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# On which of the following intervals is the function f given by f (x) is equal to x is raised to 100 plus sin x minus 1 decreasing ?

13) On which of the following intervals is the function f given by $f ( x) = x ^{100} + \sin x - 1$ decreasing ?
(A) (0,1)                      (B) $\frac{\pi}{2},\pi$                    (C)  $0,\frac{\pi}{2}$                      (D) None of these

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(A) Given function is,
$f ( x) = x ^{100} + \sin x - 1$
$f^{'}(x) = 100x^{99} + \cos x$
Now, in interval (0,1)
$f^{'}(x) > 0$
Hence, $f ( x) = x ^{100} + \sin x - 1$  is increasing function in interval  (0,1)

(B)  Now, in interval $\left ( \frac{\pi}{2},\pi \right )$
$100x^{99} > 0 \ but \ \cos x < 0$
$100x^{99} > \cos x \\ 100x^{99} - \cos x > 0$      ,       $f^{'}(x) > 0$
Hence, $f ( x) = x ^{100} + \sin x - 1$  is increasing function in interval  $\left ( \frac{\pi}{2},\pi \right )$

(C)  Now, in interval $\left ( 0,\frac{\pi}{2} \right )$
$100x^{99} > 0 \ and \ \cos x > 0$
$100x^{99} > \cos x \\ 100x^{99} - \cos x > 0$   ,  $f^{'}(x) > 0$
Hence, $f ( x) = x ^{100} + \sin x - 1$  is increasing function in interval  $\left ( 0,\frac{\pi}{2} \right )$

So,  $f ( x) = x ^{100} + \sin x - 1$ is increasing for all cases
Hence, correct answer is (D) None of these

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