Question

On which of the following intervals is the function f given by f(x)=x^100+sinx-1 strictly decreasing? (A)(0,1) (B)(π/2,π) (C)(0,π/2) (D) None of these

Answer 1

given, function $\bf{f(x)=x^{100}+sinx-1}$
differentiate with respect to x,
$\bf{\frac{dy}{dx}=100x^{99}+cosx}$

for strictly decreasing, f'(x) < 0
$\bf{100x^{99}+cosx}$ < 0

it is implicit function, we can get value of x directly , Let's check all options for getting solution.

option (a) (0,1)
100x^99 > 0 for all x belongs to (0,1)
we know, 1 < π/3
means, we can write (0,1) = (0, π/3)
but we know cosx > 0 in (0,π/3)
so, cosx > 0 in (0,1)
hence, 100x^99 + cosx = f'(x) > 0 in (0,1)
therefore , f is strictly increasing function in (0,1)

option (B) (π/2, π)
100x^99 > 0 but cosx < 0
but 100x^99 > negative value of cosx
hence, f'(x) > 0 in (π/2, π)
therefore, f is strictly increasing in (π/2, π)

option (C) (0, π/2)
100x^99 > 0 and cosx > 0
obviously , 100x^99 + cosx = f'(x) > 0 in (0,π/2)
hence, f is strictly increasing in (0,π/2)

Hence, function f is strictly decreasing on none of the intervals. option (D) is correct

Answer 2

Answer:

Refer to the attachment..

Step-by-step explanation: