43. The probability that a student will pass the final examination in both English and Hindi is 0.5 and the probability of passing neither is 0.1. If the probability of passing the English examination is 0.75, what is the probability of passing the Hindi examination?

3, 4, 5, 5

In remaining six places you have to arrange

3, 4, 5,5

So no. of ways $=\frac{6!}{2!2!2!}$  

Total no. of seven digits nos. = $\frac{7!}{2!3!2!}*1$  

Hence Req. prob. $\frac{\frac{6!}{2!2!2!}}{\frac{7!}{2!3!2!}}=\frac{6!3!}{2!7!}=\frac{3}{7}$

f (x) = x? – 4x + 1 = 0
f' (x) = 4x³ – 4
= 4 (x–1) (x²+1+x)
=> Two solution

$\tilde{z}=i{z}^2+z^2-z$

$z+\overline{Z}=z^2\left(i+1\right)$

$z+\overline{Z}=z^2\left(i+1\right)$ Let z be equal to (x + iy)

(x + iy) + (x – iy) = (x + iy)² (i + 1)

$2x=\left(x^2-y^2+2ixy\right)\left(i+1\right)$               

Equating the real & in eg part.

$\left(x^2-y^2+2ixy\right)=0.........\left(i\right)$

$\left(x^2-y^2-2xy\right)=\left(2x\right)..............\left(ii\right)$               

(i) & (ii)

 4xy = -2x Þ x = 0 or y = $\left(\frac{-1}{2}\right)$  

(for x = 0, y = 0)

For y = $\frac{-1}{2}$  

x²   $-\frac{1}{4}+2\left(\frac{-1}{2}\right)x=0$

x =   $\frac{4\pm \sqrt[]{16+16}}{2.4}$

=  $\left(\frac{1+\sqrt[]{2}}{2}\right)or\left(\frac{1-\sqrt[]{2}}{2}\right)$  

$\therefore sum$ of   ${\left|z\right|}^2={\left(\frac{1+\sqrt[]{2}}{2}\right)}^2+\frac{1}{4}+{\left(\frac{1-\sqrt[]{2}}{2}\right)}^2+\frac{1}{4}+0^2+O^2$

=   $\frac{3}{4}+\frac{\sqrt[]{2}}{2}+\frac{1}{4}+\frac{3}{4}-\frac{\sqrt[]{2}}{2}+\frac{1}{4}=\frac{3}{2}+\frac{1}{2}=2$

 $\frac{dy}{dx}=\frac{1}{1+sin2x}$

$dy=\frac{se{c}^{2}xdx}{{\left(1+tanx\right)}^2}$

$\Rightarrow y=-\frac{1}{1+tanx}+c$

When

$x=\frac{\pi }{4},y=\frac{1}{2}$ gives c = 1

So

$x+\frac{\pi }{4}=\frac{5\pi }{6}or\frac{13\pi }{6}\Rightarrow x=\frac{7\pi }{12}or\frac{23\pi }{12}$

sum of all solutions =

$\pi +\frac{7\pi }{12}+\frac{23\pi }{12}=\frac{42\pi }{12}$

Hence k = 42

Each element of ordered pair (i, j) is either present in A or in B.

So, A + B = Sum of all elements of all ordered pairs {i, j} for $1\le i\le 10$ and $1\le j\le 10$

= 20 (1 + 2 + 3 + … + 10) = 1100