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Let f (x) = max (2x + 1, 3 − 4x), where x is any real number. Then the minimum possible value of f(x) is :-
- a)1/3
- b)1/2
- c)2/3
- d)4/3
- e)5/3
Correct answer is option 'E'. Can you explain this answer?
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Let f (x) = max (2x + 1, 3 − 4x), where x is any real number. Th...
Ans.
Option (e)
To find out min of f(x) = max (2x + 1, 3 – 4x), we should be taking the point of intersection of (2x + 1) and (3 – 4x). [Since one of these equations is increasing and other one is decreasing]
2x + 1 = 3 – 4x, or, x = 1/3
2x + 1 = 3 – 4x = 5/3
Hence option (e) is the answer.
2x + 1 = 3 – 4x, or, x = 1/3
2x + 1 = 3 – 4x = 5/3
Hence option (e) is the answer.
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Let f (x) = max (2x + 1, 3 − 4x), where x is any real number. Th...
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Let f (x) = max (2x + 1, 3 − 4x), where x is any real number. Th...
f(x) is the maximum value of 2x + 1 and 3 – 4x.
Here, we see that in 1 place, x is added to something else, while in the other equation (3 – 4x), x is subtracted from something else.
So, in the 1st equation, as x goes higher, f(x) will become bigger, while in the second case, when x goes bigger, the value of f(x) goes smaller, and it will be vice-versa when x goes lower.
So, we need to find an optimum solution, which we can obtain by equating both the equations.
2x + 1 = 3 – 4x. On solving, we get x = 1/3 and f(x) = 5/3.
If we put any value more than 1/3, 2x + 1 becomes high, and f(x) increases.
If we put any value less than 1/3, 3 – 4x becomes high, and f(x) increases.
So, for x = 1/3, f(x) = 5/3 is the minimum possible value.
2x + 1 = 3 – 4x. On solving, we get x = 1/3 and f(x) = 5/3.
If we put any value more than 1/3, 2x + 1 becomes high, and f(x) increases.
If we put any value less than 1/3, 3 – 4x becomes high, and f(x) increases.
So, for x = 1/3, f(x) = 5/3 is the minimum possible value.
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