On which of the following intervals is the function f given by $f (x) = x^{100} + s i n x - 1$ strictly decreasing.

a) (0,1)

b) $(\frac{π}{2}, π)$

c) $(0, \frac{π}{2})$

d) None of these

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Solution

Given, $f (x) = x^{100} + s i n x - 1 \Rightarrow f^{'} (x) = 100 x^{99} + c o s x$
In interval $(0.1), c o s x > 0$ and $100 x^{99} > 0$
$∴ f^{'} (x) > 0 [∵ I n (0, \frac{π}{2}), c o s x > 0 i . e ., (0.157)], c o s x > 0 ∴ (0, 1), c o s x > 0$

Thus, function f is strictly increasing in interval (0,1).

In interval $(\frac{π}{2}, π), c o s x < 0$ and $100 x^{99} > 0. A l s o, 100 x^{99} > c o s x$
$∴ f^{'} (x) > 0$ in $(\frac{π}{2}, π)$
Thus, function f is strictly increasing in interval $(\frac{π}{2}, π)$ .

In interval $(0, \frac{π}{2}), c o s x > 0$ and $100 x^{99} > 0$
$∴ 100 x^{99} + c o s x > 0 \Rightarrow f^{'} (x) > 0$ in $(0, \frac{π}{2})$
Thus, f is strictly increasing in interval $(0, \frac{π}{2})$
Hence, function f is not strictly decreasing in the given intervals.
The correct answer is (d) None of these

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