retagged by
706 views

1 Answer

1 votes
1 votes

Given that,

$ 0.25 < 2^{x} < 200 $

$ \Rightarrow \frac {25}{100} < 2^{x} < 200 $

$ \Rightarrow \frac {1}{4} < 2^{x} < 200 $

$ \Rightarrow 2^{-2} < 2^{x} < 200 \quad \longrightarrow (1)$

From equation $(1)$

$ 2^{-2} < 2^{x}$

$ \Rightarrow 2^{x} > 2^{-2}$

$ \Rightarrow \boxed{x > -2}$

And, $ 2^{x} < 200 $

We know, that $ 2^{8} = 256,$ so here $ x $ should be less than $8. \Rightarrow \boxed {x < 8}$

Now, we have $ \boxed { -2 < x < 8}$

Therefore, possible value of $ x: \{ -1,0,1,2,3,4,5,6,7\}$

$ 2^{x} + 2 $ is perfectly divisible by either $3$ or $4:$

Now, we can put the values of $x$ and see, which one is satisfied.

  • $ x = -1 \Rightarrow 2^{-1} + 2 = \frac{1}{2} + 2 = \frac{5}{2} \; ($It is not divisible by either $3$ or $4)$
  • $ x = 0 \Rightarrow 2^{0} + 2 = 1 + 2 = 3\;($It is divisible by $3)$
  • $ x = 1 \Rightarrow 2^{1}  + 2 = 4\;($It is divisible by $4)$
  • $ x = 2 \Rightarrow 2^{2}  + 2 = 6\;($It is divisible by $3)$
  • $ x = 3 \Rightarrow 2^{3}  + 2 = 10 \;($It is not divisible by either $3$ or $4)$
  • $ x = 4 \Rightarrow 2^{4}  + 2 = 18\;($It is divisible by $3)$
  • $ x = 5 \Rightarrow 2^{5}  + 2 = 34 \;($It is not divisible by either $3$ or $4)$
  • $ x = 6 \Rightarrow 2^{6}  + 2 = 66 \;($It is divisible by $3)$
  • $ x = 7 \Rightarrow 2^{7}  + 2 = 130 \;($It is not divisible by either $3$ or $4)$

$ \therefore $ The number of integers $x = \{0,1,2,4,6\}$

Correct Answer $: 5 $

edited by
Answer:

Related questions

2 votes
2 votes
1 answer
1
go_editor asked Mar 19, 2020
731 views
While multiplying three real numbers, Ashok took one of the numbers as $73$ instead of $37$. As a result, the product went up by $720$. Then the minimum possible value of...
2 votes
2 votes
1 answer
4