UPSEE Solved Model Paper 7

Given: The function, f (x) = $\left(\frac{\sqrt{a+4}}{1-a}-1\right) x^5$ - 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 ⇒ $5\left(\frac{\sqrt{a+4}}{1-a}-1\right)x^4$ - 3 < 0 for all x ⇒ $\left(\frac{\sqrt{a+4}}{1-a}-1\right) x^4$ < 3/5 for all x ⇒ $\left(\frac{\sqrt{a+4}}{1-a} -1\right)$ ≤ 0 ⇒ $\frac{\sqrt{a+4}}{1-a}$ ≤ 1 This inequality is trivially true for a > 1 i.e. a ∊ (1, ∞) (i) Now. let us take a < 1 for $\sqrt{a+4}$ to be real we must have a > - 4 Thus, - 4 ≤ a < 1 For these values of a, we have $\frac{\sqrt{a+4}}{1-a}$ ≤ 1 ⇒ $\sqrt{a+4}$ < 1 - a ⇒ a + 4 ≤ $1+a^2-2a$ ⇒ 0 ≤ $a^2$ - 3a - 3 ⇒ a ≤ $\frac{3-\sqrt{21}}{2}$ or a ≥ $\frac{3+\sqrt{21}}{2}$ ⇒ - 4 ≤ a ≤ $\frac{3-\sqrt{21}}{2}$ ... (ii) using (i) and (ii) we have a ∊ $\left[-4,\frac{3-\sqrt{21}}{2}\right]$ ∪ (1 , ∞) Therefore, option B is correct