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 : 89 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^{3}+2x^{2}-7x^{2}-14x+12x+24]

=12[x^{2}(x+2)-7x(x+2)+12(x+2)]

=12(x+2)(x^{2}-7x+12)

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

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

f(x) > 0

12(x+2)(x-3)(x-4) > 0

x < -2

x > 3

x(-2,3)

So, f(x) is increasing on

(-2,3)

Go to Question:

Prev Question

Next Question