On which of the following intervals is the function f given by \(f(x)=x^{100}+sin\ x-1\) strictly decreasing?
On which of the following intervals is the function f given by \(f(x)=x^{100}+sin\ x-1\) strictly decreasing?
\((0,1)\)
\((\frac \pi2,\pi)\)
\((0,\frac \pi2)\)
\(None\ of \ these\)
The Correct Option is D
Solution and Explanation
We have,
f(x) = x100+sinx-1
f'(x) = 100x99+cosx
In interval(0,1), cosx>0 and 100x99>0.
f'(x)>0.
Thus, function f is strictly increasing in interval (0, 1).
In interval (\(\frac \pi2\),\(\pi\)), cosx<0 and 100x99>0. also, 100x100>cosx
\(\implies\)f'(x)>0 in (\(\frac \pi2\),\(\pi\)).
Thus, function f is strictly increasing in interval (\(\frac \pi2\),\(\pi\)).
In interval (0,\(\frac \pi2\)), cosx>0 and 100x99>0.
100x99+cosx>0
f'(x)>0 on (0,\(\frac \pi2\)).
Hence, function f is strictly decreasing in none of the intervals. The correct answer is (D).
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Notes on Applications of Derivatives
Concepts Used:
Increasing and Decreasing Functions
Increasing Function:
On an interval I, a function f(x) is said to be increasing, if for any two numbers x and y in I such that x < y,
⇒ f(x) ≤ f(y)
Decreasing Function:
On an interval I, a function f(x) is said to be decreasing, if for any two numbers x and y in I such that x < y,
⇒ f(x) ≥ f(y)
Strictly Increasing Function:
On an interval I, a function f(x) is said to be strictly increasing, if for any two numbers x and y in I such that x < y,
⇒ f(x) < f(y)
Strictly Decreasing Function:
On an interval I, a function f(x) is said to be strictly decreasing, if for any two numbers x and y in I such that x < y,
⇒ f(x) > f(y)
Graphical Representation of Increasing and Decreasing Functions
