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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

Answers (1)

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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 

Posted by

Gautam harsolia

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