UPSEE Solved Model Paper 7

Given:
The function, f (x) = (

√a+4

1−a

−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−

√21

2

] ∪ (1 , ∞)
Therefore, option B is correct