−1)x5 - 3x + log 5 We have to find the values of a for which '(x)is a decreasing function For f (x) to be decreasing for all x. we must have f'(x) ≤ 0 for all x ⇒ 5(
√a+4
1−a
−1)x4 - 3 < 0 for all x ⇒ (
√a+4
1−a
−1)x4 < 3/5 for all x ⇒ (
√a+4
1−a
−1) ≤ 0 ⇒
√a+4
1−a
≤ 1 This inequality is trivially true for a > 1 i.e. a ∊ (1, ∞) (i) Now. let us take a < 1 for √a+4 to be real we must have a > - 4 Thus, - 4 ≤ a < 1 For these values of a, we have
√a+4
1−a
≤ 1 ⇒ √a+4 < 1 - a ⇒ a + 4 ≤ 1+a2−2a ⇒ 0 ≤ a2 - 3a - 3 ⇒ a ≤
3−√21
2
or a ≥
3+√21
2
⇒ - 4 ≤ a ≤
3−√21
2
... (ii) using (i) and (ii) we have a ∊ [−4,3−