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Maths-

General

Easy

Question

The interval of increase of the function $f space left parenthesis x right parenthesis equals x minus e to the power of x plus t a n invisible function application open parentheses fraction numerator 2 pi over denominator 7 end fraction close parentheses$ is

$(0, \infty)$
$(- \infty, 0)$
$(1, \infty)$
$(- 1,1)$

hint Hint:

We are given a function. We have to find the interval for which it is increasing. We will take the first order derivative to find the answer

The correct answer is: $left parenthesis negative infinity comma 0 right parenthesis$

The given function is $f open parentheses x close parentheses equals x minus e to the power of x plus tan open parentheses fraction numerator 2 pi over denominator 7 end fraction close parentheses$
A function is said to be increasing when it’s first order derivative is positive.
It means f’(x) > 0
We will use this condition to find the answer.
Taking the first order derivative of the given function.
f(x) = $x minus e to the power of x plus tan open parentheses fraction numerator 2 pi over denominator 7 end fraction close parentheses$
f’(x) = 1 – e^x + 0

Now for the function to be increasing,
f'(x) > 0
1 - e^x > 0
1 > e^x
e^x< 1
Taking log_e of both the sides. It means we are taking natural log on both the sides.
xlog_ee < log_e1
x < 0 ...{ log_ee = 1 and log_e1 = 0}
x can have any value less than zero.
There is no lower limit on x.
So, the interval will be (-∞,0)

For such questions, we should know different formulas of differentiation.

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