Question

plzzzz frds explain me this solution...


i will mark as brainliest....

Answer 1

The given condition is x + 100/x > 50

$x + \frac{100}{x} > 50 \\ \\ \frac{ {x}^{2} + 100}{x} > 50 \\ \\ {x}^{2} + 100 > 50x \\ \\ {x}^{2} - 50x + 100 > 0$

$function \: will \: attain \: the \: value \: 0 \: at \\ \\ x = \frac{ - ( - 50) \pm \sqrt{ {( - 50)}^{2} - 4 \times 1 \times 100} }{2 \times 1} \\ \\ x = \frac{ 50 \pm \sqrt{2500 - 400} }{2} \\ \\ x = \frac{ 50 \pm \sqrt{2100} }{2} \\ \\ x = \frac{ 50 \pm 45.82 }{2} \\ \\ x = \frac{ 50 + 45.82 }{2} \: \: and \: \: \frac{ 50 - 45.82 }{2} \\ \\ x = 47.91 \: \: and \: \: 2.09$
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Thus, the function attains value 0 at 2.09 and 47.91

The function is concave upward. So, Between 2.09 and 47.91, the function is less than zero. You can see it by putting any value of x between 2.09 and 47.91. You will get values less than 0. for example, put x=3.

${x}^{2} - 50x + 100 = {3}^{2} - 50 \times 3 + 100 = - 41<0$

So the numbers between 2.09 and 47.91 are not the solutions.
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The numbers less than 2.09 and more than 47.91 satisfy the inequality. If you put x=1, you will see that the value of function will be more than 0.

${x}^{2} - 50x + 100 = {1}^{2} - 50 \times 1 + 100 = 51 > 0$

So numbers less than 2.09 and more than 47.91 are the solutions.

It is given that x is a natural number.(between 1 and 100)

So x = {1, 2, 48, 49, 50, 51, ... , 97, 98, 99, 100}

total number of solutions = 55

total natural numbers from 1 to 100 = 100

probability that it satisfies (x + 1/x > 50) = 55/100 = 11/20

$\boxed{ \red{ \bold{Probability = \frac{11}{20} }}}$