Test Index

UPSC NDA 2 2019 Mathematics Solved Paper

Directions (Q. Nos. 89 and 90): Read the following information and answer the two items that follow:

Consider the function

f(x)=3x^4–20x^3–12x^2+288x+1

[NDA II 2019]

Question : 90 of 120

Marks: +1, -0

Solution: 👈: Video Solution

f(x)=3x^{4}-20x^{3}-12x^{2}+288x+1

\Rightarrow f'(x)=12x^{3}-60x^{2}-24x+288

=12(x^{3}-5x^{2}-2x+24)

=12(x+2)(x-3)(x-4)

For f(x) to be decreasing, we must have

f'(x) < 0

\Rightarrow 12(x+2)(x-3)(x-4) < 0

\Rightarrow (x-3)(x-4) < 0

\Rightarrow 3 < x < 4

\Rightarrow x \in (3,4)

So, f(x) is increasing on (3, 4)

Go to Question:

Prev Question

Next Question