 2B)

+ P (4B, 4W) × P (2W and 2B)

+ P (5B, 3W) × P (2W and 2B)

+ P (6B, 2W) × P (2W and 2B)

(15 + 30 + 36 + 30 + 15)

$=\frac{36}{126}$

$=\frac{18}{63}$

$=\frac{6}{21}$

$=\frac{2}{7}$

  $?$  R is symmetric relation

⇒   (y, x) ∈ R V (x, y) ∈ R

(x, y) ∈ R ⇒ 2x = 3y and (y, x) ∈ R ⇒ 3x = 2y

Which holds only for (0, 0)

Which does not belongs to R.

∴ Value of n = 0

Take $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\lambda$  

x = 2λ + 1, y = 3λ + 2, z = 4λ + 3

  $\overrightarrow{AB}$  = (α − 2)  $\widehat{i}$ + (β − 3) $\widehat{j}$ + (γ − 4) $\widehat{k}$  

Now,

(α − 2) ⋅ 2 + (β − 3) ⋅3 + (γ − 4) ⋅ 4 = 0

2α − 4 + 3β − 9 + 4γ −16 = 0

⇒ 2α + 3β + 4γ = 29

Take e^sinx = t (t > 0)

⇒ $t-\frac{2}{t}=2$

⇒   $\frac{t^2-2}{t}=2$

->t² – 2t – 2 = 0

->t² – 2t + 1 = 3

⇒   (t −1)² = 3

⇒   t = 1 ± $\sqrt[]{3}$  

⇒   t = 1 ± 1.73

⇒   t = 2.73 or –0.73 (rejected as t > 0)

⇒   e^{sin x} = 2.73

->log_e e^{sin x} = log_e 2.73

⇒ sin x = log_e 2.73 > 1

So no solution.

f is increasing function

x < 5x < 7x

f (x) < f (5x) < f (7x)

  $\frac{f\left(x\right)}{f\left(x\right)}<\frac{f\left(5x\right)}{f\left(x\right)}<\frac{f\left(7x\right)}{f\left(x\right)}$           

$\underset{x\to \infty }{lim}\frac{f\left(x\right)}{f\left(x\right)}<\underset{x\to \infty }{lim}\frac{f\left(5x\right)}{f\left(x\right)}<\underset{x\to \infty }{lim}\frac{f\left(7x\right)}{f\left(x\right)}$             

-> $1<\underset{x\to \infty }{lim}\frac{f\left(5x\right)}{f\left(x\right)}<1$ $\underset{x\to \infty }{lim}\frac{f\left(5x\right)}{f\left(x\right)}=1$

$\underset{x\to \infty }{lim}\left(\frac{f\left(5x\right)}{f\left(x\right)}-1\right)=0$            

Let probability of tail is   $\frac{1}{3}$

⇒ Probability of getting head = $\frac{2}{3}$  

∴ Probability of getting 2 heads and 1 tail

$=\left(\frac{2}{3}\times \frac{2}{3}\times \frac{1}{3}\right)\times 3$

$=\frac{4}{27}\times 3$

$=\frac{4}{9}$                  

                  &nb

Let f (x) = x⁶ + ax⁵ + bx⁴ + ax³ + dx² + ex + f

$\underset{x\to 0}{lim}\frac{f\left(x\right)}{x^3}=1$    Non zero finite

So, d = e = f = 0

f (x)  = x⁶ + ax⁵ + bx⁴ + cx³

$\underset{x\to 0}{lim}\frac{f\left(x\right)}{x^3}=1$ Non zero finite

f’ (x) = $6x^5+5a{x}^4+4b{x}^3+3x^2$

f’ (1) = 0

6 + 5a + 4b + 3 = 0

5a + 4b = - 9 . (i)

f’ (-1) = 0

-6 + 5a – 4b + 3 = 0 . (ii)

Solving (i) and (ii)

a  -3/5, b = -3/2

$\therefore f\left(x\right)=x^6+\left(\frac{-3}{5}\right)x^5+\left(\frac{-3}{2}\right)x^4+x^3$

$\therefore 5.f\left(2\right)=144$

$L_1=\frac{x-\lambda }{1}=\frac{y-\frac{1}{2}}{\frac{1}{2}}=\frac{z}{-\frac{1}{2}}$

$\therefore SD=\frac{\left|-2\lambda +3\left(2\lambda +\frac{1}{2}\right)+\lambda \right|}{\sqrt[]{14}}=\frac{\left|5\lambda +\frac{3}{2}\right|}{\sqrt[]{14}}$

$5\lambda +\frac{3}{2}=-\frac{7}{2}\Rightarrow 5\lambda =-5\Rightarrow \lambda =-1$

$\therefore \left|\lambda \right|=1$

$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{-2}=r$

For ‘B’, x = 2r + 1, y = 3r – 1, z = -2r + 1

As AB is perpendicular to the line,

$\left\{\left(2r+1\right)-0\right\}2+\left\{\left(3r-1\right)-1\right\}3+\left\{\left(-2r+1\right)-2\right\}$

$r=\frac{2}{17}\Rightarrow B\left(\frac{2}{17},\frac{-11}{7},\frac{13}{17}\right)$     

direction ratios of AB

(2r + 1, 3r – 2, -2r – 1)

Equation of AB

$\frac{x}{-3}=\frac{y-1}{4}=\frac{z-2}{3}$

ax² + bx + c = 0

D = b² – 4ac

D = 0

b² – 4ac = 0

b² = 4ac

(i) AC = 1, b = 2 (1, 2, 1) is one way

(ii) AC = 4, b = 4

$\begin{array}{l}a=4c=1\\ a=2c=2\\ a=1c=4\end{array}\}3ways$

(iii) AC = 9, b = 6, a = 3, c = 3 is one way

1 + 3 + 1 = 5 way

Required probability = $\frac{5}{216}$